PHY6937: Difference between revisions

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To see the origins of superconductivity, it is helpful to look at a toy system, which we already know will give us superconducting behavior. This is useful because the toy system is only a simple change to a non-interacting electron gas. By adding in some small ''attractive'' interaction, we will arrive at a superconducting system! This interaction need only occur between two electrons occupying the same position in space (and necessarily having opposite spin!). Additionally, we still find the interesting behaviour regardless of the size of the interaction; the only requirement is that it be non-zero!
To see the origins of superconductivity, it is helpful to look at a toy system, which we already know will give us superconducting behavior. This is useful because the toy system is only a simple change to a non-interacting electron gas. By adding in some small ''attractive'' interaction, we will arrive at a superconducting system! This interaction need only occur between two electrons occupying the same position in space (and necessarily having opposite spin!). Additionally, we still find the interesting behaviour regardless of the size of the interaction; the only requirement is that it be non-zero!


We can write the Hamiltonian of the system as:
Our toy model will be that of free fermions with an attractive contact interaction.  We can write the Hamiltonian of the system as:


<math>H=\sum_\vec{r}[\psi_\sigma^\dagger (\vec{r})(\epsilon_\vec{p}-\mu)\psi_\sigma^\dagger (\vec{r}) +g\psi_\uparrow^\dagger (\vec{r})\psi_\downarrow^\dagger (\vec{r})\psi_\downarrow (\vec{r})\psi_\uparrow (\vec{r})]</math>
<math>H=\sum_\vec{r}[\psi_\sigma^\dagger (\vec{r})(\epsilon_\vec{p}-\mu)\psi_\sigma^\dagger (\vec{r}) +g\psi_\uparrow^\dagger (\vec{r})\psi_\downarrow^\dagger (\vec{r})\psi_\downarrow (\vec{r})\psi_\uparrow (\vec{r})]</math>
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<math>\psi^\dagger</math> and <math>\ \psi</math> are grassmann numbers.
<math>\psi^\dagger</math> and <math>\ \psi</math> are grassmann numbers.
<math>\ \Delta^*</math> and <math>\ \Delta</math> are constant.
<math>\ \Delta^*</math> and <math>\ \Delta</math> are normally commuting variables.
<math>\psi_\uparrow\psi_\downarrow</math> and <math>\psi_\downarrow\psi_\uparrow</math> behave like constant.
<math>\psi_\uparrow\psi_\downarrow</math> and <math>\psi_\downarrow\psi_\uparrow</math> behave like constant (paired fermions behaving like bosons).


Let's make a shift of the constant:
Let's make a shift of the new integration variables:


<math>\Delta \rightarrow \Delta+g\psi_\uparrow\psi_\downarrow</math>
<math>\Delta \rightarrow \Delta+g\psi_\uparrow\psi_\downarrow</math>
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<math>\Delta^*\rightarrow \Delta^*+g\psi^\dagger_\downarrow\psi^\dagger_\uparrow</math>
<math>\Delta^*\rightarrow \Delta^*+g\psi^\dagger_\downarrow\psi^\dagger_\uparrow</math>


Then, <math>S_\Delta=-\int_0^\beta d\tau \sum_{\vec{r}}{\{\frac{1}{g}\Delta^*\Delta + \Delta^*\psi_\uparrow \psi_\downarrow + \Delta\psi^\dagger_\downarrow \psi^\dagger_\uparrow+g\psi^\dagger_\downarrow \psi^\dagger_\uparrow \psi_\uparrow \psi_\downarrow}\}</math>
Then, <math>S_\Delta=-\int_0^\beta d\tau \sum_{\vec{r}}\left [\frac{1}{g}\Delta^*\Delta + \Delta^*\psi_\uparrow \psi_\downarrow + \Delta\psi^\dagger_\downarrow \psi^\dagger_\uparrow+g\psi^\dagger_\downarrow \psi^\dagger_\uparrow \psi_\uparrow \psi_\downarrow\right ]</math>


<math>\begin{align}S=&S_{BCS}+S_{\Delta}\\
<math>\begin{align}S=&S_{BCS}+S_{\Delta}\\
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then, <math>Z=\int D[\psi_{\sigma}^{*}(\tau,\mathbf{r}),\psi_{\sigma}(\tau,\mathbf{r})]D[\Delta^{*}(\tau,\mathbf{r}),\Delta(\tau,\mathbf{r})]e^{-(S_{0}+S_{int.}+S_{\Delta})}</math>.
then, <math>Z=\int D[\psi_{\sigma}^{*}(\tau,\mathbf{r}),\psi_{\sigma}(\tau,\mathbf{r})]D[\Delta^{*}(\tau,\mathbf{r}),\Delta(\tau,\mathbf{r})]e^{-(S_{0}+S_{int.}+S_{\Delta})}</math>.


<math>\left\langle e^{-S_{int.}}\right\rangle _{0}\cong exp[\frac{1}{2}\left\langle S_{int.}^{2}\right\rangle _{0}+\frac{1}{4!}(\left\langle S_{int.}^{4}\right\rangle _{0}-3\left\langle S_{int.}^{2}\right\rangle _{0}^{2})]</math> by cumulant expansion, which guarantees that until the 2nd order, it is accurate.
<math>\left\langle e^{-S_{int.}}\right\rangle _{0}\cong \exp\left [\frac{1}{2}\left\langle S_{int.}^{2}\right\rangle _{0}+\frac{1}{4!}\left (\left\langle S_{int.}^{4}\right\rangle _{0}-3\left\langle S_{int.}^{2}\right\rangle _{0}^{2}\right )\right ]</math> by cumulant expansion, which guarantees that until the 2nd order, it is accurate.


Use Matsubara's Method
Use Matsubara's Method
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<math>\Rightarrow=-\frac{1}{L^{D}}\frac{1}{\beta}\underset{\mathbf{k}}{\sum}\frac{1}{\varepsilon_{\mathbf{q}-\mathbf{k}}+\varepsilon_{\mathbf{k}}-2\mu-i\Omega_{n}}[\frac{1}{e^{\beta(\varepsilon_{\mathbf{q}}-\mu)}+1}-\frac{1}{e^{\beta(-\varepsilon_{\mathbf{q}-\mathbf{k}}+\mu)}+1}]=\int\frac{d^{D}k}{(2\pi)^{D}}\frac{1}{\varepsilon_{\mathbf{q}}+\varepsilon_{\mathbf{q}-\mathbf{k}}-2\mu-i\Omega_{n}}[1-f(\varepsilon_{\mathbf{k}})-f(\varepsilon_{\mathbf{q}-\mathbf{k}})].</math>
<math>\Rightarrow=-\frac{1}{L^{D}}\frac{1}{\beta}\underset{\mathbf{k}}{\sum}\frac{1}{\varepsilon_{\mathbf{q}-\mathbf{k}}+\varepsilon_{\mathbf{k}}-2\mu-i\Omega_{n}}[\frac{1}{e^{\beta(\varepsilon_{\mathbf{q}}-\mu)}+1}-\frac{1}{e^{\beta(-\varepsilon_{\mathbf{q}-\mathbf{k}}+\mu)}+1}]=\int\frac{d^{D}k}{(2\pi)^{D}}\frac{1}{\varepsilon_{\mathbf{q}}+\varepsilon_{\mathbf{q}-\mathbf{k}}-2\mu-i\Omega_{n}}[1-f(\varepsilon_{\mathbf{k}})-f(\varepsilon_{\mathbf{q}-\mathbf{k}})].</math>


In the static (<math>\ \Omega_{n}=0</math>) and uniform (<math>\mathbf{q}=0</math>) limit,<math> 1-2f(\varepsilon_{\mathbf{k}})=Tanh[\frac{\beta}{2}(\varepsilon_{\mathbf{k}}-\mu)]</math>.
In the static (<math>\ \Omega_{n}=0</math>) and uniform (<math>\mathbf{q}=0</math>) limit,<math> 1-2f(\varepsilon_{\mathbf{k}})=\tanh[\frac{\beta}{2}(\varepsilon_{\mathbf{k}}-\mu)]</math>.


Then <math>\chi_{p}(0,0)=\int\frac{d^{D}k}{(2\pi)^{D}}\frac{Tanh[\frac{\beta}{2}(\varepsilon_{\mathbf{k}}-\mu)]}{2(\varepsilon_{\mathbf{k}}-\mu)}</math>.
Then <math>\chi_{p}(0,0)=\int\frac{d^{D}k}{(2\pi)^{D}}\frac{\tanh[\frac{\beta}{2}(\varepsilon_{\mathbf{k}}-\mu)]}{2(\varepsilon_{\mathbf{k}}-\mu)}</math>.


In low energy, integrate the energy in the shell near Fermi energy:
In low energy, integrate the energy in the shell near Fermi energy:
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[[Image:Fig.3.jpg]]
[[Image:Fig.3.jpg]]


<math>\Rightarrow\chi_{p}(0,0)\cong N(0)\int_{\hbar\omega_{D}}^{-\hbar\omega_{D}}d\xi\frac{Tanh[\xi\beta/2]}{2\xi}\cong N(0)\int_{0}^{-\hbar\omega_{D}}d\xi\frac{Tanh[\xi\beta/2]}{\xi}=N(0)ln[\frac{2\hbar\omega_{D}e^{\gamma}}{\pi k_{B}T}].</math>
<math>\Rightarrow\chi_{p}(0,0)\cong N(0)\int_{\hbar\omega_{D}}^{-\hbar\omega_{D}}d\xi\frac{\tanh[\xi\beta/2]}{2\xi}\cong N(0)\int_{0}^{-\hbar\omega_{D}}d\xi\frac{\tanh[\xi\beta/2]}{\xi}=N(0)\ln[\frac{2\hbar\omega_{D}e^{\gamma}}{\pi k_{B}T}].</math>


Then <math>\frac{1}{2}\left\langle S_{int.}^{2}\right\rangle _{0}=L^{D}\frac{1}{\beta}\chi_{p}(0,0)\underset{\Omega_{n},\mathbf{q}}{\sum}\Delta_{\mathbf{q}}^{*}(i\Omega_{n})\Delta_{\mathbf{q}}(i\Omega_{n})</math>.
Then <math>\frac{1}{2}\left\langle S_{int.}^{2}\right\rangle _{0}=L^{D}\frac{1}{\beta}\chi_{p}(0,0)\underset{\Omega_{n},\mathbf{q}}{\sum}\Delta_{\mathbf{q}}^{*}(i\Omega_{n})\Delta_{\mathbf{q}}(i\Omega_{n})</math>.
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If we ignore the higher order in the cumulant expansion,
If we ignore the higher order in the cumulant expansion,


<math>S_{eff}=-\underset{\mathbf{r}}{\sum}\int_{0}^{\beta}d\tau\frac{1}{g}\Delta_{\mathbf{q}}^{*}(i\Omega_{n})\Delta_{\mathbf{q}}(i\Omega_{n})-\frac{1}{2}\left\langle S_{int.}^{2}\right\rangle _{0}=\underset{\mathbf{r}}{\sum}\int_{0}^{\beta}d\tau[\frac{1}{\left|g\right|}-N(0)ln(\frac{2\hbar\omega_{D}e^{\gamma}}{\pi k_{B}T})]\Delta^{*}(\tau,\mathbf{r})\Delta(\tau,\mathbf{r})</math>.
<math>S_{eff}=-\underset{\mathbf{r}}{\sum}\int_{0}^{\beta}d\tau\frac{1}{g}\Delta_{\mathbf{q}}^{*}(i\Omega_{n})\Delta_{\mathbf{q}}(i\Omega_{n})-\frac{1}{2}\left\langle S_{int.}^{2}\right\rangle _{0}=\underset{\mathbf{r}}{\sum}\int_{0}^{\beta}d\tau[\frac{1}{\left|g\right|}-N(0)\ln(\frac{2\hbar\omega_{D}e^{\gamma}}{\pi k_{B}T})]\Delta^{*}(\tau,\mathbf{r})\Delta(\tau,\mathbf{r})</math>.


Because the partition function <math>Z=\int D\Delta^{*}D\Delta e^{-S_{eff}(\Delta)}</math>, if we only consider the <math>\Delta</math> related factors.
Remember the partition function <math>Z=\int D\Delta^{*}D\Delta e^{-S_{eff}(\Delta)}</math>. To look at superconductivity, we really only must consider the <math>\Delta</math> related factors.


The superconductivity phase transition temperature is the temperature makes <math>\frac{1}{\left|g\right|}-N(0)ln(\frac{2\hbar\omega_{D}e^{\gamma}}{\pi k_{B}T})=0</math>, which is <math>T_{c}=\frac{\hbar\omega_{D}}{k_{B}}\frac{2}{\pi}e^{\gamma}e^{-\frac{1}{N(0)\left|g\right|}}=1.134\frac{\hbar\omega_{D}}{k_{B}}e^{-\frac{1}{N(0)\left|g\right|}}</math>.
The superconductivity phase transition temperature is the temperature that makes


Beyond the critical temperature, the <math>\Delta</math> related factors in the partition function is just <math>1</math>, the same as no cooper pair, which is normal state; below the critical temperature, the <math>\Delta</math> related factors in the partition function will diverge, which means superconductivity phase transition.
<math>{\frac{1}{\left|g\right|}}-N(0){\ln(\frac{2\hbar\omega_{D}e^{\gamma}}{\pi k_{B}T})}=0</math>,  
 
which happens at
 
<math>T_{c}=\frac{\hbar\omega_{D}}{k_{B}}\frac{2}{\pi}e^{\gamma}e^{-\frac{1}{N(0)\left|g\right|}}=1.134\frac{\hbar\omega_{D}}{k_{B}}e^{-\frac{1}{N(0)\left|g\right|}}</math>.
 
Above this critical temperature, the coefficient of the <math>\Delta^*\Delta</math> term becomes positive, and it is energetically unfavorable to have cooper pairs, as in the normal state. Below the critical temperature, the <math>\Delta</math> related factors in the partition function will diverge, which means superconductivity phase transition. Notice that this happens for ''any value of'' <math>|g|</math>! No matter how small the attraction between our fermions is, we can still find this funny (as of yet, it has not been shown to be specifically superconducting!) behaviour.


Finite <math>\vec{q}</math> (small) <math>\ (\Omega_n=0)</math>
Finite <math>\vec{q}</math> (small) <math>\ (\Omega_n=0)</math>
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\chi_p(q,0)-\chi_p(0,0)
\chi_p(q,0)-\chi_p(0,0)
&=\frac{1}{\beta}\sum_{i\omega_n}N(0)\int\frac{d\Omega}{\Omega_D}(q\cdot v_F)^2\frac{2\pi i}{(2i|\omega|)^3}\\
&=\frac{1}{\beta}\sum_{i\omega_n}N(0)\int\frac{d\Omega}{\Omega_D}(q\cdot v_F)^2\frac{2\pi i}{(2i|\omega|)^3}\\
&=N(0)v_F^2|\vec{q}|^2\int\frac{d\Omega}{\Omega_D}(q\cdot v_F)^2 \frac{1}{\beta}\sum_{i\omega}\frac{2\pi i}{-i8|\frac{(2n+1)\pi}{\beta}|^3}\\
&=N(0)v_F^2|\vec{q}|^2\int\frac{d\Omega}{\Omega_D}(q\cdot v_F)^2 \frac{1}{\beta}\sum_{i\omega}\frac{2\pi i}{-8i|\frac{(2n+1)\pi}{\beta}|^3}\\
&=-\frac{1}{4}N(0)v_F^2q^2(<(\hat{q}\cdot \hat{v_F})>_{F.S.})\frac{\beta^2}{\pi^2}(\sum_{N=-\infty}^{+\infty}\frac{1}{|2n+1|^3})
&=-\frac{1}{4}N(0)v_F^2q^2(<(\hat{q}\cdot \hat{v_F})>_{F.S.})\frac{\beta^2}{\pi^2}(\sum_{N=-\infty}^{+\infty}\frac{1}{|2n+1|^3})
\end{align}
\end{align}
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<math>
<math>
\int\frac{d\Omega}{\Omega_D}(\hat{q}\cdot\hat{v}_F)^2=\frac{2\pi}{4\pi}\int_{-1}^{1}dcos\theta cos^2\theta = \frac{1}{3}
\int\frac{d\Omega}{\Omega_D}(\hat{q}\cdot\hat{v}_F)^2=\frac{2\pi}{4\pi}\int_{-1}^{1}d(\cos{\theta})\,\cos^2{\theta} = \frac{1}{3}
</math>
</math>


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<math>
<math>
\int\frac{d\Omega}{\Omega_D}(\hat{q}\cdot\hat{v}_F)^2=\frac{1}{2\pi}\int_{0}^{2\pi}d\theta cos^2\theta = \frac{1}{2}
\int\frac{d\Omega}{\Omega_D}(\hat{q}\cdot\hat{v}_F)^2=\frac{1}{2\pi}\int_{0}^{2\pi}d\theta\,\cos^2{\theta} = \frac{1}{2}
</math>
</math>


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\begin{align}
\begin{align}
\frac{1}{2}\left\langle S_{int.}^{2}\right\rangle _{0}&=L^{D}\frac{1}{\beta}\underset{\Omega_{n},\mathbf{q}}{\sum}\chi_{p}(q,0)\Delta_{\mathbf{q}}^{*}(i\Omega_{n})\Delta_{\mathbf{q}}(i\Omega_{n}) \\
\frac{1}{2}\left\langle S_{int.}^{2}\right\rangle _{0}&=L^{D}\frac{1}{\beta}\underset{\Omega_{n},\mathbf{q}}{\sum}\chi_{p}(q,0)\Delta_{\mathbf{q}}^{*}(i\Omega_{n})\Delta_{\mathbf{q}}(i\Omega_{n}) \\
&=N(0)ln[\frac{2\hbar\omega_{D}e^{\gamma}}{\pi k_{B}T}]L^{D}\frac{1}{\beta}\underset{\Omega_{n},\mathbf{q}}{\sum}\Delta_{\mathbf{q}}^{*}(i\Omega_{n})\Delta_{\mathbf{q}}(i\Omega_{n})-L^{D}\frac{1}{\beta}\underset{\Omega_{n},\mathbf{q}}{\sum}N(0)q^{2}\xi^{2}\Delta_{\mathbf{q}}^{*}(i\Omega_{n})\Delta_{\mathbf{q}}(i\Omega_{n})
&=N(0)\ln[\frac{2\hbar\omega_{D}e^{\gamma}}{\pi k_{B}T}]L^{D}\frac{1}{\beta}\underset{\Omega_{n},\mathbf{q}}{\sum}\Delta_{\mathbf{q}}^{*}(i\Omega_{n})\Delta_{\mathbf{q}}(i\Omega_{n})-L^{D}\frac{1}{\beta}\underset{\Omega_{n},\mathbf{q}}{\sum}N(0)q^{2}\xi^{2}\Delta_{\mathbf{q}}^{*}(i\Omega_{n})\Delta_{\mathbf{q}}(i\Omega_{n})
\end{align}
\end{align}
</math>.
</math>.


<math>S_{eff}=\underset{\mathbf{r}}{\sum}\int_{0}^{\beta}d\tau\left[\left(\frac{1}{\left|g\right|}-N(0)ln(\frac{2\hbar\omega_{D}e^{\gamma}}{\pi k_{B}T})\right)\Delta^{*}(\tau,\mathbf{r})\Delta(\tau,\mathbf{r})-N(0)\xi^{2}(\nabla\cdot\Delta^{*}(\tau,\mathbf{r}))(\nabla\cdot\Delta(\tau,\mathbf{r}))\right]</math>.
<math>S_{eff}=\underset{\mathbf{r}}{\sum}\int_{0}^{\beta}d\tau\left[\left(\frac{1}{\left|g\right|}-N(0)\ln(\frac{2\hbar\omega_{D}e^{\gamma}}{\pi k_{B}T})\right)\Delta^{*}(\tau,\mathbf{r})\Delta(\tau,\mathbf{r})-N(0)\xi^{2}(\nabla\cdot\Delta^{*}(\tau,\mathbf{r}))(\nabla\cdot\Delta(\tau,\mathbf{r}))\right]</math>.


Note that the last term in the expression tells us that <math> S_{eff} </math> would increase if gradient of <math> \Delta </math> is not zero.
Note that the last term in the expression tells us that <math> S_{eff} </math> would increase if gradient of <math> \Delta </math> is not zero.
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[[Image:Layout48.gif]]
[[Image:Layout48.gif]]


Refer to the fig, a thin shell of superconductor with radius R is shown and a small uniform magnetic field is passing through the hollow center of the cylinder. The experiment intends to show the variation of the critical temperature with change of the magnetic field passing through the hollow superconductor cylinder.  
As shown in the above figure, a thin shell of superconductor with radius R is placed in a small uniform magnetic field passing through the hollow center of the cylinder. The experiment intends to show the variation of the critical temperature with change of the magnetic field passing through the hollow superconductor cylinder. This was [http://prl.aps.org/abstract/PRL/v9/i1/p9_1| done in 1962], by Little and Parks at Stanford University.  


Before showing it, we first have to rewrite the Giznburg-Landau functional to make it taken the presence of magnetic field into account. Hamiltonian for a free electron moving in a magnetic field can be written as:
Before showing the mathematics behind the experiment, we first have to rewrite the Giznburg-Landau functional to take the presence of magnetic field into account. The Hamiltonian for a free electron moving in a magnetic field is written as:


<math> \frac{1}{2m}(p+\frac{eA}{c})^{2}\psi + V\psi = E\psi </math>
<math> \frac{1}{2m}(p+\frac{eA}{c})^{2}\psi + V\psi = E\psi </math>


The physical observable magnetic field B would remain the same if we choose a different vector potential
The physically observable magnetic field <math>\vec{B}</math> would remain the same if we choose a different vector potential
<math> A\rightarrow A+ \nabla \chi </math> (ie perform gauge transformation). To maintain the same eigen-energy E which is observable, the wave function have to undergo a phase change:
<math> A\rightarrow A+ \nabla \chi </math> (or perform a gauge transformation). To maintain the same eigen-energy E which is observable, the wave function has to undergo a phase change:
<math> \psi \rightarrow e^{i\phi}\psi </math> where <math> \phi=\frac{e}{c\hbar}\chi </math>
<math> \psi \rightarrow e^{i\phi}\psi </math> where <math> \phi=\frac{e}{c\hbar}\chi </math>


Now in our Hamiltonian, the wave function is arranged as   
Now in our effective action, the wave function appears as  
   
<math> \Delta^{*}(\tau,\vec{r})\psi_\uparrow (\tau,\vec{r})\psi_\downarrow (\tau,\vec{r}) + \Delta(\tau,\vec{r}) \psi_\downarrow^\dagger (\tau,\vec{r})\psi_\uparrow^\dagger (\tau,\vec{r}) </math>
<math> \Delta^{*}(\tau,\vec{r})\psi_\uparrow (\tau,\vec{r})\psi_\downarrow (\tau,\vec{r}) + \Delta(\tau,\vec{r}) \psi_\downarrow^\dagger (\tau,\vec{r})\psi_\uparrow^\dagger (\tau,\vec{r}) </math>


since <math> \psi \rightarrow e^{i\phi}\psi </math>, so if we want the Hamiltonian to remind the same, <math>\ \Delta </math> has to transform as <math> \Delta \rightarrow e^{-2i\phi}\Delta  </math>
since <math> \psi \rightarrow e^{i\phi}\psi </math>, if we want the effective action to remind the same, <math>\ \Delta </math> has to transform as <math> \Delta \rightarrow e^{-2i\phi}\Delta  </math>


Since <math>\ \Delta </math> corresponds to <math>\ \Psi </math> in the Giznburg-Landau functional, so the Giznburg-Landau functional is modified as  
Since <math>\ \Delta </math> corresponds to <math>\ \Psi </math> in the Giznburg-Landau functional, so the Giznburg-Landau functional is modified as  
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<math> F=\int d^{D}r\left[ \alpha (T-T_{c}) |\Psi(\vec{r})|^{2}+\frac{1}{2m^{*}}| ( \frac{\hbar \nabla}{i} - \frac{2e}{c}A(\vec{r}) ) \Psi(\vec{r})|^{2} \right] </math>
<math> F=\int d^{D}r\left[ \alpha (T-T_{c}) |\Psi(\vec{r})|^{2}+\frac{1}{2m^{*}}| ( \frac{\hbar \nabla}{i} - \frac{2e}{c}A(\vec{r}) ) \Psi(\vec{r})|^{2} \right] </math>


choose symmetric gauge:
Note that the last term looks like a particle of charge <math>-2e</math> in a magnetic field!
 
Now, we choose the symmetric gauge:
 
<math> \vec{A}=\frac{1}{2}\vec{H}\times\vec{r}=\frac{1}{2}Hr\hat{\phi} </math>
<math> \vec{A}=\frac{1}{2}\vec{H}\times\vec{r}=\frac{1}{2}Hr\hat{\phi} </math>


In cylindrical coordinate: <math> \vec{\nabla}=\hat{r}\frac{\partial}{\partial r} + \frac{\hat{\phi}}{r}\frac{\partial}{\partial \phi} + \hat{z}\frac{\partial}{\partial z} </math>
Re-write in cylindrical coordinates:  
 
<math> \vec{\nabla}=\hat{r}\frac{\partial}{\partial r} + \frac{\hat{\phi}}{r}\frac{\partial}{\partial \phi} + \hat{z}\frac{\partial}{\partial z} </math>


define unit flux as <math> \Phi_{0}=\frac{hc}{2e} </math>  
Define the unit flux as <math> \Phi_{0}=\frac{hc}{2e} </math>  


define fluxoid as <math> \Phi(R) = \pi HR^{2}\ </math>, so we have  
Define the fluxoid as <math> \Phi(R) = \pi HR^{2}\ </math>, so we have  


<math> \begin{align} F&=\int d^{D}r\left[ \alpha (T-T_{c})|\Psi(\vec{r})|^{2}  
<math> \begin{align} F&=\int d^{D}r\left[ \alpha (T-T_{c})|\Psi(\vec{r})|^{2}  
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When <math> \Phi = N\Phi_{0}\ </math>, the critical temperature will remain the same and the phase of <math> \Psi\ </math> is changed as <math> \Psi \rightarrow e^{iN\phi} \Psi </math>. When  <math> \Phi \neq N\Phi_{0}\ </math>, the critical temperature is found to vary as
When <math> \Phi = N\Phi_{0}\ </math>, the critical temperature will remain the same and the phase of <math> \Psi\ </math> is changed as <math> \Psi \rightarrow e^{iN\phi} \Psi </math>. When  <math> \Phi \neq N\Phi_{0}\ </math>, the critical temperature is found to vary as
<math> T_{c}^{new}=T_{c}- \frac{\hbar^{2}}{2m^{*}R^{2}\alpha}\left (N-\frac{\Phi}{\Phi_{0}}\right )^{2}</math>. See the fig.
<math> T_{c}^{new}=T_{c}- \frac{\hbar^{2}}{2m^{*}R^{2}\alpha}\left (N-\frac{\Phi}{\Phi_{0}}\right )^{2}</math>. See the figure below.


[[Image:Graph49.gif]]
[[Image:Graph49.gif]]


== Microscopic derivation of the Giznburg-Landau functional ==
This experiment is significant evidence that the objects represented by <math>\Delta</math> carry a charge twice that of a single electron. As we know that this thing stands in the equations for a cooper pair, this makes perfect sense!
 
== Microscopic derivation of the Ginzburg-Landau functional ==
 
 
Let us consider the model of a metal close to the transition to the superconducting state. A complete description of its thermodynamic properties can be done through the calculation of the partition function.
 
=== Ginzburg-Landau Functional ===
 
The classical part of the Hamiltonian in the partition function, dependent of bosonic fields, may be chosen in the spirit of the Landauer theory of phase transition. However, in view of the space dependence of wave functions, Ginzberg and Landauer included in it additionally the first non vanishing term of the expansion over the gradient of the fluctuation field. Symmetry analysis shows that it should be quadratic. The weakness of the field coordinate dependence permits to omit the high-order terms of such an expansion. Therefore, the classical part of the Hamiltonian of a metal close to a superconducting transition related to the presence of the fluctuation Cooper pairs in it (the so called Ginzberg-Landauer functional)can be written as 
 
<math>F[\psi(r)]=F_{n}+\int dV\{a\mid\psi(r)\mid^{2}+\frac{b}{2}\mid\psi(r)\mid^{4}+\frac{1}{4m}\mid\nabla\psi(r)\mid^{2}\}</math>
 
The basic postulate of G-L is that if <math>\psi</math> is small and varies slowly in space, so we can expand the free energy in a series of <math>\mid\psi(r)\mid^{2}</math>.These two terms should be adequate so long as one stays near the second-order phase transition at <math>T_{c}</math>, where the order parameter <math>\mid\psi(r)\mid^{2}\rightarrow0</math>. Inspection of G-L functional shows that <math>b</math> must be positive if the theory is to be useful; otherwise the lowest free energy would occur for arbitrarily large values of <math>\mid\psi(r)\mid^{2}</math>, where the expansion is surely inadequate.
 
We already got the quadratic terms in the Ginzberg-Landauer by expanding <math><e^{-S_{int}}></math> to the second order, and we are going to go the higher order. As we discussed, we expect that this term will be a negative value to keep <math>S_{eff}</math> as a negative value under <math>T_{c}</math>. To catch this goal we start with the partition function:


<math>Z=Z_{0}< e^{-S_{int}} ></math>
<math>Z=Z_{0}< e^{-S_{int}} ></math>
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<math>Z_{0}=\int D\psi ^{*} D\psi D\Delta ^{*} D\Delta  e^{-(S_{\Delta} +S_{0})}</math>
<math>Z_{0}=\int D\psi ^{*} D\psi D\Delta ^{*} D\Delta  e^{-(S_{\Delta} +S_{0})}</math>


we can expand this average for small<math>\Delta</math> near<math>T_{c}</math>, for this perpose we can assume  asecond order phase transition  
we can expand this average for small <math>\Delta</math> near <math>T_{c}</math>, for this perpose we can assume  a second order phase transition so that it increases continiously from zero to finite number after <math>T_{c}</math>.
so that it increases continiously from zero to finite number after <math>T_{c}</math>


we need to calculate the average of <math>e^{-s_{int}}</math>which can be calculated by Tylor expansion:
we need to calculate the average of <math>e^{-s_{int}}</math>which can be calculated by Tylor expansion:


<math>e^{-S_{int}}=<-S_{int}+\frac{1}{2}S_{int}^{2}-\frac{1}{3}S_{int}^{3}+\frac{1}{4!}S_{int}^{4}+...></math>
<math>e^{-S_{int}}=<1-S_{int}+\frac{1}{2}S_{int}^{2}-\frac{1}{3}S_{int}^{3}+\frac{1}{4!}S_{int}^{4}+...></math>
 
=<math>1-<S_{int}>+\frac{1}{2} < S_{int}^{2}> -\frac{1}{3!}< S_{int}^{2}> +\frac{1}{4!}< S_{int}^{4}> +...</math>


.......................the odd power terms are zero because <math><\psi _{\uparrow}(r,\tau )\psi _{\downarrow}(r,\tau ) > =0 </math>
=<math>1-<S_{int}>+\frac{1}{2} < S_{int}^{2}> -\frac{1}{3!}< S_{int}^{3}> +\frac{1}{4!}< S_{int}^{4}> +...</math>


<math>=e^{\frac{1}{2}< S_{int}^{2}>}e^{\frac{1}{4!}< S_{int}^{4}>-\lambda }</math>
In this expansion all odd power terms go to zero because when we calculate the integral over <math>\psi</math> and its complex conjugate it will vanish since we have odd numbers of <math>\psi</math> or <math>\psi^{*}</math> in the integral. Then, we try to approximate it to
<math>e^{\frac{1}{2}< S_{int}^{2}>}e^{\frac{1}{4!}< S_{int}^{4}>-\lambda }</math> .


if we expand these two terms in to the second order the following expression can be got:
if we expand these two terms in to the second order the following expression can be got:
Line 277: Line 302:
<math>=1+\frac{1}{2} < S_{int}^{2}>+\frac{1}{8}(< S_{int}^{2}>)^{2} +...)+\frac{1}{4!}< S_{int}^{4}>-\lambda +...</math>
<math>=1+\frac{1}{2} < S_{int}^{2}>+\frac{1}{8}(< S_{int}^{2}>)^{2} +...)+\frac{1}{4!}< S_{int}^{4}>-\lambda +...</math>


<math>\lambda</math> can be choosed in such a way .......
<math>\lambda</math> can be chose in a such way that this result will match to the original expansion of <math>e^{-s_{int}}</math> up to the forth power of <math>s_{int}</math> :


so,
so,
Line 289: Line 314:
<math>S_{int}=\frac{L^{D}}{\beta ^{2}}\sum_{\omega _{n},\Omega _{n}}\sum _{k,q}[\Delta ^{*}_{q}(i\Omega _{n})\psi_{\downarrow}(i\Omega _{n}-i\omega _{n}),\vec{q}-\vec{k})\psi_{\uparrow}(i\omega _{n},k)+\Delta _{q}(i\Omega_{n})\psi_{\uparrow}^{\dagger }(i\omega _{n},k)\psi_{\downarrow}^{\dagger }(i\Omega _{n}-i\omega _{n}),\vec{q}-\vec{k})]</math>
<math>S_{int}=\frac{L^{D}}{\beta ^{2}}\sum_{\omega _{n},\Omega _{n}}\sum _{k,q}[\Delta ^{*}_{q}(i\Omega _{n})\psi_{\downarrow}(i\Omega _{n}-i\omega _{n}),\vec{q}-\vec{k})\psi_{\uparrow}(i\omega _{n},k)+\Delta _{q}(i\Omega_{n})\psi_{\uparrow}^{\dagger }(i\omega _{n},k)\psi_{\downarrow}^{\dagger }(i\Omega _{n}-i\omega _{n}),\vec{q}-\vec{k})]</math>


let's write <math>S_{int}</math> in terems od <math>a</math> for simplification. where
let's write <math>S_{int}</math> in terms of <math>a</math> for simplification. where


<math>a=\int \Delta ^{*}(1)\psi_{\downarrow}(1) \psi_{\uparrow}(1)+\Delta (1)\psi_{\downarrow}^{*}(1) \psi_{\uparrow}^{*}(1)</math>
<math>a=\int \Delta ^{*}(1)\psi_{\downarrow}(1) \psi_{\uparrow}(1)+\Delta (1)\psi_{\downarrow}^{*}(1) \psi_{\uparrow}^{*}(1)</math>


<math>a_{1}</math> is a couple grassman number, so we do not need to be worry about the sign when these terms comute with other terms.
<math>a_{1}</math> is a couple Grassman number, so we do not need to be worry about the sign when these terms commute with other terms.


<math>< S_{int}^{4}> =\int_{1234} < (a_{1}^{*}+a_{1})(a_{2}^{*}+a_{2})(a_{3}^{*}+a_{3})(a_{4}^{*}+a_{4})></math>  
<math>< S_{int}^{4}> =\int_{1234} < (a_{1}^{*}+a_{1})(a_{2}^{*}+a_{2})(a_{3}^{*}+a_{3})(a_{4}^{*}+a_{4})></math>  
Line 301: Line 326:
<math>=6< a_{1}^{*}a_{2}^{*}a_{3}a_{4}>=6\int _{1234}\Delta ^{*}(1)\Delta ^{*}(2)\Delta(3)\Delta(4)< \psi_{\downarrow}(1)\psi_{\uparrow}(1)\psi_{\downarrow}(2)\psi_{\uparrow}(2)\psi_{\downarrow}^{*}(3)\psi_{\uparrow}^{*}(3)\psi_{\downarrow}^{*}(4)\psi_{\uparrow}^{*}(4)> </math>
<math>=6< a_{1}^{*}a_{2}^{*}a_{3}a_{4}>=6\int _{1234}\Delta ^{*}(1)\Delta ^{*}(2)\Delta(3)\Delta(4)< \psi_{\downarrow}(1)\psi_{\uparrow}(1)\psi_{\downarrow}(2)\psi_{\uparrow}(2)\psi_{\downarrow}^{*}(3)\psi_{\uparrow}^{*}(3)\psi_{\downarrow}^{*}(4)\psi_{\uparrow}^{*}(4)> </math>


<math>3< S_{int}^{2}> ^{2}=3\int_{1,2}2< a_{1}^{*}a_{2}> \int_{3,4}2< a_{3}^{*}a_{4}>=12\int_{1,2,3,4}\Delta ^{*}(1)\Delta ^{*}(2)\Delta(3)\Delta(4)< \psi_{\downarrow}(1)\psi_{\uparrow}(1)\psi_{\uparrow}^{*}(3)\psi_{\downarrow}^{*}(3)>< \psi_{\downarrow}(2)\psi_{\uparrow}(2)\psi_{\uparrow}^{*}(4)\psi_{\downarrow}^{*}(4)></math>  
By Feynman diagram, <math>< S_{int}^{4}></math> has:
 
[[Image:Secondorder.png]]
 
 
[[Image:Square.png]]


<math>3< S_{int}^{2}> ^{2}=3\int_{1,2}2< a_{1}^{*}a_{2}> \int_{3,4}2< a_{3}^{*}a_{4}>=12\int_{1,2,3,4}\Delta ^{*}(1)\Delta ^{*}(2)\Delta(3)\Delta(4)< \psi_{\downarrow}(1)\psi_{\uparrow}(1)\psi_{\uparrow}^{*}(3)\times</math>


<math>\psi_{\downarrow}^{*}(3)>< \psi_{\downarrow}(2)\psi_{\uparrow}(2)\psi_{\uparrow}^{*}(4)\psi_{\downarrow}^{*}(4)></math>


<math>-2G(2-3) G(2-4)G(1-4)G(1-3)=-12\int_{1,2,3,4}\Delta ^{*}(1)\Delta _{*}(2)\Delta (3)\Delta (4)G(2-3)G(2-4)G(1-4)G(1-3)</math>
<math>-2G(2-3) G(2-4)G(1-4)G(1-3)=-12\int_{1,2,3,4}\Delta ^{*}(1)\Delta _{*}(2)\Delta (3)\Delta (4)G(2-3)G(2-4)G(1-4)G(1-3)</math>


Recall  
Recall the Fourier transform of one body Green function is:


<math>G(2-3)=< \psi (r_{2},\tau _{2})\psi ^{*}(r_{3},\tau _{3})>=\frac{1}{\beta } \sum_{\omega _{n}}\frac{1}{L^{D}}\sum_{k}e^{-i\omega _{n}(\tau _{2}-\tau _{3})}e^{ik.(r_{2}-r_{3})}\frac{1}{-i\omega _{n}+\epsilon _{k}-\mu }</math>
<math>G(2-3)=< \psi (r_{2},\tau _{2})\psi ^{*}(r_{3},\tau _{3})>=\frac{1}{\beta } \sum_{\omega _{n}}\frac{1}{L^{D}}\sum_{k}e^{-i\omega _{n}(\tau _{2}-\tau _{3})}e^{ik.(r_{2}-r_{3})}\frac{1}{-i\omega _{n}+\epsilon _{k}-\mu }</math>


see the solution which are <math>\tau</math>independent
To seek solution of which are <math>\tau</math> independent using Feynman diagram
 
<math>\frac{1}{\beta ^{4}}\sum_{\omega _{{n}_{1}}}...\sum_{\omega _{{n}_{4}}}\int_{0}^\beta {d\tau_{1}}\int_{0}^\beta {d\tau_{2}} \int_{0}^\beta {d\tau_{3}} \int_{0}^\beta {d\tau_{4}}e^{-i\omega _{{n}_{1}}(\tau _{1}-\tau _{3})} e^{-i\omega _{{n}_{2}}(\tau _{1}-\tau _{4})}e^{-i\omega _{{n}_{3}}(\tau _{2}-\tau _{3})}e^{-i\omega _{{n}_{4}}(\tau _{2}-\tau _{4})} G(i\omega _{{n}_{1}},r_{1}-r_{3})\times</math>


<math>\frac{1}{\beta ^{4}}\sum_{\omega _{{n}_{1}}}...\sum_{\omega _{{n}_{4}}}\int_{0}^\beta {d\tau_{1}}\int_{0}^\beta {d\tau_{2}} \int_{0}^\beta {d\tau_{3}} \int_{0}^\beta {d\tau_{4}}e^{-i\omega _{{n}_{1}}(\tau _{1}-\tau _{3})} e^{-i\omega _{{n}_{2}}(\tau _{1}-\tau _{4})}e^{-i\omega _{{n}_{3}}(\tau _{2}-\tau _{3})}e^{-i\omega _{{n}_{4}}(\tau _{2}-\tau _{4})} G(i\omega _{{n}_{1}},r_{1}-r_{3})G(i\omega _{{n}_{2}},r_{1}-r_{4})G(i\omega _{{n}_{3}},r_{2}-r_{3})G(i\omega _{{n}_{4}},r_{2}-r_{4})  
<math>G(i\omega _{{n}_{2}},r_{1}-r_{4})G(i\omega _{{n}_{3}},r_{2}-r_{3})G(i\omega _{{n}_{4}},r_{2}-r_{4})</math>  
</math>


after getting integration over <math>\tau_{1}</math> we will get <math>\beta \delta (\omega _{n_{1}},-\omega _{n_{2}}) </math> and similarly by getting integration over <math>\tau_{2}</math> we have <math>\beta \delta (\omega _{n_{3}},-\omega _{n_{4}}) </math>
after getting integration over <math>\tau_{1}</math> we will get <math>\beta \delta (\omega _{n_{1}},-\omega _{n_{2}}) </math> and similarly by getting integration over <math>\tau_{2}</math> we have <math>\beta \delta (\omega _{n_{3}},-\omega _{n_{4}}) </math>
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<math>r_{2}-r_{4}=\frac{1}{2}(\mu _{1,4}+\mu_{2,3})-\mu</math>
<math>r_{2}-r_{4}=\frac{1}{2}(\mu _{1,4}+\mu_{2,3})-\mu</math>


<math>\simeq -12\int d^{D}R d^{D}\mu d^{D}\mu_{1.4}d^{D}\mu_{2,3}\Delta ^{\ast }(R)\Delta^{\ast } (R)\Delta (R)\Delta (R)\sum_{\omega _n}G(i\omega _{n},\mu +\frac{1}{2}(\mu_{1,4}+\mu _{2,3}))G(-i\omega _{n},\mu _{1,4})\times</math>
<math>G(-i\omega _{n},\mu _{2,3})G(i\omega _{n},\mu +\frac{1}{2}(\mu_{1,4}+\mu _{2,3}))</math>
integrate over<math>\mu</math> gives us:  <math>L^{D}</math>        <math>\delta_{k_{1},k_{4}}</math>
and similarly <math>\mu_{1,4}</math>:  <math>L^{D}</math>        <math>\delta_{k_{1},-k_{2}}</math>
<math>\mu_{2,3}</math>  <math>l^{D}</math>        <math>\delta_{k_{1},-k_{3}}</math>
<math>-12\int d^{D}R\left | \Delta (R)^{4} \right |\sum_{\omega _{n}}\frac{1}{L^{D}}\sum_{k}\frac{1}{(-i\omega _{n}+\varepsilon_{k}-\mu)^{2}}\frac{1}{(i\omega _{n}+\varepsilon_{-k}+\mu)^{2}}</math>
<math>\epsilon _{k}=\epsilon _{-k}</math> because of timereversal symetry.
<math>\epsilon _{k}-\mu =\xi </math>
[[Image:Contour-yp.png]]
<math>-12\int d^{D}R\left | \Delta (R)\right |^{4}\sum_{\omega_{n}}\int_{-\infty }^{+\infty
}d\xi N(\xi +\mu )\frac{1}{(-i\omega _{n}+\xi )^{2}}\frac{1}{(i\omega _{n}+\xi )^{2}} </math>
<math>I(p,q)=\int_{-\infty }^{+\infty }d\xi \frac{1}{(\xi -p)^{2}}\frac{1}{(\xi +p)^{2}}={\frac{\partial }{\partial p}}\int d\xi \frac{1}{\xi -p}\frac{1}{(\xi +p)^{2}}
</math>
<math>\frac{\partial }{\partial p}2\pi i \frac{1}{(p+q)^{2}}=2\pi i\frac{-2}{(p+q)^{2}}</math>
<math>\simeq -12\int d^{D}R\left | \Delta (R) \right |^{4}\sum_{\omega _{n}}N(0)\frac{-4\pi i}{(2i\omega _{n})^{3}} </math>
<math>\simeq-12\beta\int d^{D}R\mid\triangle(R)\mid^{4}\frac{N(0)}{2\beta}\pi\frac{\beta^{3}}{\pi^{3}}\sum_{n=-\infty}^{\infty}\frac{1}{\mid2n+1\mid^{3}}</math>
We can substitute the sum using 
<math>\sum_{n=-\infty}^{\infty}\frac{1}{\mid2n+1\mid^{3}}=2\times\frac{7\xi(3)}{8}</math>
<math>=-\frac{1}{2}\frac{7\xi(3)}{8\pi^{2}}\beta^{2}N(0)\frac{1}{K_{B}T}\int d^{D}R\mid\triangle(R)\mid^{4}</math>


So, we get a term that is always negative and it corresponds to the  <math>\mid\psi(r)\mid^{4}</math> in the Ginzberg-Landauer functional.
=== Saddle Point Approximation ===




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<math>S_{eff}=\frac{1}{k_{B}T}\sum_{r}\left[\underset{A}{\underbrace{\left(\frac{1}{|g|}-N(0)In\left[\frac{2\hbar\omega_{D}e^{\gamma_{E}}}{\pi k_{B}T}\right]\right)}}|\Delta(r)|^{2}+N(0)\xi^{2}(\nabla\Delta*(r)).(\nabla\Delta(r))+\frac{1}{2}\underset{B}{\underbrace{\frac{7\zeta(3)N(0)}{8\pi^{2}k_{B}^{2}T^{2}}}}|\Delta(r)|^{4}\right]</math>
<math>S_{eff}=\frac{1}{k_{B}T}\sum_{r}\left[\underset{A}{\underbrace{\left(\frac{1}{|g|}-N(0)In\left[\frac{2\hbar\omega_{D}e^{\gamma_{E}}}{\pi k_{B}T}\right]\right)}}|\Delta(r)|^{2}+N(0)\xi^{2}(\nabla\Delta*(r)).(\nabla\Delta(r))+\frac{1}{2}\underset{B}{\underbrace{\frac{7\zeta(3)N(0)}{8\pi^{2}k_{B}^{2}T^{2}}}}|\Delta(r)|^{4}\right]</math>
We can use this expression to make quantitative experimental predictions. The path integral over <math>\Delta</math> is still imposible to carry out exactly, despite our approximations for <math>S_{eff}</math>, because <math>S_{eff}</math>contains quartic terms in <math>\Delta</math> and so we are not dealing with a Gaussian integral. The approximation strategy whic we will pursue is called saddle point approxiation, which in our contetxt means that we will expand teh integrand about a solution which minimizes <nowiki>S_{eff}</nowiki> with respect to <math>\Delta</math>. What we end up doing is replacing Z with
<math>Z\sim Z_{0}e^{-S_{eff}[\Delta_{min}]}</math>, where <math>\Delta_{min}</math> is determined from<math> \left|\frac{\delta S_{eff}}{\delta\Delta^{*}}\right|_{\Delta=\Delta_{min}}=0</math>
At this point, let's seek uniform solutions to their equations, in whcih case we can drop the gradient terms in <math>S_{eff}</math> :
<math>\frac{\delta S_{eff}}{\delta\Delta^{*}}=\frac{L^{D}}{k_{B}T}\left(A\Delta+B\Delta\Delta^{*}\Delta\right)</math>
where:
<math>A=\frac{1}{|g|}-N(0)ln\left(\frac{\hbar\omega_{D}}{k_{B}T}\frac{2e\gamma}{\pi}\right)</math> and
<math>B=\frac{7\zeta(3)}{8\pi}\frac{N(0)}{k_{B}^{2}T^{2}}</math>
Note that for :
<math>T>T_{C}\qquad A>0,B>0</math>
and
<math>T<T_{C}\qquad A<0,B>0</math>
So  <math>A\Delta_{min}+B\Delta_{min}^{3}=0</math>,
<math>\Delta_{min}=0, \;T>T_{C}</math>
and
 
<math>\Delta_{min}=\sqrt{\frac{-A}{B}},\; T<T_{C} </math>.
<math>S_{eff}[\Delta_{min}]=\frac{L^{D}}{k_{B}T}\left(A\Delta_{min}^{2}+\frac{1}{2}B\Delta_{min}^{4}\right)</math>
<math>T>T_{C};\quad S_{eff}[\Delta_{min}]=0</math>,
<math>T<T_{C};\quad S_{eff}[\Delta_{min}]=\left(A\frac{-A}{B}+\frac{1}{2}B\frac{A^{2}}{B^{2}}\right)\frac{L^{D}}{k_{B}T}=\frac{-A^{2}}{2B}\frac{L^{D}}{k_{B}T}</math>
Since, we now have the approximate expression for the partition function we can calculate thermodynamic physical properties. the one we will focus on is the specific heat. Recall that,
<math>Z=e^{-\beta F}=\sum_{n}e^{-\beta E_{n}}\Longrightarrow<E>=\frac{1}{Z}\sum_{n}E_{n}e^{-\beta E_{n}}=-\frac{\partial}{\partial\beta}lnZ =-\frac{\partial}{\partial\beta}lne^{-\beta F}=\frac{\partial}{\partial\beta}\left(\beta F\right)=F+\beta\frac{\partial F}{\partial\beta}</math>
<math>=\frac{\partial F}{\partial T}+\frac{\partial\beta}{\partial T}\frac{\partial F}{\partial\beta}+\beta\frac{\partial}{\partial T}\frac{\partial F}{\partial\beta}</math>
<math>=2\frac{\partial F}{\partial T}+\beta\frac{\partial}{\partial T}\left(\frac{\partial F}{\partial T}\frac{\partial T}{\partial\beta}\right)</math>
<math>=2\frac{\partial F}{\partial T}+\beta\frac{\partial^{2}F}{\partial T^{2}}\left(-k_{B}T^{2}\right)+\beta\frac{\partial}{\partial T}\left(-2k_{B}T\right)</math>
<math>=-T\frac{\partial^{2}F}{\partial T^{2}}</math>
if we only study the constribution to <math>c_{V}</math> from the superconducting order parameter terms in <math>S_{eff}</math>, we have
<math>c_{V}</math>
So, we see that if the double derivateive of
<math>\frac{-A^{2}}{2B}</math>
with respect to <math>T</math> is finite at <math>T_{C}</math>, then the specific heat jumps at
<math>T_{C}</math>,
since
<math>c_{V}=0</math>
for <math>T>T_{C}</math>.
We are interested in the size of this jump. Therefore, we need to simply expand
<math>\frac{-A^{2}}{2B}</math> near <math>T_{C}</math>.
Since <math>A</math> vanishes at <math>T_{C}</math>, we can simply evaluate
<math>B</math> at <math>T_{C}</math> and expand <math>A</math>:
<math>A(T)=\frac{1}{|g|}-N(0)ln\left(\frac{\hbar\omega_{D}}{k_{B}\left(T_{C}+T-T_{C}\right)}\frac{2e^{\gamma}}{\pi}\right)</math>
<math>=\frac{1}{|g|}-N(0)ln\left(\frac{2e^{\gamma}}{\pi}\frac{\hbar\omega_{D}}{k_{B}T_{C}}\left(1+\frac{T-T_{C}}{T_{C}}\right)^{-1}\right)</math>
<math>=\underset{vanishes\; by\; def.\; of\; T_{C}}{\underbrace{\frac{1}{|g|}-N(0)ln\left(\frac{2e^{\gamma}}{\pi}\frac{\hbar\omega_{D}}{k_{B}T_{C}}\right)}}by+N(0)ln\left(1+\frac{T-T_{C}}{T_{C}}\right)</math>
<math>\Rightarrow A(T)=N(0)ln\left(1+\frac{T-T_{C}}{T_{C}}\right)\simeq N(0)\frac{T-T_{C}}{T_{C}}+...</math>
<math>\Rightarrow\frac{-A^{2}(T)}{2B(T)}\simeq-\frac{1}{2}\frac{N^{2}(0)\left(\frac{T-T_{C}}{T_{C}}\right)^{2}}{\frac{7\zeta(3)}{8\pi}\frac{N(0)}{k_{B}^{2}T^{2}}}+...</math>
<math>\Delta c_{V}\simeq-\frac{T_{C}}{2}\frac{8\pi^{2}}{7\zeta(3)}k_{B}^{2}N(0)+...</math>
What is the specific heat of a non-interacting electron gas?
<math>c_{V}^{(n)}=\frac{\partial}{\partial T}\left(2(from\; spin)\int\frac{d^{D}k}{(2\pi)^{D}}\frac{\left(\epsilon_{k}-\mu\right)}{e^{\frac{\epsilon_{k}-\mu}{k_{B}T}+1}}\right)</math>
<math>=2\int\frac{d^{D}k}{(2\pi)^{D}}\frac{-\left(\epsilon_{k}-\mu\right)}{\left(e^{\frac{\epsilon_{k}-\mu}{k_{B}T}+1}\right)^{2}}\left(\frac{-\left(\epsilon_{k}-\mu\right)}{k_{B}T^{2}}\right)e^{\left(\frac{\epsilon_{k}-\mu}{k_{B}T}\right)}</math> <math>\simeq2k_{B}N(0)\int_{-\infty}^{\infty}d\xi\left(\frac{\xi}{2k_{B}T}\right)^{2}\frac{1}{cosh^{2}\left(\frac{\xi}{2k_{B}T}\right)}</math>
<math>\simeq4k_{B}^{2}TN(0)\underset{\frac{\pi^{2}}{6}}{\underbrace{\int_{-\infty}^{\infty}dx\frac{x^{2}}{cosh^{2}x}}}=\frac{2\pi^{2}}{3}k_{B}^{2}T</math>
So, if we measure the jump in the specific heat at T_c in the units of the normal state electronic contribution we find:
<math>
\frac{\Delta c_{V}}{c_{V}^{(n)}}=\frac{\frac{8\pi^{2}}{7\zeta(3)}k_{B}^{2}T_{C}N(0)}{\frac{2\pi^{2}}{3}k_{B}^{2}T_{C}N(0)}</math>
<math>=\frac{12}{7\zeta(3)}\simeq1.426</math> This is dimensionless number is a “famous” prediction of the BCS theory, although we derived it using different formalism. Let's check it with experiment:
First the '''caveats''':
''when specific is measured, all excitations contribute. Most importantly lattice vibrations (phonons) contribute as well. At low T, however, the phonon contribution drops of as <math>T^3</math> and we can neglect it if the <math>T_C</math> is sufficiently low''. In practice we have do an example:
{| border="1" align="center" style="text-align:center;"
|materials
|<math>T_C</math>
|phonon contribution at <math>T_C</math>
|-
|Al
|1.2K
|1%
|-
|Zn
|0.8K
|3%
|-
|Cd
|0.5K
|3%
|-
|Sn
|3.7K
|45%
|-
|In
|3.4K
|77%
|-
|Th
|2.4K
|83%
|-
|Pb
|7.2K
|94%
|}
Experimental data for Aluminum gives
<math>\frac{\Delta c_{V}}{c_{V}^{(n)}}\simeq 1.39</math>
This result from the saddle point approximation/mean field is very close to the experimental result. But for high Tc superconductor the behavior is very different.  The heat capacity of superconductor decrease gradually, doesn't have sharp jump like in the picture below.  The Gaussian approximation we have used is not enough to explain high Tc superconductor.
=== Limitation of Microscopic derivation of the Ginzburg-Landau functional  ===
[[image:Cv.png|thumb|400px|middle|Experimental result of heat capacity of aluminum (dotted lines), at zero magnetic field(curve peak) and at 300 Gauss when superconductivity suppressed (no peak) ,  typical curve for high Tc superconductor (additional solid blue line), (Phys. Rev., 114:676–685, 1959)]]
1)  Fluctuation is important in low dimension  (such in 2D by including Gaussian fluctuation)
2)  In 1D, beyond mean field approach is needed
3)  Mean Field give jump in second order of the derivative
4)  We need to include fluctuation in High Tc superconductor
5)  Gaussian approximation is not enough in High Tc superconductor


== Effects of an applied magnetic field; Type I and Type II superconductivity ==
== Effects of an applied magnetic field; Type I and Type II superconductivity ==
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We can now immediately write down the variational derivative, which, upon being set to zero, gives us the first GL equation,
We can now immediately write down the variational derivative, which, upon being set to zero, gives us the first GL equation,


<math>\frac{1}{2m}\left (\frac{\hbar}{i}\nabla-\frac{2e}{c}\vec{A}\right )^2\Psi+\alpha(T-T_c)\Psi+b|\Psi|^2\Psi=0.</math>
<math>\frac{1}{2m}\left (\frac{\hbar}{i}\nabla-\frac{2e}{c}\vec{A}\right )^2\Psi+\alpha(T-T_c)\Psi+b|\Psi|^2\Psi=0.</math>


We also need to minimize the free energy with respect to the magnetic field.  We have already done this for the normal case, and there is only one more term that we need to consider in the superconducting case; we will therefore only treat this term.  We can quickly write down the variation in the superconducting part of the free energy <math>F_{SC}</math>, which is
We also need to minimize the free energy with respect to the magnetic field.  We have already done this for the normal case, and there is only one more term that we need to consider in the superconducting case; we will therefore only treat this term.  We can quickly write down the variation in the superconducting part of the free energy <math>F_{SC}</math>, which is
Line 425: Line 632:
or, introducing <math>\vec{B}</math> and <math>\vec{H}</math>,
or, introducing <math>\vec{B}</math> and <math>\vec{H}</math>,


<math>\frac{e}{m}\left (\Psi^{*}\frac{\hbar}{i}\nabla\Psi-\Psi\frac{\hbar}{i}\nabla\Psi^{*}\right )-\frac{4e^2}{mc}|\Psi|^2\vec{A}=\frac{c}{4\pi}\nabla\times(\vec{B}-\vec{H}).</math>
<math>\frac{e}{m}\left (\Psi^{*}\frac{\hbar}{i}\nabla\Psi-\Psi\frac{\hbar}{i}\nabla\Psi^{*}\right )-\frac{4e^2}{mc}|\Psi|^2\vec{A}=\frac{c}{4\pi}\nabla\times(\vec{B}-\vec{H}).</math>


Given the definition of <math>\vec{H}</math> and the Maxwell equation (assuming static fields),
Given the definition of <math>\vec{H}</math> and the Maxwell equation (assuming static fields),
Line 455: Line 662:
We will find it convenient to introduce dimensionless variables when working with the GL equations.  We start by introducing a dimensionless order parameter, <math>\psi=\frac{\Psi}{\Psi_0}</math>, where
We will find it convenient to introduce dimensionless variables when working with the GL equations.  We start by introducing a dimensionless order parameter, <math>\psi=\frac{\Psi}{\Psi_0}</math>, where


<math>\Psi_0^2=\frac{\alpha(T_C-T)}{b}.</math>
<math>\Psi_0^2=\frac{\alpha(T_c-T)}{b}.</math>


We may rewrite the first GL equation in terms of this parameter as
We may rewrite the first GL equation in terms of this parameter as
Line 475: Line 682:
where <math>\lambda=\sqrt{\frac{mc^2}{16\pi e^2\Psi_0^2}}</math> is known as the penetration depth of the superconductor; we will see where this name comes from shortly.  In terms of this vector, the first GL equation becomes
where <math>\lambda=\sqrt{\frac{mc^2}{16\pi e^2\Psi_0^2}}</math> is known as the penetration depth of the superconductor; we will see where this name comes from shortly.  In terms of this vector, the first GL equation becomes


<math>\left (\frac{1}{\Psi_0\lambda\sqrt{2mb}}\tilde{\nabla}-\frac{1}{\Psi_0\sqrt{2mb}}\frac{2e}{c}\vec{A}\right )^2\psi+(\left |\psi\right |^2-1)\psi=0</math>
<math>\left (\frac{1}{\Psi_0\lambda\sqrt{2mb}}\frac{\hbar}{i}\tilde{\nabla}-\frac{1}{\Psi_0\sqrt{2mb}}\frac{2e}{c}\vec{A}\right )^2\psi+(\left |\psi\right |^2-1)\psi=0</math>


and the second becomes
and the second becomes
Line 491: Line 698:
In terms of these, the first GL equation becomes
In terms of these, the first GL equation becomes


<math>\left (-\frac{i}{\kappa}\tilde{\nabla}-\tilde{A}\right )^2\psi+(\left |\psi\right |^2-1)\psi=0</math>
<math>\left (-\frac{i}{\kappa}\tilde{\nabla}-\tilde{A}\right )^2\psi+(\left |\psi\right |^2-1)\psi=0</math>


and the second becomes
and the second becomes


<math>\frac{1}{2\kappa}\left (\psi^{\ast}\frac{\tilde{\nabla}}{i}\psi-\psi\frac{\tilde{\nabla}}{i}\psi^{\ast}\right )-\left |\psi\right |^2\vec{A}=\tilde{\nabla}\times[\tilde{\nabla}\times(\tilde{A}-\tilde{A}_0)].</math>
<math>\frac{1}{2\kappa}\left (\psi^{\ast}\frac{\tilde{\nabla}}{i}\psi-\psi\frac{\tilde{\nabla}}{i}\psi^{\ast}\right )-\left |\psi\right |^2\vec{A}=\tilde{\nabla}\times[\tilde{\nabla}\times(\tilde{A}-\tilde{A}_0)].</math>
 
We see that our theory has a dimensionless parameter in it, namely <math>\kappa</math>, which is known as the Ginzburg-Landau parameter.  We may write this parameter as
 
<math>\kappa=\frac{\lambda\sqrt{2m\alpha(T_c-T)}}{\hbar}=\frac{\lambda}{\xi_\text{GL}},</math>
 
where
 
<math>\xi_\text{GL}=\frac{\hbar}{2m\alpha(T_c-T)}</math>
 
is the GL coherence length.  This tells us that <math>\kappa</math> is the ratio of two length scales associated with the superconductor, namely the scale over which the order parameter "heals" (the coherence length <math>\xi_\text{GL}</math>) and that over which the magnetic field dies out (the penetration depth <math>\lambda</math>, as we will demonstrate shortly).  It also turns out that this parameter decides what type of superconductor we are dealing with.  If <math>\kappa < \tfrac{1}{\sqrt{2}}</math>, then we have a Type I superconductor, while, if <math>\kappa > \tfrac{1}{\sqrt{2}}</math>, then we have a Type II superconductor.
 
We may now find the value of this parameter in the microscopic model we considered earlier.  In that case, we found that
 
<math>\frac{\hbar^2}{2m}=N(\mu)\xi^2=N(\mu)\frac{7\zeta(3)}{16\pi^2 d}\ell_T^2,</math>
 
where <math>N(\mu)</math> is the density of states at the Fermi level, <math>\xi</math> is the coherence length, <math>d</math> is the number of dimensions that we are working in, and <math>l_T</math> is the thermal wavelength.  We will state the result for <math>d=3</math>.  Given that
 
<math>\ell_T=\frac{\hbar v_F}{k_B T}</math>
 
and that, in this case,
 
<math>N(\mu)=\frac{1}{2\pi^2}\frac{mk_F}{\hbar^2},</math>
 
we find that
 
<math>\kappa=\sqrt{\frac{18\pi^3}{7\zeta(3)}}\frac{k_B T_c}{\sqrt{mc^2}\sqrt{e^2 k_F}}\left (\frac{c}{v_F}\right )^2.</math>
 
Note that we set <math>T=T_c</math> in the expression for <math>l_T</math>; this is because the GL theory is only valid just below the transition temperature.  We may also express this in terms of the Fermi energy,
 
<math>E_F=\frac{p_F^2}{2m}=\tfrac{1}{2}p_F v_F.</math>
 
Doing so, we obtain
 
<math>\kappa=\sqrt{\frac{18\pi^3}{7\zeta(3)}}\frac{k_B T_c}{\sqrt{mc^2}\sqrt{\alpha}\sqrt{2E_F}}\left (\frac{c}{v_F}\right )^{3/2}=8\times 10^{-6}\cdot\frac{T_c\,[\text{K}]}{\sqrt{E_F\,[\text{eV}]}}\left (\frac{c}{v_F}\right )^{3/2}.</math>
 
In a typical metal, <math>v_F\approx 10^5 - 10^6\,\tfrac{\text{m}}{\text{s}},</math> so
 
<math>\kappa=(0.04 - 1.3)\cdot\frac{T_c\,[\text{K}]}{\sqrt{E_F\,[\text{eV}]}}.</math>
 
=== A Simple Example - The Strongly Type-I Superconductor With a Planar Surface ===
 
As a simple demonstration of the solution of the GL equations, let us consider a strongly Type I (<math>\kappa\ll 1</math>) superconductor with a planar boundary between it and an insulator.  Let us set up our coordinate system so that the boundary is at <math>x=0</math>.
 
[[Image:SC_I_Boundary.PNG]]
 
We apply a magnetic field along the <math>z</math> axis,
 
<math>\vec{H}=H\hat{z}.</math>
 
We expect by symmetry that the total magnetic field <math>\vec{B}=B(x)\hat{z}</math>.  We will choose our gauge such that
 
<math>\vec{A}=A_y(x)\hat{y}.</math>
 
We also take the order parameter to depend only on <math>x</math>.  The first GL equation becomes
 
<math>-\frac{1}{\kappa^2}\frac{d^2\psi}{d\tilde{x}^2}+A_y^2\psi-\psi+\psi^3=0.</math>
 
Since we are taking <math>\kappa</math> to be small, the derivative term dominates, and we may therefore approximate this equation as
 
<math>\frac{d^2\psi}{d\tilde{x}^2}=0,</math>
 
so that <math>\psi(\tilde{x})=c_1+c_2\tilde{x}</math>.  Our boundary condition states that
 
<math>\left. \frac{1}{i\kappa}\frac{d\psi}{d\tilde{x}}\right |_{\tilde{x}=0}=0,</math>
 
so that <math>c_2=0</math>.  Since <math>\Psi=\Psi_0</math> in the bulk, we conclude that <math>\psi(\tilde{x})=1</math> for <math>\tilde{x}<0</math>.  Similarly, <math>\Psi=0</math> deep into the insulating region, so that <math>\psi(\tilde{x})=0</math> for <math>\tilde{x}>0</math>.
 
Now we consider the second equation.  In this case, it becomes, for <math>\tilde{x}<0</math>,
 
<math>-\tilde{A}=\tilde{\nabla}\times(\tilde{B}-\tilde{H})=\tilde{\nabla}\times\tilde{B},</math>
 
or
 
<math>-\tilde{B}=\tilde{\nabla}\times(\tilde{\nabla}\times\tilde{B}).</math>
 
The right-hand side is just
 
<math>\tilde{\nabla}\times(\tilde{\nabla}\times\tilde{B})=-\frac{d^2\tilde{B}}{d\tilde{x}^2},</math>
 
so that the equation is now
 
<math>\frac{d^2\tilde{B}}{d\tilde{x}^2}=\tilde{B}.</math>
 
The solution to the equation in simply <math>\tilde{B}(\tilde{x})=\tilde{B}_{+}e^{\tilde{x}}+\tilde{B}_{-}e^{-\tilde{x}}</math>, or, in terms of dimensional quantities,
 
<math>B(x)=B_{+}e^{x/\lambda}+B_{-}e^{-x/\lambda}.</math>
 
Since our superconductor is in the region <math>x<0</math>, we must take <math>B_{-}=0</math>.  Furthermore, the field must equal the applied field at <math>x=0</math>, so
 
<math>B(x)=He^{x/\lambda}.</math>
 
For <math>x>0</math>, the second GL equation becomes
 
<math>\frac{d^2\tilde{B}}{d\tilde{x}^2}=0.</math>
 
The solution, in terms of dimensional quantities, is <math>B(x)=B_0+B_1 x</math>.  We must set <math>B_1=0</math> so that the field does not increase indefinitely as we move away from the superconductor.  Since <math>\vec{B}=\vec{H}</math> in the normal state, we conclude that <math>B(x)=H</math> for <math>x>0</math>.  The following is a plot of the full dependence of the magnetic field on position.
 
[[Image:SC_BField_PlanarInt.PNG]]
 
We have now shown why we called <math>\lambda</math> the penetration depth; it sets the length scale over which the magnetic field tends to zero inside the superconductor.  We have also illustrated the expulsion of applied magnetic fields from the bulk of a Type I superconductor; this is known as the Meissner effect.
 
=== Thermodynamics of Type-I Superconductors in Magnetic Fields ===
 
In a bulk superconductor, surface effects are unimportant; for now, we will assume that the order parameter <math>\Psi</math> is constant everywhere in the superconductor and that magnetic fields are completely expelled.  In this case, the free energy per unit volume of the superconductor is
 
<math>f_s=\alpha(T-T_c)\Psi_0^2+\tfrac{1}{2}b\Psi_0^4=-\frac{\alpha^2}{2b}(T-T_c)^2.</math>
 
This is known as the condensation energy (per unit volume).  We see that we can "save" energy by going into the superconducting state.
 
In the normal state, only the magnetic field terms are present, so that the free energy is
 
<math>F_n=\int d^d\vec{r}\,\left [\frac{1}{8\pi}(\nabla\times\vec{A})^2-\frac{1}{c}\vec{J}_\text{ext}\cdot\vec{A}\right ].</math>
 
We may substitute in
 
<math>\frac{4\pi}{c}\vec{J}_\text{ext}=\nabla\times\vec{H}</math>
 
to get
 
<math>F_n=\int d^d\vec{r}\,\left [\frac{1}{8\pi}(\nabla\times\vec{A})^2-\frac{1}{4\pi}(\nabla\times\vec{H})\cdot\vec{A}\right ]=\int d^d\vec{r}\,\left [\frac{1}{8\pi}(\nabla\times\vec{A})^2-\frac{1}{4\pi}(\nabla\times\vec{A})\cdot\vec{H}\right ]
=\int d^d\vec{r}\,\left [\frac{1}{8\pi}B^2-\frac{1}{4\pi}\vec{B}\cdot\vec{H}\right ].</math>
 
In the normal state, <math>\vec{B}=\vec{H}</math>, so
 
<math>F_n=-\int d^d\vec{r}\,\frac{1}{8\pi}H^2.</math>
 
The free energy per unit volume of the normal state is therefore
 
<math>f_n=-\frac{1}{8\pi}H^2.</math>
 
We see that, overall, we also "save" energy in the normal state.  Which state we go into depends on which "saves" more energy.  We may now define a field at which the "savings" are the same for both states; this is the (thermodynamic) critical field <math>H_c</math> (sometimes also denoted <math>H_{cm}</math>).  Equating the free energies per unit volume of each state, we obtain
 
<math>\frac{\alpha^2}{2b}(T-T_c)^2=\frac{1}{8\pi}H_c^2,</math>
 
or, solving for <math>H_c</math>,
 
<math>H_c=\sqrt{\frac{4\pi}{b}}\alpha\left |T-T_c\right |.</math>
 
We see, therefore, that GL theory predicts a linear dependence of the critical field on the temperature.  This is not what is observed experimentally, however.  The dependence of the critical field on temperature in many real superconductors can, in fact, be modeled with the following empirical law:
 
<math>H_c(T)=H_c(0)\left [1-\left (\frac{T}{T_c}\right )^2\right ].</math>
 
We plot this relation below.
 
[[Image:CriticalField_vs_T.PNG]]
 
We see that, near <math>T_c</math>, the dependence of the critical field on temperature does indeed follow the linear relation that we just derived.  However, it deviates from said relation when we go far below <math>T_c</math>.  This is not surprising; the GL theory from which we obtained the linear relation is only valid near <math>T_c</math>.  In order to obtain a more accurate relation, we require a theory for the superconductor that is valid far below <math>T_c</math>.
 
We may rewrite our dimensionless vector potential and magnetic field in terms of the critical field.  We may write our expression for <math>H_c</math> as
 
<math>H_c^2=4\pi b\left |\Psi_0\right |^4.</math>
 
Our dimensionless vector potential is then
 
<math>\tilde{A}=\frac{2e}{c}\frac{\Psi_0}{\sqrt{2mb\Psi_0^4}}\vec{A}=\frac{2e}{c}\Psi_0\sqrt{\frac{4\pi}{2mH_c^2}}\vec{A}=\sqrt{\frac{8\pi e^2\Psi_0^2}{mc^2}}\vec{A}=\frac{1}{\sqrt{2}H_c \lambda}\vec{A}.</math>
 
We may rewrite the definition of the magnetic field as
 
<math>\vec{B}=\tilde{\nabla}\times\sqrt{2}H_c\tilde{A},</math>
 
or
 
<math>\tilde{\nabla}\times\tilde{A}=\frac{1}{\sqrt{2}H_c}\vec{B}=\tilde{B}.</math>
 
=== Magnetic Properties of a Type-I Superconductor ===
 
In the approximation stated above, we find that, in the superconducting state (the applied magnetic field is below <math>H_c</math>), the total magnetic field <math>\vec{B}</math> is completely expelled, while <math>\vec{B}=\vec{H}</math> when the applied field is above <math>H_c</math>.  From the defining relation for the magnetization <math>\vec{M}</math>,
 
<math>\vec{B}=\vec{H}+4\pi\vec{M},</math>
 
we find that the magnitude of the magnetization increases linearly with, and points in the opposite direction to, <math>\vec{H}</math>, in the superconducting state, but is zero in the normal state.  We plot these two relations below.
 
[[Image:SC_BvsHandMvsH.PNG]]
 
We will now demonstrate two consequences of the total expulsion of a magnetic field from a superconductor.
 
'''1) The total field is always tangential to the surface of a superconductor.'''
 
First, recall the Maxwell equation,
 
<math>\nabla\cdot\vec{B}=0.</math>
 
Let us now consider the boundary between a superconductor and a normal region:
 
[[Image:SC_BFieldCond_PillBox.PNG]]
 
We will find the flux of a magnetic field through the "pill box" shown above.  Let the area of the circular surfaces be <math>\Delta A</math>, and let us assume that the contribution from the "tube" part of the surface is negligible (we will assume that its height is small compared to the radius of the circular surfaces).  Then the total flux may be written as
 
<math>B_{n,\bot}\cdot\Delta A-B_{sc,\bot}\cdot\Delta A,</math>
 
where <math>B_{n,\bot}</math> and <math>B_{sc,\bot}</math> are the components of the magnetic field normal to the circular surfaces of the "pill box" in the normal and superconducting regions, respectively.  The Maxwell equation stated above is equivalent to the statement that the total flux through a closed surface, such as the "pill box" we consider here, must be zero.  Therefore,
 
<math>B_{n,\bot}=B_{sc,\bot}.</math>
 
However, we know that the superconductor completely expels magnetic fields, so that <math>\vec{B}_{sc}=0</math>.  Therefore, <math>B_{n,\bot}=0</math>, thus proving that <math>\vec{B}_n</math> can only have tangential components.
 
'''2) As a direct consequence of (1), a type-I superconductor in an external magnetic field always carries an electrical current near its surface.'''
 
To show this, first recall the Maxwell equation (Ampere's Law in the case of static fields),
 
<math>\nabla\times\vec{B}=\frac{4\pi}{c}\vec{J},</math>
 
or, in integral form,
 
<math>\oint_{\partial S}\vec{B}\cdot d\vec{\ell}=\frac{4\pi}{c}I,</math>
 
where <math>S</math> is a surface with bounding curve <math>\partial S</math>, the line integral on the left is taken in the direction that would cause a right-handed screw to advance in the direction of the normal to the surface, and <math>I</math> is the total current passing through the surface.
 
Let us now, once again, consider the boundary between a superconductor and a normal region and a rectangular contour drawn around the boundary:
 
[[Image:SC_BField_Contour.PNG]]
 
In this case, because the magnetic field is zero in the superconducting region and because the magnetic field in the normal region is tangential to the surface of the superconductor, then, assuming that the lengths of segments 14 and 23 are small compared to that of segments 12 and 34.  In this case, the left-hand side of the Maxwell equation becomes
 
<math>\oint_{\partial S}\vec{B}\cdot d\vec{\ell}=B\ell_{12},</math>
 
where <math>\ell_{12}</math> is the length of segment 12.  The right-hand side, on the other hand, is
 
<math>\frac{4\pi}{c}I=\frac{4\pi}{c}K\ell_{12},</math>
 
where <math>K</math> is the surface current density (per unit length).  This implies that
 
<math>B=\frac{4\pi}{c}K,</math>
 
or, in vector form,
 
<math>\vec{K}=\frac{c}{4\pi}\hat{n}\times\vec{B},</math>
 
where <math>\hat{n}</math> is the normal to the surface pointing into the normal region.  In the case depicted above, we see that <math>\vec{K}</math> points into the page or screen.
 
=== An Example: The Magnetic Field Around a Spherical Superconductor ===
 
These observations lead to interesting conclusions for superconductors with geometries more complex than a cylinder inside a magnetic field parallel to its axis.  Consider, for example, a spherical superconductor, depicted below:
 
[[Image:MagField_SCSphere.png]]
 
As we can see, the magnetic field near the equator is stronger than the applied field, while the field at the poles vanishes.  This implies that the magnetic field near the equator may exceed <math>H_c</math> even if the applied field is less than <math>H_c</math>, simply due to the geometry of the sample.  In fact, for a certain range of fields,
 
<math>(1-\eta)H_c<H<H_c,</math>
 
the sample will enter an intermediate state, in which superconducting and normal regions coexist.  The above inequality holds for any geometry, in fact, and <math>\eta</math> is known as the demagnetizing factor of the sample.  Its value will depend on the exact geometry of the sample; we will now show that, for a sphere, <math>\eta=\tfrac{1}{3}</math>.  We will do so by solving Maxwell's equations for a spherical superconductor of radius <math>R</math> in a uniform applied magnetic field <math>\vec{H}=H\hat{z}</math>.  The boundary conditions for this problem are
 
<math>\vec{B}(\vec{r})=H\hat{z}</math>
 
for <math>|\vec{r}|\rightarrow\infty</math> and
 
<math>\hat{n}\cdot\vec{B}(\vec{r})=0</math>
 
on the surface of the sphere.  Outside the superconductor, the equations satisfied by <math>\vec{B}</math> are
 
<math>\nabla\cdot\vec{B}=0</math>
 
and
 
<math>\nabla\times\vec{B}=0.</math>
 
The second equation implies that there are no current sources outside the superconductor.  It also implies that we may write the magnetic field in terms of a scalar potential.  We therefore write
 
<math>\vec{B}=\nabla\Phi+H\vec{z}.</math>
 
Substituting this into the first equation, we get
 
<math>\nabla^2\Phi=0,</math>
 
which is just Laplace's equation.  Our boundary conditions for <math>\Phi</math> are, in spherical coordinates,
 
<math>\Phi(\vec{r}\rightarrow\infty)=0</math>
 
and
 
<math>\frac{\partial\Phi}{\partial r}+H\cos{\theta}=0</math>
 
on the surface.  Laplace's equation in spherical coordinates is
 
<math>\frac{1}{r^2}\frac{\partial}{\partial r}\left (r^2\frac{\partial\Phi}{\partial r}\right )+\frac{1}{r^2\sin{\theta}}\frac{\partial}{\partial\theta}\left (\sin{\theta}\frac{\partial\Phi}{\partial\theta}\right )+\frac{1}{r^2\sin^2{\theta}}\frac{\partial^2\Phi}{\partial\phi^2}=0.</math>
 
Because our system has azimuthal symmetry, <math>\Phi</math> should be independent of <math>\phi</math>.  Therefore, <math>\Phi(\vec{r})=\Phi(r,\theta)</math> and Laplace's equation becomes
 
<math>\frac{1}{r^2}\frac{\partial}{\partial r}\left (r^2\frac{\partial\Phi}{\partial r}\right )+\frac{1}{r^2\sin{\theta}}\frac{\partial}{\partial\theta}\left (\sin{\theta}\frac{\partial\Phi}{\partial\theta}\right )=0.</math>
 
Multiplying by <math>r^2</math>, we get
 
<math>\frac{\partial}{\partial r}\left (r^2\frac{\partial\Phi}{\partial r}\right )+\frac{1}{\sin{\theta}}\frac{\partial}{\partial\theta}\left (\sin{\theta}\frac{\partial\Phi}{\partial\theta}\right )=0.</math>
 
We will now attempt to solve this equation by separation of variables.  Let us try a solution of the form
 
<math>\Phi(r,\theta)=R(r)\chi(\theta).</math>
 
Then
 
<math>\frac{1}{R}\frac{d}{dr}\left (r^2\frac{dR}{dr}\right )=-\frac{1}{\chi}\frac{1}{\sin{\theta}}\frac{d}{d\theta}\left (\sin{\theta}\frac{d\chi}{d\theta}\right ).</math>
 
We see that we have an expression depending only on <math>r</math> on one side and one depending only on <math>\theta</math> on the other.  The only way for this equation to be satisfied is if both equal a constant, which we will call <math>-\lambda</math>.  Let us first consider the equation for <math>\chi</math>.  This equation is
 
<math>\frac{1}{\sin{\theta}}\frac{d}{d\theta}\left (\sin{\theta}\frac{d\chi}{d\theta}\right )=\lambda\chi.</math>
 
If we make the substitution, <math>x=\cos{\theta}</math>, into this equation, we get
 
<math>\frac{d}{dx}\left [(1-x^2)\frac{d\chi}{dx}\right ]=\lambda\chi.</math>
 
We recognize this as the Legendre differential equation.  The only physically interesting solutions to this equation occur when <math>\lambda=-l(l+1)</math>, in which case we find that <math>\chi</math> is a Legendre polynomial,
 
<math>\chi(\theta)=P_l(\cos{\theta}).</math>
 
The first few Legendre polynomials are <math>P_0(x)=1</math>, <math>P_1(x)=x</math>, and <math>P_2(x)=\tfrac{1}{2}(3x^2-1)</math>.
 
Now we will consider the equation for <math>R</math>.  If we take <math>\chi(\theta)=P_l(\cos{\theta})</math>, then our differential equation for <math>R</math> becomes
 
<math>\frac{d}{dr}\left (r^2\frac{dR}{dr}\right )=l(l+1)R,</math>
 
or
 
<math>\frac{d^2 R}{dr^2}+2r\frac{dR}{dr}-l(l+1)R=0.</math>
 
This is an Euler-Cauchy differential equation.  Let us assume a power law dependence for <math>R</math>,
 
<math>R(r)=r^\alpha.</math>
 
The equation becomes
 
<math>\alpha(\alpha-1)r^\alpha+2\alpha r^\alpha-l(l+1)r^\alpha=0.</math>
 
The exponent <math>\alpha</math> must therefore satisfy
 
<math>\alpha(\alpha+1)=l(l+1).</math>
 
This equation has two possible solutions, <math>\alpha=l</math> and <math>\alpha=-(l+1)</math>, so the general solution for the differential equation is
 
<math>R(r)=A_l r^l+\frac{B_l}{r^{l+1}}.</math>
 
The general solution to the original partial differential equation is then a linear combination of all possible products <math>R(r)\chi(\theta)</math>,
 
<math>\Phi(r,\theta)=\sum_{l=0}^{\infty}\left (A_l r^l+\frac{B_l}{r^{l+1}}\right )P_l(\cos{\theta}).</math>
 
Since <math>r^l\rightarrow\infty</math> for all <math>l>0</math>, we must set <math>A_l=0</math> for all such values of <math>l</math>.  We will also set <math>A_0=0</math>, so that
 
<math>\Phi(r,\theta)=\sum_{l=0}^{\infty}\frac{B_l}{r^{l+1}}P_l(\cos{\theta}).</math>
 
We now apply the boundary condition,
 
<math>\frac{\partial\Phi}{\partial r}+H\cos{\theta}=0,</math>
 
to obtain
 
<math>\sum_{l=0}^{\infty}\frac{(l+1)B_l}{R^{l+2}}P_l(\cos{\theta})=H\cos{\theta}.</math>
 
By inspection, we find that <math>B_l=0</math> for all <math>l\neq 1</math>.  For <math>l=1</math>, we have
 
<math>B_l=\tfrac{1}{2}HR^3.</math>
 
Therefore, the solution for <math>\Phi</math> is
 
<math>\Phi(r,\theta)=\frac{HR^3}{2r^2}\cos{\theta},</math>
 
and thus <math>\vec{B}</math> is
 
<math>\vec{B}=\tfrac{1}{2}HR^3\nabla\left (\frac{\cos{\theta}}{r^2}\right )+H\hat{z}.</math>
 
In spherical coordinates, the gradient operator is given by
 
<math>\nabla\Phi=\frac{\partial\Phi}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial\Phi}{\partial\theta}\hat{\theta}+\frac{1}{r\sin{\theta}}\frac{\partial\Phi}{\partial\phi}\hat{\phi},</math>
 
so
 
<math>\vec{B}=-H\left (\frac{R}{r}\right )^3(\cos{\theta}\hat{r}+\tfrac{1}{2}\sin{\theta}\hat{\theta})+H\hat{z}.</math>
 
We recognize that <math>\hat{z}=\cos{\theta}\hat{r}-\sin{\theta}\hat{\theta}</math>, so that
 
<math>\vec{B}=\left [1-\left (\frac{R}{r}\right )^3\right ]H\cos{\theta}\hat{r}-\left [1+\tfrac{1}{2}\left (\frac{R}{r}\right )^3\right ]H\sin{\theta}\hat{\theta}.</math>
 
We now want to find the point at which the magnetic field has the largest magnitude.  The magnitude of this vector is given by
 
<math>|\vec{B}|^2=H^2\left [1+\tfrac{1}{2}\left (\frac{R}{r}\right )^3\right ]^2+3\left (\frac{R}{r}\right )^3\left [\tfrac{1}{4}\left (\frac{R}{r}\right )^3-1\right ]\cos^2{\theta}</math>
 
We see that the second term, which is proportional to <math>\cos^2{\theta}</math>, is clearly negative when <math>r>R</math>.  Therefore, we must make the size of this term as small as possible.  This may be done by letting <math>\theta=\tfrac{\pi}{2}</math>.  For this value of <math>\theta</math>, we have
 
<math>|\vec{B}|^2=H^2\left [1+\tfrac{1}{2}\left (\frac{R}{r}\right )^3\right ]^2.</math>
 
This function is monotonically decreasing as we increase <math>r</math>; therefore, we choose the smallest possible value of <math>r</math>, which is <math>r=R</math>.  This gives us
 
<math>|\vec{B}|^2=\tfrac{9}{4}H^2,</math>
 
or <math>|\vec{B}|=\tfrac{3}{2}H</math>.  Therefore, the total magnetic field achieves its maximum magnitude at the equator of the sphere, where it is three halves the applied field.  Therefore, if the applied field is larger than <math>H=\tfrac{2}{3}H_c</math>, the field at the equator would become larger than <math>H_c</math>, which would destroy the sphere's superconductivity at that point.  Therefore, the sphere enters an intermediate state when <math>\tfrac{2}{3}H_c<H<H_c</math>, so that the demagnetizing factor <math>\eta=\tfrac{1}{3}</math>.
 
=== Superconductors of General Geometries ===
 
In general, the maximum magnetic field on the surface of a superconductor will be related to the applied field by
 
<math>B_{\text{max}}=\frac{H}{1-\eta}.</math>
 
We will now list the values of <math>\eta</math> for a few different geometries.
 
{| border="1" align="center" style="text-align:center;"
|'''Geometry'''
|'''Demagnetizing factor <math>\eta</math>'''
|-
|Cylinder with <math>\vec{H}</math> parallel to its axis
|<math>0</math>
|-
|Cylinder with <math>\vec{H}</math> perpendicular to its axis
|<math>\tfrac{1}{2}</math>
|-
|Sphere
|<math>\tfrac{1}{3}</math>
|-
|Infinite thin plate with <math>\vec{H}</math> perpendicular to the plate
|<math>1</math>
|}
 
The sample will enter the intermediate state at the value of <math>H=H^{\ast}</math> at which <math>B_{\text{max}}=H_c</math>, which is
 
<math>H^{\ast}=(1-\eta)H_c.</math>
 
Therefore, the superconductor will be in the intermediate state when
 
<math>(1-\eta)H_c<H<H_c.</math>
 
We will now show plots of the magnetic field of a sphere at the equator and at the pole as a function of the applied field (for details on how to derive the dependence in the intermediate state, see P. G. DeGennes, ''Superconductivity of Metals and Alloys'').
 
[[Image:BField_SCSphere.PNG]]
 
In the intermediate state, the energy per volume of both the normal and superconducting states is the same.  Therefore, the energy per unit ''area'' of a "domain wall" will be the dominant contribution.
 
=== Surface Term of the Free Energy ===
 
The free energy in the reduced units can be written as
 
<math>F=\int d^3 r \,\left[\alpha (T-T_c)|\Psi|^2 + \frac{b}{2} |\Psi|^4 + \frac{\hbar^2}{2m} |(\nabla
- \frac{2ei}{\hbar c}\vec{A})\Psi|^2 + \frac{1}{8\pi}(\nabla\times\vec{A})^2 - \frac{1}{4\pi}\vec{H}\cdot\vec{B}\right]</math>
 
<math>=\lambda^D\int d^D\tilde{r}\,\left[\alpha (T-T_c)|\Psi_o|^2 |\psi|^2 + \frac{b}{2} |\Psi_o|^4 |\psi|^4 + 
\frac{\hbar^2}{2m_{GL}}|\Psi_o|^2|(\frac{1}{\lambda}\tilde{\nabla}-\frac{2ei}{\hbar c}\sqrt{2}\lambda H_c \tilde{A})\psi|^2 + \frac{1}{8\pi \lambda^2} (\sqrt{2}\lambda H_c)^2 (\tilde{\nabla}\times\tilde{A})^2
- \frac{2\lambda^2 H_c^2}{4\pi \lambda^2} \tilde{H}\cdot\tilde{B}\,\right]</math>
 
where
 
<math>\tilde{r}=\frac{r}{\lambda}, \tilde{H}=\frac{H}{\sqrt{2}H_c}</math>
 
Using the following relations:
 
<math>b|\Psi_o|^4 = \frac{H_c^2}{4\pi}</math>
 
<math>\frac{\alpha (T-T_c)}{b|\Psi_o|^2} = -1</math>
 
<math>|\Psi_o|^2 \frac{4e^2}{mc^2} = \frac{1}{4\pi\lambda^2}</math>
 
one has
 
<math>F=\frac{\lambda^D H_c^2}{4\pi}\int d^D\tilde{r}\,\left[-|\psi|^2+\frac{1}{2}|\psi|^4+|(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})\psi|^2+\tilde{B}^2-2\tilde{H}\cdot\tilde{B}\,\right]</math>
 
Now let's consider the gradinet term:
 
<math>|(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})\psi|^2
= \,\left[ (\frac{-1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})\psi^*\,\right]\,\left[ (\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})\psi\,\right]
= (-\tilde{A}\psi^*)(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})\psi
- (\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}\psi^*)(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})\psi
</math>
 
Integration by part leads the 2nd term to the form:
 
<math>-\tilde{\nabla}\cdot(\frac{1}{\kappa}\frac{1}{i}\psi^*(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})\psi)
+ \psi^*(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})\psi)</math>
 
where the 1st term needs to be evaluated on the surface where it vanishes by the boundary condition for an insulating interface. Hence the gradient term becomes
 
<math>|(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})\psi|^2
= \psi^*(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})^2\psi</math>
 
Assuming that <math>\psi</math> and <math>\tilde{A}</math> satisfy the GL equation, namely,
 
<math>(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})^2 \psi = \psi - \psi |\psi|^2</math>
 
we have
 
<math>\psi^*(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})^2\psi = |\psi|^2-|\psi|^4</math>
 
and the free energy becomes
 
<math>F=\frac{\lambda^D H_c^2}{4\pi}\int d^D \tilde{r} \,\left[-\frac{1}{2}|\psi|^4+\tilde{B}^2-2\tilde{H}\cdot\tilde{B}\,\right]</math>
 
This is true at the saddle point. Now the free energy for the normal state and the superconducting state are given by
 
<math>F_n=\frac{\lambda^D H_c^2}{4\pi}\int d^D \tilde{r} \,\left[-\tilde{H}^2\,\right]</math>
 
and
 
<math>F_s=\frac{\lambda^D H_c^2}{4\pi}\int d^D \tilde{r} \,\left[-\frac{1}{2}|\psi|^4+\tilde{B}^2-2\tilde{B}\cdot\tilde{H}\,\right]</math>
 
Therefore, the surface free energy of the interface is
 
<math>F_s-F_n=\frac{\lambda^D H_c^2}{4\pi}\int d^D \tilde{r} \,\left[-\frac{1}{2}|\psi|^4+\tilde{B}^2-2\tilde{B}\cdot\tilde{H}+\tilde{H}^2\,\right]
= \frac{\lambda^D H_c^2}{4\pi}\int d^D \tilde{r} \,\left[-\frac{1}{2}|\psi|^4+(\tilde{B}-\tilde{H})^2\,\right]</math>
 
As shown in the figures below, type-I superconductor (left) has positive wall energy while type-II superconductor (right) negative wall energy.
[[Image:pic4.jpg]] [[Image:pic3.jpg]]
 
=== Free Energy of a Normal-Superconducting Interface ===
 
Now let's consider the free energy of a normal-superconducting interface. Recall the GL equations:
 
<math>(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})^2\psi - \psi + \psi|\psi|^2 = 0</math>
 
<math>\frac{1}{2}(\psi^*\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}\psi + c.c.)-\tilde{A}|\psi|^4
=\tilde{\nabla}\times(\tilde{\nabla}\times\tilde{A})</math>
 
Assume that the interface is along y-z plane and <math>\tilde{B}\|\hat{z}</math>. We can choose <math>\tilde{A}(\tilde{r})=\hat{y}\tilde{A}(\tilde{x})</math> which implies the order parameter depends only on <math>\tilde{x}</math>. Then the GL equation become
 
[[Image:pic5.jpg|right]]
 
<math>-\frac{1}{\kappa^2}\frac{d^2\psi}{d\tilde{x}^2}+\tilde{A}_y^2(\tilde{x})\psi -\psi + \psi^3 = 0</math>(*)
 
<math>\tilde{A}_y(\tilde{x})\psi^2 = \frac{d^2\tilde{A}_y(\tilde{x})}{d\tilde{x}^2}</math>
 
and
 
<math>\frac{1}{2}(\psi^*\frac{1}{\kappa}\frac{1}{i}\frac{d}{d\tilde{x}}\psi + c.c.) = 0</math>
 
Note that we can choose <math>\psi</math> to be real since the coefficients are real. Now let's manipulate Eq.(*):
 
<math>-\frac{1}{\kappa^2}\frac{d\psi}{d\tilde{x}}\frac{d^2\psi}{d\tilde{x}^2}+\tilde{A}_y^2\frac{d\psi}{d\tilde{x}}\psi
- \frac{d\psi}{d\tilde{x}}\psi + \frac{d\psi}{d\tilde{x}}\psi^3 = 0</math>
 
<math>-\frac{1}{2\kappa^2}\frac{d}{d\tilde{x}}(\frac{d\psi}{d\tilde{x}})^2+\frac{1}{2}\tilde{A}_y^2\frac{d}{d\tilde{x}}\psi^2 - \frac{1}{2}\frac{d}{d\tilde{x}}\psi^2 + \frac{1}{4}\frac{d}{\tilde{x}}\psi^4 = 0</math>
 
Integrating we have
 
<math>-\frac{1}{2\kappa^2}(\frac{d\psi}{d\tilde{x}})^2 + \frac{1}{2}\int d\tilde{x}\tilde{A}_y^2\frac{d}{d\tilde{x}}\psi^2
- \frac{1}{2}\psi^2 + \frac{1}{4}\psi^4 = const.</math>
 
This integral can be further manipulated:
 
<math>\int d\tilde{x}\tilde{A}_y^2\frac{d}{d\tilde{x}}\psi^2 = \int d\tilde{x}\,\left[\frac{d}{d\tilde{x}}(\tilde{A}_y^2\psi^2)-\frac{d\tilde{A}_y^2}{d\tilde{x}}\psi^2\,\right]
= \tilde{A}_y^2\psi^2 - 2\int d\tilde{x}\tilde{A}_y\frac{d\tilde{A}_y}{d\tilde{x}}\psi^2
= \tilde{A}_y^2\psi^2 - 2\int d\tilde{x}\frac{d\tilde{A}_y}{d\tilde{x}}\frac{d^2\tilde{A}_y}{d\tilde{x}^2}
= \tilde{A}_y^2\psi^2 - \int d\tilde{x}\frac{d}{d\tilde{x}}(\frac{d\tilde{A}_y}{d\tilde{x}})^2
= \tilde{A}_y^2\psi^2 - (\frac{d\tilde{A}_y}{d\tilde{x}})^2 = const.</math>
 
Thus, Eq.(*) becomes
 
<math>-\frac{1}{2\kappa^2}(\frac{d\psi}{d\tilde{x}})^2+\frac{1}{2}\tilde{A}_y^2\psi^2-\frac{1}{2}(\frac{d\tilde{A}_y}{d\tilde{x}})^2 - \frac{1}{2}\psi^2 + \frac{1}{4}\psi^4 = const.</math>
 
or,
 
<math>\frac{1}{\kappa^2}(\frac{d\psi}{d\tilde{x}})^2+(\frac{d\tilde{A}_y}{d\tilde{x}})^2+\psi^2(1-\tilde{A}_y^2)-\frac{1}{2}\psi^4 = const.</math>
 
To determine the const., note that as <math>x\rightarrow -\infty</math>, <math>\psi \rightarrow 1</math> and <math>\tilde{A}_y \rightarrow 0</math>, so
 
<math>0 + 0 + 1 - \frac{1}{2} = const.</math>
 
Therefore,
 
<math>\frac{1}{\kappa^2}(\frac{d\psi}{d\tilde{x}})^2+(\frac{d\tilde{A}_y}{d\tilde{x}})^2+\psi^2(1-\tilde{A}_y^2)-\frac{1}{2}\psi^4 = \frac{1}{2}</math>
 
Consider the case where <math>\kappa << 1</math>, that is, <math>\lambda << \xi</math>. The main contribution to the surface energy comes from the region where <math>\tilde{B}=0</math> but <math>\psi < 1</math>.
 
<math>\frac{1}{\kappa^2}(\frac{d\psi}{d\tilde{x}})^2+\psi^2-\frac{1}{2}\psi^4=\frac{1}{2}</math>
 
<math>\frac{1}{\kappa^2}(\frac{d\psi}{d\tilde{x}})^2 = \frac{1}{2}-\psi^2+\frac{1}{2}\psi^4 = \frac{1}{2}(1-\psi^2)^2</math>
 
<math>\frac{1}{\kappa}\frac{d\psi}{d\tilde{x}}=-\frac{1}{\sqrt{2}}(1-\psi^2)</math>
 
Note that the minus sign is from the experimental setup. Further,
 
<math>\frac{d\psi}{1-\psi^2}=-\frac{\kappa}{\sqrt{2}}d\tilde{x};\frac{1}{1-\psi^2}=\frac{1}{2}(\frac{1}{1-\psi}+\frac{1}{1+\psi})</math>
 
<math>ln\frac{1+\psi}{1-\psi}=-\sqrt{2}\kappa \tilde{x}+const.</math>
 
<math>\frac{1+\psi(\tilde{x})}{1-\psi(\tilde{x})}=const.\times e^{-\sqrt{2}\kappa \tilde{x}}</math>
 
[[Image:pic6.jpg|right]]
 
Requiring <math>\psi=0</math> at <math>\tilde{x}=0</math> gives <math>const.=1</math> Now we have
 
<math>\psi = \frac{e^{-\sqrt{2}\kappa \tilde{x}}-1}{e^{-\sqrt{2}\kappa \tilde{x}}+1}=-tanh\frac{\kappa \tilde{x}}{\sqrt{2}}</math>
 
whose curve is shown in the figure. Moreover,
 
<math>F_s-F_n=\frac{H_c^2}{8\pi}L^2\lambda\int_{-\infty}^{0}d\tilde{x}(1-\psi^4)</math>
 
where
 
<math>\int_{-\infty}^{0}d\tilde{x}(1-\psi^4)=\int_{-\infty}^{0}d\tilde{x}(1+\psi^2)(1-\psi^2)
=\int_{-\infty}^{0}d\tilde{x}(1+\psi^2)(-\frac{\sqrt{2}}{\kappa})\frac{d\psi}{d\tilde{x}}
=-\frac{\sqrt{2}}{\kappa}\psi |_{-\infty}^{0}-\frac{\sqrt{2}}{\kappa}\int_{-\infty}^{0}d\tilde{x}\psi^2\frac{d\psi}{d\tilde{x}}</math>
 
[[Image:pic7.jpg|right]]
 
<math>=-\frac{\sqrt{2}}{\kappa}\psi |_{-\infty}^{0}-\frac{\sqrt{2}}{3\kappa}\int_{-\infty}^{0}d\tilde{x}\frac{d\psi^3}{d\tilde{x}} = -\frac{\sqrt{2}}{\kappa}\psi |_{-\infty}^{0}-\frac{\sqrt{2}}{3\kappa}\psi^3 |_{-\infty}^{0} =
\frac{\sqrt{2}}{\kappa}+\frac{\sqrt{2}}{3\kappa} = \frac{4\sqrt{2}}{3\kappa}</math>
 
Hence one has
 
<math>F_s-F_n=\frac{H_c^2}{8\pi}L^2\lambda\cdot\frac{4\sqrt{2}}{3\kappa}</math>
 
Below we show that given <math>\psi</math> and <math>\tilde{A}_y</math> satisfying the GL equation, the surface energy of the surface vanishes at <math>\kappa=1/2</math>:
 
<math>\frac{1}{\kappa^2}(\frac{d\psi}{d\tilde{x}})^2+(\frac{d\tilde{A}_y}{d\tilde{x}})^2+\psi^2(1-\tilde{A}_y^2)-\frac{1}{2}\psi^4 = \frac{1}{2}</math>(**)
 
<math>F_s-F_n=\frac{H_c^2}{4\pi}L^2\lambda\int d\tilde{x}\,\left[(\tilde{B}-\tilde{H})^2-\frac{1}{2}\psi^4\,\right]</math>
 
Assume <math>\kappa=1/\sqrt{2}</math>.
 
<math>F_s-F_n=\frac{H_c^2}{4\pi}L^2\lambda\int d\tilde{x}\,\left[(\frac{d\tilde{A}_y}{d\tilde{x}}-\frac{1}{\sqrt{2}})^2-\frac{1}{2}\psi^4\,\right]</math>
 
Choosing
 
<math>\frac{d\tilde{A}_y}{d\tilde{x}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\psi^2</math>
 
then <math>F_s-F_n=0</math> and the 2nd GL equation becomes
 
<math>\tilde{A}_y\psi^2=\frac{d^2\tilde{A}_y}{d\tilde{x}^2}=\frac{d}{d\tilde{x}}(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\psi^2)=-\sqrt{2}\psi\frac{d\psi}{d\tilde{x}}</math>
 
<math>\frac{d\psi}{d\tilde{x}}=-\frac{1}{\sqrt{2}}\tilde{A}_y\psi</math>
 
Insert this into Eq.(**):
 
<math>\frac{1}{\kappa^2}\frac{1}{2}\tilde{A}_y^2\psi^2+(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\psi^2)^2+\psi^2(1-\tilde{A}_y^2)-\frac{1}{2}\psi^4-\frac{1}{2}=0</math>
 
<math>\frac{1}{2\kappa^2}\tilde{A}_y^2\psi^2+\frac{1}{2}-\psi^2+\frac{1}{2}\psi^4+\psi^2-\tilde{A}_y^2\psi^2-\frac{1}{2}\psi^4-\frac{1}{2}=0</math>
 
<math>(\frac{1}{2\kappa^2}-1)\tilde{A}_y^2\psi^2=0</math>
 
Therefore, <math>\kappa=1/\sqrt{2}</math> as expected.
 
In summary, for type-I superconductor (positive surface energy) <math>\kappa < 1/\sqrt{2}</math> and <math>\psi</math> is uniform in the bulk and jumps at <math>T_c</math>. On the other hand, for type-II superconductor (negative surface energy) <math>\kappa > 1/\sqrt{2}</math> and <math>\psi</math> is not uniform in the bulk but grows continuously when crossing <math>T_c</math> from above.
 
=== The "Landau-Level"-Like States above Hc and the Upper Critical Field Hc2 ===
 
Now we have the surface free energy:
 
<math>F_s-F_n=\frac{H_c^2}{4\pi}\lambda^3\int d^3\tilde{r} \,\left[-|\psi|^2+|(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})^2\psi|^2+\frac{1}{2}|\psi|^4+(\tilde{B}-\tilde{H})^2\,\right]</math>
 
where
 
<math>\tilde{B}=\frac{B}{\sqrt{2}H_c}</math> and <math>\tilde{A}=\frac{A}{\sqrt{2}\lambda H_c}</math>
 
Assuming 2nd order phase transition and <math>T=T_c</math>, then <math>\psi\rightarrow 0</math> and <math>\tilde{B}\rightarrow\tilde{H}</math>:
 
<math>F_s-F_n=\frac{H_c^2}{4\pi}\lambda^3\int d^3\tilde{r} \,\left[-|\psi|^2+|(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})^2\psi|^2\,\right]</math>
 
Using the Landau gauge:
 
<math>\vec{A}=\hat{y}Hx=\hat{y}\frac{H}{\sqrt{2}H_c}\frac{x}{\lambda}=\hat{y}\tilde{H}\tilde{x}</math>
 
Recall
 
<math>(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})^2\psi-\psi=0</math>
 
<math>(\frac{1}{\kappa}\frac{1}{i}\frac{d}{d\tilde{x}})^2\psi+(\frac{1}{\kappa}\frac{1}{i}\frac{d}{d\tilde{y}}-\tilde{H}\tilde{x})^2\psi+(\frac{1}{\kappa}\frac{1}{i}\frac{d}{d\tilde{z}})^2\psi-\psi=0</math>
 
<math>-\frac{1}{\kappa^2}\frac{d^2}{d\tilde{x}^2}\psi+(\frac{1}{\kappa}\frac{1}{i}\frac{d}{d\tilde{y}}-\tilde{H}\tilde{x})^2\psi-\frac{1}{\kappa^2}\frac{d^2}{d\tilde{z}^2}\psi-\psi=0</math>
 
Let
 
<math>\psi(\tilde{x},\tilde{y},\tilde{z})=e^{ik_z\tilde{z}}e^{ik\tilde{y}}\Phi_k(\tilde{x})</math>
 
then
 
<math>e^{ik_z\tilde{z}}e^{ik\tilde{y}}\,\left[-\frac{1}{\kappa^2}\frac{d^2}{d\tilde{x}^2}\Phi+(\frac{k}{\kappa}-\tilde{H}\tilde{x})^2\Phi+\frac{k_z^2}{\kappa^2}\Phi-\Phi\,\right]=0</math>
 
Let
 
<math>-\tilde{H}X=\frac{k}{\kappa}-\tilde{H}\tilde{x};\frac{d}{d\tilde{x}}=\frac{dX}{d\tilde{x}}\frac{d}{dX}=\frac{d}{dX}</math>
 
then
 
<math>e^{ik_z\tilde{z}}e^{ik\tilde{y}}(-\frac{1}{\kappa^2}\frac{d^2}{dX^2}+\tilde{H}^2 X^2+\frac{k_z^2}{\kappa^2}-1)\Phi=0</math>
 
Recall for a harmonic oscillator we have
 
<math>(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{1}{2}m\omega^2 x^2)\Psi=E\Psi</math>
 
and
 
<math>E=\hbar\omega(n+\frac{1}{2})</math>
 
Comparison leads to
 
<math>\frac{-\hbar^2}{2m}\rightarrow\frac{-1}{\kappa^2};\frac{1}{2}m\omega^2\rightarrow\tilde{H}^2;\hbar\omega\rightarrow\frac{2\tilde{H}}{\kappa}</math>
 
and the eigenvalues
 
<math>2\frac{\tilde{H}}{\kappa}(n+\frac{1}{2})+\frac{k_z}{\kappa}-1</math>
 
The exponent is zero when
 
<math>2\frac{\tilde{H}}{\kappa}(n+\frac{1}{2})+\frac{k_z}{\kappa}-1=0</math>
 
If <math>n=0</math> and <math>k_x=0</math> we have
 
<math>\tilde{H}_{c2}=\kappa=\frac{H_{c2}}{\sqrt{2}H_c}</math>
 
<math>H_{c2}=\sqrt{2}\kappa H_c</math>
 
For <math>\kappa < 1/\sqrt{2}</math>, <math>H_{c2}<H_c</math>, the bulk is preempted by the Meissner phase. On the other hand, <math>\kappa > 1/\sqrt{2}</math>, <math>H_{c2}>H_c</math>, the bulk undergoes a 2nd order phase transition into a mixed state in which normal state and superconducting state coexist.
 
Now let's look at the wavefunction at <math>H_{c2}</math>.
 
<math>(-\frac{1}{\kappa^2}\frac{d^2}{dX^2}+\tilde{H}_{c2}^2 X^2)\Phi=\lambda\Phi;\tilde{H}_{c2}=\kappa</math>
 
<math>(-\frac{1}{\kappa^2}\frac{d^2}{dX^2}+\kappa^2 X^2)\Phi=\lambda\Phi</math>
 
Let
 
<math>\frac{1}{\sqrt{2}}\frac{1}{\kappa}\frac{d}{dX}+\frac{1}{\sqrt{2}}\kappa X = a</math>
 
<math>\frac{-1}{\sqrt{2}}\frac{1}{\kappa}\frac{d}{dX}+\frac{1}{\sqrt{2}}\kappa X = a^+</math>
 
where <math>[a,a^{+}]=1</math> Note that
 
<math>a^+ a=\frac{-1}{2\kappa^2}\frac{d^2}{dX^2}+\frac{1}{2}\kappa^2 X^2-\frac{1}{2}</math>
 
<math>(2a^+ a+1)\Phi=\lambda\Phi</math>
 
For <math>\lambda=1</math> we have
 
<math>\frac{1}{\sqrt{2}}(\frac{1}{\kappa}\frac{d}{dX}+\kappa K)\Phi=0</math>
 
<math>\Phi=const.\times e^{-\frac{1}{2}\kappa^2 X^2}</math>
 
and the order parameter is of the form
 
<math>const.\times e^{ik\tilde{y}}e^{-\frac{\kappa^2}{2}(\tilde{x}-\frac{k}{\kappa^2})^2}</math>
 
=== Abrikosov Solution of the GL Equation for Type-II Superconductor near Hc2 ===
 
Note. In this section we will drop the "tilde(~)" on relevant quantities and recover it when necessary.
 
Since the conditions along the entire superconductor are uniform, we seek a linear combination of solutions centered through equal intervals, namely,
 
<math>\psi^{(0)}=\Sigma_{n=-\infty}^{\infty}C_n e^{ikn\hat{y}} e^{-\frac{1}{2}\kappa^2(x-\frac{kn}{\kappa^2})^2}</math>
 
or,
 
<math>\psi=\Sigma_{n=-\infty}^{\infty}C_n e^{ikny}\phi_n(x)</math>
 
with
 
<math>\phi_n(x)=e^{-\frac{1}{2}\kappa^2(x-\frac{kn}{\kappa^2})^2}</math>
 
Consider the solution to GL equation at <math>H</math> slightly less than <math>\kappa</math>.The 2nd GL equation gives
 
<math>\nabla\times(\nabla\times A^{(1)})=\frac{-i}{2\kappa}(\psi^{(0)*} \nabla\psi^{(0)}-\psi^{(0)}\nabla\psi^{(0)*})-|\psi^{(0)}|^2 A^{(0)}</math>
 
<math>\nabla\times B=\hat{x}\frac{\partial B_z}{\partial y}-\hat{y}\frac{\partial B_z}{\partial x}</math>
 
<math>A=\hat{y}A_y(x,y);\nabla\times A=\hat{z}\frac{\partial A}{\partial x}</math>
 
For the x-component, we have
 
<math>\frac{\partial B_z}{\partial y}=\frac{\partial^2 A_y}{\partial x\partial y}</math>
 
<math>=\frac{-i}{2\kappa}\Sigma_{n=-\infty}^{\infty}\Sigma_{m=-\infty}^{\infty}C_n^* e^{-ikny}\phi_n(x)(\frac{\partial}{\partial x}\phi_m(x))C_me^{ikmy}
+ \frac{i}{2\kappa}\Sigma_{n=-\infty}^{\infty}\Sigma_{m=-\infty}^{\infty}C_n^* e^{-ikny}(\frac{\partial}{\partial x}\phi_n(x))\phi_m(x)C_m e^{ikmy}</math>
 
<math>=\frac{-i}{2\kappa}\Sigma_{n,m}C_n^* C_m e^{-ik(n-m)y}\,\left[\phi_n(x)\frac{\partial}{\partial x}\phi_m(x)-\phi_m(x)\frac{\partial}{\partial x}\phi_n(x)\,\right]</math>
 
where
 
<math>[...]=-\kappa^2(x-\frac{km}{\kappa^2})\phi_n(x)\phi_m(x)+\kappa^2(x-\frac{kn}{\kappa^2})\phi_n(x)\phi_m(x)
=k(m-n)\phi_n(x)\phi_m(x)</math>
 
<math>\frac{\partial B_z}{\partial y}=\frac{-i}{2\kappa}\Sigma_{n,m}C_n^* C_m e^{-ik(n-my)}k(m-n)\phi_n(x)\phi_m(x)
=\frac{\partial}{\partial y}\,\left[\frac{1}{2\kappa}\Sigma_{n,m}C_n^* C_m e^{-ik(n-m)y}\phi_n(x)\phi_m(y)\,\right]
=\frac{\partial}{\partial y}\,\left[\frac{-1}{2\kappa}|\psi^{(0)}|^2\,\right]</math>
 
Similarly, for the y-component we have
 
<math>-\frac{\partial B_z}{\partial x}=\frac{-i}{2\kappa}\Sigma_{n,m}\,\left[C_n^* e^{-ikny}\phi_n(x)ikm e^{ikmy}\phi_m(x)C_m+C_n^* ikne^{-ikny}\phi_n(x)e^{ikmy}\phi_m C_m\,\right]-\kappa x|\psi^{(0)}|^2</math>
 
<math>=\frac{1}{2\kappa}\Sigma_{n,m}\,\left[k(n+m)-2\kappa^2x\,\right]C_n^* C_me^{-ik(n-m)y}\phi_m(x)\phi_n(x)</math>
 
Note that
 
<math>\frac{\partial}{\partial x}\,\left[\phi_m(x)\phi_m(x)\,\right]=\,\left[-\kappa^2(x-\frac{km}{\kappa^2})-\kappa^2(x-\frac{kn}{\kappa^2})\,\right]\phi_m(x)\phi_n(x)=-\,\left[2\kappa^2x-k(m+n)\,\right]\phi_m(x)\phi_n(x)</math>
 
Therefore,
 
<math>-\frac{\partial B_z}{\partial x}=\frac{\partial}{\partial x}\,\left[\frac{1}{2\kappa}|\psi^{(0)}|^2\,\right]</math>
 
Now we have
 
<math>\frac{\partial B_z}{\partial y}=-\frac{\partial}{\partial y}\,\left[\frac{1}{2\kappa}|\psi^{(0)}|^2\,\right]</math>
 
<math>\frac{\partial B_z}{\partial x}=-\frac{\partial}{\partial x}\,\left[\frac{1}{2\kappa}|\psi^{(0)}|^2\,\right]</math>
 
or,
 
<math>B_z(x,y)=f(x)-\frac{1}{2\kappa}|\psi^{(0)}|^2</math>
 
<math>B_z(x,y)=g(y)-\frac{1}{2\kappa}|\psi^{(0)}|^2</math>
 
which implies
 
<math>B_z(x,y)=const.-\frac{1}{2\kappa}|\psi^{(0)}|^2</math>
 
At points where <math>\psi^{(0)}=0</math>, <math>B_z(x,y)=H_z</math>. So
 
<math>B_z(x,y)=H_z-\frac{1}{2\kappa}|\psi^{(0)}|^2</math>
 
Note that the correction (the 2nd term) comes from induced current. Furthermore,
 
<math>A_y=xH_z-\frac{1}{2\kappa}\int dx |\psi^{(0)}|^2</math>
 
Now let's consider the 1st GL equation:
 
<math>(\frac{1}{\kappa}\frac{\nabla}{i}-A)^2\psi-\psi+\psi|\psi|^2=0</math>
 
which is solced by dropping the non-linear term and A and leads to Hc2.
 
<math>(\frac{1}{\kappa}\frac{\nabla}{i}-A^{(0)}-A^{(1)})^2(\psi^{(0)}+\psi^{(1)})-(\psi^{(0)}+\psi^{(1)})+(\psi^{(0)}+\psi^{(1)})|\psi^{(0)}+\psi^{(1)}|^2=0</math>
 
'''To 0th oeder:'''
 
<math>(\frac{1}{\kappa}\frac{\nabla}{i}-A^{(0)})^2\psi^{(0)}-\psi^{(0)}=0</math>
 
<math>A_y^{(0)}=\kappa x</math>
 
<math>\psi^{(0)}=\Sigma_{n=-\infty}^{\infty}C_ne^{ikny}\phi_n(x)</math>
 
'''To 1st order:'''
 
<math>(\frac{1}{\kappa}\frac{\nabla}{i}-A^{(0)})^2\psi^{(1)}-(\frac{1}{\kappa}\frac{\nabla}{i}-A^{(0)})\cdot A^{(1)}\psi^{(0)}-A^{(1)}\cdot(\frac{1}{\kappa}\frac{\nabla}{i}-A^{(0)})\psi^{(0)}-\psi^{(1)}+\psi^{(0)}|\psi^{(0)}|^2=0</math>
 
<math>A_y^{(1)}=(H-\kappa)x-\frac{1}{2\kappa}\int^{x}|\psi|^2</math>
 
<math>\psi^{(1)}=\Sigma_{-\infty}^{\infty}e^{ikny}\psi_n^{(1)}(x)</math>
 
[[1st term:]]
 
<math>\,\left[\frac{-1}{\kappa^2}\frac{\partial^2}{\partial x^2}+(\frac{1}{\kappa}\frac{1}{i}\frac{\partial}{\partial y}-\kappa x)^2 \,\right] \Sigma_n e^{ikny} \psi_n^{(1)}(x)= \Sigma_n e^{ikny} \,\left[\frac{-1}{\kappa^2}\frac{\partial^2}{\partial x^2}+(\frac{kn}{\kappa}-\kappa x)^2\,\right]\psi_n^{(1)}(x)</math>
 
[[2nd term:]]
 
<math>-(\frac{1}{\kappa}\frac{1}{i}\frac{\partial}{\partial y}-\kappa x)\,\left[(H-\kappa)x-\frac{1}{2\kappa}\int^x |\psi^{(0)}|^2\,\right]\Sigma_n e^{ikny} \phi_n(x)</math>
 
<math>=\Sigma_{n=-\infty}^{\infty}C_ne^{ikny}(-\frac{kn}{\kappa}+\kappa x)(H-\kappa)x\phi_n(x)
+(\frac{1}{\kappa}\frac{1}{i}\frac{\partial}{\partial y}-\kappa x)\frac{1}{2\kappa}\int^x dx'\Sigma_{m,p}C_m^*C_pe^{-ikmy}e^{ikpy}\phi_m(x')\phi_p(x')\Sigma_nC_ne^{ikny}\phi_n(x)</math>
 
<math>=\Sigma_{n=-\infty}^{\infty}C_ne^{ikmy}(-\frac{kn}{\kappa}+\kappa x)(H-\kappa)\phi_n(x)+\Sigma_{n,m,p}C_nC_m^*C_pe^{ik(n-m+p)y}(\frac{n-m+p}{\kappa}-\kappa x)\frac{1}{2\kappa}\phi_n(x)\int^x dx'\phi_m(x')\phi_p(x')</math>
 
[[3rd term:]]
 
<math>\Sigma_nC_ne^{ikny}(-\frac{kn}{\kappa}+\kappa x)(H-\kappa)x\phi_n(x)+\Sigma_{n,m,p}C_nC_m^*C_pe^{ik(n-m+p)y}(\frac{kn}{\kappa}-\kappa x)\frac{1}{2\kappa}\phi_n(x)\int^x dx'\phi_m(x')\phi_p(x')</math>
 
[[4th term:]]
 
<math>-\Sigma_ne^{ikny}\psi_n^{(1)}(x)</math>
 
[[5th term:]]
 
<math>\Sigma_{n,m,p}C_nC_m^*C_pe^{ik(n-m+p)y}\phi_n(x)\phi_m(x)\phi_p(x)</math>
 
Add them up, we have
 
<math>\Sigma_ne^{ikny}\,\left[\frac{-1}{\kappa^2}\frac{\partial^2}{\partial x^2}+(\frac{kn}{\kappa}-\kappa x)^2 + 1\,\right]\psi^{(1)}=\Sigma_ne^{ikny}C_n2x(H-\kappa)(\frac{kn}{\kappa}-\kappa x)\phi_n(x)-\Sigma_{n,m,p}e^{ik(n-m+p)y}C_nC_m^*C_p\,\left[(\frac{(2n-m+p)k}{\kappa}-2\kappa)\frac{1}{2\kappa}\phi_n(x)\int^xdx'\phi(x')\phi(x')+\phi_n(x)\phi_m(x)\phi_p(x)\,\right]</math>
 
Multiplying the whole equation by <math>e^{-ikNy}</math> and integrating over the variable y leads to
 
<math>\,\left[\frac{-1}{\kappa^2}\frac{\partial^2}{\partial x^2}+(\frac{kN}{\kappa}-\kappa)^2-1\,\right]\psi_N^{(1)}(x)=C_N2x(H-\kappa)(\frac{kx}{\kappa}-\kappa x)\phi_N(x)-\Sigma_{m,p}C_{N+m-p}C_m^*C_p\,\left[(\frac{k}{\kappa}(2N+2m-2p-m+p)-2\kappa x)\frac{1}{2\kappa}\phi_{N+m-p}(x)\int^xdx'\phi_m(x')\phi_p(x')+\phi_{N+m-p}(x)\phi_m(x)\phi_p(x)\,\right]</math>
 
where
 
<math>\phi_n(x)=e^{-\frac{1}{2}\kappa^2(x-\frac{kn}{\kappa^2})^2}</math>
 
The above equation has this (inhomogeneous) form:
 
<math>\hat{O}|\psi_N^{(1)}>=|\Phi></math>
 
Consider the homogeneous version of this equation:
 
<math>\hat{O}|\chi>=0</math>
 
The only way for the inhomogeneous equation to have a solution is if
 
<math><\chi|\Phi>=0</math>
 
Thus we need
 
<math>\,\left[\frac{-1}{\kappa^2}\frac{\partial^2}{\partial x^2}+(\frac{kN}{\kappa}-\kappa x)^2-1\,\right]\chi(x)=0</math>
 
Let
 
<math>\chi(x)=e^{-\frac{1}{2}\kappa^2(x-\frac{kN}{\kappa^2})^2}=\phi_N(x)</math>
 
then the identity is satisfied.
 
<math>0=C_N2(H-\kappa)\int_{-\infty}^{\infty}dxx(\frac{kN}{\kappa}-\kappa x)\phi_N^2(x)-\Sigma_{m,p}C_{N-p+m}C_m^*C_p
\,\left[\int_{-\infty}^{\infty}dx\phi_N(x)(\frac{k}{2\kappa^2}(2N+m-p)-x)\phi_{N+m-p}(x)\int^xdx'\phi_m(x')\phi_p(x')
+\int_{-\infty}^{\infty}dx\phi_N(x)\phi_{N+m-p}(x)\phi_m(x)\phi_p(x)\,\right]</math>
 
<math>0=\frac{\sqrt{\pi}}{\kappa}(1-\frac{H}{\kappa})C_N-\Sigma_{m,p}C_{N-p+m}C_m^*C_p(1-\frac{1}{2\kappa^2})\int_{-\infty}^{\infty}dx\phi_N(x)\phi_{N+m-p}(x)\phi_m(x)\phi_p(x)</math>
 
<math>0=\frac{\sqrt{\pi}}{\kappa}(1-\frac{H}{\kappa})C_N+\Sigma_{m,p}C_{N-p+m}C_m^*C_p(1-\frac{1}{2\kappa^2})\sqrt{\frac{\pi}{2}}\frac{1}{\kappa}e^{-\frac{k^2}{2\kappa^2}((m-p)^2+(N-p)^2)}</math>
 
Now, multiply both sides by <math>C_N^*</math> and sum over N we have
 
<math>0=(1-\frac{H}{\kappa})\Sigma_NC_N^*C_N+\frac{1}{\sqrt{2}}(\frac{1}{2\kappa^2}-1)\Sigma_{N,m,p}C_N^*C_{N-p+m}C_m^*C_pe^{-\frac{k^2}{2\kappa^2}((m-p)^2+(N-p)^2)}</math>
 
Taking the spatial average results in
 
<math>0=(1-\frac{H}{\kappa})<|\psi|^2>+(\frac{1}{2\kappa^2}-1)<|\psi|^4></math>
 
<math>1=\frac{\frac{H}{\kappa}-1}{\frac{1}{2\kappa^2}-1}\frac{<|\psi|^2>}{<|\psi|^4>}</math>
 
<math><B>=H-\frac{1}{2\kappa}<|\psi|^2>=H-\frac{1}{2\kappa}\frac{H-\kappa}{\frac{1}{2\kappa}-\kappa}\frac{<|\psi|^2>^2}{<|\psi|^4>}</math>
 
<math><B>=H-\frac{\kappa -H}{2\kappa^2-1}\frac{1}{\beta_A}</math>
 
or, recovering the "tilde(~)",
 
<math><\tilde{B}>=\tilde{H}-\frac{\kappa -\tilde{H}}{2\kappa^2-1}\frac{1}{\beta_A}</math>
 
where
 
<math>\beta_A=\frac{<|\psi|^4>}{<|\psi|^2>^2}</math>
 
is the famous Abrikosov parameter which does not depend on <math>\tilde{H}</math> or the normalization of <math>\psi</math>. It is easy to see that <math>\beta_A</math> is always greater than or equal to 1:
 
<math>|\psi|^2=<|\psi|^2>+\delta |\psi|^2</math>
 
<math><|\psi|^4>=<|\psi|^2>^2+2<|\psi|^2><\delta |\psi|^2>+<(\delta |\psi|^2)^2></math>
 
Note that
 
<math><\delta |\psi|^2>=0</math>
 
Hence
 
<math><|\psi|^4>=<|\psi|^2>^2+<(\delta |\psi|^2)^2>\ge <|\psi|^2>^2</math>
 
<math>\beta_A=\frac{<|\psi|^4>}{<|\psi|^2>^2}\ge 1</math>
 
The equality holds when <math>\psi=1</math>.
 
Recall that the surface free energy is given by
 
<math>F_s-F_n=\frac{H_c^2}{4\pi}\lambda^3\int d^3\tilde{r}\,\left[(\tilde{B}-\tilde{H})^2-\frac{1}{2}|\psi|^4\,\right]</math>
 
In order to obtain the thermodynamic potential which when differentiated with respect to H gives B we need to subtract <math>\tilde{H}^2</math> (which came from the normal term) from the above expression:
 
<math>\Omega=\frac{H_c^2}{4\pi}\lambda^3\int d^3\tilde{r}\,\left[(\tilde{B}-\tilde{H})^2-\tilde{H}^2-\frac{1}{2}|\psi|^4\,\right]</math>
 
<math>=\frac{H_c^2}{4\pi}L^3\,\left[\frac{1}{4\kappa^2}<|\psi|^4>-\tilde{H}^2-\frac{1}{2}<|\psi|^4>\,\right]</math>
 
<math>=\frac{H_c^2}{8\pi}L^3\,\left[\frac{1-2\kappa^2}{2\kappa^2}<|\psi|^4>-2\tilde{H}^2\,\right]</math>
 
<math>=\frac{H_c^2}{8\pi}L^3\,\left[\frac{2}{1-2\kappa^2}\frac{1}{\beta_A}(\tilde{H}-\kappa)^2-2\tilde{H}^2\,\right]</math>
 
Note that for type-II superconductor, <math>\kappa >1/\sqrt{2}</math>, the whole expression is negative. So, to lower the the free energy we need to minimize <math>\beta_A</math>.
 
Knowing <math>\Omega</math> we can obtain <math>\tilde{B}</math> by differentiating <math>\Omega</math> with respect to <math>\tilde{H}</math> and hence obtain the magnetization which is measurable.
 
<math>-\frac{4\pi}{L^3}\frac{\partial\Omega}{\partial H}=B</math>
 
<math>\tilde{H}=\frac{H}{\sqrt{2}H_c};\frac{\partial}{\partial H}=\frac{\partial\tilde{H}}{\partial H}\frac{\partial}{\partial H}=\frac{1}{\sqrt{2}H_c}\frac{\partial}{\partial\tilde{H}}</math>
 
<math>B=-\frac{H_c}{2\sqrt{2}}4(\frac{1}{\beta_A}\frac{\tilde{H}-\kappa}{1-2\kappa^2}-\tilde{H})
=-\frac{2H_c}{\sqrt{2}}\,\left[\frac{1}{\beta_A}\frac{1}{1-2\kappa^2}(\frac{H}{\sqrt{2}H_c}-\frac{H_{c2}}{\sqrt{2}H_c})-\frac{H}{\sqrt{2}H_c}\,\right]=\frac{1}{\beta_A}\frac{1}{1-2\kappa^2}(H-H_{c2})+H\le H</math>
 
[[Image:pic8.jpg|right]]
 
for <math>H<H_{c2}</math>.
 
<math>B=H+4\pi M</math>
 
<math>4\pi M=B-H=\frac{1}{\beta_A}\frac{1}{2\kappa^2-1}(H-H_{c2})<0</math>
 
which means the type-II superconductor behaves like a diamagnet.
 
=== Abrikosov Vortex States ===
 
Now let's follow the step of Abrikosov to derive the vortex structure in type-II superconductor.
 
<math>\psi=\Sigma_{n=-\infty}^{\infty}C_ne^{ikny}e^{-\frac{1}{2}\kappa^2(x-\frac{kn}{\kappa^2})^2}
=C\Sigma_{n=-\infty}^{\infty}e^{ikny}e^{-\frac{1}{2}\kappa^2(x-\frac{kn}{\kappa^2})^2}</math>
 
where <math>C_n=C</math>.
 
<math>|\psi|^2=|C|^2\Sigma_{m,n}e^{-ikny}e^{ikmy}e^{-\frac{1}{2}\kappa^2(\tilde{x}-\frac{kn}{\kappa^2})^2}e^{-\frac{1}{2}\kappa^2(\tilde{x}-\frac{km}{\kappa^2})^2}</math>
 
<math>\int d\tilde{x}d\tilde{y}d\tilde{z}|\psi|^2=L_yL_z\Sigma_n\int d\tilde{x}|C|^2e^{-\kappa^2(\tilde{x}-\frac{kn}{\kappa^2})^2}=L_yL_z\frac{\kappa^2}{k}\tilde{L}_x\frac{\sqrt{\pi}}{\kappa}|C|^2</math>
 
Thus
 
<math><|\psi|^2>=|C|^2\frac{\sqrt{\pi}\kappa}{k}</math>
<math>|\psi|^4=|C|^4\Sigma_{m,n}\Sigma_{s,t}e^{-ikmy}e^{ikny}e^{-iksy}e^{ikty}e^{-\frac{\kappa^2}{2}(x-\frac{km}{\kappa^2})^2}e^{-\frac{\kappa^2}{2}(x-\frac{kn}{\kappa^2})^2}e^{-\frac{\kappa^2}{2}(x-\frac{ks}{\kappa^2})^2}e^{-\frac{\kappa^2}{2}(x-\frac{kt}{\kappa^2})^2}</math>
 
<math>\int dxdydz|\psi|^4=L_yL_z|C|^4\Sigma_{m,n,s}\int dx e^{-\frac{\kappa^2}{2}(x-\frac{km}{\kappa^2})^2}e^{-\frac{\kappa^2}{2}(x-\frac{kn}{\kappa^2})^2}e^{-\frac{\kappa^2}{2}(x-\frac{ks}{\kappa^2})^2}e^{-\frac{\kappa^2}{2}(x-\frac{k(m-n+s)}{\kappa^2})^2}</math>
 
<math>=L_yL_z|C|^4\Sigma_{m,n,s}\sqrt{\frac{\pi}{2}}\frac{1}{\kappa}e^{-\frac{k^2}{2\kappa^2}((m-n)^2+(n-s)^2)}</math>
 
<math>=L_yL_z|C|^4\sqrt{\frac{\pi}{2}}\frac{1}{k}\frac{\kappa^2}{k}\tilde{L}_x\Sigma_{m,n}e^{-\frac{k^2}{2\kappa^2}(m^2+n^2)}</math>
 
<math>\beta_A=\frac{1}{\sqrt{2\pi}}\frac{k}{\kappa}(\Sigma_{n=-\infty}^{\infty}e^{-\frac{k^2n^2}{2\kappa^2}})^2</math>
 
Minimize <math>\beta_A</math> with respect to k we find that at minimum
 
<math>k=\sqrt{2\pi}\kappa</math>
 
<math>\psi=C\Sigma_{n=-\infty}^{\infty}e^{i\sqrt{2}\pi n\kappa\tilde{y}}e^{-\frac{1}{2}(\kappa\tilde{x}-\sqrt{2}\pi n)^2}</math>
 
where <math>\beta_A=1.18034</math> and
 
<math>\psi=Ce^{-\frac{1}{2}\kappa^2\tilde{x}^2}\Sigma_{n=-\infty}^{\infty}e^{i\sqrt{2\pi}n\kappa\tilde{y}}e^{\sqrt{2\pi}n\kappa\tilde{x}}e^{-\pi n^2}=Ce^{-\frac{1}{2}\kappa^2\tilde{x}^2}\Sigma_{n=-\infty}^{\infty}e^{\sqrt{2\pi}\kappa (\tilde{x}+i\tilde{y})n}e^{-\pi n^2}=Ce^{-\frac{1}{2}\kappa^2\tilde{x}^2}\theta_3 (\sqrt{\frac{\pi}{2}}\kappa (\tilde{y}-i\tilde{x}),e^{-\pi})</math>
 
where <math>\theta_3</math> is the Jacobi elliptic theta function. The <math>\psi</math> vanishes at points:
 
<math>\tilde{y}=\sqrt{\frac{\pi}{2}}\frac{1}{\kappa}(2m-1)</math>
 
and
 
<math>\tilde{x}=\sqrt{\frac{\pi}{2}}\frac{1}{\kappa}(1-2n)\rightarrow\sqrt{\frac{\pi}{2}}\frac{1}{\kappa}(2n-1)</math>
 
Obviously, the zeros form a square lattice of spacing <math>\frac{\sqrt{2\pi}}{\kappa}</math> in reduced units.
 
Now consider the current lines:
 
<math>\vec{j}=\frac{-i}{2\kappa}(\psi^*\nabla\psi-\psi\nabla\psi^*)-|\psi|^2\tilde{A}</math>
 
<math>j_x=\frac{1}{2\kappa i}\Sigma_{m,n}k(n-m)C_m^*C_ne^{-ik(m-n)\tilde{y}}\phi_n(\tilde{x})\phi_m(\tilde{x})=\frac{-1}{2\kappa}\frac{\partial}{\partial\tilde{y}}|\psi|^2</math>
 
<math>j_y=\frac{1}{2\kappa}\frac{\partial}{\partial\tilde{x}}|\psi|^2</math>
 
The lines of constant current magnitude are determined by the following condition:
 
[[Image:pic9.jpg|right]]
 
<math>\frac{dy}{dx}=\frac{j_y}{j_x}</math>
 
<math>j_xdy-j_ydx=0</math>
 
<math>\frac{-1}{2\kappa}(\frac{\partial}{\partial y}|\psi|^2)dy-\frac{1}{2\kappa}(\frac{\partial}{\partial x}|\psi|^2)dx=0</math>
 
<math>d|\psi|^2=0</math>
 
<math>|\psi|^2=const.</math>
 
which means the current flows along the contours of <math>|\psi|^2</math>.
 
However, the actual solution which minimizes <math>\beta_A</math> is a triangular lattice for which the coefficients satisfy
 
<math>C_{n+4}=C_n;C_0=C_1=C;C_2=C_3=-C</math>
 
and <math>\beta_A=1.16</math>.
 
==Going below Tc with the Saddle Point Approximation==
 
So, previously all of this work has shown us the behaviour of a superconducting system near <math>T_c</math> only. If we want to go into lower temperatures, we will have to make a careful saddle-point approximation, following Bardeen, Cooper, and Schrieffer (BCS). Once again, we can start from our [[#Pairing Hamiltonian and BCS instability|microscopic 'toy' Hamiltonian]], and gain useful information.
 
Recall that the partition function can be written,
 
<math> \mathbb{Z} = \int{d\Delta^* d\Delta} \left[ \int{D\psi^* D\psi \ e^{-S_{BCS} - S_{\Delta}}} \right] </math>
 
Where
 
<math> S_{\Delta} = \frac{-1}{g} \int_{0}^{\beta}{d\tau} \int{d^3r}\Delta^*(\vec{r},\tau) \Delta(\vec{r},\tau) </math>
 
and
 
<math> S_{BCS} = S_0 + S_{int} \;\; \text{where} \;\; S_0 = \int_{0}^{\beta}{d\tau}\int{d^3r}\left[ \psi_{\sigma}^*(\vec{r},\tau) \left( \frac{\partial}{\partial \tau} + \epsilon_p - \mu \right) \psi_{\sigma}(\vec{r},\tau) \right]  </math>
<math> \text{and} \;\; S_{int} = \int_{0}^{\beta}{d\tau}\int{d^3r} \left[\Delta^*(\vec{r},\tau)\psi_{\downarrow}(\vec{r},\tau) \psi_{\uparrow}(\vec{r},\tau) + \Delta(\vec{r},\tau)\psi_{\downarrow}(\vec{r},\tau)\psi_{\uparrow}(\vec{r}.\tau) \right]</math>
 
Previously, we used a cumulant expansion around <math>T = T_c</math> to find the Ginzburg-Landau equations, along with the assumption that <math>\Delta</math> was small. Now, if we throw away this assumption, pray that fluctuations are small, and that there is only one saddle point, we will be able to successfully describe the superconducting state deep below <math>T_c</math> by demanding:
 
<math>\frac{\partial S_{eff}[\Delta]}{\partial \Delta*} = 0 </math>
 
===Self-Consistency Equation===
 
The solution of this functional derivative equation will give the value of <math>\Delta = \Delta_{sp}</math> at the saddle point (and also the self-consistency equation for this mean-field theory.)
 
To that end:
: <math> S_{eff}[\Delta] = S_{\Delta} - \ln \left( \int{D\psi D\psi^*} \;\; e^{-S_{BCS}}\right) </math>
 
: <math> \frac{\partial S_{\Delta}}{\partial \Delta^*(\vec{r},\tau)} = \frac{-\Delta(\vec{r},\tau)}{g}</math>
 
: <math> \frac{\partial \left[ -\ln\left(\int{D\psi D\psi^*} \;\; e^{-S_{BCS}}\right) \right] }{\partial \Delta^*(\vec{r},\tau)} = \frac{-1}{\int{D\psi D\psi^*} \;\; e^{-S_{BCS}}} \int{D\psi D\psi*} \frac{\partial e^{-S_{BCS}}}{\partial \Delta^*} = \frac{\int{D\psi D\psi^*} \;\; e^{-S_{BCS}} \frac{\partial S_{BCS}}{\partial \Delta^*}}{\int{D\psi D\psi^*} \;\; e^{-S_{BCS}}} </math>
 
and, since
 
: <math>\frac{\partial S_{BCS}}{\partial \Delta^*} = \psi_{\downarrow}(\vec{r},\tau) \psi_{\uparrow}(\vec{r},\tau)</math>
 
We arrive at
 
: <math>\frac{\partial \left[ -\ln\left(\int{D\psi D\psi^*} \;\; e^{-S_{BCS}}\right) \right] }{\partial \Delta^*(\vec{r},\tau)} = \frac{\int{D\psi D\psi^*} \;\; e^{-S_{BCS}} \psi_{\downarrow}(\vec{r},\tau) \psi_{\uparrow}(\vec{r},\tau)}{\int{D\psi D\psi^*} \;\; e^{-S_{BCS}}} = \langle \psi_{\downarrow}(\vec{r},\tau) \psi_{\uparrow}(\vec{r},\tau)\rangle_{BCS} </math>
 
So that, from the saddle-point condition, we find the Self-Consistency Equation:
 
<math>{ \frac{\Delta_{sp}(\vec{r},\tau)}{g}} = {\langle\psi_{\downarrow}(\vec{r},\tau) \psi_{\uparrow}(\vec{r},\tau)\rangle_{BCS}} </math>
 
This contains the same information as the Ginzburg-Landau equations, but also more, as we can now go far below <math>T_c</math>. Evaluation of this cam be done formally, but is difficult and not terribly enlightening. Instead, we will search for a solution in which <math\psi</math> is independent of <math>\vec{r}</math> and <math>\tau</math>, similar to our solution near <math>T_c</math>. In this case, we also expect that <math>\Delta</math> will be independent of <math>\vec{r}</math> and <math>\tau</math>.
 
First, it is useful to write down the action in momentum and frequency space, before evaluating the correlator.
 
Start with <math> \psi(\vec{r},\tau) = \frac{1}{\beta}\sum_{\omega_n}\sum_{\vec{k}}\frac{1}{\sqrt{V}} e^{i\vec{k}\cdot\vec{r}}\psi_{k}(i\omega_n)</math>
 
For the three terms in <math>S_{BCS}</math>, we have:
 
<math> S_0 = \int_{0}^{\beta}{d\tau}\int{d^3r}\left[ \psi_{\sigma}^*(\vec{r},\tau) \left( \frac{\partial}{\partial \tau} + \epsilon - \mu \right) \psi_{\sigma}(\vec{r},\tau) \right] </math>
: <math> = \frac{1}{V}\sum_{\vec{k},\vec{k'}} \frac{1}{\beta^2}\sum_{\omega_n,\nu_m}\int_{0}^{\beta}{d\tau}\int{d^3r}e^{i\nu_{m}\tau}e^{-i\vec{k'}\cdot\vec{r}} \left(-i\omega_n + \epsilon_p - \mu \right) e^{-i\omega_n\tau}e^{i\vec{k}\cdot\vec{r}}\psi_{\sigma,\vec{k'}}^*(i\nu_m)\psi_{\sigma, \vec{k}}(i\omega_n)</math>
 
The integral over all space give <math>V \delta_{\vec{k},\vec{k'}}</math>, and the integral over imaginary time gives a factor of <math>\beta \delta(\omega_n - \nu_m)</math>, so that we find
 
<math>S_0 = \frac{1}{\beta}\sum_{\omega_n}\sum_{\vec{k}} \psi_{\sigma, \vec{k}}^*(i\omega_n)\left( -i\omega_n + \epsilon_p - \mu \right)\psi_{\sigma,\vec{k}}(i\omega_n) </math>
 
Now, for the 2nd (pairing) term:
 
<math> \Delta^*\int_{0}^{\beta}{d\tau} \int{d^3r} \psi_{\downarrow}(\vec{r},\tau) \psi_{\uparrow}(\vec{r},\tau) = \Delta^* \frac{1}{V}\sum_{\vec{k},\vec{k'}} \frac{1}{\beta^2}\sum_{\omega_n,\nu_m}\int_{0}^{\beta}{d\tau}\int{d^3r} e^{-i(\nu_m + \omega_n)\tau) e^{i(\vec{k} + \vec{k'})\cdot\vec{r}} \psi_\downarrow,\vec{k'}}(i\nu_m) \psi_{\uparrow,\vec{k}}(i\omega_n) </math>
 
This time, the integral over real space gives <math>V \delta(\vec{k}+\vec{k'})</math>, and the integral over imaginary time gives a factor of <math>\beta \delta(\omega_n + \nu_m)</math>, so the 2nd term becomes:
 
: <math> = \frac{\Delta^*}{\beta}\sum_{\vec{k}}\sum_{\omega_n}\psi_{\downarrow,\vec{k}}(i\omega_n)\psi_{\uparrow,-\vec{k}}(-i\omega_n)</math>
 
The hermitian conjugate of the above (the 3rd term in <math>S_{BCS}</math>) gives
 
:<math> \;\; \frac{\Delta}{\beta}\sum_{\vec{k}}\sum_{\omega_n}\psi_{\uparrow,-\vec{k}}^*(-i\omega_n)\psi_{\downarrow,\vec{k}}^*(i\omega_n)</math>
 
So that, for the entire <math>S_{BCS}</math>, we have found:
 
<math>S_{BCS} = \frac{1}{\beta}\sum_{\omega_n}\sum_{\vec{k}}\left[ \left( i\omega_n + \epsilon_{-\vec{k}} - \mu \right)\psi_{\uparrow, -\vec{k}}^*(-i\omega_n)\psi_{\uparrow,-\vec{k}}(-i\omega_n) + \left( -i\omega_n + \epsilon_{\vec{k}} - \mu \right)\psi_{\downarrow, \vec{k}}^*(i\omega_n)\psi_{\downarrow,\vec{k}}(i\omega_n) \right] </math>
::<math> + \frac{1}{\beta}\sum_{\omega_n}\sum_{\vec{k}}\left[ \Delta^*\psi_{\downarrow,\vec{k}}(i\omega_n)\psi_{\uparrow,-\vec{k}}(-i\omega_n) + \Delta\psi_{\uparrow,-\vec{k}}^*(-i\omega_n)\psi_{\downarrow,\vec{k}}^*(i\omega_n) \right] </math>
 
Now, to evaluate the Gaussian integrals in the correlator, it is extremely beneficial to write <math>S_{BCS}</math> like a matrix. To this end, we construct the so-called Nambu Spinors:
 
<math> \Psi_{\vec{k}}(i\omega_n) = \begin{pmatrix} \psi_{-\vec{k},\uparrow}(-i\omega_n) \\ \psi^*_{\vec{k},\downarrow}(i\omega_n) \end{pmatrix} </math>
 
<math> \Psi^*_{\vec{k}}(i\omega_n) = \begin{pmatrix} \psi^*_{-\vec{k},\uparrow}(-i\omega_n) & \psi^*_{\vec{k},\downarrow}(i\omega_n) \end{pmatrix} </math>
 
So now, we can write the BCS action as:
 
<math>
S_{BCS} = \frac{1}{\beta}\sum_{\omega_N}\sum_{\vec{k}}\Psi^*_{\vec{k}}(i\omega_n)
\begin{pmatrix}
i\omega_n + \epsilon_{-\vec{k}} - \mu & \Delta \\
\Delta^* & i\omega_n - \epsilon_{\vec{k}} + \mu
\end{pmatrix}
\Psi_{\vec{k}}(i\omega_n)
</math>
 
Now, we can examine the correlator we found on the right-hand-side of the self-consistency equation:
 
<math> \langle \psi_{\downarrow}(\vec{r},\tau)\psi_{\uparrow}(\vec{r},\tau)\rangle = \frac{1}{\beta^2 V}\sum_{\omega_n,\nu_m}e^{-i(\omega_n + \nu_M)\tau} \sum_{\vec{k},\vec{k'}} e^{i(\vec{k} + \vec{k'})\cdot\vec{r}}\langle \psi_{\vec{k},\downarrow}(i\omega_n) \psi_{\vec{k'},\uparrow}(i\nu_m) \rangle </math>
:<math> = \frac{1}{\beta^2 V}\sum_{\omega_n,\nu_m}e^{-i(\omega_n + \nu_M)\tau} \sum_{\vec{k},\vec{k'}} e^{i(\vec{k} + \vec{k'})\cdot\vec{r}} \langle \Psi^*_{\vec{k}}(i\omega_n)_2 \Psi_{-\vec{k'}}(-i\nu_m)_1 \rangle </math>
 
So, we need to write down a generic matrix for <math>\langle \Psi^*_{\vec{k}}(i\omega_n)_{\mu} \Psi_{-\vec{k'}}(-i\nu_m)_{\lambda} \rangle</math>, and take element <math> (2,1) </math> for our result. The gaussian integrals require that <math>\vec{k} = -\vec{k'}</math> and <math> i\omega_n = -i\nu_m </math> for convergence.
Using our [[Phy5670#Gaussian_Integrals| technology from last semester]], we can show that:
 
<math>\langle \Psi^*_{\vec{k}}(i\omega_n)_{\mu} \Psi_{-\vec{k'}}(-i\nu_m)_{\lambda} \rangle = \delta(\vec{k} + \vec{k'})\beta \delta(\omega_n + \nu_m) \left[
\begin{pmatrix}
i\omega_n + \epsilon_{\vec{k}} - \mu & \Delta \\
\Delta^* & i\omega_n - \epsilon_{\vec{k}} + \mu
\end{pmatrix}^{-1} \right]_{\mu\lambda}
</math>
 
<math>  \Psi^*_{\vec{k}}(i\omega_n)_2 \Psi_{-\vec{k'}}(-i\nu_m)_1 \rangle = \delta(\vec{k} + \vec{k'})\beta \delta(\omega_n + \nu_m)\frac{\Delta}{-\omega_n^2 - (\epsilon_{\vec{k}} - \mu)^2 - \Delta^* \Delta } </math>
 
Now, we can easily write down the correlator:
 
<math>\langle \psi_{\downarrow}(\vec{r},\tau)\psi_{\uparrow}(\vec{r},\tau)\rangle = \frac{1}{\beta}\sum_{\omega_n}\frac{1}{V}\sum_{\vec{k}}\frac{\Delta}{-\omega_n^2 - (\epsilon_{\vec{k}} - \mu)^2 - \Delta^* \Delta } </math>
 
This reshapes our consistency equation to:
 
<math>\frac{\Delta}{g} = \frac{1}{\beta}\sum_{\omega_n}\frac{1}{V}\sum_{\vec{k}}\frac{\Delta}{-\omega_n^2 - (\epsilon_{\vec{k}} - \mu)^2 - \Delta^* \Delta } </math>
 
Which has a trivial, <math> \Delta = 0 </math> solution, and far more interesting solution where
 
<math>\frac{1}{g} = \frac{1}{\beta}\sum_{\omega_n}\frac{1}{V}\sum_{\vec{k}}\frac{1}{-\omega_n^2 - (\epsilon_{\vec{k}} - \mu)^2 - |\Delta|^2 } </math>
 
Note that this is only possible if <math> g < 0 </math>!
 
-We will not get S-wave superconductivity for <math>\Delta</math> independent of <math>k</math>
 
-In cuprates, <math>\Delta(k)</math> changes sign with <math>90^{0}</math>. It changes sign like D-wave.
 
-If <math>g</math> is repulsive system does not like to make uniform gap!
 
-In Fe-Pnictides, pairing gaps on different Fermi surfaces have different signs, with repulsion.
 
 
Let us define
<math>E_{k}^{2} = (\epsilon_{k} - \mu)^{2} + |\Delta|^{2} </math>
 
So
<math>\frac{-1}{g} = \frac{1}{\beta}\sum_{\omega_n}\frac{1}{V}\sum_{\vec{k}}\frac{1}{-\omega_n^2 - (\epsilon_{\vec{k}} - \mu)^2 - |\Delta|^2 } </math>
 
<math>\frac{-1}{g} = \frac{1}{\beta}\sum_{\omega_n}\sum_{\vec{k}}\frac{1}{-\omega_n^2 - E_{\vec{k}}^{2} } </math>
<math>= \frac{1}{\beta}\sum_{\omega_{n}}\frac{1}{(i\omega_{n} - E_{\vec{k}})(i\omega_{n} + E_{\vec{k}})}</math>
 
<math>\omega_{n} = \frac{(2n + 1)\pi}{\beta}</math>
 
We can write this as a complex contour integral, with a series of pole along the imaginary axis.
 
<math>\frac{-1}{g} = \oint_{c} \frac{dz}{2\pi i}\frac{1}{e^{\beta z} + 1}f(z)</math>
 
since
<math> \frac{1}{e^{\beta(\frac{\pi}{\beta}i + \delta z)} + 1} = \frac{1}{-e^{\beta \delta z +1}} = \frac{-1}{\beta\delta z}</math>
 
<math>\frac{-1}{g}= \oint_{c}\frac{dz}{(2\pi i)}\frac{1}{e^{\beta z} + 1}\frac{1}{(z-E_{\vec{k}})(z+E_{\vec{k}})}</math>
 
There is no divergence as <math>z \to i\infty</math>
There are two poles, at <math>Z = \pm E_{\vec{k}}</math>
 
The contour can be deformed so that they exclude these two poles, and the integrand is tiny as <math>Re(z) \to \infty</math>
The only contribution comes from the the loop around the two poles. For <math>Re(z)\to -\infty</math>, the integrand falls off as <math>\frac{1}{z^{2}}</math> , which, by Jordan's lemma,is fast enough for convergence.
 
So the integrand becomes
<math> = \frac{-1}{2\pi i}2\pi i \left [\frac{1}{(e^{\beta E_{\vec{k}}} +1)}\frac{1}{2E_{\vec{k}}} + \frac{1}{(e^{\beta E_{\vec{k}}}+1)}\frac{1}{-2E_{\vec{k}}}  \right ]</math>
 
<math>= \frac{1}{2E_{\vec{k}}}\left [ \frac{1}{(e^{-\beta E_{\vec{k}}} +1)} - \frac{1}{(e^{\beta E_{\vec{k}}} +1)}\right ]</math>
 
<math>=\frac{1}{2E_{\vec{k}}}\left [ \frac{1}{e^{-\beta E_{\vec{k}}/2}2\cosh(\beta E_{\vec{k}}/2)} - \frac{1}{e^{\beta E_{\vec{k}}/2}2\cosh(\beta E_{\vec{k}}/2)}\right ]</math>
 
<math>= \frac{1}{4E_{\vec{k}}\cosh(\beta E_{\vec{k}}/2)}\left [ e^{\beta E_{\vec{k}}/2}2 - e^{-\beta E_{\vec{k}}/2}2 \right ]</math>
 
<math>= \frac{1}{2E_{\vec{k}}}\tanh(\beta E_{\vec{k}}/2)</math>
 
<math>\Rightarrow \frac{1}{g} = \frac{-1}{V}\sum_{\vec{k}}\frac{1}{2E_{\vec{k}}}\tanh(E_{\vec{k}}/2k_{b}T)</math>
 
===Solution of the Self-Consistency Equation===
 
By using our usual trick of <math>\sum_{\vec{k}} \to V\int \frac{d^{3}k}{(2\pi)^3}</math>,
 
we can convert the sum into an integral:
<math>\frac{1}{|g|} = -\frac{1}{g} = \int \frac{d^{3}k}{(2\pi)^3}\frac{\tanh(E_{\vec{k}}/2k_{b}T)}{2E_{\vec{k}}}</math>
 
At this point, we can, if we wish, solve this equation on a computer, at any temperature
 
For <math>\Delta \to 0</math>, the RHS is <math>\frac{\tanh((\epsilon_{\vec{k}}-\mu)/2k_{b}T)}{(\epsilon_{\vec{k}}-\mu)}</math> always intersects the LHS <math>\left (\frac{1}{|g|}  \right )</math>, and we are guaranteed a solution.
 
As we raise temperature, log-divergence of integer gets cut-off.
Finding T-dependence of the intercept gives T-dependence of <math>\Delta</math>
 
There can be an analytic solution as <math>T \to 0</math>
 
<math>\frac{1}{|g|} = -\frac{1}{g} = \int \frac{d^{3}k}{(2\pi)^3}\frac{1}{2E_{\vec{k}}} = \int \frac{d^{3}k}{(2\pi)^3}\frac{1}{2\sqrt{(\epsilon_{\vec{k}} - \mu)^2 + |\Delta|^2}} </math>
 
Assuming that the density of states is smooth, and dominated by <math>\xi \rightarrow 0 </math>, we can write:
 
<math>\simeq N(\mu) \int_{-\hbar \omega_D}^{\hbar \omega_D}{d\xi}\frac{1}{2\sqrt{\xi^2 + |\Delta|^2}} = N(\mu) \int_0^{\frac{\hbar \omega_D}{|\Delta|}}{d\xi}\frac{1}{\sqrt{\xi^2 + |\Delta|^2}} </math>
 
Now,
 
<math> \frac{1}{|g|} = N(\mu) \ln{\left(\frac{\hbar \omega_D}{\Delta} + \sqrt{1 + \frac{\hbar^2\omega_d^2}{|\Delta|^2}}\right)} </math>
 
In typical systems, <math>\Delta \ll \omega_D </math>! For example, in Al, <math>|\Delta| \simeq 2K</math> while <math> \omega_d \simeq 300-500K</math>.
So, we can safely assume that <math> \frac{\hbar \omega_D}{|\Delta|} \gg 1 </math>, which gives that:
 
<math>\frac{1}{|g|} = N(\mu) \ln{\left(\frac{h\hbar\omega_D} {\Delta} \right)} \Rightarrow |\Delta| = 2\hbar\omega_D e^{-\frac{1}{|g|N(\mu)}}</math> at <math> T = 0</math>.
 
This actually lowers condensation energy more than <math>\Delta = 0</math>! It is also important to note, that since <math>|g|</math> enters in a exponential, that the <math>\Delta</math> is completely flat as <math>|g| \rightarrow 0</math>, showing that this result is completely non-perturbative. Note that <math>\Delta</math> is only exactly zero when <math>|g|</math> is zero, so that ''no matter how small the interaction'', we still have a superconducting state.
 
[[Image:Deltavsg.jpeg]]
 
 
Happily, this result is also something we can compare to experimental results from tunneling measurements on suitable materials (S-wave superconductors). From the definition of <math>T_c</math>:
 
<math> \frac{|\Delta|(T = 0)}{k_B T_c} = \frac{2}{1.134} = 1.764 </math>
 
By convention, the BCS results for twice this is compared to experiment.
 
<math>\frac{2|\Delta|(T = 0)}{k_B T_c} = 3.53 </math>
 
{|border="4"
|Element || Measured <math>\frac{2\Delta(0)}{k_B T_c}</math>
|-
|Al ||<math>4.2\pm0.6</math>
|-
|  ||<math>2.5 \pm0.3</math>
|-
|  ||<math> 3.2\pm0.4 </math>
|-
|Cd ||<math>~3.2</math>
|-
|Nb ||<math>~3.84 </math>
|-
|Sn ||<math>3.49\pm0.1</math>
|-
|  ||<math>3.1\pm0.05</math>
|-
|  ||<math>3.5\pm0.18</math>
|-
|}
 
====Going between <math>T = 0</math> and <math>T_c</math>====
At zero-Temperature,
<math>\Delta = 2\hbar\omega_{D}e^{\frac{1}{gN_{0}}}</math>
 
Now lets look at how it agrees with the Ginzburg-Landau theory near <math>T_{c}</math>
 
At finite T,
<math>\frac{1}{g} = \int \frac{d^{3}k}{(2\pi)^3}\frac{\tanh(E_{\vec{k}}/2T)}{2E_{\vec{k}}}</math>
 
in units of
<math>E_{\vec{k}} = \left [ \epsilon^{2} + \Delta^2(T) \right ]^\frac{1}{2}</math>
 
<math>\Rightarrow \frac{1}{g} = N_{\mu}\int^{\omega_{D}}_{0}d\epsilon \frac{\tanh\left( \frac{\left [ (\epsilon^{2} + \Delta^2(T) \right ]^\frac{1}{2}}{2T}\right)}{ \left [ \epsilon^{2} + \Delta^2(T) \right ]^\frac{1}{2}}</math>
 
Now. let us parametrize <math>\Delta(T)</math> with <math>\Delta(T = 0)</math>
 
(Note at
<math>T =0, \frac{1}{g} = N_{\mu}\int^{\omega_{D}}_{0}d\epsilon \frac{1}{ \left [ \epsilon^{2} + \Delta^2 \right ]^\frac{1}{2}}</math>)
 
so <math> N_{0}(\mu)\int_{0}^{\omega_{D}} d\epsilon \frac{1}{\sqrt{\epsilon^2 + \Delta^{2}(T=0)}}= N_{0}(\mu)\int_{0}^{\omega_{D}} d\epsilon \frac{tanh\left( \frac{\sqrt{(\epsilon^{2} + \Delta^2(T)}}{2T}\right)}{\sqrt{\epsilon^2 + \Delta^{2}(T)}}</math>
 
Or,<math>\int_{0}^{\omega_{D}\to \infty} d\epsilon \left (\frac{1}{\sqrt{\epsilon^2 + \Delta^{2}(T=0)}} - \frac{1}{\sqrt{\epsilon^2 + \Delta^{2}(T)}}\right ) = \int_{0}^{\omega_{D\to \infty}} d\epsilon \left (\frac{\tanh\left( \frac{\sqrt{(\epsilon^{2} + \Delta^2(T)}}{2T}\right)-1}{\sqrt{\epsilon^2 + \Delta^{2}(T)}} \right )</math>
 
Now <math>\omega_{D} \sim 1000-3000 K</math> , which is very large compared to the energy scale we are working at, and so we can replace the upper limit in the integrals by <math>\infty</math> . This will cause the first term in the LHS to vanish, since it falls faster than <math>\frac{1}{\epsilon^{2}}</math>
We get,
 
<math>\ln\left(\frac{\Delta(T)}{\Delta(0)}\right ) = \int_{0}^{\omega_{D\to \infty}} d\epsilon \left (\frac{\tanh\left( \frac{\sqrt{(\epsilon^{2} + \Delta^2(T)}}{2T}\right)-1}{\sqrt{\epsilon^2 + \Delta^{2}(T)}} \right )</math>
 
Since we expect <math>\Delta(T)<\Delta(0)</math>, we can switch their order inside the logarithm:
 
<math>\ln\left(\frac{\Delta(T=0)}{\Delta(T)}\right ) = \int_{0}^{\omega_{D\to \infty}} d\epsilon \left (\frac{1-\tanh\left( \frac{\sqrt{(\epsilon^{2} + \Delta^2(T)}}{2T}\right)}{\sqrt{\epsilon^2 + \Delta^{2}(T)}} \right )</math>
We are now in a position to solve for <math>\Delta(T)</math> near <math>T = 0</math>
 
Let <math>\epsilon/\Delta(T) = \xi</math>
 
So the RHS is <math>\int_{0}^{\infty}d\xi \frac{1}{\sqrt{\xi^{2}+1}}\left(1-\tanh\left(\frac{\Delta(T)}{2T}\sqrt{\xi^{2}+1}\right)\right)</math>
 
Let us call <math>\sqrt{\xi^{2}+1} = X</math>. So
 
<math>\ln\left(\frac{\Delta(T=0)}{\Delta(T)}\right ) = \int_{1}^{\infty}dX \frac{1}{\sqrt{X^{2}-1}}\left(1-\tanh\left(\frac{\Delta(T)}{2T}X\right)\right)</math>
 
<math>\rightarrow \ln\left(\frac{\Delta(T=0)}{\Delta(T)}\right ) = \int_{1}^{\infty}dX \frac{1}{\sqrt{X^{2}-1}}\frac{2}{e^{\frac{\Delta(T)X}{T}+1}}</math>
Now as <math>T\to 0</math>, the integral clearly goes to zero
 
So the integral is
<math>2\int_{1}^{\infty}dX \frac{1}{\sqrt{X^{2}-1}}\frac{e^{-\frac{\Delta(T)X}{T}}}{e^{-\frac{\Delta(T)X}{T}}+1}</math>
 
<math>= 2\int_{1}^{\infty}dX \frac{1}{\sqrt{X^{2}-1}}e^{-\frac{\Delta(T)X}{T}}\sum_{n=0}^{\infty}(-1)^n e^{-\frac{n\Delta(T)X}{T}}</math>
 
<math>= 2\sum_{n=0}^{\infty}(-1)^n\int_{1}^{\infty}dX \frac{1}{\sqrt{X^{2}-1}} e^{-\frac{\Delta(T)X(n+1)}{T}}</math>
 
<math>= 2\sum_{n=0}^{\infty}(-1)^n K_{0}\left({\frac{\Delta(T)X(n+1)}{T}}\right)</math>
 
Where <math>K_{0}</math>  is the modified Bessel function of the 2nd kind.
 
As <math>x \to 0, K(x) \to \infty</math>, and as <math>x \to \infty, K(x) \to 0</math>
 
For small temperatures, the sum is dominated by the smallest n value. So approximately,
 
<math>\ln\left(\frac{\Delta(T=0)}{\Delta(T)}\right ) 2\simeq \sqrt{\frac{\pi T}{2\Delta(0)}}e^{-\frac{\Delta(0)}{T}}</math>
 
By noting that <math>\Delta(T)-\Delta(0)</math> is small, we can manipulate the logarithm to get
 
<math>-\ln\left(\frac{\Delta(T)-\Delta(0)+\Delta(0)}{\Delta(0)}\right )</math>
 
<math>= -\ln\left(1 + \frac{\Delta(T)-\Delta(0)}{\Delta(0)}\right )</math>
 
<math>\simeq -\frac{\Delta(T)-\Delta(0)}{\Delta(0)}</math>
 
<math>\Rightarrow 1- \frac{\Delta(T)}{\Delta(0)} \simeq 2\sqrt{\frac{\pi T}{2\Delta(0)}}e^{-\frac{\Delta(0)}{T}}</math>
 
<math>\Rightarrow \frac{\Delta(T)}{\Delta(0)} \simeq 1- 2\sqrt{\frac{\pi T}{2\Delta(0)}}e^{-\frac{\Delta(0)}{T}}</math>
 
Next,let us solve near <math>T = T_{c}</math> analytically
 
Near <math> T_{c}, \Delta(T) \to 0</math>
 
<math>-\frac{1}{gN_{0}} = \int_{0}^{\omega_{D}} d\epsilon \frac{\tanh\left(\frac{\sqrt{(\epsilon^{2} + \Delta^2(T)}}{2T}\right)}{\sqrt{\epsilon^2 + \Delta^{2}(T)}}</math>
 
At <math>T = T_{c}. \Delta = 0</math>. So,
 
<math>-\frac{1}{gN_{0}} = \int_{0}^{\omega_{D}} d\epsilon \frac{\tanh\left( \frac{\epsilon}{2T_{c}}\right)}{\epsilon}</math>
 
<math>\Rightarrow \int_{0}^{\omega_{D}} d\epsilon \frac{\tanh\left( \frac{\epsilon}{2T_{c}}\right)}{\epsilon} = \int_{0}^{\omega_{D}} d\epsilon \frac{\tanh\left(\frac{\sqrt{(\epsilon^{2} + \Delta^2(T)}}{2T}\right)}{\sqrt{\epsilon^2 + \Delta^{2}(T)}}</math>
 
Let us subtract <math>\int_{0}^{\omega_{D}} d\epsilon \frac{\tanh\left( \frac{\epsilon}{2T}\right)}{\epsilon}</math> from both sides
 
<math>\Rightarrow \int_{0}^{\omega_{D}} d\epsilon \frac{\tanh\left( \frac{\epsilon}{2T_{c}}\right)-\tanh\left( \frac{\epsilon}{2T_{c}}\right)}{\epsilon} = \int_{0}^{\omega_{D}} d\epsilon \left ( \frac{\tanh\left(\frac{\sqrt{(\epsilon^{2} + \Delta^2(T)}}{2T}\right)}{\sqrt{\epsilon^2 + \Delta^{2}(T)}} -\frac{\tanh\left( \frac{\epsilon}{2T}\right)}{\epsilon} \right )  </math>
 
Now if we let <math>\omega_{D} \to \infty</math>, the reason of which has already been argued, we can solve the LHS exactly, and we get,
 
<math>\ln \left (\frac{T}{T_{c}}\right)</math>
 
So we get
 
<math>\ln \left (\frac{T_{c}}{T}\right) = \int_{0}^{\infty} d\epsilon \left (\frac{\tanh\left( \frac{\epsilon}{2T}\right)}{\epsilon} -\frac{\tanh\left(\frac{\sqrt{(\epsilon^{2} + \Delta^2(T)}}{2T}\right)}{\sqrt{\epsilon^2 + \Delta^{2}(T)}}\right )</math>
 
The next logical step is to expand the hyperbolic tangent
 
<math>\tanh(\pi x/2) = \frac{4x}{\pi}\sum_{k = 0}^{\infty}\frac{1}{(2k+1)^2+x^2} </math>
 
So RHS
<math>=\int_{0}^{\infty} d\epsilon \left (\frac{4}{\pi^{2}T}\sum_{k = 0}^{\infty}\frac{1}{(2k+1)^2+\left ( \frac{\epsilon^2}{\pi^{2}T^{2}}\right)}-\frac{4}{\pi^{2}T}\sum_{k = 0}^{\infty}\frac{1}{(2k+1)^2+\left ( \frac{\epsilon^2}{\pi^{2}T^{2}}\right)} \right )</math>
 
<math>=\sum_{k = 0}^{\infty}\int_{0}^{\infty} d\epsilon\frac{4}{\pi^{2}T}\frac{\frac{\Delta^2}{\pi^{2}T{2}}}{\left ((2k+1)^2+\left ( \frac{\epsilon^2}{\pi^{2}T^{2}}\right)\right)^2} + O(\Delta^4)</math>
 
<math>\simeq \frac{4\Delta^{2}}{\pi^{3}T^{2}}\sum_{k = 0}^{\infty}\int_{0}^{\infty}d\epsilon\frac{1}{\left ((2k+1)^2 + \epsilon^{2} \right)}</math>
 
<math>=\frac{4\Delta^{2}}{\pi^{3}T^{2}}\sum_{k = 0}^{\infty}\frac{\pi}{4}\frac{1}{(2k +1)^{2}}</math>
 
This series converges, and we get,
 
<math>\ln \left (\frac{T_{c}}{T}\right) = \frac{\Delta^{2}(T)}{\pi^{2}T{2}}\frac{7\xi(3)}{8}</math>
Where <math>\xi(3)</math> is the order three Reimann Zeta function.
 
<math>\Rightarrow \frac{\Delta^{2}(T)}{\pi^{2}T{2}}\frac{7\xi(3)}{8}=-\ln \left (1+\frac{T-T_{c}}{T_{c}}\right) = \frac{T_{c}-T}{T{c}}</math>
 
<math>\Rightarrow \Delta^{2}(T) \simeq \frac{8\pi^{2}T_{c}(T_{c}-T)}{7\xi(3)}</math>
 
<math>\Rightarrow \Delta(T)\sim 3.063\sqrt{T_{c}(T_{c}-T)}</math>
 
===Condensation energy and T dependence of the thermodynamic field===
 
<math>Z = \int D\psi^{*}D\psi\int D\Delta^{*}D\Delta e^{-S_{BCS}-S_{\Delta}}= \int D\Delta^{*}D\Delta e^{-S_{eff}[\Delta]}</math>
 
Using the saddle point approximation
 
<math>e^{-S_{eff}[\Delta _{0}\, ]}\; =\; e^{-S_{\Delta }}\int D\psi ^{\ast }D\psi \; e^{-S_{BCS}[\Delta _{0\, }]}\; \sim e^{-\beta F}</math>
 
Then,
 
<math>e^{-\beta [F(\Delta _{0})-F(0)]}</math>
 
<math> =\frac{e^{-S_{\Delta}}\int D\psi ^{\ast }D\psi e^{-S_{BCS}[\Delta _{0}\, ]}}{\int D\psi ^{\ast }D\psi e^{-S_{BCS}[\, 0\, ]}}</math>
 
We recall that <math>S_{\Delta }=-\frac{1}{g}\int_{0}^{\beta }d\tau \sum_{k}\Delta _{k}^{\ast }(\tau )\Delta _{k}(\tau )</math>
 
<math>\Rightarrow S_{\Delta }(\Delta _{0})=-\frac{\Delta _{0}^{2}}{g}\beta N</math>
 
<math> And \int D\psi ^{\ast }D\psi e^{-S_{BCS}[\Delta _{0}]}=\int D\psi ^{\ast }D\psi \, e^{-\frac{1}{\beta }}\sum_{\omega _{n}}\sum_{k}\psi _{k}^{\ast }(i\omega _{n})\begin{pmatrix} i\omega _{n}+\epsilon _{k}-\mu  & \Delta _{0}\\\Delta _{0} &  i\omega _{n}-\epsilon _{k}+\mu \end{pmatrix}\psi _{k}(i\omega _{n}) </math>
 
<math>= const. \prod_{k}\prod_{n}\frac{1}{\beta ^{2}}\, \begin{vmatrix}i\omega _{n}+\epsilon _{k}-\mu  & \Delta _{0}\\ \Delta _{0}&  i\omega _{n}-\epsilon _{k}+\mu \end{vmatrix}</math>
 
<math>= const. \prod_{k}\prod_{n}\frac{1}{\beta ^{2}}[(i\omega _{n})^{2}-(\epsilon _{k}-\mu )^{2}-\Delta _{0}^{2}]</math>
 
<math>\Rightarrow e^{-\beta [F(\Delta _{0})-F(0)]}</math>
 
<math>=\frac{e^{-\frac{\Delta _{0}^{2}}{\left | g \right |}\beta N}const.{\prod_{k}^{}} {\prod_{n}^{}}\frac{1}{\beta ^{2}}[(i\omega _{n})^{2}-(\epsilon _{k}-\mu )^{2}-\Delta _{0}^{2}]}{const.{\prod_{k}^{}}{ \prod_{n}^{}}\frac{1}{\beta ^{2}}[(i\omega _{n})^{2}-(\epsilon _{k}-\mu )^{2}]}</math>
 
<math>=e^{-\frac{\Delta _{0}^{2}}{\left | g \right |}\beta N}{\prod_{k}^{}}{\prod_{n}^{}}(1+\frac{\Delta _{0}^{2}}{(\omega _{n}^{2})+(\epsilon _{n}-\mu )^{2}})</math>
 
<math>=e^{-\frac{\Delta _{0}^{2}}{\left | g \right |}\beta N}e^{\sum_{k}\sum_{n}ln(1+\frac{\Delta _{0}^{2}}{\omega_{n}^{2}+(\epsilon _{n}-\mu )^{2}})}</math>
 
<math>=e^{-\beta [F(\Delta _{0})-F(0)]}</math>


We see that our theory has a dimensionless parameter in it, namely <math>\kappa</math>, which is known as the Ginzburg-Landau parameter.
Then,
==Going below <math>T_c</math> with the Saddle Point Approximation==
 
<math>F(\Delta _{0})-F(0)=V\frac{\Delta _{0}}{\left | g \right |}-\frac{1}{\beta } {\sum_{\omega _{n}}^{}}{\sum_{k}^{}}ln (1+\frac{\Delta _{0}^{2}}{\omega _{n}^{2}+(\epsilon _{k}-\mu )^{2}})</math>
 
Here, the first term is positive whereas the second term should be negative.
 
In order to calculate <math>F[\Delta ]</math> we note that:
 
<math>F[\Delta ]=-\frac{1}{\beta }{\sum_{\omega _{n}}^{}}ln(1+\frac{\Delta ^{2}}{\omega _{n}^{2}+(\epsilon _{k}-\mu )^{2}})</math>
 
<math>\frac{\mathrm{d} F[\Delta ]}{\mathrm{d} \Delta }=-\frac{1}{\beta }{\sum_{\omega _{n}}^{}}\frac{1}{(1+\frac{\Delta ^{2}}{\omega _{n}^{2}+(\epsilon _{k}-\mu )^{2}})}\times \frac{2\Delta }{\omega _{n}^{2}+(\epsilon _{k}-\mu )^{2}}</math>
 
<math>-(2\Delta )\frac{1}{\beta }{\sum_{\omega _{n}}^{}}\frac{1}{\omega _{n}^{2}+(\epsilon _{k}-\mu )^{2}+\Delta ^{2}}</math>
 
Choosing <math>z=i\omega _{n}</math>
 
<math>\frac{\mathrm{d} F[\Delta ]}{\mathrm{d}\Delta }= -(2\Delta )\frac{1}{\beta }{\sum_{\omega _{n}}^{}}\frac{1}{E_{k}^{2}-z^{2}}</math>
 
<math>=(2\Delta ){\oint_{c}^{}}\frac{dz}{2\pi i}\frac{1}{e^{\beta z}+1}\frac{-1}{(z+E_{k})(z-E_{k})}</math>
 
<math>=(\Delta )(\frac{n_{F(E_{k})}}{E_{k}}+\frac{n_{F}(-E_{k})}{-E_{k}})</math>
 
<math>=(\Delta )(\frac{n_{F(E_{k})}}{E_{k}}-\frac{(1-n_{F}(E_{k}))}{E_{k}})</math>
 
<math>=\frac{\Delta }{E_{k}}(-1+2n_{F}(E_{k}))</math>
 
<math>\Rightarrow \int_{0}^{\Delta }d\Delta \frac{\mathrm{d}F }{\mathrm{d} \Delta }=\int d\Delta \frac{\Delta }{\sqrt{(\epsilon _{k}-\mu )^{2}+\Delta ^{2}}}(-1+2{\frac{1}{(e^{\frac{\sqrt{(\epsilon _{k}-\mu )^{2}+\Delta ^{2}}}{T}}+1)})}</math>
 
<math>-\sqrt{(\epsilon _{k}-\mu )^{2}+\Delta ^{2}}\, +\, 4T\, ln\, [cosh\frac{(\epsilon _{k}-\mu )^{2}+\Delta ^{2}}{2T}]+const.</math>
 
<math>\because F[0]= 0</math>
 
<math>\therefore F[\Delta ]=-E_{k}+\left | \epsilon _{k}-\mu  \right |-4T\, ln\, [\frac{1+e^{-\frac{E_{k}}{T}}}{1+e^\frac{-(\epsilon _{k}-\mu )}{T}}]</math>
 
Using the self consistency equation:
 
<math>\frac{1}{\left | g \right |}=\int \frac{d^{3}k}{\left ( 2\pi  \right )^{3}}\frac{tanh\frac{E_{k}}{2T}}{2E_{k}}</math>
 
<math>\therefore F[\Delta _{0}]-F[0]</math>
 
<math>={\sum_{k}^{}}(\frac{\Delta _{0}^{2}}{2E_{k}}-E_{k}+\left | \epsilon _{k}-\mu  \right |-\frac{\Delta _{0}^{2}(1-tanh\frac{E_{k}}{2T})}{2E_{k}})-4T\, ln\, (\frac{1+e^{-\frac{E_{k}}{T}}}{1+e^{-\frac{(\epsilon _{k}-\mu )}{T}}}) </math>
 
When T= 0, only the 1st term survives.
 
<math>\Rightarrow F[\Delta _{0}]-F[0]=-V\frac{N_{0}\Delta ^{2}}{2}</math>


== Beyond saddle-point approximation, collective modes and response in the broken symmetry state==
== Beyond saddle-point approximation, collective modes and response in the broken symmetry state==
Line 507: Line 2,199:
or
or


<math>Z=\int D(Re\Delta)D(Im\Delta)e^{-S_{eff}[Re\Delta,Im\Delta]}</math>
<math>Z=\int D(\Re e\Delta)D(\Im m\Delta)e^{-S_{eff}[\Re e\Delta,\Im m\Delta]}</math>


where
where


<math>\begin{eqnarray}S_{eff}[\Delta]&=&\frac{1}{|g|}\int_0^\beta d\tau\int d^Dr\left[(Re\Delta(r,\tau))^2+(Im\Delta(r,\tau))^2\right]\\
<math>S_{eff}[\Delta]=\frac{1}{|g|}\int_0^\beta d\tau\int d^Dr\left[(\Re e\Delta(r,\tau))^2+(\Im m\Delta(r,\tau))^2\right]-\ln\left[\int D\psi^*D\psi e^{-S_0-S_{int}}\right]</math>
&-&\ln\left[\int D\psi^*D\psi e^{-S_0-S_{int}}\right]\end{eqnarray}</math>
 
and
 
<math>\begin{align} S_0&=\int_0^\beta d\tau\int d^Dr \psi^*(r,\tau)(\frac{\partial}{\partial\tau}+\epsilon_p-\mu)\psi(r,\tau)\\
S_{int}&=\int_0^\beta d\tau\int d^Dr(\Delta^*(r,\tau)\psi_\downarrow(r,\tau)\psi_{\uparrow}(r,\tau)+\Delta(r,\tau)\psi^*_\uparrow(r,\tau)\psi^*_{\downarrow}(r,\tau))
\end{align}</math>
 
We can rewrite the interaction term in the action as
 
<math>S_{int}=\int_0^\beta d\tau\int d^Dr\left[\Re e(\Delta(r,\tau))(\psi_{\downarrow}\psi_{\uparrow}+\psi_{\uparrow}^*\psi_{\downarrow}^*)(r,\tau)+i*\Im m(\Delta(r,\tau))(\psi_{\uparrow}^*\psi_{\downarrow}^*-\psi_{\downarrow}\psi_{\uparrow})(r,\tau)\right]</math>
 
Consider now functional derivatives of <math>S_{eff}[\Delta]</math>:
 
<math>
\frac{\delta S_{eff}[\Delta]}{\delta\Re e\Delta(r,\tau)}=\frac{2}{|g|}\Re e\Delta(r,\tau)-\frac{1}{\int D\psi^*D\psi e^{-S_0-S_{int}}} \int D\psi^*D\psi e^{-S_0-S_{int}}\left(-\frac{\delta S_{int}}{\delta \Re e\Delta(r,\tau)}\right)
</math>
 
The functional derivative of <math>S_{int}</math> with <math>\Re e\Delta(r,\tau)</math> is
 
<math>\frac{\delta S_{int}}{\delta \Re e\Delta(r,\tau)}=\psi_{\downarrow}(r,\tau)\psi_{\uparrow}(r,\tau)+\psi_{\uparrow}^*(r,\tau)\psi_{\downarrow}^*(r,\tau)</math>
 
Therefore, we have
 
<math>\frac{\delta S_{eff}[\Delta]}{\delta \Re e\Delta(r,\tau)}=\frac{2}{|g|}\Re e\Delta(r,\tau)+\langle\psi_{\downarrow}(r,\tau)\psi_{\uparrow}(r,\tau)+\psi_{\uparrow}^*(r,\tau)\psi_{\downarrow}^*(r,\tau)\rangle\ \ (1)</math>
 
Similarly
 
<math>\frac{\delta S_{eff}[\Delta]}{\delta \Im m\Delta(r,\tau)}=\frac{2}{|g|}\Im m\Delta(r,\tau)+i\langle\psi_{\uparrow}^*(r,\tau)\psi_{\downarrow}^*(r,\tau)-\psi_{\downarrow}(r,\tau)\psi_{\uparrow}(r,\tau)\rangle\ \ (2)</math>
 
If we were to set the LHS of the above two equations to zero, we would obtain our self-consistency conditions.
 
The strategy is to take
 
<math>S_{eff}[\Delta(r,\tau)]=S_{eff}\left[\Delta_{sp}+(\Delta(r,\tau)-\Delta_{sp})\right]</math>
 
where <math>\Delta_{sp}</math> solves (1) and (2) with LHS set to zero and expand in  powers of <math>\Delta(r,\tau)-\Delta_{sp}</math>. So
 
<math>\begin{align}
&S_{eff}[\Delta_{sp}+(\Delta(r,\tau)-\Delta_{sp})]\\
&=S_{eff}[\Delta_{sp}]+\int_0^\beta d\tau\int d^3r\Re e(\Delta(r,\tau)-\Delta_{sp})\frac{\delta S_{eff}}{\delta\Re e\Delta(r,\tau)}|_{\Delta_{sp}}+\int_0^\beta d\tau\int d^3r\Im m(\Delta(r,\tau)-\Delta_{sp})\frac{\delta S_{eff}}{\delta\Im m\Delta(r,\tau)}|_{\Delta_{sp}}\\
&+\frac{1}{2}\int_0^\beta d\tau\int_0^\beta d\tau'\int d^3r\int d^3r'\Re e(\Delta(r,\tau)-\Delta_{sp})\Re e(\Delta(r',\tau')-\Delta_{sp})\frac{\delta^2 S_{eff}}{\delta\Re e\Delta(r,\tau)\delta\Re e\Delta(r',\tau')}|_{\Delta_{sp}}\\
&+\frac{1}{2}\int_0^\beta d\tau\int_0^\beta d\tau'\int d^3r\int d^3r'\Im m(\Delta(r,\tau)-\Delta_{sp})\Im m(\Delta(r',\tau')-\Delta_{sp})\frac{\delta^2 S_{eff}}{\delta\Im m\Delta(r,\tau)\delta\Im m\Delta(r',\tau')}|_{\Delta_{sp}}\\
&+\int_0^\beta d\tau\int_0^\beta d\tau'\int d^3r\int d^3r'\Re e(\Delta(r,\tau)-\Delta_{sp})\Im m(\Delta(r',\tau')-\Delta_{sp})\frac{\delta^2 S_{eff}}{\delta\Re e\Delta(r,\tau)\delta\Im m\Delta(r',\tau')}|_{\Delta_{sp}}+...
\end{align}</math>
 
Since by definition of <math>\Delta_{sp}</math> we have
 
<math>\frac{\delta S_{eff}}{\delta\Re e\Delta(r,\tau)}|_{\Delta_{sp}}=\frac{\delta S_{eff}}{\delta\Im m\Delta(r,\tau)}|_{\Delta_{sp}}=0</math>
 
So only the 0th and 2nd order terms contribute. The 0th order term gave us the condensation energy, and the 2nd order term will give us information about collective modes (in the broken symmetry phase).
 
<math>\begin{align}
&\frac{\delta^2 S_{eff}}{\delta\Re e\Delta(r,\tau)\delta\Re e\Delta(r',\tau')}\\
&=\frac{2}{|g|}\delta(\tau-\tau')\delta(r-r')+\frac{\delta}{\delta\Re e\Delta(r,\tau)}\left[\frac{1}{\int D\psi^*D\psi e^{-S_0-S_{int}}}\int D\psi^*D\psi e^{-S_0-S_{int}}\left(\frac{\delta S_{int}}{\delta \Re e\Delta(r',\tau')}\right)\right]\\
&=\frac{2}{|g|}\delta(\tau-\tau')\delta(r-r')-\frac{\int D\psi^*D\psi e^{-S_0-S_{int}}\left(\frac{\delta S_{int}}{\delta \Re e\Delta(r,\tau)}\right)\left(\frac{\delta S_{int}}{\delta \Re e\Delta(r',\tau')}\right)}{\int D\psi^*D\psi e^{-S_0-S_{int}}}\\
&+\left(\frac{\int D\psi^*D\psi e^{-S_0-S_{int}}\left(\frac{\delta S_{int}}{\delta \Re e\Delta(r,\tau)}\right)}{\int D\psi^*D\psi e^{-S_0-S_{int}}}\right)\left(\frac{\int D\psi^*D\psi e^{-S_0-S_{int}}\left(\frac{\delta S_{int}}{\delta \Re e\Delta(r',\tau')}\right)}{\int D\psi^*D\psi e^{-S_0-S_{int}}}\right)
\end{align}</math>
 
So
 
<math>\begin{align}
&\frac{\delta^2 S_{eff}}{\delta\Re e\Delta(r,\tau)\delta\Re e\Delta(r',\tau')}=\frac{2}{|g|}\delta(\tau-\tau')\delta(r-r')\\
&-\langle\left(\psi_{\downarrow}(r,\tau)\psi_{\uparrow}(r,\tau)+\psi^*_{\uparrow}(r,\tau)\psi^*_{\downarrow}(r,\tau)\right)\left(\psi_{\downarrow}(r',\tau')\psi_{\uparrow}(r',\tau')+\psi^*_{\uparrow}(r',\tau')\psi^*_{\downarrow}(r',\tau')\right)\rangle_{S_0+S_{int}}\\
&+\langle\psi_{\downarrow}(r,\tau)\psi_{\uparrow}(r,\tau)+\psi^*_{\uparrow}(r,\tau)\psi^*_{\downarrow}(r,\tau)\rangle\langle\psi_{\downarrow}(r',\tau')\psi_{\uparrow}(r',\tau')+\psi^*_{\uparrow}(r',\tau')\psi^*_{\downarrow}(r',\tau')\rangle_{S_0+S_{int}}
\end{align}</math>
 
i.e.
 
<math>\begin{align}
&\frac{\delta^2 S_{eff}}{\delta\Re e\Delta(r,\tau)\delta\Re e\Delta(r',\tau')}=\frac{2}{|g|}\delta(\tau-\tau')\delta(r-r')\\
&-\langle\left(\psi_{\downarrow}(r,\tau)\psi_{\uparrow}(r,\tau)+\psi^*_{\uparrow}(r,\tau)\psi^*_{\downarrow}(r,\tau)\right)\left(\psi_{\downarrow}(r',\tau')\psi_{\uparrow}(r',\tau')+\psi^*_{\uparrow}(r',\tau')\psi^*_{\downarrow}(r',\tau')\right)\rangle_{S_0+S_{int}}^{\mbox{con}}
\end{align}</math>
 
Similarly,
 
<math>\begin{align}
&\frac{\delta^2 S_{eff}}{\delta\Im m\Delta(r,\tau)\delta\Im m\Delta(r',\tau')}=\frac{2}{|g|}\delta(\tau-\tau')\delta(r-r')\\
&+\langle\left(\psi^*_{\uparrow}(r,\tau)\psi^*_{\downarrow}(r,\tau)-\psi_{\downarrow}(r,\tau)\psi_{\uparrow}(r,\tau)\right)\left(\psi^*_{\uparrow}(r',\tau')\psi^*_{\downarrow}(r',\tau')-\psi_{\downarrow}(r',\tau')\psi_{\uparrow}(r',\tau')\right)\rangle_{S_0+S_{int}}^{\mbox{con}}
\end{align}</math>
 
And
 
<math>
\frac{\delta^2 S_{eff}}{\delta\Im m\Delta(r,\tau)\delta\Re e\Delta(r',\tau')}=-i\langle\left(\psi^*_{\uparrow}(r,\tau)\psi^*_{\downarrow}(r,\tau)-\psi_{\downarrow}(r,\tau)\psi_{\uparrow}(r,\tau)\right)\left(\psi_{\downarrow}(r',\tau')\psi_{\uparrow}(r',\tau')+\psi^*_{\uparrow}(r',\tau')\psi^*_{\downarrow}(r',\tau')\right)\rangle_{S_0+S_{int}}^{\mbox{con}}
</math>
 
If we evaluate these functional derivatives at the saddle point <math>\Delta_{sp}</math> we have
 
<math>\begin{align}
S_0+S_{int}&\rightarrow\int_0^\beta d\tau\int d^3r\sum_{\sigma=\uparrow,\downarrow}\psi^*_\sigma(r,\tau)\left(\frac{\partial}{\partial\tau}+\epsilon_p-\mu\right)\psi_{\sigma}(r,\tau)\\
&+\Delta_{sp}\int_0^\beta d\tau\int d^3r\left(\psi_{\downarrow}(r,\tau)\psi_{\uparrow}(r,\tau)+\psi_{\uparrow}^*(r,\tau)\psi_{\downarrow}^*(r,\tau)\right)
\end{align}</math>
 
where we take the saddle point solution <math>\Delta_{sp}</math> to be purely real. Arranging the Grassman fields into the Nambu spinor we have:
 
<math>(\psi^*_{\uparrow}\ \psi_\downarrow)\left(\begin{align}&\frac{\partial}{\partial\tau}+\epsilon_p-\mu&\Delta_{sp}\\ &\Delta_{sp}&\frac{\partial}{\partial\tau}-\epsilon_p+\mu\end{align}\right) \left(\begin{align}&\psi_\uparrow\\&\psi^*_\downarrow\end{align}\right)</math>
 
where the Nambu spinor is defined as
 
<math>\Psi=\left(\begin{align}&\psi_\uparrow\\&\psi^*_\downarrow\end{align}\right)</math>
 
So
 
<math>\begin{align}
&\psi^*_\uparrow\psi^*_\downarrow+\psi_\downarrow\psi_\uparrow=\Psi^*\left(\begin{align}&0&1\\&1&0\end{align}\right)\Psi =\Psi^*\sigma_x\Psi\\
&\psi^*_\uparrow\psi^*_\downarrow-\psi_\downarrow\psi_\uparrow=\Psi^*\left(\begin{align}&0&1\\-&1&0\end{align}\right)\Psi =\Psi^*i\sigma_y\Psi
\end{align}</math>
 
<math>\begin{align}
\frac{\delta^2S_{eff}}{\delta\Re e\Delta(r,\tau)\delta\Re e\Delta(r',\tau')}|_{\Delta_{sp}}&=\frac{2}{|g|}\delta(\tau-\tau')\delta(r-r')-\langle\Psi^*(r,\tau)\sigma_x\Psi(r,\tau)\Psi^*(r',\tau')\sigma_x\Psi(r',\tau')\rangle^{\mbox{con}}\\
&=\frac{2}{|g|}\delta(\tau-\tau')\delta(r-r')+\mbox{Tr}\left[\sigma_xG(r-r',\tau-\tau')\sigma_xG(r'-r,\tau'-\tau)\right]
\end{align}</math>
 
where the Green's functions are 2*2 matrices,
 
<math>G^{-1}_k(i\omega_n)=\left(\begin{align}&-i\omega_n+\epsilon_k-\mu&\Delta_{sp}\\&\Delta_{sp}&-i\omega_n-\epsilon_k+\mu\end{align}\right)</math>
 
Notice that this is a function of <math>\tau-\tau'</math> and <math>r-r'</math>. Let's call it <math>\Pi_{++}(r-r',\tau-\tau')</math>. This will give rise to
 
<math>\begin{align}
&\frac{1}{2}\int_0^\beta d\tau\int_0^\beta d\tau'\int d^3r\int d^3r'\Re e(\Delta(r,\tau)-\Delta_{sp})\Pi_{++}(r-r',\tau-\tau')\Re e(\Delta(r',\tau')-\Delta_{sp})\\
&=\frac{1}{2}\int_0^\beta d\tau\int_0^\beta d\tau'\int d^3r\int d^3r'\delta\Delta_+(r,\tau)\Pi_{++}(r-r',\tau-\tau')\delta\Delta_+(r',\tau')
\end{align}</math>
 
Fourier transforming we find the contribution to <math>S_{eff}</math>
 
<math>\frac{1}{2}\frac{1}{\beta}\sum_{\Omega_n}\sum_{q}\delta\Delta_+(-q,-i\Omega_n)\Pi_{++}(q,i\Omega_n)\delta\Delta_+(q,i\Omega_n)</math>
 
Similarly
 
<math>\begin{align}
\Pi_{--}(r-r',\tau-\tau')&=\frac{2}{|g|}\delta(\tau-\tau')\delta(r-r')+\mbox{Tr}\left[\sigma_yG(r-r',\tau-\tau')\sigma_yG(r'-r,\tau'-\tau)\right]\\
\Pi_{-+}(r-r',\tau-\tau')&=\langle\Psi^*\sigma_y\Psi(r,\tau)\Psi^*\sigma_x\Psi(r',\tau')\rangle^{\mbox{con}}=-\mbox{Tr}\left[\sigma_yG(r-r',\tau-\tau')\sigma_xG(r'-r,\tau'-\tau)\right]
\end{align}</math>
 
Then
 
<math>\begin{align}
&S_{eff}[\Delta]\approx S_{eff}[\Delta_{sp}]+\\
&\frac{1}{2}\frac{1}{\beta}\sum_{\Omega_n}\sum_q\left[\delta\Delta_+(-q,-i\Omega_n)\Pi_{++}(q,i\Omega_n)\delta\Delta_+(q,i\Omega_n)+\delta\Delta_-(-q,-i\Omega_n)\Pi_{--}(q,i\Omega_n)\delta\Delta_-(q,i\Omega_n)\right.\\
&\left.+2\delta\Delta_-(-q,-i\Omega_n)\Pi_{-+}(q,i\Omega_n)\delta\Delta_+(q,i\Omega_n)\right]+...
\end{align}</math>
 
To proceed with the evaluation of <math>\Pi</math>'s, note that
 
<math>\begin{align}
&G^{-1}_k(i\omega_n)=-i\omega_n\mathbb{I}+(\epsilon_k-\mu)\sigma_3+\Delta_{sp}\sigma_1\\
\Rightarrow&G_k(i\omega_n)=\frac{i\omega_n\mathbb{I}+(\epsilon_k-\mu)\sigma_3+\Delta_{sp}\sigma_1}{\omega_n^2+(\epsilon_k-\mu)^2+\Delta_{sp}^2}
\end{align}</math>
 
where <math>\omega_n=\frac{(2n+1)\pi}{\beta}</math>. After Fourier transform,
 
<math>\begin{align}
\Pi_{++}(q,i\Omega_n)&=\frac{2}{|g|}+\frac{1}{\beta}\sum_{\omega_n}\int\frac{d^3k}{(2\pi)^3}\mbox{Tr}\left[\sigma_1G_{k+q}(i\omega_n+i\Omega_n)\sigma_1G_k(i\omega_n)\right]\\
\Pi_{--}(q,i\Omega_n)&=\frac{2}{|g|}+\frac{1}{\beta}\sum_{\omega_n}\int\frac{d^3k}{(2\pi)^3}\mbox{Tr}\left[\sigma_2G_{k+q}(i\omega_n+i\Omega_n)\sigma_2G_k(i\omega_n)\right]\\
\Pi_{-+}(q,i\Omega_n)&=-\frac{1}{\beta}\sum_{\omega_n}\int\frac{d^3k}{(2\pi)^3}\mbox{Tr}\left[\sigma_2G_{k+q}(i\omega_n+i\Omega_n)\sigma_1G_k(i\omega_n)\right]\\
\end{align}</math>
 
Consider now
 
<math>\begin{align}
&\mbox{Tr}\left[\sigma_{1,2}\frac{(i\omega_n+i\Omega_n)\mathbb{I}+(\epsilon_{k+q}-\mu)\sigma_3+\Delta_{sp}\sigma_1}{(\omega_n+\Omega_n)^2+(\epsilon_{k+q}-\mu)^2+\Delta_{sp}^2}\sigma_{1,2}\frac{i\omega_n\mathbb{I}+(\epsilon_k-\mu)\sigma_3+\Delta_{sp}\sigma_1}{\omega_n^2+(\epsilon_k-\mu)^2+\Delta_{sp}^2}\right]\\
=&2\frac{(i\omega_n+i\Omega_n)i\omega_n-(\epsilon_{k+q}-\mu)(\epsilon_k-\mu)\pm\Delta^2_{sp}}{\left[(\omega_n+\Omega_n)^2+(\epsilon_{k+q}-\mu)^2+\Delta_{sp}^2\right]\left[\omega_n^2+(\epsilon_k-\mu)^2+\Delta_{sp}^2\right]}
\end{align}</math>


and
and


<math> S_0&=&\int_0^\beta d\tau\int d^Dr \psi^*(r,\tau)(\frac{\partial}{\partial\tau}+\epsilon_p-\mu)\psi(r,\tau)\\
<math>\begin{align}
S_{int}&=&\int_0^\beta d\tau\int d^Dr(\Delta^*(r,\tau)\psi_\downarrow(r,\tau)\psi_{\uparrow}(r,\tau)+\Delta(r,\tau)\psi^*_\downarrow(r,\tau)\psi^*_{\uparrow}(r,\tau))</math>
&\mbox{Tr}\left[\sigma_{2}\frac{(i\omega_n+i\Omega_n)\mathbb{I}+(\epsilon_{k+q}-\mu)\sigma_3+\Delta_{sp}\sigma_1}{(\omega_n+\Omega_n)^2+(\epsilon_{k+q}-\mu)^2+\Delta_{sp}^2}\sigma_{1}\frac{i\omega_n\mathbb{I}+(\epsilon_k-\mu)\sigma_3+\Delta_{sp}\sigma_1}{\omega_n^2+(\epsilon_k-\mu)^2+\Delta_{sp}^2}\right]\\
=&\frac{(i\omega_n+i\Omega_n)(\epsilon_k-\mu)\mbox{Tr}[\sigma_2\sigma_1\sigma_3]+i\omega_n(\epsilon_{k+q}-\mu)\mbox{Tr}[\sigma_2\sigma_3\sigma_1]}{\left[(\omega_n+\Omega_n)^2+(\epsilon_{k+q}-\mu)^2+\Delta_{sp}^2\right]\left[\omega_n^2+(\epsilon_k-\mu)^2+\Delta_{sp}^2\right]}\\
=&\frac{(i\omega_n+i\Omega_n)(\epsilon_k-\mu)(-2i)+i\omega_n(\epsilon_{k+q}-\mu)2i}{\left[(\omega_n+\Omega_n)^2+(\epsilon_{k+q}-\mu)^2+\Delta_{sp}^2\right]\left[\omega_n^2+(\epsilon_k-\mu)^2+\Delta_{sp}^2\right]}
\end{align}</math>
 
Now, note that at <math>q=0,\Omega_n=0</math>,
 
<math>\Pi_{-+}(0,0)=0</math>
 
because the numerator vanishes for all <math>k</math> and <math>\omega_n</math>. Also note that by self-consistency condition
 
<math>\begin{align}
\Pi_{--}(0,0)&=\frac{2}{|g|}-\frac{1}{\beta}\sum_{\omega_n}\int\frac{d^3k}{(2\pi)^3}2\frac{1}{[\omega_n^2+(\epsilon_k-\mu)^2+\Delta_{sp}^2]}\\
&=\frac{2}{|g|}+2\int\frac{d^3k}{(2\pi)^3}\frac{1}{\beta}\sum_{\omega_n}\frac{1}{(i\omega_n-E_k)(i\omega_n+E_k)}\\
&=\frac{2}{|g|}+2\int\frac{d^3k}{(2\pi)^3}\left(\frac{n_F(E_k)}{2E_k}-\frac{n_F(-E_k)}{2E_k}\right)\\
&=2\left(\frac{1}{|g|}-\int\frac{d^3k}{(2\pi)^3}\frac{1-2n_F(E_k)}{2E_k}\right)\\
&=0
\end{align}</math>
 
However,
 
<math>\begin{align}
\Pi_{++}(0,0)&=\frac{2}{|g|}+\frac{1}{\beta}\sum_{\omega_n}\int\frac{d^3k}{(2\pi)^3}2\frac{-\omega_n^2-(\epsilon_k-\mu)^2+\Delta_{sp}^2}{[\omega_n^2+(\epsilon_k-\mu)^2+\Delta_{sp}^2]^2}\\
&=4\Delta_{sp}^2\frac{1}{\beta}\sum_{\omega_n}\int\frac{d^3k}{(2\pi)^3}\frac{1}{[\omega_n^2+(\epsilon_k-\mu)^2+\Delta_{sp}^2]^2}\\
&=4\Delta_{sp}^2N_0\frac{1}{\beta}\sum_{\omega_n}\int_{-\infty}^\infty\frac{d\xi}{[\xi^2+\omega_n^2+\Delta_{sp}^2]^2}\\
&=4\Delta_{sp}^2N_0\frac{1}{\beta}\sum_{\omega_n}\frac{\pi}{2}\frac{1}{(\omega_n^2+\Delta_{sp}^2)^{\frac{3}{2}}}\\
&=2\pi N_0\Delta_{sp}^2\frac{1}{\pi^3T^2}\sum_{n=-\infty}^\infty\frac{1}{[(2n+1)^2+(\frac{\Delta_{sp}}{\pi T})^2]^{\frac{3}{2}}}
\end{align}</math>
 
This sum is slowly convergent. To evaluate it efficiently, we note that
 
<math>\frac{1}{a^{\frac{3}{2}}}=\frac{2}{\sqrt{\pi}}\int_0^\infty d\lambda\sqrt{\lambda}e^{-\lambda a}\ \ \mbox{for }a>0</math>
 
So
 
<math>\begin{align}
\sum_{n=-\infty}^\infty\frac{1}{[(2n+1)^2+(\frac{\Delta_{sp}}{\pi T})^2]^\frac{3}{2}}&=\frac{2}{\sqrt{\pi}} \int_0^\infty d\lambda\sqrt{\lambda}e^{-\lambda(\frac{\Delta_{sp}}{\pi T})^2}\left(\sum_{n=-\infty}^\infty e^{-\lambda(2n+1)^2}\right)\\
&=\frac{2}{\sqrt{\pi}} \int_0^\infty d\lambda\sqrt{\lambda}e^{-\lambda(\frac{\Delta_{sp}}{\pi T})^2}\Theta_2(0,e^{-4\lambda})\\
&=\frac{1}{4\sqrt{\pi}} \int_0^\infty d\xi\sqrt{\xi}e^{-\xi(\frac{\Delta_{sp}}{2\pi T})^2}\Theta_2(0,e^{-\xi})
\end{align}</math>
 
where <math>\Theta_2</math> is Jacobi elliptic theta function. Then
 
<math>\begin{align}
\Pi_{++}(0,0)&=2\pi N_0\Delta^2_{sp}\frac{1}{\pi^3T^2}\sum_{n=-\infty}^\infty\frac{1}{[(2n+1)^2+(\frac{\Delta_{sp}}{\pi T})^2]^\frac{3}{2}}\\
&=2N_0\Phi(\frac{\Delta_{sp}}{2\pi T})
\end{align}</math>
 
where
 
<math>\Phi(x)=x^2\frac{1}{\sqrt{\pi}}\int_0^\infty d\xi\sqrt{\xi}e^{-\xi x^2}\Theta_2(0,e^{-\xi})</math>
 
[[Image:Phi.jpg]]
 
Note that
 
<math>\Pi_{++}(0,0)\rightarrow 2N_0\frac{7\zeta(3)}{4\pi^2T^2}\Delta^2_{sp}</math>
 
as <math>\Delta_{sp}\rightarrow 0</math>, i.e. as <math>T\rightarrow T_C</math>.
 
That is precisely the curvature of the new minimum in the Ginzburg-Landau free energy we found before. So, at 2nd order our effective action corresponds to the action for two real free bosons, <math>\delta\Delta_+</math> and <math>\delta\Delta_-</math>. At <math>q=0, i\Omega_n=0</math>, <math>\delta\Delta_+</math> mode is gapped (massive), but <math>\delta\Delta_-</math> mode is not gapped (massless). Physically, <math>\delta\Delta_+</math> corresponds to the fluctuations of the order parameter amplitude (because we chose <math>\Delta_{sp}</math> to be real).
 
Recall our discussion from many-body course:
 
Amplitude fluctuations are not hydrodynamic modes since they do not correspond to either conserved or to broken symmetry variable. We have to extend our approach to higher order in <math>\delta\Delta_\pm</math> to describe its (rapid) decay.
 
<math>\delta\Delta_-</math> corresponds to fluctuations along the direction of the minimum of the double well potential, where there is no barrier. It corresponds to (part of) a "phase" mode.
 
To determine the kinematics (of our collective modes) we need to expand <math>\Pi_{\mu\nu}(q, i\Omega_n)</math> in powers of <math>\vec{q}</math> and <math>\Omega_n</math>. Our small expansion parameters are
 
<math>\Omega_n \ll \Delta_{sp}</math> and <math>q \ll \frac{\Delta_{sp}}{\hbar v_F}</math>
 
(Obviously <math>q \ll k_F</math>)
 
Start with <math>\Pi_{++}</math> :
What we need is
 
<math>\Pi_{++}\left(q, i\Omega_n\right) - \Pi_{++}\left(0, 0\right)</math>
 
<math>= \frac{2}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\left[\frac{1}{\omega_n^2 + E_k^2} - \frac{\omega_n\left(\omega_n + \Omega_n\right) + \left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \mu\right) - \Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)\left(\left(\omega_n + \Omega_n\right)^2 + E_{k + q}^2\right)} - \frac{2\Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)^2}\right]}}</math>
 
<math>= \frac{2}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\left[\frac{\omega_n^2 + \left(\epsilon_k - \mu\right)^2 - \Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)^2} - \frac{\omega_n\left(\omega_n + \Omega_n\right) + \left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \mu\right) - \Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)\left(\left(\omega_n + \Omega_n\right)^2 + E_{k + q}^2\right)}\right]}}</math>
 
<math>= \frac{2}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\left[\frac{\omega_n^2 + E_k^2 - 2\Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)^2} - \frac{\omega_n^2 + E_k^2 + \omega_n\Omega_n + \left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \epsilon_k\right) - 2\Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)\left(\omega_n^2 + E_k^2 + 2\omega_n\Omega_n + \Omega_n^2 + \left(\epsilon_{k + q} - \mu\right)^2 - \left(\epsilon_k - \mu\right)\right)}\right]}}</math>
 
The second term in the brackets:
 
<math>\frac{\omega_n^2 + E_k^2 - 2\Delta_{sp}^2 + \overbrace{\omega_n\Omega_n + \left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \epsilon_k\right)}^{\text{small}}}{\left(\omega_n^2 + E_k^2\right)\left(\omega_n^2 + E_k^2 + \underbrace{2\omega_n\Omega_n + \Omega_n^2 + \left(\epsilon_{k + q} - \mu\right)^2 - \left(\epsilon_k - \mu\right)}_{\text{small}}\right)}</math>
 
<math>= \frac{\omega_n^2 + E_k^2 - 2\Delta_{sp}^2 + \omega_n\Omega_n + \left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \epsilon_k\right)}{\left(\omega_n^2 + E_k^2\right)^2}\left[1 + \frac{2\omega_n\Omega_n + \Omega_n^2 + \left(\epsilon_{k + q} - \epsilon_k\right)\left(\epsilon_{k + q} + \epsilon_k - 2\mu\right)}{\omega_n^2 + E_k^2}\right]^{-1}</math>
 
Now, by Taylor expansion,
 
<math>\epsilon_{k + q} \simeq \epsilon_k + q\cdot \nabla\epsilon_k + \ldots</math>
 
and near the Fermi level we have
 
<math>q\cdot \nabla\epsilon_k \simeq q\cdot v_F</math>
 
Also note that after integrating over <math>\vec{k}</math> and <math>\omega_n</math>, terms with odd power of <math>\omega_n</math> and <math>q\cdot v_F</math> vanish. Keeping only terms of order <math>q^2</math> and <math>\Omega_n^2</math> we have
 
<math>\frac{\omega_n^2 + E_k^2 - 2\Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)^2}\left(1 - \frac{\Omega_n^2 + \left(q\cdot v_F\right)^2}{\omega_n^2 + E_k^2}\right) - \frac{2\omega_n^2\Omega_n^2 + 2\left(\epsilon_k - \mu\right)^2\left(q\cdot v_F\right)^2}{\left(\omega_n^2 + E_k^2\right)^3} + \frac{\omega_n^2 + E_k^2 - 2\Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)^2}\cdot\frac{4\omega_n^2\Omega_n^2 + 4\left(\epsilon_k - \mu\right)^2\left(q\cdot v_F\right)^2}{\left(\omega_n^2 + E_k^2\right)^2}</math>
 
The terms which we dropped are down by powers of <math>q/k_F</math> and/or vanish assuming particle-hole symmetry.
 
<math>\begin{align} \Pi_{++}\left(q, i\Omega_n\right) - \Pi_{++}\left(0, 0\right) \simeq & \quad \frac{2}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\frac{\omega_n^2 + E_k^2 - 2\Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)^3}\left(\Omega_n^2 + \left(q\cdot v_F\right)^2\right)}} \\
 
& + \frac{2}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\frac{2\omega_n^2\Omega_n^2 + 2\left(\epsilon_k - \mu\right)^2\left(q\cdot v_F\right)^2}{\left(\omega_n^2 + E_k^2\right)^3}}} \\
 
& - \frac{2}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3} 4 \cdot \frac{\omega_n^2 + E_k^2 - 2\Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)^4}\left(\omega_n^2\Omega_n^2 + \left(\epsilon_k - \mu\right)^2\left(q\cdot v_F\right)^2\right)}} \\ \end{align} </math>
 
Let's focus on <math>T = 0</math>. Then <math>\frac{1}{\beta}\sum_{\omega_n} \longrightarrow \int_{-\infty}^{\infty}{\frac{d\omega}{2\pi}}</math> <math>\Longrightarrow</math>
 
<math>\begin{align} \Pi_{++}\left(q, i\Omega_n\right) - \Pi_{++}\left(0, 0\right) \simeq &-2N_0\int_{-\infty}^{\infty}{\frac{d\omega}{2\pi}\int_{-\infty}^{\infty}{d\xi\frac{\omega^2 + \xi^2 - \Delta_{sp}^2}{\left(\omega^2 + \xi^2 + \Delta_{sp}^2\right)^3}\left(\Omega_n^2 + q^2\langle \left(\hat{q}\cdot v_F\right)^2 \rangle_{FS}\right)}} \\
 
&+ 4N_0\int_{-\infty}^{\infty}{\frac{d\omega}{2\pi}\int_{-\infty}^{\infty}{d\xi\frac{\omega^2\Omega_n^2 + \xi^2q^2\langle\left(\hat{q}\cdot v_F\right)^2\rangle_{FS}}{\left(\omega^2 + \xi^2 + \Delta_{sp}^2\right)^3}}} \\
 
&- 8N_0\int_{-\infty}^{\infty}{\frac{d\omega}{2\pi}\int_{-\infty}^{\infty}{d\xi\frac{\omega^2 + \xi^2 - \Delta_{sp}^2}{\left(\omega^2 + \xi^2 + \Delta_{sp}^2\right)^4}\left(\omega^2\Omega_n^2 + \xi^2q^2\langle \left(\hat{q}\cdot v_F\right)^2 \rangle_{FS}\right)}} \\ \end{align}</math>
 
So at <math>T = 0</math> :
 
<math>\begin{align} \Pi_{++}\left(q, i\Omega_n\right) - \Pi_{++}\left(0, 0\right) \simeq & \quad 0 \\
&+ \frac{N_0}{2\Delta_{sp}^2}\left(\Omega_n^2 + q^2\langle \left(\hat{q}\cdot v_F\right)^2 \rangle_{FS}\right) \\
&- \frac{N_0}{3\Delta_{sp}^2}\left(\Omega_n^2 + q^2\langle \left(\hat{q}\cdot v_F\right)^2 \rangle_{FS}\right) \\ \end{align}</math>
 
in 3D and for a spherical FS
 
<math> \begin{align} \langle \left(\hat{q}\cdot v_F\right)^2 \rangle_{FS} &= \frac{v_F^2}{2}\int_{-1}^{1}{d\left(\cos\theta\right)\cos^2\theta} \\
&= \frac{v_F^2}{2}\frac{1}{3}\left(1 - \left(-1\right)\right) \\
&= \frac{v_F^2}{3} \\ \end{align}</math>
 
<math>\Longrightarrow</math>
<math>\Pi_{++}\left(q, i\Omega_n\right) - \Pi_{++}\left(0, 0\right) \simeq \frac{N_0}{6\Delta_{sp}^2}\left(\Delta_n^2 + \frac{1}{3}v_F^2q^2\right) + \ldots</math>
 
Next
 
<math> \begin{align} \Pi_{--}\left(q, i\Omega_n\right) &= \frac{2}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\left[\frac{1}{\omega_n^2 + E_k^2} - \frac{\omega_n\left(\omega_n + \Omega_n\right) + \left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \mu\right) + \Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)\left(\left(\omega_n + \Omega_n\right)^2 + E_{k + q}^2\right)}\right]}} \\
 
&= \frac{2}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\left[\frac{1}{\omega_n^2 + E_k^2} - \frac{\omega_n^2 + \left(\epsilon_k - \mu\right)^2 + \Delta_{sp}^2 + \omega_n\Omega_n + \left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \epsilon_k\right)}{\left(\omega_n^2 + E_k^2\right)\left(\omega_n^2 + 2\omega_n\Omega_n + \Omega_n^2 + E_k^2 + \left(\epsilon_{k + q} - \mu\right)^2 - \left(\epsilon_k - \mu\right)^2\right)}\right]}} \\
 
&= \frac{2}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\left[\frac{1}{\omega_n^2 + E_k^2} - \frac{\omega_n^2 + E_k^2 + \omega_n\Omega_n + \left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \epsilon_k\right)}{\left(\omega_n^2 + E_k^2\right)^2\left(1 + \frac{\Omega_n^2 + 2\omega_n\Omega_n + \left(\epsilon_{k + q} - \epsilon_k\right)\left(\epsilon_{k + q} + \epsilon_k - 2\mu\right)}{\omega_n^2 + E_k^2}\right)}\right]}} \\ \end{align}</math>
 
Now, the second term of the integrand:
 
<math>\left[\frac{1}{\omega_n^2 + E_k^2} + \frac{\omega_n\Omega_n + \left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \epsilon_k\right)}{\left(\omega_n^2 + E_k^2\right)^2}\right]\left(1 + \frac{\Omega_n^2 + 2\omega_n\Omega_n + \left(\epsilon_{k + q} - \epsilon_k\right)\left(\epsilon_{k + q} + \epsilon_k - 2\mu\right)}{\omega_n^2 + E_k^2}\right)^{-1}</math>
 
<math>\begin{align} = & \frac{1}{\omega_n^2 + E_k^2} - \frac{\Omega_n^2 + \left(\epsilon_{k + q} - \epsilon_k\right)^2}{\left(\omega_n^2 + E_k^2\right)^2} - \frac{2\omega_n^2\Omega_n^2 + 2\left(\epsilon_k - \mu\right)^2\left(\epsilon_{k + q} - \epsilon_k\right)^2}{\left(\omega_n^2 + E_k^2\right)^3} + \\
 
& \frac{1}{\omega_n^2 + E_k^2}\frac{4\omega_n^2\Omega_n^2 + 4\left(\epsilon_{k + q} - \epsilon_k\right)^2\left(\epsilon_k - \mu\right)^2}{\left(\omega_n^2 + E_k^2\right)^2} + \ldots \\ \end{align}</math>
 
Therefore,
 
<math>\Pi_{--}\left(q, i\Omega_n\right) \simeq \frac{2}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\left[\frac{\Omega_n^2 + \left(\epsilon_{k + q} - \epsilon_k\right)^2}{\left(\omega_n^2 + E_k^2\right)^2} - \frac{2\omega_n^2\Omega_n^2 + 2\left(\epsilon_k - \mu\right)^2\left(\epsilon_{k + q} - \epsilon_k\right)^2}{\left(\omega_n^2 + E_k^2\right)^3}\right]}}</math>
 
Again at <math>T = 0</math> ,
 
<math>\Pi_{--}\left(q, i\Omega_n\right) \simeq 2N_0\int_{-\infty}^{\infty}{\frac{d\omega}{2\pi}\int_{-\infty}^{\infty}{d\xi\left[\frac{\Omega_n^2 + \frac{1}{3}v_F^2q^2}{\left(\omega^2 + \xi^2 + \Delta_{sp}^2\right)^2} - 2\frac{\omega^2\Omega_n^2 + \xi^2\frac{1}{3}v_F^2q^2}{\left(\omega^2 + \xi^2 + \Delta_{sp}^2\right)^3} + \ldots \right]}}</math>
 
<math>\Pi_{--}\left(q, i\Omega_n\right) \simeq \frac{N_0}{2\Delta_{sp}^2}\left(\Omega_n^2 + \frac{1}{3}v_F^2q^2\right) + \ldots</math>
 
Finally,
 
<math>\begin{align} \Pi_{-+}\left(q, i\Omega_n\right) = & -\frac{1}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\frac{\left(i\omega_n + i\Omega_n\right)\left(\epsilon_k - \mu\right)\left(-2i\right) + i\omega_n\left(\epsilon_{k + q} - \mu\right)\cdot 2i}{\left[\left(\omega_n + \Omega_n\right)^2 + \left(\epsilon_{k + q} - \mu\right)^2 + \Delta_{sp}^2\right]\left[\omega_n^2 + \left(\epsilon_k - \mu\right)^2 + \Delta_{sp}^2\right]}}} \\
 
& = -\frac{1}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\frac{2\Omega_n\left(\epsilon_k - \mu\right) - 2\omega_n\left(\epsilon_{k + q} - \epsilon_k\right)}{\omega_n^2 + 2\omega_n\Omega_n + \Omega_n^2 + E_k^2 + \left(\epsilon_{k + q} - \epsilon_k\right)\left(\epsilon_{k + q} + \epsilon_k - 2\mu\right)} \cdot \frac{1}{\omega_n^2 + E_k^2}}} \\
 
& = -\frac{1}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\frac{2\Omega_n\left(\epsilon_k - \mu\right) - 2\omega_n\left(\epsilon_{k + q} - \epsilon_k\right)}{\left(\omega_n^2 + E_k^2\right)^2\left(1 + \frac{2\omega_n\Omega_n + \Omega_n^2 + \left(\epsilon_{k + q} - \epsilon_k\right)\left(2\left(\epsilon_k - \mu\right) + \epsilon_{k + q} - \epsilon_k\right)}{\omega_n^2 + E_k^2}\right)}}} \\ \end{align}</math>
 
Now, expanding:
 
<math>\left[1 + \frac{2\omega_n\Omega_n + 2\left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \epsilon_k\right) + \left(\epsilon_{k + q} - \epsilon_k\right)^2 + \Omega_n^2}{\omega_n^2 + E_k^2}\right]^{-1}</math>
 
<math>\begin{align} = 1 & - \frac{2\omega_n\Omega_n + 2\left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \epsilon_k\right) + \left(\epsilon_{k + q} - \epsilon_k\right)^2 + \Omega_n^2}{\omega_n^2 + E_k^2} \\
 
& + \frac{\left[2\omega_n\Omega_n + 2\left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \epsilon_k\right) + \left(\epsilon_{k + q} - \epsilon_k\right)^2 + \Omega_n^2\right]^2}{\left(\omega_n^2 + E_k^2\right)^2} \\
 
& - \frac{\left[2\omega_n\Omega_n + 2\left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \epsilon_k\right) + \left(\epsilon_{k + q} - \epsilon_k\right)^2 + \Omega_n^2\right]^3}{\left(\omega_n^2 + E_k^2\right)^3} + \ldots \\ \end{align}</math>
 
When multiplied by <math>\Omega_n\left(\epsilon_k - \mu\right) - \omega_n\left(\epsilon_{k + q} - \epsilon_k\right)</math> and integrated over <math>\vec{k}</math> or summed over <math>\omega_n</math> all terms vanish if we assume particle-hole symmetry and <math>q/k_F \rightarrow 0</math>. That is because <math>\Omega_n\left(\epsilon_k - \mu\right) - \omega_n\left(\epsilon_{k + q} - \epsilon_k\right)</math> needs to be multiplied either by a term which is odd in <math>\epsilon_k - \mu</math> and even in <math>\epsilon_{k + q} - \epsilon_k</math> and in <math>\omega_n</math>, or by a term which is odd in <math>\omega_n</math> and <math>\epsilon_{k + q} - \epsilon_k</math> as well as even in <math>\epsilon_k - \mu</math>. But there is no such term to any order. That means that the coupling term is given by particle-hole symmetry breaking terms or terms down by <math>q/k_F</math>.
With such terms present, we expect it to be <math>\sim q^2\Omega_n</math>. Although, there is a term with <math>\Omega_nN_0^'</math>
 
<math>\left[\begin{array}{cc} \delta\Delta_+\left(-q, -i\Omega_n\right) & \delta\Delta_-\left(-q, -i\Omega_n\right) \end{array}\right]\left[\begin{array}{cc} m^2 + \frac{1}{3}\rho\left(\Omega_n^2 + c^2q^2\right) & \alpha\Omega_n \\ \alpha\Omega_n & \rho\left(\Omega_n^2 + c^2q^2\right) \end{array}\right]\left[\begin{array}{cc} \delta\Delta_+\left(q, i\Omega_n\right) \\ \delta\Delta_-\left(q, i\Omega_n\right) \end{array}\right]</math>
 
<math>\Longrightarrow</math>
 
<math>\begin{align} & \frac{1}{2}m^2 + \frac{2}{3}\rho\left(\Omega_n^2 + c^2q^2\right) & \pm \sqrt{\left[\frac{1}{2}m^2 - \frac{1}{3}\rho\left(\Omega_n^2 + c^2q^2\right)\right]^2 + \alpha^2\Omega_n^2} \\
 
\simeq & \frac{1}{2}m^2 + \frac{2}{3}\rho\left(\Omega_n^2 + c^2q^2\right) & \underbrace{\pm \left(\frac{1}{4}m^4- \frac{m^2}{3}\rho\left(\eta\Omega_n^2 + c^2q^2\right) + \ldots \right)^{\frac{1}{2}}}_{\begin{array}{clr} \Downarrow \\ \pm \frac{1}{2}m^2\left(1 - \frac{1}{3}\frac{\rho}{m^2}\left(\eta\Omega_n^2 + c^2q^2\right) + \ldots\right) \end{array}} \\
 
\simeq & \left\{\begin{array}{rcl} m^2 + \frac{1}{3}\left(\eta'\Omega_n^2 + c^2q^2\right) \\ \rho\left(\eta'\Omega_n^2 + c^2q^2\right) \end{array}\right. \end{align}</math>
 
<math>\Longrightarrow</math> we have a massive propagating mode AND a massless propagating mode with velocity <math>c = \frac{v_F}{\sqrt{3}}</math>. Such a mode would have to show up in specific-heat for instance as a correction to <math>T^3</math> but it does not. Rather, we have found a (thermodynamic) sound mode of a super-fluid (which is neutral), not superconductor which is charged.
 
==Anderson-Higgs "mechanism"==
 
In order to properly describe the collective modes of a (charged) superconductor, we must include the effects of long-range Coulomb repulsion. That means that in addition to including the short-range attraction we must include the long-range repulsion.

Latest revision as of 18:07, 3 May 2011

Welcome to Phy 6937 Superconductivity and superfluidity

PHY6937 is a one semester advanced graduate level course. Its aim is to introduce concepts and theoretical techniques for the description of superconductors and superfluids. This course is a natural continuation of the "many-body" course PHY5670 and will build on the logical framework introduced therein, i.e. broken symmetry and adiabatic continuity. The course will cover a range of topics, such as the connection between the phenomenological Ginzburg-Landau and the microscpic BCS theory, Migdal-Eliashberg treatment of phonon mediated superconductivity, unconventional superconductivity, superfluidity in He-4 and He-3, and Kosterlitz-Thouless theory of two dimensional superfluids.


The key component of the course is the collaborative student contribution to the course Wiki-textbook. Each team of students is responsible for BOTH writing the assigned chapter AND editing chapters of others.

Team assignments: Spring 2011 student teams


Outline of the course:

Pairing Hamiltonian and BCS instability

To see the origins of superconductivity, it is helpful to look at a toy system, which we already know will give us superconducting behavior. This is useful because the toy system is only a simple change to a non-interacting electron gas. By adding in some small attractive interaction, we will arrive at a superconducting system! This interaction need only occur between two electrons occupying the same position in space (and necessarily having opposite spin!). Additionally, we still find the interesting behaviour regardless of the size of the interaction; the only requirement is that it be non-zero!

Our toy model will be that of free fermions with an attractive contact interaction. We can write the Hamiltonian of the system as:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H=\sum_\vec{r}[\psi_\sigma^\dagger (\vec{r})(\epsilon_\vec{p}-\mu)\psi_\sigma^\dagger (\vec{r}) +g\psi_\uparrow^\dagger (\vec{r})\psi_\downarrow^\dagger (\vec{r})\psi_\downarrow (\vec{r})\psi_\uparrow (\vec{r})]}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ g<0} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ |g|<<\epsilon_{F}} .

For this system, the partition function is:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z=\int D[\psi_\sigma ^{*} (\tau, \vec{r}), \psi_\sigma (\tau, \vec{r})]e^{-S_{BCS}}}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{BCS}=\int_0^\beta d\tau \sum_\vec{r}[\psi_\sigma^\dagger (\tau, \vec{r})(\partial _\tau+\epsilon_\vec{p}-\mu)\psi_\sigma^\dagger (\vec{r}) +g\psi_\uparrow^\dagger (\vec{r})\psi_\downarrow^\dagger (\vec{r})\psi_\downarrow (\vec{r})\psi_\uparrow (\vec{r})]}

It doesn't matter to multiply partition function by a constant:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z\rightarrow Z=\int D[\psi_\sigma ^{*} (\tau, \vec{r}), \psi_\sigma (\tau, \vec{r})] D[\Delta^{*}(\tau, \vec{r}),\Delta (\tau, \vec{r})] e^{-S_{BCS}-S_{\Delta}}}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_\Delta=-\int_0^\beta d\tau\sum_{\vec{r}}\frac{1}{g}\Delta^*(\tau,\vec{r})\Delta(\tau,\vec{r})}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi^\dagger} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \psi} are grassmann numbers. Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \Delta^*} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \Delta} are normally commuting variables. Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_\uparrow\psi_\downarrow} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_\downarrow\psi_\uparrow} behave like constant (paired fermions behaving like bosons).

Let's make a shift of the new integration variables:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta \rightarrow \Delta+g\psi_\uparrow\psi_\downarrow}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta^*\rightarrow \Delta^*+g\psi^\dagger_\downarrow\psi^\dagger_\uparrow}

Then, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_\Delta=-\int_0^\beta d\tau \sum_{\vec{r}}\left [\frac{1}{g}\Delta^*\Delta + \Delta^*\psi_\uparrow \psi_\downarrow + \Delta\psi^\dagger_\downarrow \psi^\dagger_\uparrow+g\psi^\dagger_\downarrow \psi^\dagger_\uparrow \psi_\uparrow \psi_\downarrow\right ]}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align}S=&S_{BCS}+S_{\Delta}\\ =&\int_0^\beta d\tau \sum_{\vec{r}}\{ \psi_\sigma^\dagger(\tau, \vec{r})(\partial _\tau+\epsilon_\vec{p}-\mu)\psi_\sigma^\dagger (\tau, \vec{r}) \ \ \ \ \ \ \ \ \ \ \rightarrow S_0 \\ &+\Delta^*(\tau, \vec{r})\psi_\uparrow (\tau, \vec{r})\psi_\downarrow (\tau, \vec{r}) \Delta (\tau, \vec{r})\psi^\dagger_\downarrow (\tau, \vec{r})\psi^\dagger_\uparrow (\tau, \vec{r}) \rightarrow S_{int}\\ &-\frac{1}{g}\Delta^* (\tau, \vec{r})\Delta (\tau, \vec{r}) \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \rightarrow S_{\Delta} \end{align}}

then, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z=\int D[\psi_{\sigma}^{*}(\tau,\mathbf{r}),\psi_{\sigma}(\tau,\mathbf{r})]D[\Delta^{*}(\tau,\mathbf{r}),\Delta(\tau,\mathbf{r})]e^{-(S_{0}+S_{int.}+S_{\Delta})}} .

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\langle e^{-S_{int.}}\right\rangle _{0}\cong \exp\left [\frac{1}{2}\left\langle S_{int.}^{2}\right\rangle _{0}+\frac{1}{4!}\left (\left\langle S_{int.}^{4}\right\rangle _{0}-3\left\langle S_{int.}^{2}\right\rangle _{0}^{2}\right )\right ]} by cumulant expansion, which guarantees that until the 2nd order, it is accurate.

Use Matsubara's Method

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_{\sigma}(\tau,\mathbf{r})=\frac{1}{\beta}\underset{\omega_{n}}{\sum}\underset{\mathbf{k}}{\sum}e^{i\mathbf{k}\cdot\mathbf{r}}e^{-i\omega_{n}\tau}\psi_{\sigma}(i\omega_{n},\mathbf{k}), \omega_{n}=(2n+1)\frac{\pi}{\beta};}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta(\tau,\mathbf{r})=\frac{1}{\beta}\underset{\Omega_{n}}{\sum}\underset{\mathbf{k}}{\sum}e^{i\mathbf{k}\cdot\mathbf{r}}e^{-i\Omega_{n}\tau}\Delta_{\mathbf{k}}(i\Omega_{n}), \omega_{n}=2n\frac{\pi}{\beta}.}

Then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{0}=\frac{L^{D}}{\beta}\underset{\omega_{n}}{\sum}\underset{\mathbf{k}}{\sum}[-i\omega_{n}+\varepsilon_{\mathbf{k}}-\mu]\psi_{\sigma}^{\dagger}(i\omega_{n},\mathbf{k})\psi_{\sigma}(i\omega_{n},\mathbf{k}).}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{int.}=\frac{L^{D}}{\beta^{2}}\underset{\omega_{n},\Omega_{n}}{\sum}\underset{\mathbf{k},\mathbf{q}}{\sum}[\Delta_{\mathbf{q}}^{*}(i\Omega_{n})\psi_{\uparrow}(i\Omega_{n}-i\omega_{n},\mathbf{\mathbf{q}-k})\psi_{\downarrow}(i\omega_{n},\mathbf{k})+\Delta_{\mathbf{q}}(i\Omega_{n})\psi_{\downarrow}^{\dagger}(i\omega_{n},\mathbf{k})\psi_{\uparrow}^{\dagger}(i\Omega_{n}-i\omega_{n},\mathbf{\mathbf{q}-k})].}

The Fourier transform of 1 body Green's function is (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i=1,2} mean Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\mathbf{r}_{i},\tau_{i}}} ) Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G(1-2)=\left\langle \psi(1)\psi^{*}(2)\right\rangle =\frac{1}{\beta}\underset{\omega_{n}}{\sum}\frac{1}{L^{D}}\underset{\mathbf{k}}{\sum}e^{-i\omega_{n}(\tau_{1}-\tau_{2})}e^{i\mathbf{k}\cdot(\mathbf{r}_{1}-\mathbf{r}_{2})}\frac{1}{-i\omega_{n}+\varepsilon_{\mathbf{k}}-\mu}} ,

so Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G_{\sigma}^{0}(i\omega_{n},\mathbf{k})=\left\langle \psi_{\sigma}(i\omega_{n},\mathbf{k})\psi_{\sigma}^{\dagger}(i\omega_{n},\mathbf{k})\right\rangle _{0}=\frac{\beta}{L^{D}}\frac{1}{-i\omega_{n}+\varepsilon_{\mathbf{k}}-\mu}} .

Then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\langle S_{int.}^{2}\right\rangle _{0}=\frac{2L^{2D}}{\beta^{4}}\underset{\omega_{n},\Omega_{n}}{\sum}\underset{\mathbf{k},\mathbf{q}}{\sum}[G_{\uparrow}^{0}(i\omega_{n},\mathbf{k})G_{\downarrow}^{0}(i\Omega_{n}-i\omega_{n},\mathbf{q}-\mathbf{k})]\Delta_{\mathbf{q}}^{*}(i\Omega_{n})\Delta_{\mathbf{q}}(i\Omega_{n})=L^{D}\frac{2}{\beta}\underset{\Omega_{n},\mathbf{q}}{\sum}\chi_{p}(\mathbf{q},i\Omega_{n})\Delta_{\mathbf{q}}^{*}(i\Omega_{n})\Delta_{\mathbf{q}}(i\Omega_{n})} ,

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \chi_{p}(\mathbf{q},i\Omega_{n})=\frac{L^{D}}{\beta^{3}}\underset{\omega_{n},\mathbf{k}}{\sum}G_{\uparrow}^{0}(i\omega_{n},\mathbf{k})G_{\downarrow}^{0}(i\Omega_{n}-i\omega_{n},\mathbf{q}-\mathbf{k})} is called pairing susceptibility.

Let's calculate it:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \chi_{p}(\mathbf{q},i\Omega_{n})=\frac{L^{D}}{\beta^{3}}\underset{\omega_{n},\mathbf{k}}{\sum}G_{\uparrow}^{0}(i\omega_{n},\mathbf{k})G_{\downarrow}^{0}(i\Omega_{n}-i\omega_{n},\mathbf{q}-\mathbf{k})=\frac{1}{L^{D}}\frac{1}{\beta}\underset{\omega_{n},\mathbf{k}}{\sum}\frac{-1}{i\omega_{n}-\varepsilon_{\mathbf{k}}+\mu}\times\frac{1}{i\omega_{n}-i\Omega_{n}+\varepsilon_{\mathbf{q}-\mathbf{k}}-\mu}} ,

Fig.1.jpg

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow=\frac{1}{L^{D}}\frac{1}{\beta}\underset{\mathbf{k}}{\sum}\oint_{c}\frac{dz}{2\pi i}\frac{-1}{z-\varepsilon_{\mathbf{k}}+\mu}\times\frac{1}{z-i\Omega_{n}+\varepsilon_{\mathbf{q}-\mathbf{k}}-\mu}\frac{1}{e^{\beta z}+1}} .

Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{-1}{z-\varepsilon_{\mathbf{k}}+\mu}\times\frac{1}{z-i\Omega_{n}+\varepsilon_{\mathbf{q}-\mathbf{k}}-\mu}=\frac{1}{\varepsilon_{\mathbf{q}-\mathbf{k}}+\varepsilon_{\mathbf{k}}-2\mu-i\Omega_{n}}[\frac{1}{z-\varepsilon_{\mathbf{q}}+\mu}-\frac{1}{z-i\Omega_{n}+\varepsilon_{\mathbf{q}-\mathbf{k}}-\mu}]} ,

and change the integral path to

Fig.2.jpg

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow=-\frac{1}{L^{D}}\frac{1}{\beta}\underset{\mathbf{k}}{\sum}\frac{1}{\varepsilon_{\mathbf{q}-\mathbf{k}}+\varepsilon_{\mathbf{k}}-2\mu-i\Omega_{n}}[\frac{1}{e^{\beta(\varepsilon_{\mathbf{q}}-\mu)}+1}-\frac{1}{e^{\beta(-\varepsilon_{\mathbf{q}-\mathbf{k}}+\mu)}+1}]=\int\frac{d^{D}k}{(2\pi)^{D}}\frac{1}{\varepsilon_{\mathbf{q}}+\varepsilon_{\mathbf{q}-\mathbf{k}}-2\mu-i\Omega_{n}}[1-f(\varepsilon_{\mathbf{k}})-f(\varepsilon_{\mathbf{q}-\mathbf{k}})].}

In the static (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \Omega_{n}=0} ) and uniform (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{q}=0} ) limit,Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1-2f(\varepsilon_{\mathbf{k}})=\tanh[\frac{\beta}{2}(\varepsilon_{\mathbf{k}}-\mu)]} .

Then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \chi_{p}(0,0)=\int\frac{d^{D}k}{(2\pi)^{D}}\frac{\tanh[\frac{\beta}{2}(\varepsilon_{\mathbf{k}}-\mu)]}{2(\varepsilon_{\mathbf{k}}-\mu)}} .

In low energy, integrate the energy in the shell near Fermi energy:

Fig.3.jpg

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow\chi_{p}(0,0)\cong N(0)\int_{\hbar\omega_{D}}^{-\hbar\omega_{D}}d\xi\frac{\tanh[\xi\beta/2]}{2\xi}\cong N(0)\int_{0}^{-\hbar\omega_{D}}d\xi\frac{\tanh[\xi\beta/2]}{\xi}=N(0)\ln[\frac{2\hbar\omega_{D}e^{\gamma}}{\pi k_{B}T}].}

Then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2}\left\langle S_{int.}^{2}\right\rangle _{0}=L^{D}\frac{1}{\beta}\chi_{p}(0,0)\underset{\Omega_{n},\mathbf{q}}{\sum}\Delta_{\mathbf{q}}^{*}(i\Omega_{n})\Delta_{\mathbf{q}}(i\Omega_{n})} .

If we ignore the higher order in the cumulant expansion,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{eff}=-\underset{\mathbf{r}}{\sum}\int_{0}^{\beta}d\tau\frac{1}{g}\Delta_{\mathbf{q}}^{*}(i\Omega_{n})\Delta_{\mathbf{q}}(i\Omega_{n})-\frac{1}{2}\left\langle S_{int.}^{2}\right\rangle _{0}=\underset{\mathbf{r}}{\sum}\int_{0}^{\beta}d\tau[\frac{1}{\left|g\right|}-N(0)\ln(\frac{2\hbar\omega_{D}e^{\gamma}}{\pi k_{B}T})]\Delta^{*}(\tau,\mathbf{r})\Delta(\tau,\mathbf{r})} .

Remember the partition function Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z=\int D\Delta^{*}D\Delta e^{-S_{eff}(\Delta)}} . To look at superconductivity, we really only must consider the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta} related factors.

The superconductivity phase transition temperature is the temperature that makes 

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\frac{1}{\left|g\right|}}-N(0){\ln(\frac{2\hbar\omega_{D}e^{\gamma}}{\pi k_{B}T})}=0}
,

which happens at 

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_{c}=\frac{\hbar\omega_{D}}{k_{B}}\frac{2}{\pi}e^{\gamma}e^{-\frac{1}{N(0)\left|g\right|}}=1.134\frac{\hbar\omega_{D}}{k_{B}}e^{-\frac{1}{N(0)\left|g\right|}}}
.

Above this critical temperature, the coefficient of the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta^*\Delta} term becomes positive, and it is energetically unfavorable to have cooper pairs, as in the normal state. Below the critical temperature, the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta} related factors in the partition function will diverge, which means superconductivity phase transition. Notice that this happens for any value of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |g|} ! No matter how small the attraction between our fermions is, we can still find this funny (as of yet, it has not been shown to be specifically superconducting!) behaviour.

Finite Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{q}} (small) Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ (\Omega_n=0)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \chi_p (q,0)-\chi_p (0,0)=\frac{1}{L^D} \sum_k \frac{1}{\beta} \sum_{i\omega_n}\frac{-1}{i\omega_n-\epsilon_k+\mu}(\frac{1}{i\omega_n+\epsilon_{q-k}-\mu}-\frac{1}{i\omega_n+\epsilon_{-k}-\mu}) }

for small ,

and

Thus,

Consider the states near the shell near fermi surface, we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \chi_p(q,0)-\chi_p(0,0) =\frac{1}{\beta}\sum_{i\omega_n}\int\frac{d\Omega_{F.S.}}{\Omega_D}(q\cdot v_F)^2\int_{-\infty}^{+\infty} d\xi N(\xi+\mu)\frac{1}{(\xi-i\omega_n)(\xi+i\omega_n)^3}}

where, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \xi=\epsilon_k-\mu}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int_{-\infty}^{+\infty} d\xi \frac{1}{(\xi-i\omega_n)(\xi+i\omega_n)^3} &=\frac{2\pi i}{(2i\omega_n)^3}\theta(\omega_n)-\frac{2\pi i}{(2i\omega_n)^3}\theta(-\omega_n)\\ &=\frac{2\pi i}{(2i|\omega|)^3} \end{align} }

So,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \chi_p(q,0)-\chi_p(0,0) &=\frac{1}{\beta}\sum_{i\omega_n}N(0)\int\frac{d\Omega}{\Omega_D}(q\cdot v_F)^2\frac{2\pi i}{(2i|\omega|)^3}\\ &=N(0)v_F^2|\vec{q}|^2\int\frac{d\Omega}{\Omega_D}(q\cdot v_F)^2 \frac{1}{\beta}\sum_{i\omega}\frac{2\pi i}{-8i|\frac{(2n+1)\pi}{\beta}|^3}\\ &=-\frac{1}{4}N(0)v_F^2q^2(<(\hat{q}\cdot \hat{v_F})>_{F.S.})\frac{\beta^2}{\pi^2}(\sum_{N=-\infty}^{+\infty}\frac{1}{|2n+1|^3}) \end{align} }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{N=-\infty}^{+\infty}\frac{1}{|2n+1|^3}=\sum_{n=0}^\infty\frac{2}{(2n+1)^3}=\frac{2}{\pi}\frac{7\zeta(3)}{8} }

where, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \zeta(3)} is Riemann zeta function.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <(\hat{q}\cdot \hat{v}_F)^2>_{F.S.}=\frac{1}{D} }

For spherical F.S. in 3D,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{d\Omega}{\Omega_D}(\hat{q}\cdot\hat{v}_F)^2=\frac{2\pi}{4\pi}\int_{-1}^{1}d(\cos{\theta})\,\cos^2{\theta} = \frac{1}{3} }

For circular F.S. in 2D,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{d\Omega}{\Omega_D}(\hat{q}\cdot\hat{v}_F)^2=\frac{1}{2\pi}\int_{0}^{2\pi}d\theta\,\cos^2{\theta} = \frac{1}{2} }

Then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \chi_p(q,0)-\chi_p(0,0) &=-\frac{1}{4}N(0)v_{F}^{2}q^{2}\frac{1}{D}\frac{\beta^{2}}{\pi^{2}}\frac{2}{\pi}\frac{7\zeta(3)}{8} \\ &=-N(0)\frac{7\zeta(3)}{16D\pi^{2}}q^{2}\frac{1}{\pi \hbar^{2}}\left(\frac{\hbar v_{F}}{k_{B}T}\right)^{2} \\ &\equiv-N(0)q^{2}\xi^{2} \end{align} }


So

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{1}{2}\left\langle S_{int.}^{2}\right\rangle _{0}&=L^{D}\frac{1}{\beta}\underset{\Omega_{n},\mathbf{q}}{\sum}\chi_{p}(q,0)\Delta_{\mathbf{q}}^{*}(i\Omega_{n})\Delta_{\mathbf{q}}(i\Omega_{n}) \\ &=N(0)\ln[\frac{2\hbar\omega_{D}e^{\gamma}}{\pi k_{B}T}]L^{D}\frac{1}{\beta}\underset{\Omega_{n},\mathbf{q}}{\sum}\Delta_{\mathbf{q}}^{*}(i\Omega_{n})\Delta_{\mathbf{q}}(i\Omega_{n})-L^{D}\frac{1}{\beta}\underset{\Omega_{n},\mathbf{q}}{\sum}N(0)q^{2}\xi^{2}\Delta_{\mathbf{q}}^{*}(i\Omega_{n})\Delta_{\mathbf{q}}(i\Omega_{n}) \end{align} } .

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{eff}=\underset{\mathbf{r}}{\sum}\int_{0}^{\beta}d\tau\left[\left(\frac{1}{\left|g\right|}-N(0)\ln(\frac{2\hbar\omega_{D}e^{\gamma}}{\pi k_{B}T})\right)\Delta^{*}(\tau,\mathbf{r})\Delta(\tau,\mathbf{r})-N(0)\xi^{2}(\nabla\cdot\Delta^{*}(\tau,\mathbf{r}))(\nabla\cdot\Delta(\tau,\mathbf{r}))\right]} .

Note that the last term in the expression tells us that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{eff} } would increase if gradient of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta } is not zero.

Note that the above expression has a one-one correspondant to the Giznburg-Landau functional:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F=\int d^{D}r\left[ \alpha (T-T_{c}) |\Psi(\vec{r})|^{2}+\frac{\hbar^{2}}{2m^{*}}|\nabla \Psi(\vec{r})|^{2} \right] } ,

here Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \Psi(\vec{r}) } corresponds to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \frac{\Delta(\tau,\vec{r})}{|g|N(0)a_{0}} } in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{eff}} .


Little Parks experiment

Layout48.gif

As shown in the above figure, a thin shell of superconductor with radius R is placed in a small uniform magnetic field passing through the hollow center of the cylinder. The experiment intends to show the variation of the critical temperature with change of the magnetic field passing through the hollow superconductor cylinder. This was done in 1962, by Little and Parks at Stanford University.

Before showing the mathematics behind the experiment, we first have to rewrite the Giznburg-Landau functional to take the presence of magnetic field into account. The Hamiltonian for a free electron moving in a magnetic field is written as:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2m}(p+\frac{eA}{c})^{2}\psi + V\psi = E\psi }

The physically observable magnetic field Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}} would remain the same if we choose a different vector potential Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A\rightarrow A+ \nabla \chi } (or perform a gauge transformation). To maintain the same eigen-energy E which is observable, the wave function has to undergo a phase change: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi \rightarrow e^{i\phi}\psi } where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi=\frac{e}{c\hbar}\chi }

Now in our effective action, the wave function appears as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta^{*}(\tau,\vec{r})\psi_\uparrow (\tau,\vec{r})\psi_\downarrow (\tau,\vec{r}) + \Delta(\tau,\vec{r}) \psi_\downarrow^\dagger (\tau,\vec{r})\psi_\uparrow^\dagger (\tau,\vec{r}) }

since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi \rightarrow e^{i\phi}\psi } , if we want the effective action to remind the same, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \Delta } has to transform as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta \rightarrow e^{-2i\phi}\Delta }

Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \Delta } corresponds to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \Psi } in the Giznburg-Landau functional, so the Giznburg-Landau functional is modified as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F=\int d^{D}r\left[ \alpha (T-T_{c}) |\Psi(\vec{r})|^{2}+\frac{1}{2m^{*}}| ( \frac{\hbar \nabla}{i} - \frac{2e}{c}A(\vec{r}) ) \Psi(\vec{r})|^{2} \right] }

Note that the last term looks like a particle of charge Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -2e} in a magnetic field!

Now, we choose the symmetric gauge:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A}=\frac{1}{2}\vec{H}\times\vec{r}=\frac{1}{2}Hr\hat{\phi} }

Re-write in cylindrical coordinates:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{\nabla}=\hat{r}\frac{\partial}{\partial r} + \frac{\hat{\phi}}{r}\frac{\partial}{\partial \phi} + \hat{z}\frac{\partial}{\partial z} }

Define the unit flux as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi_{0}=\frac{hc}{2e} }

Define the fluxoid as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi(R) = \pi HR^{2}\ } , so we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} F&=\int d^{D}r\left[ \alpha (T-T_{c})|\Psi(\vec{r})|^{2} +\frac{\hbar^{2}}{2m^{*}}| (\frac{1}{R}\frac{\partial}{\partial \phi} - \frac{ie}{\hbar c} HR )\Psi(\vec{r}) |^{2}+ \frac{\hbar^{2}}{2m^{*}}|\frac{\partial}{\partial z} \Psi(\vec{r}) |^{2} \right] \\ &=\int d^{D}r\left[ \alpha (T-T_{c})|\Psi(\vec{r})|^{2} +\frac{\hbar^{2}}{2m^{*}R^{2}}| (\frac{\partial}{\partial \phi} - \frac{i\Phi}{\Phi_{0}} )\Psi(\vec{r}) |^{2} \right] \\ \end{align} }

When Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi = N\Phi_{0}\ } , the critical temperature will remain the same and the phase of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi\ } is changed as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi \rightarrow e^{iN\phi} \Psi } . When Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi \neq N\Phi_{0}\ } , the critical temperature is found to vary as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_{c}^{new}=T_{c}- \frac{\hbar^{2}}{2m^{*}R^{2}\alpha}\left (N-\frac{\Phi}{\Phi_{0}}\right )^{2}} . See the figure below.

Graph49.gif

This experiment is significant evidence that the objects represented by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta} carry a charge twice that of a single electron. As we know that this thing stands in the equations for a cooper pair, this makes perfect sense!

Microscopic derivation of the Ginzburg-Landau functional

Let us consider the model of a metal close to the transition to the superconducting state. A complete description of its thermodynamic properties can be done through the calculation of the partition function.

Ginzburg-Landau Functional

The classical part of the Hamiltonian in the partition function, dependent of bosonic fields, may be chosen in the spirit of the Landauer theory of phase transition. However, in view of the space dependence of wave functions, Ginzberg and Landauer included in it additionally the first non vanishing term of the expansion over the gradient of the fluctuation field. Symmetry analysis shows that it should be quadratic. The weakness of the field coordinate dependence permits to omit the high-order terms of such an expansion. Therefore, the classical part of the Hamiltonian of a metal close to a superconducting transition related to the presence of the fluctuation Cooper pairs in it (the so called Ginzberg-Landauer functional)can be written as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F[\psi(r)]=F_{n}+\int dV\{a\mid\psi(r)\mid^{2}+\frac{b}{2}\mid\psi(r)\mid^{4}+\frac{1}{4m}\mid\nabla\psi(r)\mid^{2}\}}

The basic postulate of G-L is that if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi} is small and varies slowly in space, so we can expand the free energy in a series of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mid\psi(r)\mid^{2}} .These two terms should be adequate so long as one stays near the second-order phase transition at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_{c}} , where the order parameter Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mid\psi(r)\mid^{2}\rightarrow0} . Inspection of G-L functional shows that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b} must be positive if the theory is to be useful; otherwise the lowest free energy would occur for arbitrarily large values of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mid\psi(r)\mid^{2}} , where the expansion is surely inadequate.

We already got the quadratic terms in the Ginzberg-Landauer by expanding Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <e^{-S_{int}}>} to the second order, and we are going to go the higher order. As we discussed, we expect that this term will be a negative value to keep Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{eff}} as a negative value under Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_{c}} . To catch this goal we start with the partition function:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z=Z_{0}< e^{-S_{int}} >}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z_{0}=\int D\psi ^{*} D\psi D\Delta ^{*} D\Delta e^{-(S_{\Delta} +S_{0})}}

we can expand this average for small Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta} near Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_{c}} , for this perpose we can assume a second order phase transition so that it increases continiously from zero to finite number after Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_{c}} .

we need to calculate the average of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{-s_{int}}} which can be calculated by Tylor expansion:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{-S_{int}}=<1-S_{int}+\frac{1}{2}S_{int}^{2}-\frac{1}{3}S_{int}^{3}+\frac{1}{4!}S_{int}^{4}+...>}

=Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1-<S_{int}>+\frac{1}{2} < S_{int}^{2}> -\frac{1}{3!}< S_{int}^{3}> +\frac{1}{4!}< S_{int}^{4}> +...}

In this expansion all odd power terms go to zero because when we calculate the integral over Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi} and its complex conjugate it will vanish since we have odd numbers of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi} or Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi^{*}} in the integral. Then, we try to approximate it to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{\frac{1}{2}< S_{int}^{2}>}e^{\frac{1}{4!}< S_{int}^{4}>-\lambda }} .

if we expand these two terms in to the second order the following expression can be got:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (1+\frac{1}{2} < S_{int}^{2}>+\frac{1}{2}(<\frac{1}{2} S_{int}^{2}>)^{2} +...)(1+\frac{1}{4!}< S_{int}^{4}>+...)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =1+\frac{1}{2} < S_{int}^{2}>+\frac{1}{8}(< S_{int}^{2}>)^{2} +...)+\frac{1}{4!}< S_{int}^{4}>-\lambda +...}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda} can be chose in a such way that this result will match to the original expansion of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{-s_{int}}} up to the forth power of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_{int}}  :

so, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda =\frac{1}{8}< S_{int}^{2}> ^{2} }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =e^{\frac{1}{2} < S_{int}^{2}>+\frac{1}{4!}(< S_{int}^{4}>-3<S_{int}^{2}>^{2})+...}}

according to the expression we got before:


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{int}=\frac{L^{D}}{\beta ^{2}}\sum_{\omega _{n},\Omega _{n}}\sum _{k,q}[\Delta ^{*}_{q}(i\Omega _{n})\psi_{\downarrow}(i\Omega _{n}-i\omega _{n}),\vec{q}-\vec{k})\psi_{\uparrow}(i\omega _{n},k)+\Delta _{q}(i\Omega_{n})\psi_{\uparrow}^{\dagger }(i\omega _{n},k)\psi_{\downarrow}^{\dagger }(i\Omega _{n}-i\omega _{n}),\vec{q}-\vec{k})]}

let's write Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{int}} in terms of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} for simplification. where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a=\int \Delta ^{*}(1)\psi_{\downarrow}(1) \psi_{\uparrow}(1)+\Delta (1)\psi_{\downarrow}^{*}(1) \psi_{\uparrow}^{*}(1)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{1}} is a couple Grassman number, so we do not need to be worry about the sign when these terms commute with other terms.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle < S_{int}^{4}> =\int_{1234} < (a_{1}^{*}+a_{1})(a_{2}^{*}+a_{2})(a_{3}^{*}+a_{3})(a_{4}^{*}+a_{4})>}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =(< a_{1}^{*}a_{2}^{*}a_{3}a_{4}>+ < a_{1}^{*}a_{2}a_{3}^{*}a_{4}>+< a_{1}^{*}a_{2}a_{3}a_{4}^{*}>+< a_{1}a_{2}^{*}a_{3}^{*}a_{4}> +< a_{1}a_{2}^{*}a_{3}a_{4}^{*}>)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =6< a_{1}^{*}a_{2}^{*}a_{3}a_{4}>=6\int _{1234}\Delta ^{*}(1)\Delta ^{*}(2)\Delta(3)\Delta(4)< \psi_{\downarrow}(1)\psi_{\uparrow}(1)\psi_{\downarrow}(2)\psi_{\uparrow}(2)\psi_{\downarrow}^{*}(3)\psi_{\uparrow}^{*}(3)\psi_{\downarrow}^{*}(4)\psi_{\uparrow}^{*}(4)> }

By Feynman diagram, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle < S_{int}^{4}>} has:

Secondorder.png


Square.png

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3< S_{int}^{2}> ^{2}=3\int_{1,2}2< a_{1}^{*}a_{2}> \int_{3,4}2< a_{3}^{*}a_{4}>=12\int_{1,2,3,4}\Delta ^{*}(1)\Delta ^{*}(2)\Delta(3)\Delta(4)< \psi_{\downarrow}(1)\psi_{\uparrow}(1)\psi_{\uparrow}^{*}(3)\times}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_{\downarrow}^{*}(3)>< \psi_{\downarrow}(2)\psi_{\uparrow}(2)\psi_{\uparrow}^{*}(4)\psi_{\downarrow}^{*}(4)>}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -2G(2-3) G(2-4)G(1-4)G(1-3)=-12\int_{1,2,3,4}\Delta ^{*}(1)\Delta _{*}(2)\Delta (3)\Delta (4)G(2-3)G(2-4)G(1-4)G(1-3)}

Recall the Fourier transform of one body Green function is:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G(2-3)=< \psi (r_{2},\tau _{2})\psi ^{*}(r_{3},\tau _{3})>=\frac{1}{\beta } \sum_{\omega _{n}}\frac{1}{L^{D}}\sum_{k}e^{-i\omega _{n}(\tau _{2}-\tau _{3})}e^{ik.(r_{2}-r_{3})}\frac{1}{-i\omega _{n}+\epsilon _{k}-\mu }}

To seek solution of which are Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tau} independent using Feynman diagram

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\beta ^{4}}\sum_{\omega _{{n}_{1}}}...\sum_{\omega _{{n}_{4}}}\int_{0}^\beta {d\tau_{1}}\int_{0}^\beta {d\tau_{2}} \int_{0}^\beta {d\tau_{3}} \int_{0}^\beta {d\tau_{4}}e^{-i\omega _{{n}_{1}}(\tau _{1}-\tau _{3})} e^{-i\omega _{{n}_{2}}(\tau _{1}-\tau _{4})}e^{-i\omega _{{n}_{3}}(\tau _{2}-\tau _{3})}e^{-i\omega _{{n}_{4}}(\tau _{2}-\tau _{4})} G(i\omega _{{n}_{1}},r_{1}-r_{3})\times}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G(i\omega _{{n}_{2}},r_{1}-r_{4})G(i\omega _{{n}_{3}},r_{2}-r_{3})G(i\omega _{{n}_{4}},r_{2}-r_{4})}

after getting integration over Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tau_{1}} we will get Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta \delta (\omega _{n_{1}},-\omega _{n_{2}}) } and similarly by getting integration over Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tau_{2}} we have Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta \delta (\omega _{n_{3}},-\omega _{n_{4}}) }

So, the final result can be written: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i\omega _{n}}G(i\omega _{n},r_{1}-r_{3}G(-i\omega _{n},r_{2}-r_{3})G(-i\omega _{n},r_{1}-r_{4})G(i\omega _{n},r_{2}-r_{4}) }

Now, we wish to perform gradiant expansion:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta ^{\ast }(1)=\Delta ^{\ast }(\frac{r_{1}+r_{4}}{2}+\frac{r_{1}-r_{4}}{2})=\Delta ^{\ast }(\frac{r_{1}+r_{4}}{2})+(\frac{r_{1}-r_{4}}{2})\nabla\Delta^{\ast }(\frac{r_{1}+r_{4}}{2})+... }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta ^{\ast }(2)=\Delta ^{\ast }(\frac{r_{2}+r_{3}}{2}+\frac{r_{2}-r_{3}}{2})=\Delta ^{\ast }(\frac{r_{2}+r_{3}}{2})+(\frac{r_{2}-r_{3}}{2})\nabla\Delta^{\ast }(\frac{r_{2}+r_{3}}{2})+... } Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -12\int d^{D}R_{1,4}d^{D}R_{2,3}d^{D}\mu _{1,4}d^{D}\mu_{2,3}\Delta ^{\ast }(R_{1,4})\Delta ^{\ast }(R_{2,3})\Delta (R_{2,3}) \Delta (R_{1,4})\sum_{\omega_{n}}G(i\omega_{n},R_{1,4}-R_{2,3}+\frac{1}{2}(\mu _{1,4}+\mu _{2,3}))}

where:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R=\frac{1}{2}(R_{1,4}+R_{2,3})}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu =R_{1,4}-R_{2,3} }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r_{1}-r_{3}=\frac{1}{2}(\mu _{1,4}+\mu_{2,3})+\mu}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r_{2}-r_{4}=\frac{1}{2}(\mu _{1,4}+\mu_{2,3})-\mu}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \simeq -12\int d^{D}R d^{D}\mu d^{D}\mu_{1.4}d^{D}\mu_{2,3}\Delta ^{\ast }(R)\Delta^{\ast } (R)\Delta (R)\Delta (R)\sum_{\omega _n}G(i\omega _{n},\mu +\frac{1}{2}(\mu_{1,4}+\mu _{2,3}))G(-i\omega _{n},\mu _{1,4})\times}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G(-i\omega _{n},\mu _{2,3})G(i\omega _{n},\mu +\frac{1}{2}(\mu_{1,4}+\mu _{2,3}))}

integrate overFailed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu} gives us: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L^{D}} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta_{k_{1},k_{4}}}

and similarly Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu_{1,4}} : Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L^{D}} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta_{k_{1},-k_{2}}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu_{2,3}} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l^{D}} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta_{k_{1},-k_{3}}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -12\int d^{D}R\left | \Delta (R)^{4} \right |\sum_{\omega _{n}}\frac{1}{L^{D}}\sum_{k}\frac{1}{(-i\omega _{n}+\varepsilon_{k}-\mu)^{2}}\frac{1}{(i\omega _{n}+\varepsilon_{-k}+\mu)^{2}}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon _{k}=\epsilon _{-k}} because of timereversal symetry.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon _{k}-\mu =\xi }

Contour-yp.png


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -12\int d^{D}R\left | \Delta (R)\right |^{4}\sum_{\omega_{n}}\int_{-\infty }^{+\infty }d\xi N(\xi +\mu )\frac{1}{(-i\omega _{n}+\xi )^{2}}\frac{1}{(i\omega _{n}+\xi )^{2}} }


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I(p,q)=\int_{-\infty }^{+\infty }d\xi \frac{1}{(\xi -p)^{2}}\frac{1}{(\xi +p)^{2}}={\frac{\partial }{\partial p}}\int d\xi \frac{1}{\xi -p}\frac{1}{(\xi +p)^{2}} }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial }{\partial p}2\pi i \frac{1}{(p+q)^{2}}=2\pi i\frac{-2}{(p+q)^{2}}} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \simeq -12\int d^{D}R\left | \Delta (R) \right |^{4}\sum_{\omega _{n}}N(0)\frac{-4\pi i}{(2i\omega _{n})^{3}} }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \simeq-12\beta\int d^{D}R\mid\triangle(R)\mid^{4}\frac{N(0)}{2\beta}\pi\frac{\beta^{3}}{\pi^{3}}\sum_{n=-\infty}^{\infty}\frac{1}{\mid2n+1\mid^{3}}}

We can substitute the sum using Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=-\infty}^{\infty}\frac{1}{\mid2n+1\mid^{3}}=2\times\frac{7\xi(3)}{8}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =-\frac{1}{2}\frac{7\xi(3)}{8\pi^{2}}\beta^{2}N(0)\frac{1}{K_{B}T}\int d^{D}R\mid\triangle(R)\mid^{4}}

So, we get a term that is always negative and it corresponds to the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mid\psi(r)\mid^{4}} in the Ginzberg-Landauer functional.

Saddle Point Approximation

Starting from the microscopic model, we found that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z\backsimeq Z_{0}\int D\Delta*D\Delta e^{-S_{eff}} } , where the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 4^{th}} order in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta} , and keeping only quadratic qradient terms, we have:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{eff}=\frac{1}{k_{B}T}\sum_{r}\left[\underset{A}{\underbrace{\left(\frac{1}{|g|}-N(0)In\left[\frac{2\hbar\omega_{D}e^{\gamma_{E}}}{\pi k_{B}T}\right]\right)}}|\Delta(r)|^{2}+N(0)\xi^{2}(\nabla\Delta*(r)).(\nabla\Delta(r))+\frac{1}{2}\underset{B}{\underbrace{\frac{7\zeta(3)N(0)}{8\pi^{2}k_{B}^{2}T^{2}}}}|\Delta(r)|^{4}\right]}

We can use this expression to make quantitative experimental predictions. The path integral over Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta} is still imposible to carry out exactly, despite our approximations for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{eff}} , because Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{eff}} contains quartic terms in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta} and so we are not dealing with a Gaussian integral. The approximation strategy whic we will pursue is called saddle point approxiation, which in our contetxt means that we will expand teh integrand about a solution which minimizes S_{eff} with respect to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta} . What we end up doing is replacing Z with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z\sim Z_{0}e^{-S_{eff}[\Delta_{min}]}} , where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta_{min}} is determined fromFailed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|\frac{\delta S_{eff}}{\delta\Delta^{*}}\right|_{\Delta=\Delta_{min}}=0} At this point, let's seek uniform solutions to their equations, in whcih case we can drop the gradient terms in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{eff}}  : Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\delta S_{eff}}{\delta\Delta^{*}}=\frac{L^{D}}{k_{B}T}\left(A\Delta+B\Delta\Delta^{*}\Delta\right)} where: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A=\frac{1}{|g|}-N(0)ln\left(\frac{\hbar\omega_{D}}{k_{B}T}\frac{2e\gamma}{\pi}\right)} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B=\frac{7\zeta(3)}{8\pi}\frac{N(0)}{k_{B}^{2}T^{2}}}

Note that for : Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T>T_{C}\qquad A>0,B>0} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T<T_{C}\qquad A<0,B>0}

So Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A\Delta_{min}+B\Delta_{min}^{3}=0} ,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta_{min}=0, \;T>T_{C}} and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta_{min}=\sqrt{\frac{-A}{B}},\; T<T_{C} } .


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{eff}[\Delta_{min}]=\frac{L^{D}}{k_{B}T}\left(A\Delta_{min}^{2}+\frac{1}{2}B\Delta_{min}^{4}\right)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T>T_{C};\quad S_{eff}[\Delta_{min}]=0} ,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T<T_{C};\quad S_{eff}[\Delta_{min}]=\left(A\frac{-A}{B}+\frac{1}{2}B\frac{A^{2}}{B^{2}}\right)\frac{L^{D}}{k_{B}T}=\frac{-A^{2}}{2B}\frac{L^{D}}{k_{B}T}}

Since, we now have the approximate expression for the partition function we can calculate thermodynamic physical properties. the one we will focus on is the specific heat. Recall that,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{\partial F}{\partial T}+\frac{\partial\beta}{\partial T}\frac{\partial F}{\partial\beta}+\beta\frac{\partial}{\partial T}\frac{\partial F}{\partial\beta}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =2\frac{\partial F}{\partial T}+\beta\frac{\partial}{\partial T}\left(\frac{\partial F}{\partial T}\frac{\partial T}{\partial\beta}\right)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =2\frac{\partial F}{\partial T}+\beta\frac{\partial^{2}F}{\partial T^{2}}\left(-k_{B}T^{2}\right)+\beta\frac{\partial}{\partial T}\left(-2k_{B}T\right)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =-T\frac{\partial^{2}F}{\partial T^{2}}}

if we only study the constribution to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{V}} from the superconducting order parameter terms in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{eff}} , we have Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{V}} So, we see that if the double derivateive of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{-A^{2}}{2B}} with respect to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T} is finite at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_{C}} , then the specific heat jumps at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_{C}} , since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{V}=0} for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T>T_{C}} . We are interested in the size of this jump. Therefore, we need to simply expand Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{-A^{2}}{2B}} near Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_{C}} . Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} vanishes at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_{C}} , we can simply evaluate Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B} at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_{C}} and expand Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A(T)=\frac{1}{|g|}-N(0)ln\left(\frac{\hbar\omega_{D}}{k_{B}\left(T_{C}+T-T_{C}\right)}\frac{2e^{\gamma}}{\pi}\right)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{1}{|g|}-N(0)ln\left(\frac{2e^{\gamma}}{\pi}\frac{\hbar\omega_{D}}{k_{B}T_{C}}\left(1+\frac{T-T_{C}}{T_{C}}\right)^{-1}\right)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\underset{vanishes\; by\; def.\; of\; T_{C}}{\underbrace{\frac{1}{|g|}-N(0)ln\left(\frac{2e^{\gamma}}{\pi}\frac{\hbar\omega_{D}}{k_{B}T_{C}}\right)}}by+N(0)ln\left(1+\frac{T-T_{C}}{T_{C}}\right)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow A(T)=N(0)ln\left(1+\frac{T-T_{C}}{T_{C}}\right)\simeq N(0)\frac{T-T_{C}}{T_{C}}+...}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow\frac{-A^{2}(T)}{2B(T)}\simeq-\frac{1}{2}\frac{N^{2}(0)\left(\frac{T-T_{C}}{T_{C}}\right)^{2}}{\frac{7\zeta(3)}{8\pi}\frac{N(0)}{k_{B}^{2}T^{2}}}+...}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta c_{V}\simeq-\frac{T_{C}}{2}\frac{8\pi^{2}}{7\zeta(3)}k_{B}^{2}N(0)+...}

What is the specific heat of a non-interacting electron gas?

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{V}^{(n)}=\frac{\partial}{\partial T}\left(2(from\; spin)\int\frac{d^{D}k}{(2\pi)^{D}}\frac{\left(\epsilon_{k}-\mu\right)}{e^{\frac{\epsilon_{k}-\mu}{k_{B}T}+1}}\right)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =2\int\frac{d^{D}k}{(2\pi)^{D}}\frac{-\left(\epsilon_{k}-\mu\right)}{\left(e^{\frac{\epsilon_{k}-\mu}{k_{B}T}+1}\right)^{2}}\left(\frac{-\left(\epsilon_{k}-\mu\right)}{k_{B}T^{2}}\right)e^{\left(\frac{\epsilon_{k}-\mu}{k_{B}T}\right)}} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \simeq2k_{B}N(0)\int_{-\infty}^{\infty}d\xi\left(\frac{\xi}{2k_{B}T}\right)^{2}\frac{1}{cosh^{2}\left(\frac{\xi}{2k_{B}T}\right)}} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \simeq4k_{B}^{2}TN(0)\underset{\frac{\pi^{2}}{6}}{\underbrace{\int_{-\infty}^{\infty}dx\frac{x^{2}}{cosh^{2}x}}}=\frac{2\pi^{2}}{3}k_{B}^{2}T}

So, if we measure the jump in the specific heat at T_c in the units of the normal state electronic contribution we find: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\Delta c_{V}}{c_{V}^{(n)}}=\frac{\frac{8\pi^{2}}{7\zeta(3)}k_{B}^{2}T_{C}N(0)}{\frac{2\pi^{2}}{3}k_{B}^{2}T_{C}N(0)}} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{12}{7\zeta(3)}\simeq1.426} This is dimensionless number is a “famous” prediction of the BCS theory, although we derived it using different formalism. Let's check it with experiment:

First the caveats:

when specific is measured, all excitations contribute. Most importantly lattice vibrations (phonons) contribute as well. At low T, however, the phonon contribution drops of as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T^3} and we can neglect it if the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_C} is sufficiently low. In practice we have do an example:

materials Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_C} phonon contribution at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_C}
Al 1.2K 1%
Zn 0.8K 3%
Cd 0.5K 3%
Sn 3.7K 45%
In 3.4K 77%
Th 2.4K 83%
Pb 7.2K 94%


Experimental data for Aluminum gives Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\Delta c_{V}}{c_{V}^{(n)}}\simeq 1.39}

This result from the saddle point approximation/mean field is very close to the experimental result. But for high Tc superconductor the behavior is very different. The heat capacity of superconductor decrease gradually, doesn't have sharp jump like in the picture below. The Gaussian approximation we have used is not enough to explain high Tc superconductor.

Limitation of Microscopic derivation of the Ginzburg-Landau functional

Experimental result of heat capacity of aluminum (dotted lines), at zero magnetic field(curve peak) and at 300 Gauss when superconductivity suppressed (no peak) , typical curve for high Tc superconductor (additional solid blue line), (Phys. Rev., 114:676–685, 1959)

1) Fluctuation is important in low dimension (such in 2D by including Gaussian fluctuation)

2) In 1D, beyond mean field approach is needed

3) Mean Field give jump in second order of the derivative

4) We need to include fluctuation in High Tc superconductor

5) Gaussian approximation is not enough in High Tc superconductor

Effects of an applied magnetic field; Type I and Type II superconductivity

Derivation of the Ginzburg-Landau equations

Our starting point will be the Ginzburg-Landau (GL) free energy in the presence of an external magnetic field,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F=\int d^d \vec{r} \left [\alpha(T-T_c)|\Psi(\vec{r})|^2+\tfrac{1}{2}b|\Psi(\vec{r})|^4+\frac{\hbar^2}{2m}\left |\left (\nabla-i\frac{2e}{\hbar c}\vec{A}(\vec{r})\right )\Psi(\vec{r})\right |^2+\frac{1}{8\pi}(\nabla\times\vec{A}(\vec{r}))^2-\frac{1}{c}\vec{J}_{\text{ext}}(\vec{r})\cdot\vec{A}(\vec{r})\right ],}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A}} is the total vector potential and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{J}_{\text{ext}}} is an external current density, assumed to be controlled experimentally. This current satisfies

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla\times\vec{H}=\frac{4\pi}{c}\vec{J}_{\text{ext}},}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{H}} is the external magnetic field. The expression is the sum of the energy due to the superconducting order parameter, with the magnetic field introduced via the gauge invariance argument given above, the energy of the magnetic field alone, and the work done by the superconductor to maintain the external current at a constant value.

Let us first derive the "saddle point" equations satisfied by the magnetic field in the normal state. In this case, we set Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi} to zero everywhere and set

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left. \frac{\delta F}{\delta\vec{A}(\vec{r})}\right |_{\vec{A}=\vec{A}_\text{min}}=0.}

We will find this derivative by first finding the variation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta F} in the free energy for this case, which is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta F=\int d^d \vec{r}' \left [\frac{1}{4\pi}(\nabla\times\vec{A}(\vec{r}'))\cdot(\nabla\times\delta\vec{A}(\vec{r}'))-\frac{1}{c}\vec{J}_{\text{ext}}(\vec{r}')\cdot\delta\vec{A}(\vec{r}')\right ],}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta\vec{A}} is a small variation in the vector potential; we assume that it vanishes on the "surface" of our system. We now transform the first term using the identity,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\nabla\times\vec{A})\cdot(\nabla\times\vec{B})=\nabla\cdot[\vec{A}\times(\nabla\times\vec{B})]+\vec{A}\cdot[\nabla\times(\nabla\times\vec{B})],}

obtaining

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta F=\int d^d \vec{r}' \left [\frac{1}{4\pi}\nabla\cdot[\delta\vec{A}(\vec{r}')\times(\nabla\times\vec{A}(\vec{r}'))]+\frac{1}{4\pi}\delta\vec{A}(\vec{r}')\cdot[\nabla\times(\nabla\times\vec{A}(\vec{r}'))]-\frac{1}{c}\vec{J}_{\text{ext}}(\vec{r}')\cdot\delta\vec{A}(\vec{r}')\right ].}

The first term is a "surface" term; since we assumed that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta\vec{A}} vanishes everywhere on the "surface", we are left with just

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta F=\int d^d \vec{r} \left [\frac{1}{4\pi}[\nabla\times(\nabla\times\vec{A}(\vec{r}'))]-\frac{1}{c}\vec{J}_{\text{ext}}(\vec{r}')\right ]\cdot\delta\vec{A}(\vec{r}').}

We conclude that the variational derivative that we are interested in is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\delta F}{\delta\vec{A}(\vec{r})}=\frac{1}{4\pi}[\nabla\times(\nabla\times\vec{A}(\vec{r}'))]-\frac{1}{c}\vec{J}_{\text{ext}}(\vec{r}').}

At the "saddle point", this derivative is zero, so we obtain the equation,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla\times(\nabla\times\vec{A})=\frac{4\pi}{c}\vec{J}_{\text{ext}}.}

We may introduce the total magnetic field Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}=\nabla\times\vec{A}} , thus obtaining

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla\times\vec{B}=\frac{4\pi}{c}\vec{J}_{\text{ext}}.}

Comparing this to the definition of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{J}_{\text{ext}}} given above, we conclude that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}=\vec{H}} in the normal state. In reality, this will only be approximately true due to para- or diamagnetic effects in the metal, but these effects will be small in comparison to those due to superconductivity, which we will now derive.

First, we will apply the "saddle point" condition for the superconducting order parameter, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi} , which is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left. \frac{\delta F}{\delta\Psi^{*}(\vec{r})}\right |_{\Psi=\Psi_{\text{min}}}=0.}

Again, we start by finding the variation in the free energy in terms of a small variation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta\Psi^{*}} in the order parameter:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta F=\int d^d \vec{r}' \left [\alpha(T-T_c)\Psi(\vec{r}')\,\delta\Psi^{*}(\vec{r}')+b|\Psi(\vec{r}')|^2\Psi(\vec{r}')\,\delta\Psi^{*}(\vec{r}')-\frac{e}{mc}\vec{A}(\vec{r}')\cdot\left (\frac{\hbar}{i}\nabla-\frac{2e}{c}\vec{A}(\vec{r}')\right )\Psi(\vec{r}')\,\delta\Psi^{*}(\vec{r}')-\frac{1}{2m}\frac{\hbar}{i}\nabla\delta\Psi^{*}(\vec{r}')\cdot\left (\frac{\hbar}{i}\nabla-\frac{2e}{c}\vec{A}(\vec{r}')\right )\Psi(\vec{r}')\right ]}

The last term is equal to

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{1}{2m}\left\{\frac{\hbar}{i}\nabla\cdot\left [\left (\frac{\hbar}{i}\nabla-\frac{2e}{c}\vec{A}(\vec{r}')\right )\Psi(\vec{r}')\,\delta\Psi^{*}(\vec{r}')\right ]-\left [\frac{\hbar}{i}\nabla\cdot\left (\frac{\hbar}{i}\nabla-\frac{2e}{c}\vec{A}(\vec{r}')\right )\Psi(\vec{r} ')\right ]\delta\Psi^{*}(\vec{r}')\right\}.}

The second term in this expression is a "surface" term. If we assume that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta\Psi^{*}} is zero on the "surface", then this term vanishes, leaving us with

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta F=\int d^d \vec{r}' \left\{\alpha(T-T_c)\Psi(\vec{r}')+b|\Psi(\vec{r}')|^2\Psi(\vec{r}')-\frac{e}{mc}\vec{A}(\vec{r}')\cdot\left (\frac{\hbar}{i}\nabla-\frac{2e}{c}\vec{A}(\vec{r}')\right )\Psi(\vec{r}')+\frac{1}{2m}\left [\frac{\hbar}{i}\nabla\cdot\left (\frac{\hbar}{i}\nabla-\frac{2e}{c}\vec{A}(\vec{r}')\right )\Psi(\vec{r}')\right ]\right\}\delta\Psi^{*}(\vec{r}').}

We can now immediately write down the variational derivative, which, upon being set to zero, gives us the first GL equation, 

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2m}\left (\frac{\hbar}{i}\nabla-\frac{2e}{c}\vec{A}\right )^2\Psi+\alpha(T-T_c)\Psi+b|\Psi|^2\Psi=0.}

We also need to minimize the free energy with respect to the magnetic field. We have already done this for the normal case, and there is only one more term that we need to consider in the superconducting case; we will therefore only treat this term. We can quickly write down the variation in the superconducting part of the free energy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_{SC}} , which is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta F_{SC}=i\frac{e\hbar}{mc}\int d^d \vec{r}' \left [\Psi^{*}(\vec{r}')\left (\nabla-i\frac{2e}{\hbar c}\vec{A}(\vec{r}')\right )\Psi(\vec{r}')-\Psi(\vec{r}')\left (\nabla+i\frac{2e}{\hbar c}\vec{A}(\vec{r}')\right )\Psi^{*}(\vec{r}')\right ]\cdot\delta\vec{A}(\vec{r}').}

Combining this result with the previous result for the normal metal, we obtain the second GL equation,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{4\pi}\nabla\times(\nabla\times\vec{A})-\frac{1}{c}\vec{J}_{\text{ext}}-\frac{e}{mc}\left (\Psi^{*}\frac{\hbar}{i}\nabla\Psi-\Psi\frac{\hbar}{i}\nabla\Psi^{*}\right )+\frac{4e^2}{mc^2}|\Psi|^2\vec{A}=0,}

or, introducing Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{H}} , 

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{e}{m}\left (\Psi^{*}\frac{\hbar}{i}\nabla\Psi-\Psi\frac{\hbar}{i}\nabla\Psi^{*}\right )-\frac{4e^2}{mc}|\Psi|^2\vec{A}=\frac{c}{4\pi}\nabla\times(\vec{B}-\vec{H}).}

Given the definition of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{H}} and the Maxwell equation (assuming static fields),

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla\times\vec{B}=\frac{4\pi}{c}\vec{J},}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{J}} is the total current density, we conclude that the left-hand side of this equation is the current density induced inside the superconductor.

Let us now suppose that we do not assume that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta\Psi^{*}} vanishes on the surface. It may then be shown that the following boundary condition holds on the surface (see P. G. de Gennes, Superconductivity in Metals and Alloys):

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{n}\cdot\left (\frac{\hbar}{i}\nabla-\frac{2e}{c}\vec{A}\right )\Psi=\frac{i\hbar}{b_{dG}}\Psi.}

This relation holds for a superconductor-metal interface; for a superconductor-insulator interface, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b_{dG}\rightarrow\infty} . We may show that this condition implies that the normal component of the current density on the surface vanishes. If we multiply the above condition by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi^{*}} on both sides, we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{n}\cdot\Psi^{*}\left (\frac{\hbar}{i}\nabla-\frac{2e}{c}\vec{A}\right )\Psi=\frac{i\hbar}{b_{dG}}\Psi^{*}\Psi.}

Taking the complex conjugate of both sides gives us

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{n}\cdot\Psi\left (-\frac{\hbar}{i}\nabla-\frac{2e}{c}\vec{A}\right )\Psi^{*}=-\frac{i\hbar}{b_{dG}}\Psi^{*}\Psi.}

Adding these two equations together gives us

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{n}\cdot\left [\left (\Psi^{*}\frac{\hbar}{i}\nabla\Psi-\Psi\frac{\hbar}{i}\nabla\Psi^{*}\right )-\frac{4e}{c}|\Psi|^2\vec{A}\right ]=0.}

The left-hand side is proportional to the normal component of the current density inside the superconductor.

The GL Equations in Dimensionless Form

We will find it convenient to introduce dimensionless variables when working with the GL equations. We start by introducing a dimensionless order parameter, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi=\frac{\Psi}{\Psi_0}} , where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi_0^2=\frac{\alpha(T_c-T)}{b}.}

We may rewrite the first GL equation in terms of this parameter as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2mb\Psi_0^2}\left (\frac{\hbar}{i}\nabla-\frac{2e}{c}\vec{A}\right )^2\psi-(\left |\psi\right |^2-1)\psi=0,}

and the second as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{e\Psi_0^2}{mc}\left (\psi^{\ast}\frac{\hbar}{i}\nabla\psi-\psi\frac{\hbar}{i}\nabla\psi^{\ast}\right )-\frac{4e^2\Psi_0^2}{mc^2}\left |\psi\right |^2\vec{A}=\frac{1}{4\pi}\nabla\times[\nabla\times(\vec{A}-\vec{A}_0)],}

where we re-introduced Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A}} into the right-hand side and also introduced Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A}_0} , defined as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{H}=\nabla\times\vec{A}_0.}

Next, we introduce a dimensionless position vector,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{r}=\frac{1}{\lambda}\vec{r},}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda=\sqrt{\frac{mc^2}{16\pi e^2\Psi_0^2}}} is known as the penetration depth of the superconductor; we will see where this name comes from shortly. In terms of this vector, the first GL equation becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left (\frac{1}{\Psi_0\lambda\sqrt{2mb}}\frac{\hbar}{i}\tilde{\nabla}-\frac{1}{\Psi_0\sqrt{2mb}}\frac{2e}{c}\vec{A}\right )^2\psi+(\left |\psi\right |^2-1)\psi=0}

and the second becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{4\pi e\lambda\Psi_0^2}{mc}\left (\psi^{\ast}\frac{\hbar}{i}\tilde{\nabla}\psi-\psi\frac{\hbar}{i}\tilde{\nabla}\psi^{\ast}\right )-\left |\psi\right |^2\vec{A}=\tilde{\nabla}\times[\tilde{\nabla}\times(\vec{A}-\vec{A}_0)].}

Finally, we introduce a dimensionless vector potential,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{A}=\frac{1}{\Psi_0\sqrt{2mb}}\frac{2e}{c}\vec{A}}

and the dimensionless parameter,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa=\frac{\Psi_0\lambda\sqrt{2mb}}{\hbar}.}

In terms of these, the first GL equation becomes 

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left (-\frac{i}{\kappa}\tilde{\nabla}-\tilde{A}\right )^2\psi+(\left |\psi\right |^2-1)\psi=0}

and the second becomes 

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2\kappa}\left (\psi^{\ast}\frac{\tilde{\nabla}}{i}\psi-\psi\frac{\tilde{\nabla}}{i}\psi^{\ast}\right )-\left |\psi\right |^2\vec{A}=\tilde{\nabla}\times[\tilde{\nabla}\times(\tilde{A}-\tilde{A}_0)].}

We see that our theory has a dimensionless parameter in it, namely Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa} , which is known as the Ginzburg-Landau parameter. We may write this parameter as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa=\frac{\lambda\sqrt{2m\alpha(T_c-T)}}{\hbar}=\frac{\lambda}{\xi_\text{GL}},}

where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \xi_\text{GL}=\frac{\hbar}{2m\alpha(T_c-T)}}

is the GL coherence length. This tells us that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa} is the ratio of two length scales associated with the superconductor, namely the scale over which the order parameter "heals" (the coherence length Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \xi_\text{GL}} ) and that over which the magnetic field dies out (the penetration depth Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda} , as we will demonstrate shortly). It also turns out that this parameter decides what type of superconductor we are dealing with. If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa < \tfrac{1}{\sqrt{2}}} , then we have a Type I superconductor, while, if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa > \tfrac{1}{\sqrt{2}}} , then we have a Type II superconductor.

We may now find the value of this parameter in the microscopic model we considered earlier. In that case, we found that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\hbar^2}{2m}=N(\mu)\xi^2=N(\mu)\frac{7\zeta(3)}{16\pi^2 d}\ell_T^2,}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N(\mu)} is the density of states at the Fermi level, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \xi} is the coherence length, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d} is the number of dimensions that we are working in, and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l_T} is the thermal wavelength. We will state the result for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d=3} . Given that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ell_T=\frac{\hbar v_F}{k_B T}}

and that, in this case,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N(\mu)=\frac{1}{2\pi^2}\frac{mk_F}{\hbar^2},}

we find that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa=\sqrt{\frac{18\pi^3}{7\zeta(3)}}\frac{k_B T_c}{\sqrt{mc^2}\sqrt{e^2 k_F}}\left (\frac{c}{v_F}\right )^2.}

Note that we set Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T=T_c} in the expression for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l_T} ; this is because the GL theory is only valid just below the transition temperature. We may also express this in terms of the Fermi energy,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_F=\frac{p_F^2}{2m}=\tfrac{1}{2}p_F v_F.}

Doing so, we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa=\sqrt{\frac{18\pi^3}{7\zeta(3)}}\frac{k_B T_c}{\sqrt{mc^2}\sqrt{\alpha}\sqrt{2E_F}}\left (\frac{c}{v_F}\right )^{3/2}=8\times 10^{-6}\cdot\frac{T_c\,[\text{K}]}{\sqrt{E_F\,[\text{eV}]}}\left (\frac{c}{v_F}\right )^{3/2}.}

In a typical metal, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v_F\approx 10^5 - 10^6\,\tfrac{\text{m}}{\text{s}},} so

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa=(0.04 - 1.3)\cdot\frac{T_c\,[\text{K}]}{\sqrt{E_F\,[\text{eV}]}}.}

A Simple Example - The Strongly Type-I Superconductor With a Planar Surface

As a simple demonstration of the solution of the GL equations, let us consider a strongly Type I (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa\ll 1} ) superconductor with a planar boundary between it and an insulator. Let us set up our coordinate system so that the boundary is at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=0} .

SC I Boundary.PNG

We apply a magnetic field along the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z} axis,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{H}=H\hat{z}.}

We expect by symmetry that the total magnetic field Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}=B(x)\hat{z}} . We will choose our gauge such that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A}=A_y(x)\hat{y}.}

We also take the order parameter to depend only on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} . The first GL equation becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{1}{\kappa^2}\frac{d^2\psi}{d\tilde{x}^2}+A_y^2\psi-\psi+\psi^3=0.}

Since we are taking Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa} to be small, the derivative term dominates, and we may therefore approximate this equation as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d^2\psi}{d\tilde{x}^2}=0,}

so that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(\tilde{x})=c_1+c_2\tilde{x}} . Our boundary condition states that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left. \frac{1}{i\kappa}\frac{d\psi}{d\tilde{x}}\right |_{\tilde{x}=0}=0,}

so that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_2=0} . Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi=\Psi_0} in the bulk, we conclude that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(\tilde{x})=1} for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{x}<0} . Similarly, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi=0} deep into the insulating region, so that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(\tilde{x})=0} for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{x}>0} .

Now we consider the second equation. In this case, it becomes, for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{x}<0} ,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\tilde{A}=\tilde{\nabla}\times(\tilde{B}-\tilde{H})=\tilde{\nabla}\times\tilde{B},}

or

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\tilde{B}=\tilde{\nabla}\times(\tilde{\nabla}\times\tilde{B}).}

The right-hand side is just

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{\nabla}\times(\tilde{\nabla}\times\tilde{B})=-\frac{d^2\tilde{B}}{d\tilde{x}^2},}

so that the equation is now

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d^2\tilde{B}}{d\tilde{x}^2}=\tilde{B}.}

The solution to the equation in simply Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{B}(\tilde{x})=\tilde{B}_{+}e^{\tilde{x}}+\tilde{B}_{-}e^{-\tilde{x}}} , or, in terms of dimensional quantities,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B(x)=B_{+}e^{x/\lambda}+B_{-}e^{-x/\lambda}.}

Since our superconductor is in the region Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x<0} , we must take Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_{-}=0} . Furthermore, the field must equal the applied field at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=0} , so

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B(x)=He^{x/\lambda}.}

For Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x>0} , the second GL equation becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d^2\tilde{B}}{d\tilde{x}^2}=0.}

The solution, in terms of dimensional quantities, is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B(x)=B_0+B_1 x} . We must set Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_1=0} so that the field does not increase indefinitely as we move away from the superconductor. Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}=\vec{H}} in the normal state, we conclude that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B(x)=H} for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x>0} . The following is a plot of the full dependence of the magnetic field on position.

SC BField PlanarInt.PNG

We have now shown why we called Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda} the penetration depth; it sets the length scale over which the magnetic field tends to zero inside the superconductor. We have also illustrated the expulsion of applied magnetic fields from the bulk of a Type I superconductor; this is known as the Meissner effect.

Thermodynamics of Type-I Superconductors in Magnetic Fields

In a bulk superconductor, surface effects are unimportant; for now, we will assume that the order parameter Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi} is constant everywhere in the superconductor and that magnetic fields are completely expelled. In this case, the free energy per unit volume of the superconductor is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_s=\alpha(T-T_c)\Psi_0^2+\tfrac{1}{2}b\Psi_0^4=-\frac{\alpha^2}{2b}(T-T_c)^2.}

This is known as the condensation energy (per unit volume). We see that we can "save" energy by going into the superconducting state.

In the normal state, only the magnetic field terms are present, so that the free energy is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_n=\int d^d\vec{r}\,\left [\frac{1}{8\pi}(\nabla\times\vec{A})^2-\frac{1}{c}\vec{J}_\text{ext}\cdot\vec{A}\right ].}

We may substitute in

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{4\pi}{c}\vec{J}_\text{ext}=\nabla\times\vec{H}}

to get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_n=\int d^d\vec{r}\,\left [\frac{1}{8\pi}(\nabla\times\vec{A})^2-\frac{1}{4\pi}(\nabla\times\vec{H})\cdot\vec{A}\right ]=\int d^d\vec{r}\,\left [\frac{1}{8\pi}(\nabla\times\vec{A})^2-\frac{1}{4\pi}(\nabla\times\vec{A})\cdot\vec{H}\right ] =\int d^d\vec{r}\,\left [\frac{1}{8\pi}B^2-\frac{1}{4\pi}\vec{B}\cdot\vec{H}\right ].}

In the normal state, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}=\vec{H}} , so

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_n=-\int d^d\vec{r}\,\frac{1}{8\pi}H^2.}

The free energy per unit volume of the normal state is therefore

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_n=-\frac{1}{8\pi}H^2.}

We see that, overall, we also "save" energy in the normal state. Which state we go into depends on which "saves" more energy. We may now define a field at which the "savings" are the same for both states; this is the (thermodynamic) critical field Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_c} (sometimes also denoted Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_{cm}} ). Equating the free energies per unit volume of each state, we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\alpha^2}{2b}(T-T_c)^2=\frac{1}{8\pi}H_c^2,}

or, solving for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_c} ,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_c=\sqrt{\frac{4\pi}{b}}\alpha\left |T-T_c\right |.}

We see, therefore, that GL theory predicts a linear dependence of the critical field on the temperature. This is not what is observed experimentally, however. The dependence of the critical field on temperature in many real superconductors can, in fact, be modeled with the following empirical law:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_c(T)=H_c(0)\left [1-\left (\frac{T}{T_c}\right )^2\right ].}

We plot this relation below.

CriticalField vs T.PNG

We see that, near Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_c} , the dependence of the critical field on temperature does indeed follow the linear relation that we just derived. However, it deviates from said relation when we go far below Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_c} . This is not surprising; the GL theory from which we obtained the linear relation is only valid near Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_c} . In order to obtain a more accurate relation, we require a theory for the superconductor that is valid far below Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_c} .

We may rewrite our dimensionless vector potential and magnetic field in terms of the critical field. We may write our expression for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_c} as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_c^2=4\pi b\left |\Psi_0\right |^4.}

Our dimensionless vector potential is then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{A}=\frac{2e}{c}\frac{\Psi_0}{\sqrt{2mb\Psi_0^4}}\vec{A}=\frac{2e}{c}\Psi_0\sqrt{\frac{4\pi}{2mH_c^2}}\vec{A}=\sqrt{\frac{8\pi e^2\Psi_0^2}{mc^2}}\vec{A}=\frac{1}{\sqrt{2}H_c \lambda}\vec{A}.}

We may rewrite the definition of the magnetic field as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}=\tilde{\nabla}\times\sqrt{2}H_c\tilde{A},}

or

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{\nabla}\times\tilde{A}=\frac{1}{\sqrt{2}H_c}\vec{B}=\tilde{B}.}

Magnetic Properties of a Type-I Superconductor

In the approximation stated above, we find that, in the superconducting state (the applied magnetic field is below Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_c} ), the total magnetic field Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}} is completely expelled, while Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}=\vec{H}} when the applied field is above Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_c} . From the defining relation for the magnetization Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{M}} ,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}=\vec{H}+4\pi\vec{M},}

we find that the magnitude of the magnetization increases linearly with, and points in the opposite direction to, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{H}} , in the superconducting state, but is zero in the normal state. We plot these two relations below.

SC BvsHandMvsH.PNG

We will now demonstrate two consequences of the total expulsion of a magnetic field from a superconductor.

1) The total field is always tangential to the surface of a superconductor.

First, recall the Maxwell equation,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla\cdot\vec{B}=0.}

Let us now consider the boundary between a superconductor and a normal region:

SC BFieldCond PillBox.PNG

We will find the flux of a magnetic field through the "pill box" shown above. Let the area of the circular surfaces be Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta A} , and let us assume that the contribution from the "tube" part of the surface is negligible (we will assume that its height is small compared to the radius of the circular surfaces). Then the total flux may be written as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_{n,\bot}\cdot\Delta A-B_{sc,\bot}\cdot\Delta A,}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_{n,\bot}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_{sc,\bot}} are the components of the magnetic field normal to the circular surfaces of the "pill box" in the normal and superconducting regions, respectively. The Maxwell equation stated above is equivalent to the statement that the total flux through a closed surface, such as the "pill box" we consider here, must be zero. Therefore,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_{n,\bot}=B_{sc,\bot}.}

However, we know that the superconductor completely expels magnetic fields, so that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}_{sc}=0} . Therefore, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_{n,\bot}=0} , thus proving that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}_n} can only have tangential components.

2) As a direct consequence of (1), a type-I superconductor in an external magnetic field always carries an electrical current near its surface.

To show this, first recall the Maxwell equation (Ampere's Law in the case of static fields),

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla\times\vec{B}=\frac{4\pi}{c}\vec{J},}

or, in integral form,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \oint_{\partial S}\vec{B}\cdot d\vec{\ell}=\frac{4\pi}{c}I,}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S} is a surface with bounding curve Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \partial S} , the line integral on the left is taken in the direction that would cause a right-handed screw to advance in the direction of the normal to the surface, and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I} is the total current passing through the surface.

Let us now, once again, consider the boundary between a superconductor and a normal region and a rectangular contour drawn around the boundary:

SC BField Contour.PNG

In this case, because the magnetic field is zero in the superconducting region and because the magnetic field in the normal region is tangential to the surface of the superconductor, then, assuming that the lengths of segments 14 and 23 are small compared to that of segments 12 and 34. In this case, the left-hand side of the Maxwell equation becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \oint_{\partial S}\vec{B}\cdot d\vec{\ell}=B\ell_{12},}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ell_{12}} is the length of segment 12. The right-hand side, on the other hand, is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{4\pi}{c}I=\frac{4\pi}{c}K\ell_{12},}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K} is the surface current density (per unit length). This implies that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B=\frac{4\pi}{c}K,}

or, in vector form,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{K}=\frac{c}{4\pi}\hat{n}\times\vec{B},}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{n}} is the normal to the surface pointing into the normal region. In the case depicted above, we see that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{K}} points into the page or screen.

An Example: The Magnetic Field Around a Spherical Superconductor

These observations lead to interesting conclusions for superconductors with geometries more complex than a cylinder inside a magnetic field parallel to its axis. Consider, for example, a spherical superconductor, depicted below:

MagField SCSphere.png

As we can see, the magnetic field near the equator is stronger than the applied field, while the field at the poles vanishes. This implies that the magnetic field near the equator may exceed Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_c} even if the applied field is less than Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_c} , simply due to the geometry of the sample. In fact, for a certain range of fields,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (1-\eta)H_c<H<H_c,}

the sample will enter an intermediate state, in which superconducting and normal regions coexist. The above inequality holds for any geometry, in fact, and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \eta} is known as the demagnetizing factor of the sample. Its value will depend on the exact geometry of the sample; we will now show that, for a sphere, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \eta=\tfrac{1}{3}} . We will do so by solving Maxwell's equations for a spherical superconductor of radius Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R} in a uniform applied magnetic field Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{H}=H\hat{z}} . The boundary conditions for this problem are

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}(\vec{r})=H\hat{z}}

for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\vec{r}|\rightarrow\infty} and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{n}\cdot\vec{B}(\vec{r})=0}

on the surface of the sphere. Outside the superconductor, the equations satisfied by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}} are

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla\cdot\vec{B}=0}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla\times\vec{B}=0.}

The second equation implies that there are no current sources outside the superconductor. It also implies that we may write the magnetic field in terms of a scalar potential. We therefore write

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}=\nabla\Phi+H\vec{z}.}

Substituting this into the first equation, we get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla^2\Phi=0,}

which is just Laplace's equation. Our boundary conditions for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi} are, in spherical coordinates,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi(\vec{r}\rightarrow\infty)=0}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial\Phi}{\partial r}+H\cos{\theta}=0}

on the surface. Laplace's equation in spherical coordinates is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{r^2}\frac{\partial}{\partial r}\left (r^2\frac{\partial\Phi}{\partial r}\right )+\frac{1}{r^2\sin{\theta}}\frac{\partial}{\partial\theta}\left (\sin{\theta}\frac{\partial\Phi}{\partial\theta}\right )+\frac{1}{r^2\sin^2{\theta}}\frac{\partial^2\Phi}{\partial\phi^2}=0.}

Because our system has azimuthal symmetry, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi} should be independent of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi} . Therefore, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi(\vec{r})=\Phi(r,\theta)} and Laplace's equation becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{r^2}\frac{\partial}{\partial r}\left (r^2\frac{\partial\Phi}{\partial r}\right )+\frac{1}{r^2\sin{\theta}}\frac{\partial}{\partial\theta}\left (\sin{\theta}\frac{\partial\Phi}{\partial\theta}\right )=0.}

Multiplying by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r^2} , we get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial r}\left (r^2\frac{\partial\Phi}{\partial r}\right )+\frac{1}{\sin{\theta}}\frac{\partial}{\partial\theta}\left (\sin{\theta}\frac{\partial\Phi}{\partial\theta}\right )=0.}

We will now attempt to solve this equation by separation of variables. Let us try a solution of the form

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi(r,\theta)=R(r)\chi(\theta).}

Then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{R}\frac{d}{dr}\left (r^2\frac{dR}{dr}\right )=-\frac{1}{\chi}\frac{1}{\sin{\theta}}\frac{d}{d\theta}\left (\sin{\theta}\frac{d\chi}{d\theta}\right ).}

We see that we have an expression depending only on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r} on one side and one depending only on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta} on the other. The only way for this equation to be satisfied is if both equal a constant, which we will call Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\lambda} . Let us first consider the equation for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \chi} . This equation is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\sin{\theta}}\frac{d}{d\theta}\left (\sin{\theta}\frac{d\chi}{d\theta}\right )=\lambda\chi.}

If we make the substitution, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=\cos{\theta}} , into this equation, we get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx}\left [(1-x^2)\frac{d\chi}{dx}\right ]=\lambda\chi.}

We recognize this as the Legendre differential equation. The only physically interesting solutions to this equation occur when Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda=-l(l+1)} , in which case we find that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \chi} is a Legendre polynomial,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \chi(\theta)=P_l(\cos{\theta}).}

The first few Legendre polynomials are Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_0(x)=1} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_1(x)=x} , and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_2(x)=\tfrac{1}{2}(3x^2-1)} .

Now we will consider the equation for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R} . If we take Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \chi(\theta)=P_l(\cos{\theta})} , then our differential equation for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R} becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dr}\left (r^2\frac{dR}{dr}\right )=l(l+1)R,}

or

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d^2 R}{dr^2}+2r\frac{dR}{dr}-l(l+1)R=0.}

This is an Euler-Cauchy differential equation. Let us assume a power law dependence for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R} ,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R(r)=r^\alpha.}

The equation becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha(\alpha-1)r^\alpha+2\alpha r^\alpha-l(l+1)r^\alpha=0.}

The exponent Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha} must therefore satisfy

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha(\alpha+1)=l(l+1).}

This equation has two possible solutions, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha=l} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha=-(l+1)} , so the general solution for the differential equation is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R(r)=A_l r^l+\frac{B_l}{r^{l+1}}.}

The general solution to the original partial differential equation is then a linear combination of all possible products Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R(r)\chi(\theta)} ,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi(r,\theta)=\sum_{l=0}^{\infty}\left (A_l r^l+\frac{B_l}{r^{l+1}}\right )P_l(\cos{\theta}).}

Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r^l\rightarrow\infty} for all Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l>0} , we must set Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_l=0} for all such values of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l} . We will also set Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_0=0} , so that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi(r,\theta)=\sum_{l=0}^{\infty}\frac{B_l}{r^{l+1}}P_l(\cos{\theta}).}

We now apply the boundary condition,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial\Phi}{\partial r}+H\cos{\theta}=0,}

to obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{l=0}^{\infty}\frac{(l+1)B_l}{R^{l+2}}P_l(\cos{\theta})=H\cos{\theta}.}

By inspection, we find that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_l=0} for all Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l\neq 1} . For Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l=1} , we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_l=\tfrac{1}{2}HR^3.}

Therefore, the solution for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi} is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi(r,\theta)=\frac{HR^3}{2r^2}\cos{\theta},}

and thus Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}} is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}=\tfrac{1}{2}HR^3\nabla\left (\frac{\cos{\theta}}{r^2}\right )+H\hat{z}.}

In spherical coordinates, the gradient operator is given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla\Phi=\frac{\partial\Phi}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial\Phi}{\partial\theta}\hat{\theta}+\frac{1}{r\sin{\theta}}\frac{\partial\Phi}{\partial\phi}\hat{\phi},}

so

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}=-H\left (\frac{R}{r}\right )^3(\cos{\theta}\hat{r}+\tfrac{1}{2}\sin{\theta}\hat{\theta})+H\hat{z}.}

We recognize that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{z}=\cos{\theta}\hat{r}-\sin{\theta}\hat{\theta}} , so that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}=\left [1-\left (\frac{R}{r}\right )^3\right ]H\cos{\theta}\hat{r}-\left [1+\tfrac{1}{2}\left (\frac{R}{r}\right )^3\right ]H\sin{\theta}\hat{\theta}.}

We now want to find the point at which the magnetic field has the largest magnitude. The magnitude of this vector is given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\vec{B}|^2=H^2\left [1+\tfrac{1}{2}\left (\frac{R}{r}\right )^3\right ]^2+3\left (\frac{R}{r}\right )^3\left [\tfrac{1}{4}\left (\frac{R}{r}\right )^3-1\right ]\cos^2{\theta}}

We see that the second term, which is proportional to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos^2{\theta}} , is clearly negative when Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r>R} . Therefore, we must make the size of this term as small as possible. This may be done by letting Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta=\tfrac{\pi}{2}} . For this value of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta} , we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\vec{B}|^2=H^2\left [1+\tfrac{1}{2}\left (\frac{R}{r}\right )^3\right ]^2.}

This function is monotonically decreasing as we increase Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r} ; therefore, we choose the smallest possible value of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r} , which is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r=R} . This gives us

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\vec{B}|^2=\tfrac{9}{4}H^2,}

or Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\vec{B}|=\tfrac{3}{2}H} . Therefore, the total magnetic field achieves its maximum magnitude at the equator of the sphere, where it is three halves the applied field. Therefore, if the applied field is larger than Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H=\tfrac{2}{3}H_c} , the field at the equator would become larger than Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_c} , which would destroy the sphere's superconductivity at that point. Therefore, the sphere enters an intermediate state when Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tfrac{2}{3}H_c<H<H_c} , so that the demagnetizing factor Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \eta=\tfrac{1}{3}} .

Superconductors of General Geometries

In general, the maximum magnetic field on the surface of a superconductor will be related to the applied field by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_{\text{max}}=\frac{H}{1-\eta}.}

We will now list the values of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \eta} for a few different geometries.

Geometry Demagnetizing factor Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \eta}
Cylinder with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{H}} parallel to its axis Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0}
Cylinder with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{H}} perpendicular to its axis Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tfrac{1}{2}}
Sphere Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tfrac{1}{3}}
Infinite thin plate with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{H}} perpendicular to the plate Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1}

The sample will enter the intermediate state at the value of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H=H^{\ast}} at which Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_{\text{max}}=H_c} , which is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H^{\ast}=(1-\eta)H_c.}

Therefore, the superconductor will be in the intermediate state when

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (1-\eta)H_c<H<H_c.}

We will now show plots of the magnetic field of a sphere at the equator and at the pole as a function of the applied field (for details on how to derive the dependence in the intermediate state, see P. G. DeGennes, Superconductivity of Metals and Alloys).

BField SCSphere.PNG

In the intermediate state, the energy per volume of both the normal and superconducting states is the same. Therefore, the energy per unit area of a "domain wall" will be the dominant contribution.

Surface Term of the Free Energy

The free energy in the reduced units can be written as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F=\int d^3 r \,\left[\alpha (T-T_c)|\Psi|^2 + \frac{b}{2} |\Psi|^4 + \frac{\hbar^2}{2m} |(\nabla - \frac{2ei}{\hbar c}\vec{A})\Psi|^2 + \frac{1}{8\pi}(\nabla\times\vec{A})^2 - \frac{1}{4\pi}\vec{H}\cdot\vec{B}\right]}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\lambda^D\int d^D\tilde{r}\,\left[\alpha (T-T_c)|\Psi_o|^2 |\psi|^2 + \frac{b}{2} |\Psi_o|^4 |\psi|^4 + \frac{\hbar^2}{2m_{GL}}|\Psi_o|^2|(\frac{1}{\lambda}\tilde{\nabla}-\frac{2ei}{\hbar c}\sqrt{2}\lambda H_c \tilde{A})\psi|^2 + \frac{1}{8\pi \lambda^2} (\sqrt{2}\lambda H_c)^2 (\tilde{\nabla}\times\tilde{A})^2 - \frac{2\lambda^2 H_c^2}{4\pi \lambda^2} \tilde{H}\cdot\tilde{B}\,\right]}

where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{r}=\frac{r}{\lambda}, \tilde{H}=\frac{H}{\sqrt{2}H_c}}

Using the following relations:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b|\Psi_o|^4 = \frac{H_c^2}{4\pi}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\alpha (T-T_c)}{b|\Psi_o|^2} = -1}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi_o|^2 \frac{4e^2}{mc^2} = \frac{1}{4\pi\lambda^2}}

one has

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F=\frac{\lambda^D H_c^2}{4\pi}\int d^D\tilde{r}\,\left[-|\psi|^2+\frac{1}{2}|\psi|^4+|(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})\psi|^2+\tilde{B}^2-2\tilde{H}\cdot\tilde{B}\,\right]}

Now let's consider the gradinet term:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})\psi|^2 = \,\left[ (\frac{-1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})\psi^*\,\right]\,\left[ (\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})\psi\,\right] = (-\tilde{A}\psi^*)(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})\psi - (\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}\psi^*)(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})\psi }

Integration by part leads the 2nd term to the form:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\tilde{\nabla}\cdot(\frac{1}{\kappa}\frac{1}{i}\psi^*(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})\psi) + \psi^*(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})\psi)}

where the 1st term needs to be evaluated on the surface where it vanishes by the boundary condition for an insulating interface. Hence the gradient term becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})\psi|^2 = \psi^*(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})^2\psi}

Assuming that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{A}} satisfy the GL equation, namely,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})^2 \psi = \psi - \psi |\psi|^2}

we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi^*(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})^2\psi = |\psi|^2-|\psi|^4}

and the free energy becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F=\frac{\lambda^D H_c^2}{4\pi}\int d^D \tilde{r} \,\left[-\frac{1}{2}|\psi|^4+\tilde{B}^2-2\tilde{H}\cdot\tilde{B}\,\right]}

This is true at the saddle point. Now the free energy for the normal state and the superconducting state are given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_n=\frac{\lambda^D H_c^2}{4\pi}\int d^D \tilde{r} \,\left[-\tilde{H}^2\,\right]}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_s=\frac{\lambda^D H_c^2}{4\pi}\int d^D \tilde{r} \,\left[-\frac{1}{2}|\psi|^4+\tilde{B}^2-2\tilde{B}\cdot\tilde{H}\,\right]}

Therefore, the surface free energy of the interface is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_s-F_n=\frac{\lambda^D H_c^2}{4\pi}\int d^D \tilde{r} \,\left[-\frac{1}{2}|\psi|^4+\tilde{B}^2-2\tilde{B}\cdot\tilde{H}+\tilde{H}^2\,\right] = \frac{\lambda^D H_c^2}{4\pi}\int d^D \tilde{r} \,\left[-\frac{1}{2}|\psi|^4+(\tilde{B}-\tilde{H})^2\,\right]}

As shown in the figures below, type-I superconductor (left) has positive wall energy while type-II superconductor (right) negative wall energy.

Pic4.jpg Pic3.jpg

Free Energy of a Normal-Superconducting Interface

Now let's consider the free energy of a normal-superconducting interface. Recall the GL equations:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})^2\psi - \psi + \psi|\psi|^2 = 0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2}(\psi^*\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}\psi + c.c.)-\tilde{A}|\psi|^4 =\tilde{\nabla}\times(\tilde{\nabla}\times\tilde{A})}

Assume that the interface is along y-z plane and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{B}\|\hat{z}} . We can choose Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{A}(\tilde{r})=\hat{y}\tilde{A}(\tilde{x})} which implies the order parameter depends only on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{x}} . Then the GL equation become

Pic5.jpg

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{1}{\kappa^2}\frac{d^2\psi}{d\tilde{x}^2}+\tilde{A}_y^2(\tilde{x})\psi -\psi + \psi^3 = 0} (*)

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{A}_y(\tilde{x})\psi^2 = \frac{d^2\tilde{A}_y(\tilde{x})}{d\tilde{x}^2}}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2}(\psi^*\frac{1}{\kappa}\frac{1}{i}\frac{d}{d\tilde{x}}\psi + c.c.) = 0}

Note that we can choose Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi} to be real since the coefficients are real. Now let's manipulate Eq.(*):

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{1}{\kappa^2}\frac{d\psi}{d\tilde{x}}\frac{d^2\psi}{d\tilde{x}^2}+\tilde{A}_y^2\frac{d\psi}{d\tilde{x}}\psi - \frac{d\psi}{d\tilde{x}}\psi + \frac{d\psi}{d\tilde{x}}\psi^3 = 0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{1}{2\kappa^2}\frac{d}{d\tilde{x}}(\frac{d\psi}{d\tilde{x}})^2+\frac{1}{2}\tilde{A}_y^2\frac{d}{d\tilde{x}}\psi^2 - \frac{1}{2}\frac{d}{d\tilde{x}}\psi^2 + \frac{1}{4}\frac{d}{\tilde{x}}\psi^4 = 0}

Integrating we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{1}{2\kappa^2}(\frac{d\psi}{d\tilde{x}})^2 + \frac{1}{2}\int d\tilde{x}\tilde{A}_y^2\frac{d}{d\tilde{x}}\psi^2 - \frac{1}{2}\psi^2 + \frac{1}{4}\psi^4 = const.}

This integral can be further manipulated:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int d\tilde{x}\tilde{A}_y^2\frac{d}{d\tilde{x}}\psi^2 = \int d\tilde{x}\,\left[\frac{d}{d\tilde{x}}(\tilde{A}_y^2\psi^2)-\frac{d\tilde{A}_y^2}{d\tilde{x}}\psi^2\,\right] = \tilde{A}_y^2\psi^2 - 2\int d\tilde{x}\tilde{A}_y\frac{d\tilde{A}_y}{d\tilde{x}}\psi^2 = \tilde{A}_y^2\psi^2 - 2\int d\tilde{x}\frac{d\tilde{A}_y}{d\tilde{x}}\frac{d^2\tilde{A}_y}{d\tilde{x}^2} = \tilde{A}_y^2\psi^2 - \int d\tilde{x}\frac{d}{d\tilde{x}}(\frac{d\tilde{A}_y}{d\tilde{x}})^2 = \tilde{A}_y^2\psi^2 - (\frac{d\tilde{A}_y}{d\tilde{x}})^2 = const.}

Thus, Eq.(*) becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{1}{2\kappa^2}(\frac{d\psi}{d\tilde{x}})^2+\frac{1}{2}\tilde{A}_y^2\psi^2-\frac{1}{2}(\frac{d\tilde{A}_y}{d\tilde{x}})^2 - \frac{1}{2}\psi^2 + \frac{1}{4}\psi^4 = const.}

or,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\kappa^2}(\frac{d\psi}{d\tilde{x}})^2+(\frac{d\tilde{A}_y}{d\tilde{x}})^2+\psi^2(1-\tilde{A}_y^2)-\frac{1}{2}\psi^4 = const.}

To determine the const., note that as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\rightarrow -\infty} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi \rightarrow 1} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{A}_y \rightarrow 0} , so

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0 + 0 + 1 - \frac{1}{2} = const.}

Therefore,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\kappa^2}(\frac{d\psi}{d\tilde{x}})^2+(\frac{d\tilde{A}_y}{d\tilde{x}})^2+\psi^2(1-\tilde{A}_y^2)-\frac{1}{2}\psi^4 = \frac{1}{2}}

Consider the case where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa << 1} , that is, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda << \xi} . The main contribution to the surface energy comes from the region where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{B}=0} but Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi < 1} .

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\kappa^2}(\frac{d\psi}{d\tilde{x}})^2+\psi^2-\frac{1}{2}\psi^4=\frac{1}{2}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\kappa^2}(\frac{d\psi}{d\tilde{x}})^2 = \frac{1}{2}-\psi^2+\frac{1}{2}\psi^4 = \frac{1}{2}(1-\psi^2)^2}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\kappa}\frac{d\psi}{d\tilde{x}}=-\frac{1}{\sqrt{2}}(1-\psi^2)}

Note that the minus sign is from the experimental setup. Further,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d\psi}{1-\psi^2}=-\frac{\kappa}{\sqrt{2}}d\tilde{x};\frac{1}{1-\psi^2}=\frac{1}{2}(\frac{1}{1-\psi}+\frac{1}{1+\psi})}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ln\frac{1+\psi}{1-\psi}=-\sqrt{2}\kappa \tilde{x}+const.}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1+\psi(\tilde{x})}{1-\psi(\tilde{x})}=const.\times e^{-\sqrt{2}\kappa \tilde{x}}}

Pic6.jpg

Requiring Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi=0} at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{x}=0} gives Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle const.=1} Now we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi = \frac{e^{-\sqrt{2}\kappa \tilde{x}}-1}{e^{-\sqrt{2}\kappa \tilde{x}}+1}=-tanh\frac{\kappa \tilde{x}}{\sqrt{2}}}

whose curve is shown in the figure. Moreover,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_s-F_n=\frac{H_c^2}{8\pi}L^2\lambda\int_{-\infty}^{0}d\tilde{x}(1-\psi^4)}

where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{-\infty}^{0}d\tilde{x}(1-\psi^4)=\int_{-\infty}^{0}d\tilde{x}(1+\psi^2)(1-\psi^2) =\int_{-\infty}^{0}d\tilde{x}(1+\psi^2)(-\frac{\sqrt{2}}{\kappa})\frac{d\psi}{d\tilde{x}} =-\frac{\sqrt{2}}{\kappa}\psi |_{-\infty}^{0}-\frac{\sqrt{2}}{\kappa}\int_{-\infty}^{0}d\tilde{x}\psi^2\frac{d\psi}{d\tilde{x}}}

Pic7.jpg

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =-\frac{\sqrt{2}}{\kappa}\psi |_{-\infty}^{0}-\frac{\sqrt{2}}{3\kappa}\int_{-\infty}^{0}d\tilde{x}\frac{d\psi^3}{d\tilde{x}} = -\frac{\sqrt{2}}{\kappa}\psi |_{-\infty}^{0}-\frac{\sqrt{2}}{3\kappa}\psi^3 |_{-\infty}^{0} = \frac{\sqrt{2}}{\kappa}+\frac{\sqrt{2}}{3\kappa} = \frac{4\sqrt{2}}{3\kappa}}

Hence one has

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_s-F_n=\frac{H_c^2}{8\pi}L^2\lambda\cdot\frac{4\sqrt{2}}{3\kappa}}

Below we show that given Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{A}_y} satisfying the GL equation, the surface energy of the surface vanishes at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa=1/2} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\kappa^2}(\frac{d\psi}{d\tilde{x}})^2+(\frac{d\tilde{A}_y}{d\tilde{x}})^2+\psi^2(1-\tilde{A}_y^2)-\frac{1}{2}\psi^4 = \frac{1}{2}} (**)

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_s-F_n=\frac{H_c^2}{4\pi}L^2\lambda\int d\tilde{x}\,\left[(\tilde{B}-\tilde{H})^2-\frac{1}{2}\psi^4\,\right]}

Assume Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa=1/\sqrt{2}} .

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_s-F_n=\frac{H_c^2}{4\pi}L^2\lambda\int d\tilde{x}\,\left[(\frac{d\tilde{A}_y}{d\tilde{x}}-\frac{1}{\sqrt{2}})^2-\frac{1}{2}\psi^4\,\right]}

Choosing

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d\tilde{A}_y}{d\tilde{x}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\psi^2}

then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_s-F_n=0} and the 2nd GL equation becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{A}_y\psi^2=\frac{d^2\tilde{A}_y}{d\tilde{x}^2}=\frac{d}{d\tilde{x}}(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\psi^2)=-\sqrt{2}\psi\frac{d\psi}{d\tilde{x}}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d\psi}{d\tilde{x}}=-\frac{1}{\sqrt{2}}\tilde{A}_y\psi}

Insert this into Eq.(**):

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\kappa^2}\frac{1}{2}\tilde{A}_y^2\psi^2+(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\psi^2)^2+\psi^2(1-\tilde{A}_y^2)-\frac{1}{2}\psi^4-\frac{1}{2}=0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2\kappa^2}\tilde{A}_y^2\psi^2+\frac{1}{2}-\psi^2+\frac{1}{2}\psi^4+\psi^2-\tilde{A}_y^2\psi^2-\frac{1}{2}\psi^4-\frac{1}{2}=0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\frac{1}{2\kappa^2}-1)\tilde{A}_y^2\psi^2=0}

Therefore, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa=1/\sqrt{2}} as expected.

In summary, for type-I superconductor (positive surface energy) Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa < 1/\sqrt{2}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi} is uniform in the bulk and jumps at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_c} . On the other hand, for type-II superconductor (negative surface energy) Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa > 1/\sqrt{2}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi} is not uniform in the bulk but grows continuously when crossing Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_c} from above.

The "Landau-Level"-Like States above Hc and the Upper Critical Field Hc2

Now we have the surface free energy:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_s-F_n=\frac{H_c^2}{4\pi}\lambda^3\int d^3\tilde{r} \,\left[-|\psi|^2+|(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})^2\psi|^2+\frac{1}{2}|\psi|^4+(\tilde{B}-\tilde{H})^2\,\right]}

where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{B}=\frac{B}{\sqrt{2}H_c}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{A}=\frac{A}{\sqrt{2}\lambda H_c}}

Assuming 2nd order phase transition and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T=T_c} , then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi\rightarrow 0} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{B}\rightarrow\tilde{H}} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_s-F_n=\frac{H_c^2}{4\pi}\lambda^3\int d^3\tilde{r} \,\left[-|\psi|^2+|(\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})^2\psi|^2\,\right]}

Using the Landau gauge:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A}=\hat{y}Hx=\hat{y}\frac{H}{\sqrt{2}H_c}\frac{x}{\lambda}=\hat{y}\tilde{H}\tilde{x}}

Recall

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\frac{1}{\kappa}\frac{\tilde{\nabla}}{i}-\tilde{A})^2\psi-\psi=0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\frac{1}{\kappa}\frac{1}{i}\frac{d}{d\tilde{x}})^2\psi+(\frac{1}{\kappa}\frac{1}{i}\frac{d}{d\tilde{y}}-\tilde{H}\tilde{x})^2\psi+(\frac{1}{\kappa}\frac{1}{i}\frac{d}{d\tilde{z}})^2\psi-\psi=0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{1}{\kappa^2}\frac{d^2}{d\tilde{x}^2}\psi+(\frac{1}{\kappa}\frac{1}{i}\frac{d}{d\tilde{y}}-\tilde{H}\tilde{x})^2\psi-\frac{1}{\kappa^2}\frac{d^2}{d\tilde{z}^2}\psi-\psi=0}

Let

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(\tilde{x},\tilde{y},\tilde{z})=e^{ik_z\tilde{z}}e^{ik\tilde{y}}\Phi_k(\tilde{x})}

then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{ik_z\tilde{z}}e^{ik\tilde{y}}\,\left[-\frac{1}{\kappa^2}\frac{d^2}{d\tilde{x}^2}\Phi+(\frac{k}{\kappa}-\tilde{H}\tilde{x})^2\Phi+\frac{k_z^2}{\kappa^2}\Phi-\Phi\,\right]=0}

Let

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\tilde{H}X=\frac{k}{\kappa}-\tilde{H}\tilde{x};\frac{d}{d\tilde{x}}=\frac{dX}{d\tilde{x}}\frac{d}{dX}=\frac{d}{dX}}

then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{ik_z\tilde{z}}e^{ik\tilde{y}}(-\frac{1}{\kappa^2}\frac{d^2}{dX^2}+\tilde{H}^2 X^2+\frac{k_z^2}{\kappa^2}-1)\Phi=0}

Recall for a harmonic oscillator we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{1}{2}m\omega^2 x^2)\Psi=E\Psi}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E=\hbar\omega(n+\frac{1}{2})}

Comparison leads to

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{-\hbar^2}{2m}\rightarrow\frac{-1}{\kappa^2};\frac{1}{2}m\omega^2\rightarrow\tilde{H}^2;\hbar\omega\rightarrow\frac{2\tilde{H}}{\kappa}}

and the eigenvalues

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2\frac{\tilde{H}}{\kappa}(n+\frac{1}{2})+\frac{k_z}{\kappa}-1}

The exponent is zero when

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2\frac{\tilde{H}}{\kappa}(n+\frac{1}{2})+\frac{k_z}{\kappa}-1=0}

If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n=0} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_x=0} we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{H}_{c2}=\kappa=\frac{H_{c2}}{\sqrt{2}H_c}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_{c2}=\sqrt{2}\kappa H_c}

For Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa < 1/\sqrt{2}} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_{c2}<H_c} , the bulk is preempted by the Meissner phase. On the other hand, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa > 1/\sqrt{2}} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_{c2}>H_c} , the bulk undergoes a 2nd order phase transition into a mixed state in which normal state and superconducting state coexist.

Now let's look at the wavefunction at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_{c2}} .

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (-\frac{1}{\kappa^2}\frac{d^2}{dX^2}+\tilde{H}_{c2}^2 X^2)\Phi=\lambda\Phi;\tilde{H}_{c2}=\kappa}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (-\frac{1}{\kappa^2}\frac{d^2}{dX^2}+\kappa^2 X^2)\Phi=\lambda\Phi}

Let

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\sqrt{2}}\frac{1}{\kappa}\frac{d}{dX}+\frac{1}{\sqrt{2}}\kappa X = a}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{-1}{\sqrt{2}}\frac{1}{\kappa}\frac{d}{dX}+\frac{1}{\sqrt{2}}\kappa X = a^+}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,a^{+}]=1} Note that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^+ a=\frac{-1}{2\kappa^2}\frac{d^2}{dX^2}+\frac{1}{2}\kappa^2 X^2-\frac{1}{2}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (2a^+ a+1)\Phi=\lambda\Phi}

For Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda=1} we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\sqrt{2}}(\frac{1}{\kappa}\frac{d}{dX}+\kappa K)\Phi=0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi=const.\times e^{-\frac{1}{2}\kappa^2 X^2}}

and the order parameter is of the form

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle const.\times e^{ik\tilde{y}}e^{-\frac{\kappa^2}{2}(\tilde{x}-\frac{k}{\kappa^2})^2}}

Abrikosov Solution of the GL Equation for Type-II Superconductor near Hc2

Note. In this section we will drop the "tilde(~)" on relevant quantities and recover it when necessary.

Since the conditions along the entire superconductor are uniform, we seek a linear combination of solutions centered through equal intervals, namely,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi^{(0)}=\Sigma_{n=-\infty}^{\infty}C_n e^{ikn\hat{y}} e^{-\frac{1}{2}\kappa^2(x-\frac{kn}{\kappa^2})^2}}

or,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi=\Sigma_{n=-\infty}^{\infty}C_n e^{ikny}\phi_n(x)}

with

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi_n(x)=e^{-\frac{1}{2}\kappa^2(x-\frac{kn}{\kappa^2})^2}}

Consider the solution to GL equation at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H} slightly less than Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa} .The 2nd GL equation gives

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla\times(\nabla\times A^{(1)})=\frac{-i}{2\kappa}(\psi^{(0)*} \nabla\psi^{(0)}-\psi^{(0)}\nabla\psi^{(0)*})-|\psi^{(0)}|^2 A^{(0)}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla\times B=\hat{x}\frac{\partial B_z}{\partial y}-\hat{y}\frac{\partial B_z}{\partial x}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A=\hat{y}A_y(x,y);\nabla\times A=\hat{z}\frac{\partial A}{\partial x}}

For the x-component, we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial B_z}{\partial y}=\frac{\partial^2 A_y}{\partial x\partial y}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{-i}{2\kappa}\Sigma_{n=-\infty}^{\infty}\Sigma_{m=-\infty}^{\infty}C_n^* e^{-ikny}\phi_n(x)(\frac{\partial}{\partial x}\phi_m(x))C_me^{ikmy} + \frac{i}{2\kappa}\Sigma_{n=-\infty}^{\infty}\Sigma_{m=-\infty}^{\infty}C_n^* e^{-ikny}(\frac{\partial}{\partial x}\phi_n(x))\phi_m(x)C_m e^{ikmy}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{-i}{2\kappa}\Sigma_{n,m}C_n^* C_m e^{-ik(n-m)y}\,\left[\phi_n(x)\frac{\partial}{\partial x}\phi_m(x)-\phi_m(x)\frac{\partial}{\partial x}\phi_n(x)\,\right]}

where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [...]=-\kappa^2(x-\frac{km}{\kappa^2})\phi_n(x)\phi_m(x)+\kappa^2(x-\frac{kn}{\kappa^2})\phi_n(x)\phi_m(x) =k(m-n)\phi_n(x)\phi_m(x)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial B_z}{\partial y}=\frac{-i}{2\kappa}\Sigma_{n,m}C_n^* C_m e^{-ik(n-my)}k(m-n)\phi_n(x)\phi_m(x) =\frac{\partial}{\partial y}\,\left[\frac{1}{2\kappa}\Sigma_{n,m}C_n^* C_m e^{-ik(n-m)y}\phi_n(x)\phi_m(y)\,\right] =\frac{\partial}{\partial y}\,\left[\frac{-1}{2\kappa}|\psi^{(0)}|^2\,\right]}

Similarly, for the y-component we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\partial B_z}{\partial x}=\frac{-i}{2\kappa}\Sigma_{n,m}\,\left[C_n^* e^{-ikny}\phi_n(x)ikm e^{ikmy}\phi_m(x)C_m+C_n^* ikne^{-ikny}\phi_n(x)e^{ikmy}\phi_m C_m\,\right]-\kappa x|\psi^{(0)}|^2}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{1}{2\kappa}\Sigma_{n,m}\,\left[k(n+m)-2\kappa^2x\,\right]C_n^* C_me^{-ik(n-m)y}\phi_m(x)\phi_n(x)}

Note that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial x}\,\left[\phi_m(x)\phi_m(x)\,\right]=\,\left[-\kappa^2(x-\frac{km}{\kappa^2})-\kappa^2(x-\frac{kn}{\kappa^2})\,\right]\phi_m(x)\phi_n(x)=-\,\left[2\kappa^2x-k(m+n)\,\right]\phi_m(x)\phi_n(x)}

Therefore,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\partial B_z}{\partial x}=\frac{\partial}{\partial x}\,\left[\frac{1}{2\kappa}|\psi^{(0)}|^2\,\right]}

Now we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial B_z}{\partial y}=-\frac{\partial}{\partial y}\,\left[\frac{1}{2\kappa}|\psi^{(0)}|^2\,\right]}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial B_z}{\partial x}=-\frac{\partial}{\partial x}\,\left[\frac{1}{2\kappa}|\psi^{(0)}|^2\,\right]}

or,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_z(x,y)=f(x)-\frac{1}{2\kappa}|\psi^{(0)}|^2}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_z(x,y)=g(y)-\frac{1}{2\kappa}|\psi^{(0)}|^2}

which implies

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_z(x,y)=const.-\frac{1}{2\kappa}|\psi^{(0)}|^2}

At points where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi^{(0)}=0} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_z(x,y)=H_z} . So

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_z(x,y)=H_z-\frac{1}{2\kappa}|\psi^{(0)}|^2}

Note that the correction (the 2nd term) comes from induced current. Furthermore,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_y=xH_z-\frac{1}{2\kappa}\int dx |\psi^{(0)}|^2}

Now let's consider the 1st GL equation:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\frac{1}{\kappa}\frac{\nabla}{i}-A)^2\psi-\psi+\psi|\psi|^2=0}

which is solced by dropping the non-linear term and A and leads to Hc2.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\frac{1}{\kappa}\frac{\nabla}{i}-A^{(0)}-A^{(1)})^2(\psi^{(0)}+\psi^{(1)})-(\psi^{(0)}+\psi^{(1)})+(\psi^{(0)}+\psi^{(1)})|\psi^{(0)}+\psi^{(1)}|^2=0}

To 0th oeder:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\frac{1}{\kappa}\frac{\nabla}{i}-A^{(0)})^2\psi^{(0)}-\psi^{(0)}=0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_y^{(0)}=\kappa x}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi^{(0)}=\Sigma_{n=-\infty}^{\infty}C_ne^{ikny}\phi_n(x)}

To 1st order:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\frac{1}{\kappa}\frac{\nabla}{i}-A^{(0)})^2\psi^{(1)}-(\frac{1}{\kappa}\frac{\nabla}{i}-A^{(0)})\cdot A^{(1)}\psi^{(0)}-A^{(1)}\cdot(\frac{1}{\kappa}\frac{\nabla}{i}-A^{(0)})\psi^{(0)}-\psi^{(1)}+\psi^{(0)}|\psi^{(0)}|^2=0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_y^{(1)}=(H-\kappa)x-\frac{1}{2\kappa}\int^{x}|\psi|^2}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi^{(1)}=\Sigma_{-\infty}^{\infty}e^{ikny}\psi_n^{(1)}(x)}

1st term:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \,\left[\frac{-1}{\kappa^2}\frac{\partial^2}{\partial x^2}+(\frac{1}{\kappa}\frac{1}{i}\frac{\partial}{\partial y}-\kappa x)^2 \,\right] \Sigma_n e^{ikny} \psi_n^{(1)}(x)= \Sigma_n e^{ikny} \,\left[\frac{-1}{\kappa^2}\frac{\partial^2}{\partial x^2}+(\frac{kn}{\kappa}-\kappa x)^2\,\right]\psi_n^{(1)}(x)}

2nd term:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -(\frac{1}{\kappa}\frac{1}{i}\frac{\partial}{\partial y}-\kappa x)\,\left[(H-\kappa)x-\frac{1}{2\kappa}\int^x |\psi^{(0)}|^2\,\right]\Sigma_n e^{ikny} \phi_n(x)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\Sigma_{n=-\infty}^{\infty}C_ne^{ikny}(-\frac{kn}{\kappa}+\kappa x)(H-\kappa)x\phi_n(x) +(\frac{1}{\kappa}\frac{1}{i}\frac{\partial}{\partial y}-\kappa x)\frac{1}{2\kappa}\int^x dx'\Sigma_{m,p}C_m^*C_pe^{-ikmy}e^{ikpy}\phi_m(x')\phi_p(x')\Sigma_nC_ne^{ikny}\phi_n(x)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\Sigma_{n=-\infty}^{\infty}C_ne^{ikmy}(-\frac{kn}{\kappa}+\kappa x)(H-\kappa)\phi_n(x)+\Sigma_{n,m,p}C_nC_m^*C_pe^{ik(n-m+p)y}(\frac{n-m+p}{\kappa}-\kappa x)\frac{1}{2\kappa}\phi_n(x)\int^x dx'\phi_m(x')\phi_p(x')}

3rd term:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Sigma_nC_ne^{ikny}(-\frac{kn}{\kappa}+\kappa x)(H-\kappa)x\phi_n(x)+\Sigma_{n,m,p}C_nC_m^*C_pe^{ik(n-m+p)y}(\frac{kn}{\kappa}-\kappa x)\frac{1}{2\kappa}\phi_n(x)\int^x dx'\phi_m(x')\phi_p(x')}

4th term:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\Sigma_ne^{ikny}\psi_n^{(1)}(x)}

5th term:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Sigma_{n,m,p}C_nC_m^*C_pe^{ik(n-m+p)y}\phi_n(x)\phi_m(x)\phi_p(x)}

Add them up, we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Sigma_ne^{ikny}\,\left[\frac{-1}{\kappa^2}\frac{\partial^2}{\partial x^2}+(\frac{kn}{\kappa}-\kappa x)^2 + 1\,\right]\psi^{(1)}=\Sigma_ne^{ikny}C_n2x(H-\kappa)(\frac{kn}{\kappa}-\kappa x)\phi_n(x)-\Sigma_{n,m,p}e^{ik(n-m+p)y}C_nC_m^*C_p\,\left[(\frac{(2n-m+p)k}{\kappa}-2\kappa)\frac{1}{2\kappa}\phi_n(x)\int^xdx'\phi(x')\phi(x')+\phi_n(x)\phi_m(x)\phi_p(x)\,\right]}

Multiplying the whole equation by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{-ikNy}} and integrating over the variable y leads to

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \,\left[\frac{-1}{\kappa^2}\frac{\partial^2}{\partial x^2}+(\frac{kN}{\kappa}-\kappa)^2-1\,\right]\psi_N^{(1)}(x)=C_N2x(H-\kappa)(\frac{kx}{\kappa}-\kappa x)\phi_N(x)-\Sigma_{m,p}C_{N+m-p}C_m^*C_p\,\left[(\frac{k}{\kappa}(2N+2m-2p-m+p)-2\kappa x)\frac{1}{2\kappa}\phi_{N+m-p}(x)\int^xdx'\phi_m(x')\phi_p(x')+\phi_{N+m-p}(x)\phi_m(x)\phi_p(x)\,\right]}

where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi_n(x)=e^{-\frac{1}{2}\kappa^2(x-\frac{kn}{\kappa^2})^2}}

The above equation has this (inhomogeneous) form:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{O}|\psi_N^{(1)}>=|\Phi>}

Consider the homogeneous version of this equation:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{O}|\chi>=0}

The only way for the inhomogeneous equation to have a solution is if

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <\chi|\Phi>=0}

Thus we need

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \,\left[\frac{-1}{\kappa^2}\frac{\partial^2}{\partial x^2}+(\frac{kN}{\kappa}-\kappa x)^2-1\,\right]\chi(x)=0}

Let

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \chi(x)=e^{-\frac{1}{2}\kappa^2(x-\frac{kN}{\kappa^2})^2}=\phi_N(x)}

then the identity is satisfied.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0=C_N2(H-\kappa)\int_{-\infty}^{\infty}dxx(\frac{kN}{\kappa}-\kappa x)\phi_N^2(x)-\Sigma_{m,p}C_{N-p+m}C_m^*C_p \,\left[\int_{-\infty}^{\infty}dx\phi_N(x)(\frac{k}{2\kappa^2}(2N+m-p)-x)\phi_{N+m-p}(x)\int^xdx'\phi_m(x')\phi_p(x') +\int_{-\infty}^{\infty}dx\phi_N(x)\phi_{N+m-p}(x)\phi_m(x)\phi_p(x)\,\right]}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0=\frac{\sqrt{\pi}}{\kappa}(1-\frac{H}{\kappa})C_N-\Sigma_{m,p}C_{N-p+m}C_m^*C_p(1-\frac{1}{2\kappa^2})\int_{-\infty}^{\infty}dx\phi_N(x)\phi_{N+m-p}(x)\phi_m(x)\phi_p(x)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0=\frac{\sqrt{\pi}}{\kappa}(1-\frac{H}{\kappa})C_N+\Sigma_{m,p}C_{N-p+m}C_m^*C_p(1-\frac{1}{2\kappa^2})\sqrt{\frac{\pi}{2}}\frac{1}{\kappa}e^{-\frac{k^2}{2\kappa^2}((m-p)^2+(N-p)^2)}}

Now, multiply both sides by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C_N^*} and sum over N we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0=(1-\frac{H}{\kappa})\Sigma_NC_N^*C_N+\frac{1}{\sqrt{2}}(\frac{1}{2\kappa^2}-1)\Sigma_{N,m,p}C_N^*C_{N-p+m}C_m^*C_pe^{-\frac{k^2}{2\kappa^2}((m-p)^2+(N-p)^2)}}

Taking the spatial average results in

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0=(1-\frac{H}{\kappa})<|\psi|^2>+(\frac{1}{2\kappa^2}-1)<|\psi|^4>}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1=\frac{\frac{H}{\kappa}-1}{\frac{1}{2\kappa^2}-1}\frac{<|\psi|^2>}{<|\psi|^4>}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <B>=H-\frac{1}{2\kappa}<|\psi|^2>=H-\frac{1}{2\kappa}\frac{H-\kappa}{\frac{1}{2\kappa}-\kappa}\frac{<|\psi|^2>^2}{<|\psi|^4>}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <B>=H-\frac{\kappa -H}{2\kappa^2-1}\frac{1}{\beta_A}}

or, recovering the "tilde(~)",

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <\tilde{B}>=\tilde{H}-\frac{\kappa -\tilde{H}}{2\kappa^2-1}\frac{1}{\beta_A}}

where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta_A=\frac{<|\psi|^4>}{<|\psi|^2>^2}}

is the famous Abrikosov parameter which does not depend on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{H}} or the normalization of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi} . It is easy to see that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta_A} is always greater than or equal to 1:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\psi|^2=<|\psi|^2>+\delta |\psi|^2}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <|\psi|^4>=<|\psi|^2>^2+2<|\psi|^2><\delta |\psi|^2>+<(\delta |\psi|^2)^2>}

Note that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <\delta |\psi|^2>=0}

Hence

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <|\psi|^4>=<|\psi|^2>^2+<(\delta |\psi|^2)^2>\ge <|\psi|^2>^2}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta_A=\frac{<|\psi|^4>}{<|\psi|^2>^2}\ge 1}

The equality holds when Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi=1} .

Recall that the surface free energy is given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_s-F_n=\frac{H_c^2}{4\pi}\lambda^3\int d^3\tilde{r}\,\left[(\tilde{B}-\tilde{H})^2-\frac{1}{2}|\psi|^4\,\right]}

In order to obtain the thermodynamic potential which when differentiated with respect to H gives B we need to subtract Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{H}^2} (which came from the normal term) from the above expression:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Omega=\frac{H_c^2}{4\pi}\lambda^3\int d^3\tilde{r}\,\left[(\tilde{B}-\tilde{H})^2-\tilde{H}^2-\frac{1}{2}|\psi|^4\,\right]}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{H_c^2}{4\pi}L^3\,\left[\frac{1}{4\kappa^2}<|\psi|^4>-\tilde{H}^2-\frac{1}{2}<|\psi|^4>\,\right]}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{H_c^2}{8\pi}L^3\,\left[\frac{1-2\kappa^2}{2\kappa^2}<|\psi|^4>-2\tilde{H}^2\,\right]}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{H_c^2}{8\pi}L^3\,\left[\frac{2}{1-2\kappa^2}\frac{1}{\beta_A}(\tilde{H}-\kappa)^2-2\tilde{H}^2\,\right]}

Note that for type-II superconductor, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa >1/\sqrt{2}} , the whole expression is negative. So, to lower the the free energy we need to minimize Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta_A} .

Knowing Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Omega} we can obtain Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{B}} by differentiating Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Omega} with respect to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{H}} and hence obtain the magnetization which is measurable.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{4\pi}{L^3}\frac{\partial\Omega}{\partial H}=B}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{H}=\frac{H}{\sqrt{2}H_c};\frac{\partial}{\partial H}=\frac{\partial\tilde{H}}{\partial H}\frac{\partial}{\partial H}=\frac{1}{\sqrt{2}H_c}\frac{\partial}{\partial\tilde{H}}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B=-\frac{H_c}{2\sqrt{2}}4(\frac{1}{\beta_A}\frac{\tilde{H}-\kappa}{1-2\kappa^2}-\tilde{H}) =-\frac{2H_c}{\sqrt{2}}\,\left[\frac{1}{\beta_A}\frac{1}{1-2\kappa^2}(\frac{H}{\sqrt{2}H_c}-\frac{H_{c2}}{\sqrt{2}H_c})-\frac{H}{\sqrt{2}H_c}\,\right]=\frac{1}{\beta_A}\frac{1}{1-2\kappa^2}(H-H_{c2})+H\le H}

Pic8.jpg

for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H<H_{c2}} .

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B=H+4\pi M}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 4\pi M=B-H=\frac{1}{\beta_A}\frac{1}{2\kappa^2-1}(H-H_{c2})<0}

which means the type-II superconductor behaves like a diamagnet.

Abrikosov Vortex States

Now let's follow the step of Abrikosov to derive the vortex structure in type-II superconductor.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi=\Sigma_{n=-\infty}^{\infty}C_ne^{ikny}e^{-\frac{1}{2}\kappa^2(x-\frac{kn}{\kappa^2})^2} =C\Sigma_{n=-\infty}^{\infty}e^{ikny}e^{-\frac{1}{2}\kappa^2(x-\frac{kn}{\kappa^2})^2}}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C_n=C} .

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\psi|^2=|C|^2\Sigma_{m,n}e^{-ikny}e^{ikmy}e^{-\frac{1}{2}\kappa^2(\tilde{x}-\frac{kn}{\kappa^2})^2}e^{-\frac{1}{2}\kappa^2(\tilde{x}-\frac{km}{\kappa^2})^2}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int d\tilde{x}d\tilde{y}d\tilde{z}|\psi|^2=L_yL_z\Sigma_n\int d\tilde{x}|C|^2e^{-\kappa^2(\tilde{x}-\frac{kn}{\kappa^2})^2}=L_yL_z\frac{\kappa^2}{k}\tilde{L}_x\frac{\sqrt{\pi}}{\kappa}|C|^2}

Thus

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <|\psi|^2>=|C|^2\frac{\sqrt{\pi}\kappa}{k}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\psi|^4=|C|^4\Sigma_{m,n}\Sigma_{s,t}e^{-ikmy}e^{ikny}e^{-iksy}e^{ikty}e^{-\frac{\kappa^2}{2}(x-\frac{km}{\kappa^2})^2}e^{-\frac{\kappa^2}{2}(x-\frac{kn}{\kappa^2})^2}e^{-\frac{\kappa^2}{2}(x-\frac{ks}{\kappa^2})^2}e^{-\frac{\kappa^2}{2}(x-\frac{kt}{\kappa^2})^2}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int dxdydz|\psi|^4=L_yL_z|C|^4\Sigma_{m,n,s}\int dx e^{-\frac{\kappa^2}{2}(x-\frac{km}{\kappa^2})^2}e^{-\frac{\kappa^2}{2}(x-\frac{kn}{\kappa^2})^2}e^{-\frac{\kappa^2}{2}(x-\frac{ks}{\kappa^2})^2}e^{-\frac{\kappa^2}{2}(x-\frac{k(m-n+s)}{\kappa^2})^2}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =L_yL_z|C|^4\Sigma_{m,n,s}\sqrt{\frac{\pi}{2}}\frac{1}{\kappa}e^{-\frac{k^2}{2\kappa^2}((m-n)^2+(n-s)^2)}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =L_yL_z|C|^4\sqrt{\frac{\pi}{2}}\frac{1}{k}\frac{\kappa^2}{k}\tilde{L}_x\Sigma_{m,n}e^{-\frac{k^2}{2\kappa^2}(m^2+n^2)}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta_A=\frac{1}{\sqrt{2\pi}}\frac{k}{\kappa}(\Sigma_{n=-\infty}^{\infty}e^{-\frac{k^2n^2}{2\kappa^2}})^2}

Minimize Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta_A} with respect to k we find that at minimum

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k=\sqrt{2\pi}\kappa}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi=C\Sigma_{n=-\infty}^{\infty}e^{i\sqrt{2}\pi n\kappa\tilde{y}}e^{-\frac{1}{2}(\kappa\tilde{x}-\sqrt{2}\pi n)^2}}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta_A=1.18034} and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi=Ce^{-\frac{1}{2}\kappa^2\tilde{x}^2}\Sigma_{n=-\infty}^{\infty}e^{i\sqrt{2\pi}n\kappa\tilde{y}}e^{\sqrt{2\pi}n\kappa\tilde{x}}e^{-\pi n^2}=Ce^{-\frac{1}{2}\kappa^2\tilde{x}^2}\Sigma_{n=-\infty}^{\infty}e^{\sqrt{2\pi}\kappa (\tilde{x}+i\tilde{y})n}e^{-\pi n^2}=Ce^{-\frac{1}{2}\kappa^2\tilde{x}^2}\theta_3 (\sqrt{\frac{\pi}{2}}\kappa (\tilde{y}-i\tilde{x}),e^{-\pi})}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta_3} is the Jacobi elliptic theta function. The Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi} vanishes at points:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{y}=\sqrt{\frac{\pi}{2}}\frac{1}{\kappa}(2m-1)}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{x}=\sqrt{\frac{\pi}{2}}\frac{1}{\kappa}(1-2n)\rightarrow\sqrt{\frac{\pi}{2}}\frac{1}{\kappa}(2n-1)}

Obviously, the zeros form a square lattice of spacing Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\sqrt{2\pi}}{\kappa}} in reduced units.

Now consider the current lines:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{j}=\frac{-i}{2\kappa}(\psi^*\nabla\psi-\psi\nabla\psi^*)-|\psi|^2\tilde{A}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j_x=\frac{1}{2\kappa i}\Sigma_{m,n}k(n-m)C_m^*C_ne^{-ik(m-n)\tilde{y}}\phi_n(\tilde{x})\phi_m(\tilde{x})=\frac{-1}{2\kappa}\frac{\partial}{\partial\tilde{y}}|\psi|^2}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j_y=\frac{1}{2\kappa}\frac{\partial}{\partial\tilde{x}}|\psi|^2}

The lines of constant current magnitude are determined by the following condition:

Pic9.jpg

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx}=\frac{j_y}{j_x}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j_xdy-j_ydx=0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{-1}{2\kappa}(\frac{\partial}{\partial y}|\psi|^2)dy-\frac{1}{2\kappa}(\frac{\partial}{\partial x}|\psi|^2)dx=0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d|\psi|^2=0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\psi|^2=const.}

which means the current flows along the contours of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\psi|^2} .

However, the actual solution which minimizes Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta_A} is a triangular lattice for which the coefficients satisfy

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C_{n+4}=C_n;C_0=C_1=C;C_2=C_3=-C}

and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta_A=1.16} .

Going below Tc with the Saddle Point Approximation

So, previously all of this work has shown us the behaviour of a superconducting system near Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_c} only. If we want to go into lower temperatures, we will have to make a careful saddle-point approximation, following Bardeen, Cooper, and Schrieffer (BCS). Once again, we can start from our microscopic 'toy' Hamiltonian, and gain useful information.

Recall that the partition function can be written,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{Z} = \int{d\Delta^* d\Delta} \left[ \int{D\psi^* D\psi \ e^{-S_{BCS} - S_{\Delta}}} \right] }

Where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{\Delta} = \frac{-1}{g} \int_{0}^{\beta}{d\tau} \int{d^3r}\Delta^*(\vec{r},\tau) \Delta(\vec{r},\tau) }

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{BCS} = S_0 + S_{int} \;\; \text{where} \;\; S_0 = \int_{0}^{\beta}{d\tau}\int{d^3r}\left[ \psi_{\sigma}^*(\vec{r},\tau) \left( \frac{\partial}{\partial \tau} + \epsilon_p - \mu \right) \psi_{\sigma}(\vec{r},\tau) \right] } Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{and} \;\; S_{int} = \int_{0}^{\beta}{d\tau}\int{d^3r} \left[\Delta^*(\vec{r},\tau)\psi_{\downarrow}(\vec{r},\tau) \psi_{\uparrow}(\vec{r},\tau) + \Delta(\vec{r},\tau)\psi_{\downarrow}(\vec{r},\tau)\psi_{\uparrow}(\vec{r}.\tau) \right]}

Previously, we used a cumulant expansion around Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T = T_c} to find the Ginzburg-Landau equations, along with the assumption that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta} was small. Now, if we throw away this assumption, pray that fluctuations are small, and that there is only one saddle point, we will be able to successfully describe the superconducting state deep below Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_c} by demanding:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial S_{eff}[\Delta]}{\partial \Delta*} = 0 }

Self-Consistency Equation

The solution of this functional derivative equation will give the value of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta = \Delta_{sp}} at the saddle point (and also the self-consistency equation for this mean-field theory.)

To that end:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{eff}[\Delta] = S_{\Delta} - \ln \left( \int{D\psi D\psi^*} \;\; e^{-S_{BCS}}\right) }
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial S_{\Delta}}{\partial \Delta^*(\vec{r},\tau)} = \frac{-\Delta(\vec{r},\tau)}{g}}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial \left[ -\ln\left(\int{D\psi D\psi^*} \;\; e^{-S_{BCS}}\right) \right] }{\partial \Delta^*(\vec{r},\tau)} = \frac{-1}{\int{D\psi D\psi^*} \;\; e^{-S_{BCS}}} \int{D\psi D\psi*} \frac{\partial e^{-S_{BCS}}}{\partial \Delta^*} = \frac{\int{D\psi D\psi^*} \;\; e^{-S_{BCS}} \frac{\partial S_{BCS}}{\partial \Delta^*}}{\int{D\psi D\psi^*} \;\; e^{-S_{BCS}}} }

and, since

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial S_{BCS}}{\partial \Delta^*} = \psi_{\downarrow}(\vec{r},\tau) \psi_{\uparrow}(\vec{r},\tau)}

We arrive at

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial \left[ -\ln\left(\int{D\psi D\psi^*} \;\; e^{-S_{BCS}}\right) \right] }{\partial \Delta^*(\vec{r},\tau)} = \frac{\int{D\psi D\psi^*} \;\; e^{-S_{BCS}} \psi_{\downarrow}(\vec{r},\tau) \psi_{\uparrow}(\vec{r},\tau)}{\int{D\psi D\psi^*} \;\; e^{-S_{BCS}}} = \langle \psi_{\downarrow}(\vec{r},\tau) \psi_{\uparrow}(\vec{r},\tau)\rangle_{BCS} }

So that, from the saddle-point condition, we find the Self-Consistency Equation: 

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle { \frac{\Delta_{sp}(\vec{r},\tau)}{g}} = {\langle\psi_{\downarrow}(\vec{r},\tau) \psi_{\uparrow}(\vec{r},\tau)\rangle_{BCS}} }

This contains the same information as the Ginzburg-Landau equations, but also more, as we can now go far below Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_c} . Evaluation of this cam be done formally, but is difficult and not terribly enlightening. Instead, we will search for a solution in which <math\psi</math> is independent of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{r}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tau} , similar to our solution near Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_c} . In this case, we also expect that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta} will be independent of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{r}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tau} .

First, it is useful to write down the action in momentum and frequency space, before evaluating the correlator.

Start with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(\vec{r},\tau) = \frac{1}{\beta}\sum_{\omega_n}\sum_{\vec{k}}\frac{1}{\sqrt{V}} e^{i\vec{k}\cdot\vec{r}}\psi_{k}(i\omega_n)}

For the three terms in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{BCS}} , we have:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_0 = \int_{0}^{\beta}{d\tau}\int{d^3r}\left[ \psi_{\sigma}^*(\vec{r},\tau) \left( \frac{\partial}{\partial \tau} + \epsilon - \mu \right) \psi_{\sigma}(\vec{r},\tau) \right] }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \frac{1}{V}\sum_{\vec{k},\vec{k'}} \frac{1}{\beta^2}\sum_{\omega_n,\nu_m}\int_{0}^{\beta}{d\tau}\int{d^3r}e^{i\nu_{m}\tau}e^{-i\vec{k'}\cdot\vec{r}} \left(-i\omega_n + \epsilon_p - \mu \right) e^{-i\omega_n\tau}e^{i\vec{k}\cdot\vec{r}}\psi_{\sigma,\vec{k'}}^*(i\nu_m)\psi_{\sigma, \vec{k}}(i\omega_n)}

The integral over all space give Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V \delta_{\vec{k},\vec{k'}}} , and the integral over imaginary time gives a factor of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta \delta(\omega_n - \nu_m)} , so that we find

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_0 = \frac{1}{\beta}\sum_{\omega_n}\sum_{\vec{k}} \psi_{\sigma, \vec{k}}^*(i\omega_n)\left( -i\omega_n + \epsilon_p - \mu \right)\psi_{\sigma,\vec{k}}(i\omega_n) }

Now, for the 2nd (pairing) term:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta^*\int_{0}^{\beta}{d\tau} \int{d^3r} \psi_{\downarrow}(\vec{r},\tau) \psi_{\uparrow}(\vec{r},\tau) = \Delta^* \frac{1}{V}\sum_{\vec{k},\vec{k'}} \frac{1}{\beta^2}\sum_{\omega_n,\nu_m}\int_{0}^{\beta}{d\tau}\int{d^3r} e^{-i(\nu_m + \omega_n)\tau) e^{i(\vec{k} + \vec{k'})\cdot\vec{r}} \psi_\downarrow,\vec{k'}}(i\nu_m) \psi_{\uparrow,\vec{k}}(i\omega_n) }

This time, the integral over real space gives Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V \delta(\vec{k}+\vec{k'})} , and the integral over imaginary time gives a factor of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta \delta(\omega_n + \nu_m)} , so the 2nd term becomes:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \frac{\Delta^*}{\beta}\sum_{\vec{k}}\sum_{\omega_n}\psi_{\downarrow,\vec{k}}(i\omega_n)\psi_{\uparrow,-\vec{k}}(-i\omega_n)}

The hermitian conjugate of the above (the 3rd term in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{BCS}} ) gives

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \;\; \frac{\Delta}{\beta}\sum_{\vec{k}}\sum_{\omega_n}\psi_{\uparrow,-\vec{k}}^*(-i\omega_n)\psi_{\downarrow,\vec{k}}^*(i\omega_n)}

So that, for the entire Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{BCS}} , we have found:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{BCS} = \frac{1}{\beta}\sum_{\omega_n}\sum_{\vec{k}}\left[ \left( i\omega_n + \epsilon_{-\vec{k}} - \mu \right)\psi_{\uparrow, -\vec{k}}^*(-i\omega_n)\psi_{\uparrow,-\vec{k}}(-i\omega_n) + \left( -i\omega_n + \epsilon_{\vec{k}} - \mu \right)\psi_{\downarrow, \vec{k}}^*(i\omega_n)\psi_{\downarrow,\vec{k}}(i\omega_n) \right] }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle + \frac{1}{\beta}\sum_{\omega_n}\sum_{\vec{k}}\left[ \Delta^*\psi_{\downarrow,\vec{k}}(i\omega_n)\psi_{\uparrow,-\vec{k}}(-i\omega_n) + \Delta\psi_{\uparrow,-\vec{k}}^*(-i\omega_n)\psi_{\downarrow,\vec{k}}^*(i\omega_n) \right] }

Now, to evaluate the Gaussian integrals in the correlator, it is extremely beneficial to write Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{BCS}} like a matrix. To this end, we construct the so-called Nambu Spinors:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi_{\vec{k}}(i\omega_n) = \begin{pmatrix} \psi_{-\vec{k},\uparrow}(-i\omega_n) \\ \psi^*_{\vec{k},\downarrow}(i\omega_n) \end{pmatrix} }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi^*_{\vec{k}}(i\omega_n) = \begin{pmatrix} \psi^*_{-\vec{k},\uparrow}(-i\omega_n) & \psi^*_{\vec{k},\downarrow}(i\omega_n) \end{pmatrix} }

So now, we can write the BCS action as:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{BCS} = \frac{1}{\beta}\sum_{\omega_N}\sum_{\vec{k}}\Psi^*_{\vec{k}}(i\omega_n) \begin{pmatrix} i\omega_n + \epsilon_{-\vec{k}} - \mu & \Delta \\ \Delta^* & i\omega_n - \epsilon_{\vec{k}} + \mu \end{pmatrix} \Psi_{\vec{k}}(i\omega_n) }

Now, we can examine the correlator we found on the right-hand-side of the self-consistency equation:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle \psi_{\downarrow}(\vec{r},\tau)\psi_{\uparrow}(\vec{r},\tau)\rangle = \frac{1}{\beta^2 V}\sum_{\omega_n,\nu_m}e^{-i(\omega_n + \nu_M)\tau} \sum_{\vec{k},\vec{k'}} e^{i(\vec{k} + \vec{k'})\cdot\vec{r}}\langle \psi_{\vec{k},\downarrow}(i\omega_n) \psi_{\vec{k'},\uparrow}(i\nu_m) \rangle }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \frac{1}{\beta^2 V}\sum_{\omega_n,\nu_m}e^{-i(\omega_n + \nu_M)\tau} \sum_{\vec{k},\vec{k'}} e^{i(\vec{k} + \vec{k'})\cdot\vec{r}} \langle \Psi^*_{\vec{k}}(i\omega_n)_2 \Psi_{-\vec{k'}}(-i\nu_m)_1 \rangle }

So, we need to write down a generic matrix for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle \Psi^*_{\vec{k}}(i\omega_n)_{\mu} \Psi_{-\vec{k'}}(-i\nu_m)_{\lambda} \rangle} , and take element Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (2,1) } for our result. The gaussian integrals require that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{k} = -\vec{k'}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i\omega_n = -i\nu_m } for convergence. Using our technology from last semester, we can show that:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle \Psi^*_{\vec{k}}(i\omega_n)_{\mu} \Psi_{-\vec{k'}}(-i\nu_m)_{\lambda} \rangle = \delta(\vec{k} + \vec{k'})\beta \delta(\omega_n + \nu_m) \left[ \begin{pmatrix} i\omega_n + \epsilon_{\vec{k}} - \mu & \Delta \\ \Delta^* & i\omega_n - \epsilon_{\vec{k}} + \mu \end{pmatrix}^{-1} \right]_{\mu\lambda} }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi^*_{\vec{k}}(i\omega_n)_2 \Psi_{-\vec{k'}}(-i\nu_m)_1 \rangle = \delta(\vec{k} + \vec{k'})\beta \delta(\omega_n + \nu_m)\frac{\Delta}{-\omega_n^2 - (\epsilon_{\vec{k}} - \mu)^2 - \Delta^* \Delta } }

Now, we can easily write down the correlator:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle \psi_{\downarrow}(\vec{r},\tau)\psi_{\uparrow}(\vec{r},\tau)\rangle = \frac{1}{\beta}\sum_{\omega_n}\frac{1}{V}\sum_{\vec{k}}\frac{\Delta}{-\omega_n^2 - (\epsilon_{\vec{k}} - \mu)^2 - \Delta^* \Delta } }

This reshapes our consistency equation to: 

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\Delta}{g} = \frac{1}{\beta}\sum_{\omega_n}\frac{1}{V}\sum_{\vec{k}}\frac{\Delta}{-\omega_n^2 - (\epsilon_{\vec{k}} - \mu)^2 - \Delta^* \Delta } }

Which has a trivial, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta = 0 } solution, and far more interesting solution where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{g} = \frac{1}{\beta}\sum_{\omega_n}\frac{1}{V}\sum_{\vec{k}}\frac{1}{-\omega_n^2 - (\epsilon_{\vec{k}} - \mu)^2 - |\Delta|^2 } }

Note that this is only possible if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g < 0 } !

-We will not get S-wave superconductivity for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta} independent of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k}

-In cuprates, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta(k)} changes sign with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 90^{0}} . It changes sign like D-wave.

-If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} is repulsive system does not like to make uniform gap!

-In Fe-Pnictides, pairing gaps on different Fermi surfaces have different signs, with repulsion.


Let us define Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{k}^{2} = (\epsilon_{k} - \mu)^{2} + |\Delta|^{2} }

So Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{-1}{g} = \frac{1}{\beta}\sum_{\omega_n}\frac{1}{V}\sum_{\vec{k}}\frac{1}{-\omega_n^2 - (\epsilon_{\vec{k}} - \mu)^2 - |\Delta|^2 } }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{-1}{g} = \frac{1}{\beta}\sum_{\omega_n}\sum_{\vec{k}}\frac{1}{-\omega_n^2 - E_{\vec{k}}^{2} } } Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \frac{1}{\beta}\sum_{\omega_{n}}\frac{1}{(i\omega_{n} - E_{\vec{k}})(i\omega_{n} + E_{\vec{k}})}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega_{n} = \frac{(2n + 1)\pi}{\beta}}

We can write this as a complex contour integral, with a series of pole along the imaginary axis.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{-1}{g} = \oint_{c} \frac{dz}{2\pi i}\frac{1}{e^{\beta z} + 1}f(z)}

since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{e^{\beta(\frac{\pi}{\beta}i + \delta z)} + 1} = \frac{1}{-e^{\beta \delta z +1}} = \frac{-1}{\beta\delta z}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{-1}{g}= \oint_{c}\frac{dz}{(2\pi i)}\frac{1}{e^{\beta z} + 1}\frac{1}{(z-E_{\vec{k}})(z+E_{\vec{k}})}}

There is no divergence as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z \to i\infty} There are two poles, at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z = \pm E_{\vec{k}}}

The contour can be deformed so that they exclude these two poles, and the integrand is tiny as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Re(z) \to \infty} The only contribution comes from the the loop around the two poles. For Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Re(z)\to -\infty} , the integrand falls off as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{z^{2}}} , which, by Jordan's lemma,is fast enough for convergence.

So the integrand becomes Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \frac{-1}{2\pi i}2\pi i \left [\frac{1}{(e^{\beta E_{\vec{k}}} +1)}\frac{1}{2E_{\vec{k}}} + \frac{1}{(e^{\beta E_{\vec{k}}}+1)}\frac{1}{-2E_{\vec{k}}} \right ]}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \frac{1}{2E_{\vec{k}}}\left [ \frac{1}{(e^{-\beta E_{\vec{k}}} +1)} - \frac{1}{(e^{\beta E_{\vec{k}}} +1)}\right ]}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{1}{2E_{\vec{k}}}\left [ \frac{1}{e^{-\beta E_{\vec{k}}/2}2\cosh(\beta E_{\vec{k}}/2)} - \frac{1}{e^{\beta E_{\vec{k}}/2}2\cosh(\beta E_{\vec{k}}/2)}\right ]}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \frac{1}{4E_{\vec{k}}\cosh(\beta E_{\vec{k}}/2)}\left [ e^{\beta E_{\vec{k}}/2}2 - e^{-\beta E_{\vec{k}}/2}2 \right ]}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \frac{1}{2E_{\vec{k}}}\tanh(\beta E_{\vec{k}}/2)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \frac{1}{g} = \frac{-1}{V}\sum_{\vec{k}}\frac{1}{2E_{\vec{k}}}\tanh(E_{\vec{k}}/2k_{b}T)}

Solution of the Self-Consistency Equation

By using our usual trick of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{\vec{k}} \to V\int \frac{d^{3}k}{(2\pi)^3}} ,

we can convert the sum into an integral: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{|g|} = -\frac{1}{g} = \int \frac{d^{3}k}{(2\pi)^3}\frac{\tanh(E_{\vec{k}}/2k_{b}T)}{2E_{\vec{k}}}}

At this point, we can, if we wish, solve this equation on a computer, at any temperature

For Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta \to 0} , the RHS is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\tanh((\epsilon_{\vec{k}}-\mu)/2k_{b}T)}{(\epsilon_{\vec{k}}-\mu)}} always intersects the LHS Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left (\frac{1}{|g|} \right )} , and we are guaranteed a solution.

As we raise temperature, log-divergence of integer gets cut-off. Finding T-dependence of the intercept gives T-dependence of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta}

There can be an analytic solution as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T \to 0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{|g|} = -\frac{1}{g} = \int \frac{d^{3}k}{(2\pi)^3}\frac{1}{2E_{\vec{k}}} = \int \frac{d^{3}k}{(2\pi)^3}\frac{1}{2\sqrt{(\epsilon_{\vec{k}} - \mu)^2 + |\Delta|^2}} }

Assuming that the density of states is smooth, and dominated by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \xi \rightarrow 0 } , we can write:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \simeq N(\mu) \int_{-\hbar \omega_D}^{\hbar \omega_D}{d\xi}\frac{1}{2\sqrt{\xi^2 + |\Delta|^2}} = N(\mu) \int_0^{\frac{\hbar \omega_D}{|\Delta|}}{d\xi}\frac{1}{\sqrt{\xi^2 + |\Delta|^2}} }

Now,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{|g|} = N(\mu) \ln{\left(\frac{\hbar \omega_D}{\Delta} + \sqrt{1 + \frac{\hbar^2\omega_d^2}{|\Delta|^2}}\right)} }

In typical systems, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta \ll \omega_D } ! For example, in Al, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Delta| \simeq 2K} while Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega_d \simeq 300-500K} . So, we can safely assume that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\hbar \omega_D}{|\Delta|} \gg 1 } , which gives that:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{|g|} = N(\mu) \ln{\left(\frac{h\hbar\omega_D} {\Delta} \right)} \Rightarrow |\Delta| = 2\hbar\omega_D e^{-\frac{1}{|g|N(\mu)}}} at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T = 0} .

This actually lowers condensation energy more than Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta = 0} ! It is also important to note, that since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |g|} enters in a exponential, that the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta} is completely flat as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |g| \rightarrow 0} , showing that this result is completely non-perturbative. Note that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta} is only exactly zero when Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |g|} is zero, so that no matter how small the interaction, we still have a superconducting state.

Deltavsg.jpeg


Happily, this result is also something we can compare to experimental results from tunneling measurements on suitable materials (S-wave superconductors). From the definition of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_c} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{|\Delta|(T = 0)}{k_B T_c} = \frac{2}{1.134} = 1.764 }

By convention, the BCS results for twice this is compared to experiment.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{2|\Delta|(T = 0)}{k_B T_c} = 3.53 }

Element Measured Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{2\Delta(0)}{k_B T_c}}
Al Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 4.2\pm0.6}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2.5 \pm0.3}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3.2\pm0.4 }
Cd Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ~3.2}
Nb Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ~3.84 }
Sn Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3.49\pm0.1}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3.1\pm0.05}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3.5\pm0.18}

Going between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T = 0} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_c}

At zero-Temperature, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta = 2\hbar\omega_{D}e^{\frac{1}{gN_{0}}}}

Now lets look at how it agrees with the Ginzburg-Landau theory near Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_{c}}

At finite T, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{g} = \int \frac{d^{3}k}{(2\pi)^3}\frac{\tanh(E_{\vec{k}}/2T)}{2E_{\vec{k}}}}

in units of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{\vec{k}} = \left [ \epsilon^{2} + \Delta^2(T) \right ]^\frac{1}{2}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \frac{1}{g} = N_{\mu}\int^{\omega_{D}}_{0}d\epsilon \frac{\tanh\left( \frac{\left [ (\epsilon^{2} + \Delta^2(T) \right ]^\frac{1}{2}}{2T}\right)}{ \left [ \epsilon^{2} + \Delta^2(T) \right ]^\frac{1}{2}}}

Now. let us parametrize Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta(T)} with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta(T = 0)}

(Note at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T =0, \frac{1}{g} = N_{\mu}\int^{\omega_{D}}_{0}d\epsilon \frac{1}{ \left [ \epsilon^{2} + \Delta^2 \right ]^\frac{1}{2}}} )

so Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N_{0}(\mu)\int_{0}^{\omega_{D}} d\epsilon \frac{1}{\sqrt{\epsilon^2 + \Delta^{2}(T=0)}}= N_{0}(\mu)\int_{0}^{\omega_{D}} d\epsilon \frac{tanh\left( \frac{\sqrt{(\epsilon^{2} + \Delta^2(T)}}{2T}\right)}{\sqrt{\epsilon^2 + \Delta^{2}(T)}}}

Or,Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{0}^{\omega_{D}\to \infty} d\epsilon \left (\frac{1}{\sqrt{\epsilon^2 + \Delta^{2}(T=0)}} - \frac{1}{\sqrt{\epsilon^2 + \Delta^{2}(T)}}\right ) = \int_{0}^{\omega_{D\to \infty}} d\epsilon \left (\frac{\tanh\left( \frac{\sqrt{(\epsilon^{2} + \Delta^2(T)}}{2T}\right)-1}{\sqrt{\epsilon^2 + \Delta^{2}(T)}} \right )}

Now Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega_{D} \sim 1000-3000 K} , which is very large compared to the energy scale we are working at, and so we can replace the upper limit in the integrals by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \infty} . This will cause the first term in the LHS to vanish, since it falls faster than Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\epsilon^{2}}} We get,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ln\left(\frac{\Delta(T)}{\Delta(0)}\right ) = \int_{0}^{\omega_{D\to \infty}} d\epsilon \left (\frac{\tanh\left( \frac{\sqrt{(\epsilon^{2} + \Delta^2(T)}}{2T}\right)-1}{\sqrt{\epsilon^2 + \Delta^{2}(T)}} \right )}

Since we expect Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta(T)<\Delta(0)} , we can switch their order inside the logarithm:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ln\left(\frac{\Delta(T=0)}{\Delta(T)}\right ) = \int_{0}^{\omega_{D\to \infty}} d\epsilon \left (\frac{1-\tanh\left( \frac{\sqrt{(\epsilon^{2} + \Delta^2(T)}}{2T}\right)}{\sqrt{\epsilon^2 + \Delta^{2}(T)}} \right )} We are now in a position to solve for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta(T)} near Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T = 0}

Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon/\Delta(T) = \xi}

So the RHS is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{0}^{\infty}d\xi \frac{1}{\sqrt{\xi^{2}+1}}\left(1-\tanh\left(\frac{\Delta(T)}{2T}\sqrt{\xi^{2}+1}\right)\right)}

Let us call Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{\xi^{2}+1} = X} . So

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ln\left(\frac{\Delta(T=0)}{\Delta(T)}\right ) = \int_{1}^{\infty}dX \frac{1}{\sqrt{X^{2}-1}}\left(1-\tanh\left(\frac{\Delta(T)}{2T}X\right)\right)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rightarrow \ln\left(\frac{\Delta(T=0)}{\Delta(T)}\right ) = \int_{1}^{\infty}dX \frac{1}{\sqrt{X^{2}-1}}\frac{2}{e^{\frac{\Delta(T)X}{T}+1}}}

Now as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T\to 0} , the integral clearly goes to zero

So the integral is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2\int_{1}^{\infty}dX \frac{1}{\sqrt{X^{2}-1}}\frac{e^{-\frac{\Delta(T)X}{T}}}{e^{-\frac{\Delta(T)X}{T}}+1}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = 2\int_{1}^{\infty}dX \frac{1}{\sqrt{X^{2}-1}}e^{-\frac{\Delta(T)X}{T}}\sum_{n=0}^{\infty}(-1)^n e^{-\frac{n\Delta(T)X}{T}}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = 2\sum_{n=0}^{\infty}(-1)^n\int_{1}^{\infty}dX \frac{1}{\sqrt{X^{2}-1}} e^{-\frac{\Delta(T)X(n+1)}{T}}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = 2\sum_{n=0}^{\infty}(-1)^n K_{0}\left({\frac{\Delta(T)X(n+1)}{T}}\right)}

Where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K_{0}} is the modified Bessel function of the 2nd kind.

As Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \to 0, K(x) \to \infty} , and as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \to \infty, K(x) \to 0}

For small temperatures, the sum is dominated by the smallest n value. So approximately,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ln\left(\frac{\Delta(T=0)}{\Delta(T)}\right ) 2\simeq \sqrt{\frac{\pi T}{2\Delta(0)}}e^{-\frac{\Delta(0)}{T}}}

By noting that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta(T)-\Delta(0)} is small, we can manipulate the logarithm to get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\ln\left(\frac{\Delta(T)-\Delta(0)+\Delta(0)}{\Delta(0)}\right )}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = -\ln\left(1 + \frac{\Delta(T)-\Delta(0)}{\Delta(0)}\right )}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \simeq -\frac{\Delta(T)-\Delta(0)}{\Delta(0)}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow 1- \frac{\Delta(T)}{\Delta(0)} \simeq 2\sqrt{\frac{\pi T}{2\Delta(0)}}e^{-\frac{\Delta(0)}{T}}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \frac{\Delta(T)}{\Delta(0)} \simeq 1- 2\sqrt{\frac{\pi T}{2\Delta(0)}}e^{-\frac{\Delta(0)}{T}}}

Next,let us solve near Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T = T_{c}} analytically

Near Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_{c}, \Delta(T) \to 0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{1}{gN_{0}} = \int_{0}^{\omega_{D}} d\epsilon \frac{\tanh\left(\frac{\sqrt{(\epsilon^{2} + \Delta^2(T)}}{2T}\right)}{\sqrt{\epsilon^2 + \Delta^{2}(T)}}}

At Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T = T_{c}. \Delta = 0} . So,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{1}{gN_{0}} = \int_{0}^{\omega_{D}} d\epsilon \frac{\tanh\left( \frac{\epsilon}{2T_{c}}\right)}{\epsilon}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \int_{0}^{\omega_{D}} d\epsilon \frac{\tanh\left( \frac{\epsilon}{2T_{c}}\right)}{\epsilon} = \int_{0}^{\omega_{D}} d\epsilon \frac{\tanh\left(\frac{\sqrt{(\epsilon^{2} + \Delta^2(T)}}{2T}\right)}{\sqrt{\epsilon^2 + \Delta^{2}(T)}}}

Let us subtract Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{0}^{\omega_{D}} d\epsilon \frac{\tanh\left( \frac{\epsilon}{2T}\right)}{\epsilon}} from both sides

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \int_{0}^{\omega_{D}} d\epsilon \frac{\tanh\left( \frac{\epsilon}{2T_{c}}\right)-\tanh\left( \frac{\epsilon}{2T_{c}}\right)}{\epsilon} = \int_{0}^{\omega_{D}} d\epsilon \left ( \frac{\tanh\left(\frac{\sqrt{(\epsilon^{2} + \Delta^2(T)}}{2T}\right)}{\sqrt{\epsilon^2 + \Delta^{2}(T)}} -\frac{\tanh\left( \frac{\epsilon}{2T}\right)}{\epsilon} \right ) }

Now if we let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega_{D} \to \infty} , the reason of which has already been argued, we can solve the LHS exactly, and we get,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ln \left (\frac{T}{T_{c}}\right)}

So we get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ln \left (\frac{T_{c}}{T}\right) = \int_{0}^{\infty} d\epsilon \left (\frac{\tanh\left( \frac{\epsilon}{2T}\right)}{\epsilon} -\frac{\tanh\left(\frac{\sqrt{(\epsilon^{2} + \Delta^2(T)}}{2T}\right)}{\sqrt{\epsilon^2 + \Delta^{2}(T)}}\right )}

The next logical step is to expand the hyperbolic tangent

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tanh(\pi x/2) = \frac{4x}{\pi}\sum_{k = 0}^{\infty}\frac{1}{(2k+1)^2+x^2} }

So RHS Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\int_{0}^{\infty} d\epsilon \left (\frac{4}{\pi^{2}T}\sum_{k = 0}^{\infty}\frac{1}{(2k+1)^2+\left ( \frac{\epsilon^2}{\pi^{2}T^{2}}\right)}-\frac{4}{\pi^{2}T}\sum_{k = 0}^{\infty}\frac{1}{(2k+1)^2+\left ( \frac{\epsilon^2}{\pi^{2}T^{2}}\right)} \right )}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\sum_{k = 0}^{\infty}\int_{0}^{\infty} d\epsilon\frac{4}{\pi^{2}T}\frac{\frac{\Delta^2}{\pi^{2}T{2}}}{\left ((2k+1)^2+\left ( \frac{\epsilon^2}{\pi^{2}T^{2}}\right)\right)^2} + O(\Delta^4)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \simeq \frac{4\Delta^{2}}{\pi^{3}T^{2}}\sum_{k = 0}^{\infty}\int_{0}^{\infty}d\epsilon\frac{1}{\left ((2k+1)^2 + \epsilon^{2} \right)}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{4\Delta^{2}}{\pi^{3}T^{2}}\sum_{k = 0}^{\infty}\frac{\pi}{4}\frac{1}{(2k +1)^{2}}}

This series converges, and we get,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ln \left (\frac{T_{c}}{T}\right) = \frac{\Delta^{2}(T)}{\pi^{2}T{2}}\frac{7\xi(3)}{8}}

Where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \xi(3)} is the order three Reimann Zeta function.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \frac{\Delta^{2}(T)}{\pi^{2}T{2}}\frac{7\xi(3)}{8}=-\ln \left (1+\frac{T-T_{c}}{T_{c}}\right) = \frac{T_{c}-T}{T{c}}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \Delta^{2}(T) \simeq \frac{8\pi^{2}T_{c}(T_{c}-T)}{7\xi(3)}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \Delta(T)\sim 3.063\sqrt{T_{c}(T_{c}-T)}}

Condensation energy and T dependence of the thermodynamic field

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z = \int D\psi^{*}D\psi\int D\Delta^{*}D\Delta e^{-S_{BCS}-S_{\Delta}}= \int D\Delta^{*}D\Delta e^{-S_{eff}[\Delta]}}

Using the saddle point approximation

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{-S_{eff}[\Delta _{0}\, ]}\; =\; e^{-S_{\Delta }}\int D\psi ^{\ast }D\psi \; e^{-S_{BCS}[\Delta _{0\, }]}\; \sim e^{-\beta F}}

Then,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{-\beta [F(\Delta _{0})-F(0)]}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{e^{-S_{\Delta}}\int D\psi ^{\ast }D\psi e^{-S_{BCS}[\Delta _{0}\, ]}}{\int D\psi ^{\ast }D\psi e^{-S_{BCS}[\, 0\, ]}}}

We recall that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{\Delta }=-\frac{1}{g}\int_{0}^{\beta }d\tau \sum_{k}\Delta _{k}^{\ast }(\tau )\Delta _{k}(\tau )}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow S_{\Delta }(\Delta _{0})=-\frac{\Delta _{0}^{2}}{g}\beta N}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle And \int D\psi ^{\ast }D\psi e^{-S_{BCS}[\Delta _{0}]}=\int D\psi ^{\ast }D\psi \, e^{-\frac{1}{\beta }}\sum_{\omega _{n}}\sum_{k}\psi _{k}^{\ast }(i\omega _{n})\begin{pmatrix} i\omega _{n}+\epsilon _{k}-\mu & \Delta _{0}\\\Delta _{0} & i\omega _{n}-\epsilon _{k}+\mu \end{pmatrix}\psi _{k}(i\omega _{n}) }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = const. \prod_{k}\prod_{n}\frac{1}{\beta ^{2}}\, \begin{vmatrix}i\omega _{n}+\epsilon _{k}-\mu & \Delta _{0}\\ \Delta _{0}& i\omega _{n}-\epsilon _{k}+\mu \end{vmatrix}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = const. \prod_{k}\prod_{n}\frac{1}{\beta ^{2}}[(i\omega _{n})^{2}-(\epsilon _{k}-\mu )^{2}-\Delta _{0}^{2}]}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow e^{-\beta [F(\Delta _{0})-F(0)]}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{e^{-\frac{\Delta _{0}^{2}}{\left | g \right |}\beta N}const.{\prod_{k}^{}} {\prod_{n}^{}}\frac{1}{\beta ^{2}}[(i\omega _{n})^{2}-(\epsilon _{k}-\mu )^{2}-\Delta _{0}^{2}]}{const.{\prod_{k}^{}}{ \prod_{n}^{}}\frac{1}{\beta ^{2}}[(i\omega _{n})^{2}-(\epsilon _{k}-\mu )^{2}]}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =e^{-\frac{\Delta _{0}^{2}}{\left | g \right |}\beta N}{\prod_{k}^{}}{\prod_{n}^{}}(1+\frac{\Delta _{0}^{2}}{(\omega _{n}^{2})+(\epsilon _{n}-\mu )^{2}})}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =e^{-\frac{\Delta _{0}^{2}}{\left | g \right |}\beta N}e^{\sum_{k}\sum_{n}ln(1+\frac{\Delta _{0}^{2}}{\omega_{n}^{2}+(\epsilon _{n}-\mu )^{2}})}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =e^{-\beta [F(\Delta _{0})-F(0)]}}

Then,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F(\Delta _{0})-F(0)=V\frac{\Delta _{0}}{\left | g \right |}-\frac{1}{\beta } {\sum_{\omega _{n}}^{}}{\sum_{k}^{}}ln (1+\frac{\Delta _{0}^{2}}{\omega _{n}^{2}+(\epsilon _{k}-\mu )^{2}})}

Here, the first term is positive whereas the second term should be negative.

In order to calculate Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F[\Delta ]} we note that:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F[\Delta ]=-\frac{1}{\beta }{\sum_{\omega _{n}}^{}}ln(1+\frac{\Delta ^{2}}{\omega _{n}^{2}+(\epsilon _{k}-\mu )^{2}})}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\mathrm{d} F[\Delta ]}{\mathrm{d} \Delta }=-\frac{1}{\beta }{\sum_{\omega _{n}}^{}}\frac{1}{(1+\frac{\Delta ^{2}}{\omega _{n}^{2}+(\epsilon _{k}-\mu )^{2}})}\times \frac{2\Delta }{\omega _{n}^{2}+(\epsilon _{k}-\mu )^{2}}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -(2\Delta )\frac{1}{\beta }{\sum_{\omega _{n}}^{}}\frac{1}{\omega _{n}^{2}+(\epsilon _{k}-\mu )^{2}+\Delta ^{2}}}

Choosing Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z=i\omega _{n}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\mathrm{d} F[\Delta ]}{\mathrm{d}\Delta }= -(2\Delta )\frac{1}{\beta }{\sum_{\omega _{n}}^{}}\frac{1}{E_{k}^{2}-z^{2}}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =(2\Delta ){\oint_{c}^{}}\frac{dz}{2\pi i}\frac{1}{e^{\beta z}+1}\frac{-1}{(z+E_{k})(z-E_{k})}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =(\Delta )(\frac{n_{F(E_{k})}}{E_{k}}+\frac{n_{F}(-E_{k})}{-E_{k}})}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =(\Delta )(\frac{n_{F(E_{k})}}{E_{k}}-\frac{(1-n_{F}(E_{k}))}{E_{k}})}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{\Delta }{E_{k}}(-1+2n_{F}(E_{k}))}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \int_{0}^{\Delta }d\Delta \frac{\mathrm{d}F }{\mathrm{d} \Delta }=\int d\Delta \frac{\Delta }{\sqrt{(\epsilon _{k}-\mu )^{2}+\Delta ^{2}}}(-1+2{\frac{1}{(e^{\frac{\sqrt{(\epsilon _{k}-\mu )^{2}+\Delta ^{2}}}{T}}+1)})}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\sqrt{(\epsilon _{k}-\mu )^{2}+\Delta ^{2}}\, +\, 4T\, ln\, [cosh\frac{(\epsilon _{k}-\mu )^{2}+\Delta ^{2}}{2T}]+const.}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \because F[0]= 0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \therefore F[\Delta ]=-E_{k}+\left | \epsilon _{k}-\mu \right |-4T\, ln\, [\frac{1+e^{-\frac{E_{k}}{T}}}{1+e^\frac{-(\epsilon _{k}-\mu )}{T}}]}

Using the self consistency equation:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\left | g \right |}=\int \frac{d^{3}k}{\left ( 2\pi \right )^{3}}\frac{tanh\frac{E_{k}}{2T}}{2E_{k}}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \therefore F[\Delta _{0}]-F[0]}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ={\sum_{k}^{}}(\frac{\Delta _{0}^{2}}{2E_{k}}-E_{k}+\left | \epsilon _{k}-\mu \right |-\frac{\Delta _{0}^{2}(1-tanh\frac{E_{k}}{2T})}{2E_{k}})-4T\, ln\, (\frac{1+e^{-\frac{E_{k}}{T}}}{1+e^{-\frac{(\epsilon _{k}-\mu )}{T}}}) }

When T= 0, only the 1st term survives.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow F[\Delta _{0}]-F[0]=-V\frac{N_{0}\Delta ^{2}}{2}}

Beyond saddle-point approximation, collective modes and response in the broken symmetry state

Recall we can write our partition function as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z=\int D\Delta^*D\Delta e^{-S_{eff}[\Delta]}}

or

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z=\int D(\Re e\Delta)D(\Im m\Delta)e^{-S_{eff}[\Re e\Delta,\Im m\Delta]}}

where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{eff}[\Delta]=\frac{1}{|g|}\int_0^\beta d\tau\int d^Dr\left[(\Re e\Delta(r,\tau))^2+(\Im m\Delta(r,\tau))^2\right]-\ln\left[\int D\psi^*D\psi e^{-S_0-S_{int}}\right]}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} S_0&=\int_0^\beta d\tau\int d^Dr \psi^*(r,\tau)(\frac{\partial}{\partial\tau}+\epsilon_p-\mu)\psi(r,\tau)\\ S_{int}&=\int_0^\beta d\tau\int d^Dr(\Delta^*(r,\tau)\psi_\downarrow(r,\tau)\psi_{\uparrow}(r,\tau)+\Delta(r,\tau)\psi^*_\uparrow(r,\tau)\psi^*_{\downarrow}(r,\tau)) \end{align}}

We can rewrite the interaction term in the action as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{int}=\int_0^\beta d\tau\int d^Dr\left[\Re e(\Delta(r,\tau))(\psi_{\downarrow}\psi_{\uparrow}+\psi_{\uparrow}^*\psi_{\downarrow}^*)(r,\tau)+i*\Im m(\Delta(r,\tau))(\psi_{\uparrow}^*\psi_{\downarrow}^*-\psi_{\downarrow}\psi_{\uparrow})(r,\tau)\right]}

Consider now functional derivatives of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{eff}[\Delta]} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\delta S_{eff}[\Delta]}{\delta\Re e\Delta(r,\tau)}=\frac{2}{|g|}\Re e\Delta(r,\tau)-\frac{1}{\int D\psi^*D\psi e^{-S_0-S_{int}}} \int D\psi^*D\psi e^{-S_0-S_{int}}\left(-\frac{\delta S_{int}}{\delta \Re e\Delta(r,\tau)}\right) }

The functional derivative of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{int}} with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Re e\Delta(r,\tau)} is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\delta S_{int}}{\delta \Re e\Delta(r,\tau)}=\psi_{\downarrow}(r,\tau)\psi_{\uparrow}(r,\tau)+\psi_{\uparrow}^*(r,\tau)\psi_{\downarrow}^*(r,\tau)}

Therefore, we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\delta S_{eff}[\Delta]}{\delta \Re e\Delta(r,\tau)}=\frac{2}{|g|}\Re e\Delta(r,\tau)+\langle\psi_{\downarrow}(r,\tau)\psi_{\uparrow}(r,\tau)+\psi_{\uparrow}^*(r,\tau)\psi_{\downarrow}^*(r,\tau)\rangle\ \ (1)}

Similarly

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\delta S_{eff}[\Delta]}{\delta \Im m\Delta(r,\tau)}=\frac{2}{|g|}\Im m\Delta(r,\tau)+i\langle\psi_{\uparrow}^*(r,\tau)\psi_{\downarrow}^*(r,\tau)-\psi_{\downarrow}(r,\tau)\psi_{\uparrow}(r,\tau)\rangle\ \ (2)}

If we were to set the LHS of the above two equations to zero, we would obtain our self-consistency conditions.

The strategy is to take

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{eff}[\Delta(r,\tau)]=S_{eff}\left[\Delta_{sp}+(\Delta(r,\tau)-\Delta_{sp})\right]}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta_{sp}} solves (1) and (2) with LHS set to zero and expand in powers of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta(r,\tau)-\Delta_{sp}} . So

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} &S_{eff}[\Delta_{sp}+(\Delta(r,\tau)-\Delta_{sp})]\\ &=S_{eff}[\Delta_{sp}]+\int_0^\beta d\tau\int d^3r\Re e(\Delta(r,\tau)-\Delta_{sp})\frac{\delta S_{eff}}{\delta\Re e\Delta(r,\tau)}|_{\Delta_{sp}}+\int_0^\beta d\tau\int d^3r\Im m(\Delta(r,\tau)-\Delta_{sp})\frac{\delta S_{eff}}{\delta\Im m\Delta(r,\tau)}|_{\Delta_{sp}}\\ &+\frac{1}{2}\int_0^\beta d\tau\int_0^\beta d\tau'\int d^3r\int d^3r'\Re e(\Delta(r,\tau)-\Delta_{sp})\Re e(\Delta(r',\tau')-\Delta_{sp})\frac{\delta^2 S_{eff}}{\delta\Re e\Delta(r,\tau)\delta\Re e\Delta(r',\tau')}|_{\Delta_{sp}}\\ &+\frac{1}{2}\int_0^\beta d\tau\int_0^\beta d\tau'\int d^3r\int d^3r'\Im m(\Delta(r,\tau)-\Delta_{sp})\Im m(\Delta(r',\tau')-\Delta_{sp})\frac{\delta^2 S_{eff}}{\delta\Im m\Delta(r,\tau)\delta\Im m\Delta(r',\tau')}|_{\Delta_{sp}}\\ &+\int_0^\beta d\tau\int_0^\beta d\tau'\int d^3r\int d^3r'\Re e(\Delta(r,\tau)-\Delta_{sp})\Im m(\Delta(r',\tau')-\Delta_{sp})\frac{\delta^2 S_{eff}}{\delta\Re e\Delta(r,\tau)\delta\Im m\Delta(r',\tau')}|_{\Delta_{sp}}+... \end{align}}

Since by definition of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta_{sp}} we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\delta S_{eff}}{\delta\Re e\Delta(r,\tau)}|_{\Delta_{sp}}=\frac{\delta S_{eff}}{\delta\Im m\Delta(r,\tau)}|_{\Delta_{sp}}=0}

So only the 0th and 2nd order terms contribute. The 0th order term gave us the condensation energy, and the 2nd order term will give us information about collective modes (in the broken symmetry phase).

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} &\frac{\delta^2 S_{eff}}{\delta\Re e\Delta(r,\tau)\delta\Re e\Delta(r',\tau')}\\ &=\frac{2}{|g|}\delta(\tau-\tau')\delta(r-r')+\frac{\delta}{\delta\Re e\Delta(r,\tau)}\left[\frac{1}{\int D\psi^*D\psi e^{-S_0-S_{int}}}\int D\psi^*D\psi e^{-S_0-S_{int}}\left(\frac{\delta S_{int}}{\delta \Re e\Delta(r',\tau')}\right)\right]\\ &=\frac{2}{|g|}\delta(\tau-\tau')\delta(r-r')-\frac{\int D\psi^*D\psi e^{-S_0-S_{int}}\left(\frac{\delta S_{int}}{\delta \Re e\Delta(r,\tau)}\right)\left(\frac{\delta S_{int}}{\delta \Re e\Delta(r',\tau')}\right)}{\int D\psi^*D\psi e^{-S_0-S_{int}}}\\ &+\left(\frac{\int D\psi^*D\psi e^{-S_0-S_{int}}\left(\frac{\delta S_{int}}{\delta \Re e\Delta(r,\tau)}\right)}{\int D\psi^*D\psi e^{-S_0-S_{int}}}\right)\left(\frac{\int D\psi^*D\psi e^{-S_0-S_{int}}\left(\frac{\delta S_{int}}{\delta \Re e\Delta(r',\tau')}\right)}{\int D\psi^*D\psi e^{-S_0-S_{int}}}\right) \end{align}}

So

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} &\frac{\delta^2 S_{eff}}{\delta\Re e\Delta(r,\tau)\delta\Re e\Delta(r',\tau')}=\frac{2}{|g|}\delta(\tau-\tau')\delta(r-r')\\ &-\langle\left(\psi_{\downarrow}(r,\tau)\psi_{\uparrow}(r,\tau)+\psi^*_{\uparrow}(r,\tau)\psi^*_{\downarrow}(r,\tau)\right)\left(\psi_{\downarrow}(r',\tau')\psi_{\uparrow}(r',\tau')+\psi^*_{\uparrow}(r',\tau')\psi^*_{\downarrow}(r',\tau')\right)\rangle_{S_0+S_{int}}\\ &+\langle\psi_{\downarrow}(r,\tau)\psi_{\uparrow}(r,\tau)+\psi^*_{\uparrow}(r,\tau)\psi^*_{\downarrow}(r,\tau)\rangle\langle\psi_{\downarrow}(r',\tau')\psi_{\uparrow}(r',\tau')+\psi^*_{\uparrow}(r',\tau')\psi^*_{\downarrow}(r',\tau')\rangle_{S_0+S_{int}} \end{align}}

i.e.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} &\frac{\delta^2 S_{eff}}{\delta\Re e\Delta(r,\tau)\delta\Re e\Delta(r',\tau')}=\frac{2}{|g|}\delta(\tau-\tau')\delta(r-r')\\ &-\langle\left(\psi_{\downarrow}(r,\tau)\psi_{\uparrow}(r,\tau)+\psi^*_{\uparrow}(r,\tau)\psi^*_{\downarrow}(r,\tau)\right)\left(\psi_{\downarrow}(r',\tau')\psi_{\uparrow}(r',\tau')+\psi^*_{\uparrow}(r',\tau')\psi^*_{\downarrow}(r',\tau')\right)\rangle_{S_0+S_{int}}^{\mbox{con}} \end{align}}

Similarly,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} &\frac{\delta^2 S_{eff}}{\delta\Im m\Delta(r,\tau)\delta\Im m\Delta(r',\tau')}=\frac{2}{|g|}\delta(\tau-\tau')\delta(r-r')\\ &+\langle\left(\psi^*_{\uparrow}(r,\tau)\psi^*_{\downarrow}(r,\tau)-\psi_{\downarrow}(r,\tau)\psi_{\uparrow}(r,\tau)\right)\left(\psi^*_{\uparrow}(r',\tau')\psi^*_{\downarrow}(r',\tau')-\psi_{\downarrow}(r',\tau')\psi_{\uparrow}(r',\tau')\right)\rangle_{S_0+S_{int}}^{\mbox{con}} \end{align}}

And

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\delta^2 S_{eff}}{\delta\Im m\Delta(r,\tau)\delta\Re e\Delta(r',\tau')}=-i\langle\left(\psi^*_{\uparrow}(r,\tau)\psi^*_{\downarrow}(r,\tau)-\psi_{\downarrow}(r,\tau)\psi_{\uparrow}(r,\tau)\right)\left(\psi_{\downarrow}(r',\tau')\psi_{\uparrow}(r',\tau')+\psi^*_{\uparrow}(r',\tau')\psi^*_{\downarrow}(r',\tau')\right)\rangle_{S_0+S_{int}}^{\mbox{con}} }

If we evaluate these functional derivatives at the saddle point Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta_{sp}} we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} S_0+S_{int}&\rightarrow\int_0^\beta d\tau\int d^3r\sum_{\sigma=\uparrow,\downarrow}\psi^*_\sigma(r,\tau)\left(\frac{\partial}{\partial\tau}+\epsilon_p-\mu\right)\psi_{\sigma}(r,\tau)\\ &+\Delta_{sp}\int_0^\beta d\tau\int d^3r\left(\psi_{\downarrow}(r,\tau)\psi_{\uparrow}(r,\tau)+\psi_{\uparrow}^*(r,\tau)\psi_{\downarrow}^*(r,\tau)\right) \end{align}}

where we take the saddle point solution Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta_{sp}} to be purely real. Arranging the Grassman fields into the Nambu spinor we have:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\psi^*_{\uparrow}\ \psi_\downarrow)\left(\begin{align}&\frac{\partial}{\partial\tau}+\epsilon_p-\mu&\Delta_{sp}\\ &\Delta_{sp}&\frac{\partial}{\partial\tau}-\epsilon_p+\mu\end{align}\right) \left(\begin{align}&\psi_\uparrow\\&\psi^*_\downarrow\end{align}\right)}

where the Nambu spinor is defined as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi=\left(\begin{align}&\psi_\uparrow\\&\psi^*_\downarrow\end{align}\right)}

So

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} &\psi^*_\uparrow\psi^*_\downarrow+\psi_\downarrow\psi_\uparrow=\Psi^*\left(\begin{align}&0&1\\&1&0\end{align}\right)\Psi =\Psi^*\sigma_x\Psi\\ &\psi^*_\uparrow\psi^*_\downarrow-\psi_\downarrow\psi_\uparrow=\Psi^*\left(\begin{align}&0&1\\-&1&0\end{align}\right)\Psi =\Psi^*i\sigma_y\Psi \end{align}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{\delta^2S_{eff}}{\delta\Re e\Delta(r,\tau)\delta\Re e\Delta(r',\tau')}|_{\Delta_{sp}}&=\frac{2}{|g|}\delta(\tau-\tau')\delta(r-r')-\langle\Psi^*(r,\tau)\sigma_x\Psi(r,\tau)\Psi^*(r',\tau')\sigma_x\Psi(r',\tau')\rangle^{\mbox{con}}\\ &=\frac{2}{|g|}\delta(\tau-\tau')\delta(r-r')+\mbox{Tr}\left[\sigma_xG(r-r',\tau-\tau')\sigma_xG(r'-r,\tau'-\tau)\right] \end{align}}

where the Green's functions are 2*2 matrices,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G^{-1}_k(i\omega_n)=\left(\begin{align}&-i\omega_n+\epsilon_k-\mu&\Delta_{sp}\\&\Delta_{sp}&-i\omega_n-\epsilon_k+\mu\end{align}\right)}

Notice that this is a function of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tau-\tau'} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r-r'} . Let's call it Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Pi_{++}(r-r',\tau-\tau')} . This will give rise to

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} &\frac{1}{2}\int_0^\beta d\tau\int_0^\beta d\tau'\int d^3r\int d^3r'\Re e(\Delta(r,\tau)-\Delta_{sp})\Pi_{++}(r-r',\tau-\tau')\Re e(\Delta(r',\tau')-\Delta_{sp})\\ &=\frac{1}{2}\int_0^\beta d\tau\int_0^\beta d\tau'\int d^3r\int d^3r'\delta\Delta_+(r,\tau)\Pi_{++}(r-r',\tau-\tau')\delta\Delta_+(r',\tau') \end{align}}

Fourier transforming we find the contribution to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{eff}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2}\frac{1}{\beta}\sum_{\Omega_n}\sum_{q}\delta\Delta_+(-q,-i\Omega_n)\Pi_{++}(q,i\Omega_n)\delta\Delta_+(q,i\Omega_n)}

Similarly

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \Pi_{--}(r-r',\tau-\tau')&=\frac{2}{|g|}\delta(\tau-\tau')\delta(r-r')+\mbox{Tr}\left[\sigma_yG(r-r',\tau-\tau')\sigma_yG(r'-r,\tau'-\tau)\right]\\ \Pi_{-+}(r-r',\tau-\tau')&=\langle\Psi^*\sigma_y\Psi(r,\tau)\Psi^*\sigma_x\Psi(r',\tau')\rangle^{\mbox{con}}=-\mbox{Tr}\left[\sigma_yG(r-r',\tau-\tau')\sigma_xG(r'-r,\tau'-\tau)\right] \end{align}}

Then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} &S_{eff}[\Delta]\approx S_{eff}[\Delta_{sp}]+\\ &\frac{1}{2}\frac{1}{\beta}\sum_{\Omega_n}\sum_q\left[\delta\Delta_+(-q,-i\Omega_n)\Pi_{++}(q,i\Omega_n)\delta\Delta_+(q,i\Omega_n)+\delta\Delta_-(-q,-i\Omega_n)\Pi_{--}(q,i\Omega_n)\delta\Delta_-(q,i\Omega_n)\right.\\ &\left.+2\delta\Delta_-(-q,-i\Omega_n)\Pi_{-+}(q,i\Omega_n)\delta\Delta_+(q,i\Omega_n)\right]+... \end{align}}

To proceed with the evaluation of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Pi} 's, note that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} &G^{-1}_k(i\omega_n)=-i\omega_n\mathbb{I}+(\epsilon_k-\mu)\sigma_3+\Delta_{sp}\sigma_1\\ \Rightarrow&G_k(i\omega_n)=\frac{i\omega_n\mathbb{I}+(\epsilon_k-\mu)\sigma_3+\Delta_{sp}\sigma_1}{\omega_n^2+(\epsilon_k-\mu)^2+\Delta_{sp}^2} \end{align}}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega_n=\frac{(2n+1)\pi}{\beta}} . After Fourier transform,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \Pi_{++}(q,i\Omega_n)&=\frac{2}{|g|}+\frac{1}{\beta}\sum_{\omega_n}\int\frac{d^3k}{(2\pi)^3}\mbox{Tr}\left[\sigma_1G_{k+q}(i\omega_n+i\Omega_n)\sigma_1G_k(i\omega_n)\right]\\ \Pi_{--}(q,i\Omega_n)&=\frac{2}{|g|}+\frac{1}{\beta}\sum_{\omega_n}\int\frac{d^3k}{(2\pi)^3}\mbox{Tr}\left[\sigma_2G_{k+q}(i\omega_n+i\Omega_n)\sigma_2G_k(i\omega_n)\right]\\ \Pi_{-+}(q,i\Omega_n)&=-\frac{1}{\beta}\sum_{\omega_n}\int\frac{d^3k}{(2\pi)^3}\mbox{Tr}\left[\sigma_2G_{k+q}(i\omega_n+i\Omega_n)\sigma_1G_k(i\omega_n)\right]\\ \end{align}}

Consider now

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} &\mbox{Tr}\left[\sigma_{1,2}\frac{(i\omega_n+i\Omega_n)\mathbb{I}+(\epsilon_{k+q}-\mu)\sigma_3+\Delta_{sp}\sigma_1}{(\omega_n+\Omega_n)^2+(\epsilon_{k+q}-\mu)^2+\Delta_{sp}^2}\sigma_{1,2}\frac{i\omega_n\mathbb{I}+(\epsilon_k-\mu)\sigma_3+\Delta_{sp}\sigma_1}{\omega_n^2+(\epsilon_k-\mu)^2+\Delta_{sp}^2}\right]\\ =&2\frac{(i\omega_n+i\Omega_n)i\omega_n-(\epsilon_{k+q}-\mu)(\epsilon_k-\mu)\pm\Delta^2_{sp}}{\left[(\omega_n+\Omega_n)^2+(\epsilon_{k+q}-\mu)^2+\Delta_{sp}^2\right]\left[\omega_n^2+(\epsilon_k-\mu)^2+\Delta_{sp}^2\right]} \end{align}}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} &\mbox{Tr}\left[\sigma_{2}\frac{(i\omega_n+i\Omega_n)\mathbb{I}+(\epsilon_{k+q}-\mu)\sigma_3+\Delta_{sp}\sigma_1}{(\omega_n+\Omega_n)^2+(\epsilon_{k+q}-\mu)^2+\Delta_{sp}^2}\sigma_{1}\frac{i\omega_n\mathbb{I}+(\epsilon_k-\mu)\sigma_3+\Delta_{sp}\sigma_1}{\omega_n^2+(\epsilon_k-\mu)^2+\Delta_{sp}^2}\right]\\ =&\frac{(i\omega_n+i\Omega_n)(\epsilon_k-\mu)\mbox{Tr}[\sigma_2\sigma_1\sigma_3]+i\omega_n(\epsilon_{k+q}-\mu)\mbox{Tr}[\sigma_2\sigma_3\sigma_1]}{\left[(\omega_n+\Omega_n)^2+(\epsilon_{k+q}-\mu)^2+\Delta_{sp}^2\right]\left[\omega_n^2+(\epsilon_k-\mu)^2+\Delta_{sp}^2\right]}\\ =&\frac{(i\omega_n+i\Omega_n)(\epsilon_k-\mu)(-2i)+i\omega_n(\epsilon_{k+q}-\mu)2i}{\left[(\omega_n+\Omega_n)^2+(\epsilon_{k+q}-\mu)^2+\Delta_{sp}^2\right]\left[\omega_n^2+(\epsilon_k-\mu)^2+\Delta_{sp}^2\right]} \end{align}}

Now, note that at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q=0,\Omega_n=0} ,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Pi_{-+}(0,0)=0}

because the numerator vanishes for all Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega_n} . Also note that by self-consistency condition

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \Pi_{--}(0,0)&=\frac{2}{|g|}-\frac{1}{\beta}\sum_{\omega_n}\int\frac{d^3k}{(2\pi)^3}2\frac{1}{[\omega_n^2+(\epsilon_k-\mu)^2+\Delta_{sp}^2]}\\ &=\frac{2}{|g|}+2\int\frac{d^3k}{(2\pi)^3}\frac{1}{\beta}\sum_{\omega_n}\frac{1}{(i\omega_n-E_k)(i\omega_n+E_k)}\\ &=\frac{2}{|g|}+2\int\frac{d^3k}{(2\pi)^3}\left(\frac{n_F(E_k)}{2E_k}-\frac{n_F(-E_k)}{2E_k}\right)\\ &=2\left(\frac{1}{|g|}-\int\frac{d^3k}{(2\pi)^3}\frac{1-2n_F(E_k)}{2E_k}\right)\\ &=0 \end{align}}

However,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \Pi_{++}(0,0)&=\frac{2}{|g|}+\frac{1}{\beta}\sum_{\omega_n}\int\frac{d^3k}{(2\pi)^3}2\frac{-\omega_n^2-(\epsilon_k-\mu)^2+\Delta_{sp}^2}{[\omega_n^2+(\epsilon_k-\mu)^2+\Delta_{sp}^2]^2}\\ &=4\Delta_{sp}^2\frac{1}{\beta}\sum_{\omega_n}\int\frac{d^3k}{(2\pi)^3}\frac{1}{[\omega_n^2+(\epsilon_k-\mu)^2+\Delta_{sp}^2]^2}\\ &=4\Delta_{sp}^2N_0\frac{1}{\beta}\sum_{\omega_n}\int_{-\infty}^\infty\frac{d\xi}{[\xi^2+\omega_n^2+\Delta_{sp}^2]^2}\\ &=4\Delta_{sp}^2N_0\frac{1}{\beta}\sum_{\omega_n}\frac{\pi}{2}\frac{1}{(\omega_n^2+\Delta_{sp}^2)^{\frac{3}{2}}}\\ &=2\pi N_0\Delta_{sp}^2\frac{1}{\pi^3T^2}\sum_{n=-\infty}^\infty\frac{1}{[(2n+1)^2+(\frac{\Delta_{sp}}{\pi T})^2]^{\frac{3}{2}}} \end{align}}

This sum is slowly convergent. To evaluate it efficiently, we note that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{a^{\frac{3}{2}}}=\frac{2}{\sqrt{\pi}}\int_0^\infty d\lambda\sqrt{\lambda}e^{-\lambda a}\ \ \mbox{for }a>0}

So

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \sum_{n=-\infty}^\infty\frac{1}{[(2n+1)^2+(\frac{\Delta_{sp}}{\pi T})^2]^\frac{3}{2}}&=\frac{2}{\sqrt{\pi}} \int_0^\infty d\lambda\sqrt{\lambda}e^{-\lambda(\frac{\Delta_{sp}}{\pi T})^2}\left(\sum_{n=-\infty}^\infty e^{-\lambda(2n+1)^2}\right)\\ &=\frac{2}{\sqrt{\pi}} \int_0^\infty d\lambda\sqrt{\lambda}e^{-\lambda(\frac{\Delta_{sp}}{\pi T})^2}\Theta_2(0,e^{-4\lambda})\\ &=\frac{1}{4\sqrt{\pi}} \int_0^\infty d\xi\sqrt{\xi}e^{-\xi(\frac{\Delta_{sp}}{2\pi T})^2}\Theta_2(0,e^{-\xi}) \end{align}}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Theta_2} is Jacobi elliptic theta function. Then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \Pi_{++}(0,0)&=2\pi N_0\Delta^2_{sp}\frac{1}{\pi^3T^2}\sum_{n=-\infty}^\infty\frac{1}{[(2n+1)^2+(\frac{\Delta_{sp}}{\pi T})^2]^\frac{3}{2}}\\ &=2N_0\Phi(\frac{\Delta_{sp}}{2\pi T}) \end{align}}

where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi(x)=x^2\frac{1}{\sqrt{\pi}}\int_0^\infty d\xi\sqrt{\xi}e^{-\xi x^2}\Theta_2(0,e^{-\xi})}

Phi.jpg

Note that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Pi_{++}(0,0)\rightarrow 2N_0\frac{7\zeta(3)}{4\pi^2T^2}\Delta^2_{sp}}

as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta_{sp}\rightarrow 0} , i.e. as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T\rightarrow T_C} .

That is precisely the curvature of the new minimum in the Ginzburg-Landau free energy we found before. So, at 2nd order our effective action corresponds to the action for two real free bosons, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta\Delta_+} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta\Delta_-} . At Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q=0, i\Omega_n=0} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta\Delta_+} mode is gapped (massive), but Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta\Delta_-} mode is not gapped (massless). Physically, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta\Delta_+} corresponds to the fluctuations of the order parameter amplitude (because we chose Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta_{sp}} to be real).

Recall our discussion from many-body course:

Amplitude fluctuations are not hydrodynamic modes since they do not correspond to either conserved or to broken symmetry variable. We have to extend our approach to higher order in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta\Delta_\pm} to describe its (rapid) decay.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta\Delta_-} corresponds to fluctuations along the direction of the minimum of the double well potential, where there is no barrier. It corresponds to (part of) a "phase" mode.

To determine the kinematics (of our collective modes) we need to expand Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Pi_{\mu\nu}(q, i\Omega_n)} in powers of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{q}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Omega_n} . Our small expansion parameters are

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Omega_n \ll \Delta_{sp}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q \ll \frac{\Delta_{sp}}{\hbar v_F}}

(Obviously Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q \ll k_F} )

Start with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Pi_{++}}  : What we need is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Pi_{++}\left(q, i\Omega_n\right) - \Pi_{++}\left(0, 0\right)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \frac{2}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\left[\frac{1}{\omega_n^2 + E_k^2} - \frac{\omega_n\left(\omega_n + \Omega_n\right) + \left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \mu\right) - \Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)\left(\left(\omega_n + \Omega_n\right)^2 + E_{k + q}^2\right)} - \frac{2\Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)^2}\right]}}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \frac{2}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\left[\frac{\omega_n^2 + \left(\epsilon_k - \mu\right)^2 - \Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)^2} - \frac{\omega_n\left(\omega_n + \Omega_n\right) + \left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \mu\right) - \Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)\left(\left(\omega_n + \Omega_n\right)^2 + E_{k + q}^2\right)}\right]}}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \frac{2}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\left[\frac{\omega_n^2 + E_k^2 - 2\Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)^2} - \frac{\omega_n^2 + E_k^2 + \omega_n\Omega_n + \left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \epsilon_k\right) - 2\Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)\left(\omega_n^2 + E_k^2 + 2\omega_n\Omega_n + \Omega_n^2 + \left(\epsilon_{k + q} - \mu\right)^2 - \left(\epsilon_k - \mu\right)\right)}\right]}}}

The second term in the brackets:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\omega_n^2 + E_k^2 - 2\Delta_{sp}^2 + \overbrace{\omega_n\Omega_n + \left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \epsilon_k\right)}^{\text{small}}}{\left(\omega_n^2 + E_k^2\right)\left(\omega_n^2 + E_k^2 + \underbrace{2\omega_n\Omega_n + \Omega_n^2 + \left(\epsilon_{k + q} - \mu\right)^2 - \left(\epsilon_k - \mu\right)}_{\text{small}}\right)}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \frac{\omega_n^2 + E_k^2 - 2\Delta_{sp}^2 + \omega_n\Omega_n + \left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \epsilon_k\right)}{\left(\omega_n^2 + E_k^2\right)^2}\left[1 + \frac{2\omega_n\Omega_n + \Omega_n^2 + \left(\epsilon_{k + q} - \epsilon_k\right)\left(\epsilon_{k + q} + \epsilon_k - 2\mu\right)}{\omega_n^2 + E_k^2}\right]^{-1}}

Now, by Taylor expansion,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_{k + q} \simeq \epsilon_k + q\cdot \nabla\epsilon_k + \ldots}

and near the Fermi level we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q\cdot \nabla\epsilon_k \simeq q\cdot v_F}

Also note that after integrating over Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{k}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega_n} , terms with odd power of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega_n} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q\cdot v_F} vanish. Keeping only terms of order Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q^2} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Omega_n^2} we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\omega_n^2 + E_k^2 - 2\Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)^2}\left(1 - \frac{\Omega_n^2 + \left(q\cdot v_F\right)^2}{\omega_n^2 + E_k^2}\right) - \frac{2\omega_n^2\Omega_n^2 + 2\left(\epsilon_k - \mu\right)^2\left(q\cdot v_F\right)^2}{\left(\omega_n^2 + E_k^2\right)^3} + \frac{\omega_n^2 + E_k^2 - 2\Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)^2}\cdot\frac{4\omega_n^2\Omega_n^2 + 4\left(\epsilon_k - \mu\right)^2\left(q\cdot v_F\right)^2}{\left(\omega_n^2 + E_k^2\right)^2}}

The terms which we dropped are down by powers of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q/k_F} and/or vanish assuming particle-hole symmetry.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \Pi_{++}\left(q, i\Omega_n\right) - \Pi_{++}\left(0, 0\right) \simeq & \quad \frac{2}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\frac{\omega_n^2 + E_k^2 - 2\Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)^3}\left(\Omega_n^2 + \left(q\cdot v_F\right)^2\right)}} \\ & + \frac{2}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\frac{2\omega_n^2\Omega_n^2 + 2\left(\epsilon_k - \mu\right)^2\left(q\cdot v_F\right)^2}{\left(\omega_n^2 + E_k^2\right)^3}}} \\ & - \frac{2}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3} 4 \cdot \frac{\omega_n^2 + E_k^2 - 2\Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)^4}\left(\omega_n^2\Omega_n^2 + \left(\epsilon_k - \mu\right)^2\left(q\cdot v_F\right)^2\right)}} \\ \end{align} }

Let's focus on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T = 0} . Then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\beta}\sum_{\omega_n} \longrightarrow \int_{-\infty}^{\infty}{\frac{d\omega}{2\pi}}} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Longrightarrow}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \Pi_{++}\left(q, i\Omega_n\right) - \Pi_{++}\left(0, 0\right) \simeq &-2N_0\int_{-\infty}^{\infty}{\frac{d\omega}{2\pi}\int_{-\infty}^{\infty}{d\xi\frac{\omega^2 + \xi^2 - \Delta_{sp}^2}{\left(\omega^2 + \xi^2 + \Delta_{sp}^2\right)^3}\left(\Omega_n^2 + q^2\langle \left(\hat{q}\cdot v_F\right)^2 \rangle_{FS}\right)}} \\ &+ 4N_0\int_{-\infty}^{\infty}{\frac{d\omega}{2\pi}\int_{-\infty}^{\infty}{d\xi\frac{\omega^2\Omega_n^2 + \xi^2q^2\langle\left(\hat{q}\cdot v_F\right)^2\rangle_{FS}}{\left(\omega^2 + \xi^2 + \Delta_{sp}^2\right)^3}}} \\ &- 8N_0\int_{-\infty}^{\infty}{\frac{d\omega}{2\pi}\int_{-\infty}^{\infty}{d\xi\frac{\omega^2 + \xi^2 - \Delta_{sp}^2}{\left(\omega^2 + \xi^2 + \Delta_{sp}^2\right)^4}\left(\omega^2\Omega_n^2 + \xi^2q^2\langle \left(\hat{q}\cdot v_F\right)^2 \rangle_{FS}\right)}} \\ \end{align}}

So at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T = 0}  :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \Pi_{++}\left(q, i\Omega_n\right) - \Pi_{++}\left(0, 0\right) \simeq & \quad 0 \\ &+ \frac{N_0}{2\Delta_{sp}^2}\left(\Omega_n^2 + q^2\langle \left(\hat{q}\cdot v_F\right)^2 \rangle_{FS}\right) \\ &- \frac{N_0}{3\Delta_{sp}^2}\left(\Omega_n^2 + q^2\langle \left(\hat{q}\cdot v_F\right)^2 \rangle_{FS}\right) \\ \end{align}}

in 3D and for a spherical FS

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \langle \left(\hat{q}\cdot v_F\right)^2 \rangle_{FS} &= \frac{v_F^2}{2}\int_{-1}^{1}{d\left(\cos\theta\right)\cos^2\theta} \\ &= \frac{v_F^2}{2}\frac{1}{3}\left(1 - \left(-1\right)\right) \\ &= \frac{v_F^2}{3} \\ \end{align}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Longrightarrow} 

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Pi_{++}\left(q, i\Omega_n\right) - \Pi_{++}\left(0, 0\right) \simeq \frac{N_0}{6\Delta_{sp}^2}\left(\Delta_n^2 + \frac{1}{3}v_F^2q^2\right) + \ldots}

Next

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \Pi_{--}\left(q, i\Omega_n\right) &= \frac{2}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\left[\frac{1}{\omega_n^2 + E_k^2} - \frac{\omega_n\left(\omega_n + \Omega_n\right) + \left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \mu\right) + \Delta_{sp}^2}{\left(\omega_n^2 + E_k^2\right)\left(\left(\omega_n + \Omega_n\right)^2 + E_{k + q}^2\right)}\right]}} \\ &= \frac{2}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\left[\frac{1}{\omega_n^2 + E_k^2} - \frac{\omega_n^2 + \left(\epsilon_k - \mu\right)^2 + \Delta_{sp}^2 + \omega_n\Omega_n + \left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \epsilon_k\right)}{\left(\omega_n^2 + E_k^2\right)\left(\omega_n^2 + 2\omega_n\Omega_n + \Omega_n^2 + E_k^2 + \left(\epsilon_{k + q} - \mu\right)^2 - \left(\epsilon_k - \mu\right)^2\right)}\right]}} \\ &= \frac{2}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\left[\frac{1}{\omega_n^2 + E_k^2} - \frac{\omega_n^2 + E_k^2 + \omega_n\Omega_n + \left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \epsilon_k\right)}{\left(\omega_n^2 + E_k^2\right)^2\left(1 + \frac{\Omega_n^2 + 2\omega_n\Omega_n + \left(\epsilon_{k + q} - \epsilon_k\right)\left(\epsilon_{k + q} + \epsilon_k - 2\mu\right)}{\omega_n^2 + E_k^2}\right)}\right]}} \\ \end{align}}

Now, the second term of the integrand:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[\frac{1}{\omega_n^2 + E_k^2} + \frac{\omega_n\Omega_n + \left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \epsilon_k\right)}{\left(\omega_n^2 + E_k^2\right)^2}\right]\left(1 + \frac{\Omega_n^2 + 2\omega_n\Omega_n + \left(\epsilon_{k + q} - \epsilon_k\right)\left(\epsilon_{k + q} + \epsilon_k - 2\mu\right)}{\omega_n^2 + E_k^2}\right)^{-1}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} = & \frac{1}{\omega_n^2 + E_k^2} - \frac{\Omega_n^2 + \left(\epsilon_{k + q} - \epsilon_k\right)^2}{\left(\omega_n^2 + E_k^2\right)^2} - \frac{2\omega_n^2\Omega_n^2 + 2\left(\epsilon_k - \mu\right)^2\left(\epsilon_{k + q} - \epsilon_k\right)^2}{\left(\omega_n^2 + E_k^2\right)^3} + \\ & \frac{1}{\omega_n^2 + E_k^2}\frac{4\omega_n^2\Omega_n^2 + 4\left(\epsilon_{k + q} - \epsilon_k\right)^2\left(\epsilon_k - \mu\right)^2}{\left(\omega_n^2 + E_k^2\right)^2} + \ldots \\ \end{align}}

Therefore,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Pi_{--}\left(q, i\Omega_n\right) \simeq \frac{2}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\left[\frac{\Omega_n^2 + \left(\epsilon_{k + q} - \epsilon_k\right)^2}{\left(\omega_n^2 + E_k^2\right)^2} - \frac{2\omega_n^2\Omega_n^2 + 2\left(\epsilon_k - \mu\right)^2\left(\epsilon_{k + q} - \epsilon_k\right)^2}{\left(\omega_n^2 + E_k^2\right)^3}\right]}}}

Again at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T = 0} ,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Pi_{--}\left(q, i\Omega_n\right) \simeq 2N_0\int_{-\infty}^{\infty}{\frac{d\omega}{2\pi}\int_{-\infty}^{\infty}{d\xi\left[\frac{\Omega_n^2 + \frac{1}{3}v_F^2q^2}{\left(\omega^2 + \xi^2 + \Delta_{sp}^2\right)^2} - 2\frac{\omega^2\Omega_n^2 + \xi^2\frac{1}{3}v_F^2q^2}{\left(\omega^2 + \xi^2 + \Delta_{sp}^2\right)^3} + \ldots \right]}}} 

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Pi_{--}\left(q, i\Omega_n\right) \simeq \frac{N_0}{2\Delta_{sp}^2}\left(\Omega_n^2 + \frac{1}{3}v_F^2q^2\right) + \ldots}

Finally,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \Pi_{-+}\left(q, i\Omega_n\right) = & -\frac{1}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\frac{\left(i\omega_n + i\Omega_n\right)\left(\epsilon_k - \mu\right)\left(-2i\right) + i\omega_n\left(\epsilon_{k + q} - \mu\right)\cdot 2i}{\left[\left(\omega_n + \Omega_n\right)^2 + \left(\epsilon_{k + q} - \mu\right)^2 + \Delta_{sp}^2\right]\left[\omega_n^2 + \left(\epsilon_k - \mu\right)^2 + \Delta_{sp}^2\right]}}} \\ & = -\frac{1}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\frac{2\Omega_n\left(\epsilon_k - \mu\right) - 2\omega_n\left(\epsilon_{k + q} - \epsilon_k\right)}{\omega_n^2 + 2\omega_n\Omega_n + \Omega_n^2 + E_k^2 + \left(\epsilon_{k + q} - \epsilon_k\right)\left(\epsilon_{k + q} + \epsilon_k - 2\mu\right)} \cdot \frac{1}{\omega_n^2 + E_k^2}}} \\ & = -\frac{1}{\beta}\sum_{\omega_n}{\int{\frac{d^3k}{(2\pi)^3}\frac{2\Omega_n\left(\epsilon_k - \mu\right) - 2\omega_n\left(\epsilon_{k + q} - \epsilon_k\right)}{\left(\omega_n^2 + E_k^2\right)^2\left(1 + \frac{2\omega_n\Omega_n + \Omega_n^2 + \left(\epsilon_{k + q} - \epsilon_k\right)\left(2\left(\epsilon_k - \mu\right) + \epsilon_{k + q} - \epsilon_k\right)}{\omega_n^2 + E_k^2}\right)}}} \\ \end{align}}

Now, expanding:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[1 + \frac{2\omega_n\Omega_n + 2\left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \epsilon_k\right) + \left(\epsilon_{k + q} - \epsilon_k\right)^2 + \Omega_n^2}{\omega_n^2 + E_k^2}\right]^{-1}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} = 1 & - \frac{2\omega_n\Omega_n + 2\left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \epsilon_k\right) + \left(\epsilon_{k + q} - \epsilon_k\right)^2 + \Omega_n^2}{\omega_n^2 + E_k^2} \\ & + \frac{\left[2\omega_n\Omega_n + 2\left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \epsilon_k\right) + \left(\epsilon_{k + q} - \epsilon_k\right)^2 + \Omega_n^2\right]^2}{\left(\omega_n^2 + E_k^2\right)^2} \\ & - \frac{\left[2\omega_n\Omega_n + 2\left(\epsilon_k - \mu\right)\left(\epsilon_{k + q} - \epsilon_k\right) + \left(\epsilon_{k + q} - \epsilon_k\right)^2 + \Omega_n^2\right]^3}{\left(\omega_n^2 + E_k^2\right)^3} + \ldots \\ \end{align}}

When multiplied by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Omega_n\left(\epsilon_k - \mu\right) - \omega_n\left(\epsilon_{k + q} - \epsilon_k\right)} and integrated over Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{k}} or summed over Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega_n} all terms vanish if we assume particle-hole symmetry and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q/k_F \rightarrow 0} . That is because Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Omega_n\left(\epsilon_k - \mu\right) - \omega_n\left(\epsilon_{k + q} - \epsilon_k\right)} needs to be multiplied either by a term which is odd in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_k - \mu} and even in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_{k + q} - \epsilon_k} and in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega_n} , or by a term which is odd in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega_n} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_{k + q} - \epsilon_k} as well as even in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_k - \mu} . But there is no such term to any order. That means that the coupling term is given by particle-hole symmetry breaking terms or terms down by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q/k_F} . With such terms present, we expect it to be Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sim q^2\Omega_n} . Although, there is a term with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Omega_nN_0^'}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[\begin{array}{cc} \delta\Delta_+\left(-q, -i\Omega_n\right) & \delta\Delta_-\left(-q, -i\Omega_n\right) \end{array}\right]\left[\begin{array}{cc} m^2 + \frac{1}{3}\rho\left(\Omega_n^2 + c^2q^2\right) & \alpha\Omega_n \\ \alpha\Omega_n & \rho\left(\Omega_n^2 + c^2q^2\right) \end{array}\right]\left[\begin{array}{cc} \delta\Delta_+\left(q, i\Omega_n\right) \\ \delta\Delta_-\left(q, i\Omega_n\right) \end{array}\right]}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Longrightarrow}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} & \frac{1}{2}m^2 + \frac{2}{3}\rho\left(\Omega_n^2 + c^2q^2\right) & \pm \sqrt{\left[\frac{1}{2}m^2 - \frac{1}{3}\rho\left(\Omega_n^2 + c^2q^2\right)\right]^2 + \alpha^2\Omega_n^2} \\ \simeq & \frac{1}{2}m^2 + \frac{2}{3}\rho\left(\Omega_n^2 + c^2q^2\right) & \underbrace{\pm \left(\frac{1}{4}m^4- \frac{m^2}{3}\rho\left(\eta\Omega_n^2 + c^2q^2\right) + \ldots \right)^{\frac{1}{2}}}_{\begin{array}{clr} \Downarrow \\ \pm \frac{1}{2}m^2\left(1 - \frac{1}{3}\frac{\rho}{m^2}\left(\eta\Omega_n^2 + c^2q^2\right) + \ldots\right) \end{array}} \\ \simeq & \left\{\begin{array}{rcl} m^2 + \frac{1}{3}\left(\eta'\Omega_n^2 + c^2q^2\right) \\ \rho\left(\eta'\Omega_n^2 + c^2q^2\right) \end{array}\right. \end{align}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Longrightarrow} we have a massive propagating mode AND a massless propagating mode with velocity Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c = \frac{v_F}{\sqrt{3}}} . Such a mode would have to show up in specific-heat for instance as a correction to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T^3} but it does not. Rather, we have found a (thermodynamic) sound mode of a super-fluid (which is neutral), not superconductor which is charged.

Anderson-Higgs "mechanism"

In order to properly describe the collective modes of a (charged) superconductor, we must include the effects of long-range Coulomb repulsion. That means that in addition to including the short-range attraction we must include the long-range repulsion.

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