Talk:Phy5670

Collective Modes

Sound waves propagating through air or water are examples for collective modes. Such modes arise without broken symmetry of the system as ordinary gases and fluids do not brake any symmetry. Such waves have wavelengths that are very large compared to the distance between neighboring particles, which is the reason why we will look at very small wave vectors. An example for a collective mode in due to broken symmetry is a shear mode in a solid.

Remember that the partition function can be written as the Feynman path integral

$Z=\int D(c^{*},c)e^{-S},$

where

$S_{0}={\frac {1}{\beta }}\sum _{\omega _{n}}\sum _{k}(-i\omega _{n}+\epsilon _{n}-\mu )c_{k,\sigma }^{*}c_{k,\sigma }(\omega _{n})$

is the action of the non-interacting system; while the action of the interacting system is $S_{int}={\frac {1}{\beta }}\sum _{\Omega _{n}}{\frac {1}{L^{3}}}\sum _{q}V_{q}n_{-q}(-i\Omega _{n})n_{q}(i\Omega _{n})$

with $n_{q}(i\Omega _{n})={\frac {1}{\beta }}\sum _{\omega _{n}}\sum _{k}c_{k+q,\sigma }^{*}(i\omega _{n}+i\Omega _{n})c_{k,\sigma }(i\omega _{n})$

Hubbard-Stratonovich Transformation

The partition function can be calculated via the Hubbard-Stratonovich Transformation, which can be motivated by the one-dimensional integral

$1={\frac {1}{\sqrt {\pi a}}}\int \limits _{-\infty }^{\infty }dxe^{-{\frac {1}{a}}x^{2}}.$

The pre-factor ${\sqrt {\pi a}}$ is simply the constant number that equals to the value of the integral.
In the language of path integrals this implies

$1={\frac {1}{N}}\int D(\phi (r,\tau ))e^{-\int \limits _{0}^{\beta }d\tau \int d^{3}r\phi (r)V^{-1}(r-r')\phi (r')}.$

Here the number $N$ resembles the pre-factor ${\sqrt {\pi a}}$ in the previous example. Rewriting the potential in momentum space, this identity allows to convert the partition function. This is done by a shift of the field $\Phi _{q}(i\Omega _{n})$

$\phi (i\Omega _{n})={\tilde {\phi }}(i\Omega _{n})+iV_{q}n_{q}(i\Omega _{n}).$

The partition function, as introduced at the beginning of this chapter, simplifies thereby to

${\begin{aligned}Z&=1\cdot \int D(c^{*},c)e^{-S}\\&={\frac {1}{N}}\int D\phi _{q}(i\Omega _{n})e^{-{\frac {1}{\beta }}\sum _{\Omega _{n}}{\frac {1}{L^{3}}}\sum _{q}\phi _{-q}(-i\Omega _{n}){\frac {1}{V_{q}}}\Phi _{q}(i\Omega _{n})}\cdot \int D(c^{*},c)e^{-S}\\&=Z_{0}\int D{\tilde {\phi }}_{q}(i\Omega _{n})\left[\int D(c^{*},c)e^{-S_{0}}e^{-{\frac {1}{\beta }}\sum _{\Omega _{n}}{\frac {1}{L^{3}}}\sum _{q}2in_{-q}(-i\Omega _{n}){\tilde {\phi }}_{q}(i\Omega _{n})}{\frac {1}{Z_{0}}}\right]e^{-{\frac {1}{\beta }}\sum {\frac {1}{L^{3}}}\sum _{q}{\frac {{\tilde {\phi }}_{-q}(-i\Omega _{n}){\tilde {\phi }}_{q}(i\Omega _{n})}{V_{q}}}}\\&=\left\langle e^{-{\frac {1}{\beta }}\sum _{\Omega _{n}}{\frac {1}{L^{3}}}\sum _{q}2in_{-q}{\tilde {\phi }}_{q}(i\Omega _{n})}\right\rangle \\&\equiv \left\langle e^{A}\right\rangle \end{aligned}}$

Cumulant Expansion

This can be calculated by a cumulant expansion, such that

${\begin{aligned}Z&=\left\langle e^{A}\right\rangle _{0}\\&=e^{\langle A\rangle _{0}+{\frac {1}{2}}\left(\langle A^{2}\rangle _{0}-0\langle A\rangle ^{2}\right)+\ldots }\end{aligned}}$

Now, since the the average of the density $n_{q}(i\Omega _{n})$ is equal to zero, $\left\langle n_{q}(i\Omega _{n})\right\rangle =0,\quad \forall q\neq 0\land \Omega _{n}\neq 0,$

we only have to calculate the term with $\langle A^{2}\rangle$ in the exponent. This term contains the average of two density operators $n_{q}(i\Omega _{n})$, which can be evaluated using the Wick's theorem.

$\langle n_{-q}(-i\Omega _{n})n_{q'}(i\Omega _{n}')\rangle ={\frac {1}{\beta ^{2}}}\sum _{\omega _{n},\omega _{n}'}\sum _{k,k'}\langle c_{k-q,\sigma }^{*}(i\omega _{n}-i\Omega _{n})c_{k,\sigma }(i\omega _{n})c_{k'+q',\sigma '}^{*}(i\omega _{n}'+i\Omega _{n}')c_{k',\sigma '}(i\omega _{n}')\rangle _{0}$

Since we have two creation operators and two annihilation operators, two possible contractions are possible. However, one of these contractions is a non-connected Feynman diagram and so, according to the linked cluster theorem, we only have to compute the connected one. Due to Kronecker-delta symbols (arising from the contractions) we will be left with only one sum over $k$ and one sum over $\omega$. The resulting sum over $k$ can be written as an integral over a continuum (by taking the appropriate limit 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\rightarrow\infin} ), while the sum 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} is written as a contour integral. The result of this exercise 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 Z{Z_0} \simeq \int D\phi e^{S_{eff}(\phi)}, }

where the effective action 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 S_{eff}(\phi) = \frac1{\beta}\sum_{\Omega_n}\frac1{L^3} \sum_q\left(\frac1{V_q}-2 \mathcal N\Pi_q(i\Omega_n) \right) \phi_{-q}(i\Omega_n) \phi_q(i\Omega_n) + \mathcal O(\phi^4).}

Here we use the degeneracy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{N}} of each level. This action is expressed in terms of an integral that we are about to solve, 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 \Pi_q(i\Omega_n) = \int \frac{d^3k}{(2\pi)^3} \frac{n_F(\epsilon_k)-n_F(\epsilon_{k-q})} {\epsilon_k - \epsilon_{k-q}-i\Omega_n}. }

In this 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 n_F(E)=\frac1{e^{\frac{E-\mu}{kT}+1}}} is the Fermi-Dirac distribution function, together with the chemical 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 \mu} , the Boltzmann 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 k_B} and the 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 T} .

Plasmons and Zero Sound

As said before, our special interest is pointed towards density waves in an electron gas. So finally, the degeneracy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{N}} will be set equal to 2. Notice 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_q(i\Omega_n)} couples to density.
We also said in the beginning of the chapter that wavelengths of such excitations are typically very long compared to the distance between molecules; or, in case of a plasma, the distance between electrons. Therefore, when we 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 \Pi_q(i\Omega_n)} we look at the limit of very small wave vectors Failed to parse (SVG (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} , and we will also consider small temperatures. For an electron gas the energy dispersion is, of course, quadratic 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 k} , being Failed to parse (SVG (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=\frac{\hbar^2k^2}{2m}} .

Therefore, let us 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 |q|\ll k_F} 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\ll \mu} .
In that case the difference

Failed to parse (SVG (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} n_F(\epsilon_k)-n_F(\epsilon_{k-q}) \end{align}}

in the integrand 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_q(i\Omega_n)} has no contribution 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 \epsilon_k} being far away from Failed to parse (SVG (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_F} . Hence we can expand the numerator of the integrand by writing

Failed to parse (SVG (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} n_F(\epsilon_{k-q}) &= n_F(\epsilon_{k}+\epsilon_{k-q}-\epsilon_{k})\\ &\simeq n_F(\epsilon_k) + n_F'(\epsilon_k) (\epsilon_{k-q} - \epsilon_k). \end{align} }