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

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

where the prefactor ${\sqrt {\pi a}}$ is simply the constant number that equals to the value of the integral. <\br> 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

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

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 $L\rightarrow \infty$), while the sum over $\omega$ is written as a contour integral. The result of this exercise is