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Question: is there any fermionic version of the fluctuation–dissipation theorem?

Context:

In classical physics, the fluctuation–dissipation theorem is roughly written as

$$\langle x(t)x(t')\rangle = 2\frac{k_{\rm B}T}{\omega} \operatorname{Im}[\chi(\omega)]\delta(t-t')$$

where $x$ is some variable of interest that changes in time $t$, $\omega$ is the frequency, $k_{\rm B}$ Boltzmann constant, $T$ is temperature and $\chi(\omega)$ is just the linear response function (in frequency space).

When passing to quantum mechanics, we just replace $k_{\rm B}T $ by $\hbar \omega (n_{\rm BE}+1)$, where $n_{\rm BE}=1/(\exp[\hbar\omega/k_{\rm B}T]-1)$ is the Bose–Einstein occupation distribution.

If I understand it correctly, usually the systems of interest are mostly concerned by bosonic-like fluctuations (radiation, vibrations, etc.), that's why bosonic statitics are used. But, are there any cases of interests where $k_{\rm B}T$ is replaced by something proportional to $1/(\exp(\hbar \omega/k_{\rm B}T)+1)$ (Fermi–Dirac statistics)?

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There is a version for the FDT in the Keldysh formalism, which is general for treating non-equilibrium systems in the so called steady state limit. In this approach, one assumes that the system is initially non-interacting and characterised by some equilibrium distribution function $n(\omega)$, which will depend on the particle's statistics. Then interactions are switched on adiabatically at later times. Using Green's functions formalism, one can work out a relation between the different time-ordered Green's functions in the Keldysh contour:

https://en.wikipedia.org/wiki/Keldysh_formalism

One can see that in the Keldysh contour, four different contour orderings take place. The four different contour ordered Green's functions are not independent from each other, and they are related by a causality constraint. This allows to "rotate" the components of the Green's functions into the usual advanced and retarded components $G^{A},G^{R}$, plus another component that is termed the Keldysh Green function $G^{K}$. In the special case where the system is found in thermodynamic equilibrium (and thus there is time-translation invariance), these three components are not independent from each other: they relate by a FDT relation of the form: \begin{eqnarray} G^{K}(\omega)=(1+ \eta 2n(\omega))\left(G^{R}(\omega)-G^{A}(\omega)\right), \end{eqnarray} where $\eta=\pm$ is for bosons $+$ and $-$ for fermions. Here $n(\omega)$ can be either bosonic (which would be the Bose distribution function) or fermionic (the Fermi-Dirac distribution). These components of the Green's functions appear in the solution of Dyson's equation: \begin{eqnarray} G^{-1}(\omega)=G^{-1}_0(\omega)-\Sigma(\omega), \end{eqnarray} where $G_0(\omega)$ corresponds to the non-interacting Green's functions, and $\Sigma(\omega)$ is the self-energy. Dyson's equation is completely general for bosons or fermions, and the FDT equation above, which is only valid in equilibrium, is also general independent of the particle's statistics.

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  • $\begingroup$ Conclusion: the FDT theorem (equilibrium condition) does not change for fermions. Is that right? $\endgroup$
    – Mauricio
    Commented Dec 12, 2021 at 15:09
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    $\begingroup$ Notice there is a minus sign change just in front of the particle distribution $n(\omega)$, so it kind of changes a bit as one must always be sure signs are properly checked. Apart from that, the fundamental relation stays. $\endgroup$ Commented Dec 12, 2021 at 15:19

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