In Kamenev's book Field Theory of Non-Equilibrium Systems (he also has lecture notes online here, which contains the relevant statement on pg. 17), he states that the following equation $$G^K(\epsilon) = \coth\left(\frac{\epsilon-\mu}{2T}\right) \left[G^R(\epsilon) - G^A(\epsilon)\right]$$ is a statement of the fluctuation-dissipation theorem, where $G^{(K,R,A)}(\epsilon)$ are the Keldysh, retarded, and advanced propagators, respectively. I have only ever seen the FDT stated in terms of structure factors and susceptibilities. While I can see the superficial connection (since $G^A(\epsilon) = G^R(\epsilon)^\dagger$, the RHS should resemble something like $\text{Im}\chi$), I'm having difficulty rigorously connecting the two. Can someone help me understand the connection between these statements?


2 Answers 2


What you are referring to is the form of fluctuation-dissipation theorem (FDT) that relates the dynamical structure factor to some retarded susceptibilty. The equation you wrote down holds for bosonic systems, in which case the RHS can be interpreted as a susceptibility while the LHS is related to the dynamical structure factor through the relation $G^{<} = G^{K}+\frac{1}{2}\left(G^A-G^R\right)$. This leads to $G^{<}(\epsilon) = n_{B}(\epsilon)\mbox{Im}\left[G^R(\epsilon)\right]$, where $n_{B}(\epsilon)$ is the Bose distribution function.

For a fermionic system, however, $n_{B}(\epsilon)$ must be replaced by $n_{F}(\epsilon)$ - the Fermi distribution function - in the equation above. This gives a fermionic FDT. The familiar bosonic FDT can be recovered in this case by considering the two-particle excitations, which can be expressed as the product of single particle excitations using Wick's theorem.

$\Pi^{R}(t,t^{'}) = G^{R}(t,t^{'})G^{K}(t^{'},t) + G^{K}(t,t^{'})G^{A}(t^{'},t)$ is the retarded susceptibility, and similarly one can also write an expression for $\Pi^{<}$ in terms of $G^{R,A,K}$

At equilibrium, one can show that: $\Pi^{<}(\epsilon) = n_{B}(\epsilon)\mbox{Im}\left[\Pi^{R}(\epsilon)\right]$. This is the familiar form of FDT. You will find a detailed discussion in Kamenev's book ch. 9.


Linear response and Green's functions
In linear response theory, if we are given Hamiltonian, $H=H_0+\lambda (t)X$, the response of variable $Y(t)$ is given by $$\langle Y^h(t)\rangle = \langle Y(t)\rangle+ \int_{-\infty}^{+\infty} dt_1\left\langle\frac{-i}{\hbar}[Y(t),X(t_1)]\right\rangle\theta(t-t_1)\lambda(t_1), $$ where the response function is just a retarded function for operators $Y(t), X(t_1)$ (operators without subscript has time evolution governed only by $H_0$): $$ G_{YX}^r(t,t_1)= \left\langle\frac{-i}{\hbar}[Y(t),X(t_1)]\right\rangle\theta(t-t_1)=\left[G_{YX}^>(t,t_1) - G_{YX}^<(t,t_1)\right]\theta(t-t_1).$$ The susceptibility is just the Fourier transform of this function, whereas the advance Green's function is defined as $$ G_{YX}^a(t,t_1)= \left\langle\frac{i}{\hbar}[Y(t),X(t_1)]\right\rangle\theta(t-t_1)=\left[G_{YX}^>(t,t_1) - G_{YX}^<(t,t_1)\right]\theta(t_1-t),$$ so that $$ G_{YX}^r(t,t_1) - G_{YX}^a(t,t_1) = G_{YX}^>(t,t_1) - G_{YX}^<(t,t_1) = \left\langle\frac{-i}{\hbar}[Y(t),X(t_1)]\right\rangle $$ Note also that the frequency space (i.e., for Fourier transforms): $$ G_{YX}^r(\omega) - G_{YX}^a(\omega) = G_{YX}^>(\omega) - G_{YX}^<(\omega), $$ as a simple consequence of the definitions.

Lehmann representation
In the eigenbasis of the unperturbed Hamiltonian, $H_0|n\rangle=E_n|n\rangle$ the greater Green's function has the following representation $$ G_{YX}^>(t,t_1)=\frac{-i}{\hbar}\left\langle Y(t)X(t_1)\right\rangle= \frac{-i}{\hbar}\sum_{n,m}e^{-\beta E_n}e^{-i(E_m-E_n)(t-t_1)/\hbar}Y_{nm}X_{mn},\\ G_{YX}^>(\omega) = \frac{-2\pi i}{\hbar}\sum_{n,m}e^{-\beta E_n}Y_{nm}X_{mn}\delta\left(\omega -\frac{E_m-E_n}{\hbar}\right) $$ Similarly $$ G_{YX}^<(\omega) = \frac{-2\pi i}{\hbar}\sum_{n,m}e^{-\beta E_m}Y_{nm}X_{mn}\delta\left(\omega -\frac{E_m-E_n}{\hbar}\right)=\\ \frac{-2\pi i}{\hbar}\sum_{n,m}e^{-\beta (E_n+\hbar\omega)}Y_{nm}X_{mn}\delta\left(\omega -\frac{E_m-E_n}{\hbar}\right)=e^{-\beta\hbar\omega}G_{YX}^>(\omega) $$ We thus have $$ G_{YX}^>(\omega)\pm G_{YX}^<(\omega)=\left(1\pm e^{-\beta\hbar\omega}\right)G_{YX}^>(\omega)\Rightarrow\\ G_{YX}^>(\omega)+ G_{YX}^<(\omega)=\frac{1+e^{-\beta\hbar\omega}}{1-e^{-\beta\hbar\omega}}\left[G_{YX}^>(\omega)- G_{YX}^<(\omega)\right]=\\ \coth\left(\frac{\beta\hbar\omega}{2}\right)\left[G_{YX}^>(\omega)- G_{YX}^<(\omega)\right]. $$ Recognizing the left part of this expression as the definition of the Keldysh Green-s function we thus have $$ G_{YX}^K(\omega)= \coth\left(\frac{\beta\hbar\omega}{2}\right)\left[G_{YX}^r(\omega)- G_{YX}^a(\omega)\right] $$

Is this the FDT?

  • Note that the Keldysh function that we defined is given by $$ G_{YX}^K(t,t_1) - G_{YX}^a(t,t_1) = \frac{-2i}{\hbar}\left\langle\frac{1}{2}\{Y(t),X(t_1)\}\right\rangle, $$ that is it is simply the correlation function (up to a factor), whose Fourier transform is the noise intensity. We thus have the statement of the Fluctuation-dissipation theorem.
  • This relationship will also hold, if $X$ and $Y$ are creation and annihilation operators, in which the Green's functions take the more familiar form. However, its interpretation as FDT becomes less reliable, unless we introduce generalized susceptibilities. More conventionally the difference between the advanced and the retarded Green's functions corresponds to the density-of-states, whereas the Keldysh function is the particle distribution function.
  • Although the relationship holds beyond the linear response, it is still a statement about the linear response coefficients! This is in contradistinction to the non-linear FDT, which usually implies statements about higher-order response (higher order in $\lambda(t)$).

Old answer
The difference of the retarded and the advanced Green's functions in the right-had-side of this equation is actually the density-of-states, i.e. what you might call a structure factor, whereas $G^K$ tests the possibilities of adding/removing a particle, i.e. the susceptibility.

What makes me personally skeptical about interpreting this equation is that formulating it in terms of Keldysh formalism gives superficial illusion that FDT can be applied out of equilibrium (or at least that it has such a simple form out of equilibrium), whereas this is not the case.


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