3
$\begingroup$

I learned from a lecture that $G^{<}=-f(\epsilon)(G^r-G^a)$ is a type of fluctuation-dissipation theorem. However as far as I know the fluctuation-dissipation theorem is stated as $S(\omega)=2\hbar(1+n(\omega))\chi''(\omega)$ where $S$ is a correlation function, $\chi''$ is the imaginary part of response function and $n$ is the Bose distribution. Now here I doubt:

  1. What are those Green's functions in the formula $G^{<}=-f(\epsilon)(G^r-G^a)$? Are they one-body or many-body ones?
  2. What's the relation between the two FDTs?

It would be most helpful if references on this topic could be listed.

$\endgroup$
1

1 Answer 1

2
$\begingroup$

To your first point of "are they one-body or many-body ones?", I am not quite sure what you mean. This quantity involves all of the many-body eigenstates, but it is a one-body quantity in the sense that it only accounts for processes of adding/removing one particle.

From the definition of the retarded Green's functions we have

$$ G^{r} = -i\Theta(t-t')\frac{1}{Z}\text{Tr}\Big[e^{-\beta(\hat{H}-\mu \hat{N})}\{\hat{\psi}(xt),\hat{\psi}^{\dagger}(x't') \} \Big] $$ where $\{\cdot,\cdot\}$ is the anti-commutator and $ Z = \text{Tr}\exp(-\beta(\hat{H}-\mu\hat{N}))$, and the trace is over all the many-body eigenstates. The advanced is defined analogously: $$ G^{a} = i\Theta(t'-t)\frac{1}{Z}\text{Tr}\Big[e^{-\beta(\hat{H}-\mu \hat{N})}\{\hat{\psi}(xt),\hat{\psi}^{\dagger}(x't') \} \Big] $$

Assuming a time-independent $\hat{H}$, we fourier transform to $t-t' \to \omega$, from the Lehmann representation we have that: $$ G^{r}(\omega) - G^{a}(\omega) = 2i\ \text{Im}G^{r}(\omega) $$

Substituting this relation back into your expression gives,

$$ G^{<}(\omega) = - 2if(\omega)\text{Im}G^{r}(\omega), $$ which shows a much more immediate analogy to your standard fluctuation-dissipation relation. That is, an analog that connects a correlation function and the imaginary part of a response function. I am unaware of a deeper physical intuition in this expression, so if someone else has a better sense, please feel free to correct my answer.

A great source is this chapter submitted to the Autumn School on Correlated Electrons: https://arxiv.org/abs/1907.11302

$\endgroup$
1
  • $\begingroup$ Thanks, this is exactly what I've been looking for. $\endgroup$
    – user835469
    Apr 16, 2023 at 13:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.