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The time-ordered and retarded-Green-functions are defined as

\begin{align} G_{ \alpha \alpha^{\prime}} (t) &= - \mathrm{i} \langle T_{t} \, a_{ \alpha } ( t ) a_{ \alpha^{\prime}}^{ \dagger } ( 0 ) \rangle \\ G_{ \alpha \alpha^{\prime}}^{ \text{ret}} (t) &= - \mathrm{i} \theta ( t ) \langle \{ a_{ \alpha } ( t ) , \, a_{ \alpha^{\prime}}^{ \dagger } ( 0 ) \} \rangle \end{align}

with grand-canonical-time-evolution-operators

$$ U(t) = \mathrm{e}^{- \mathrm{i} (H - \mu N) t / \hbar } $$

The e.o.m.-technique yields

\begin{align} G_{ \alpha \alpha^{\prime}} (\omega) &= \big[ \omega - \frac{ \operatorname{mat} H - \mu \mathbb{1}}{\hbar} \big]_{ \alpha \alpha^{\prime}}^{\operatorname{inv}} \\ G_{ \alpha \alpha^{\prime}}^{ \text{ret}} (\omega + \mathrm{i} \delta ) &= \big[ \omega + \mathrm{i} \delta - \frac{ \operatorname{mat} H - \mu \mathbb{1}}{\hbar} \big]_{ \alpha \alpha^{\prime}}^{\operatorname{inv}} \end{align}

However... In the book from Alexander Altland, in chapter 7, I find the identities

\begin{align} \operatorname{Re} G ( \omega ) &= \operatorname{Re} G^{ \text{ret}} (\omega) \\ \operatorname{Im} G( \omega ) &= \operatorname{Im} G^{ \text{ret}} (\omega) \cdot \tanh \frac{\hbar \beta \omega}{2} \end{align}

following from the Lehmann-representation.

This seems to me a little bit illogical due to the appearance of the tangens hyperbolicus in the lower equation. Why do these two sets of equations agree with each other?

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  • $\begingroup$ I think in the case, that the Hamilton-matrix is real-valued, the two identities should perfectly coincide... It's still a little bit confusing... $\endgroup$
    – Antihero
    Commented Jul 12, 2022 at 8:17
  • $\begingroup$ See this answer to a possible duplicate. $\endgroup$
    – Roger V.
    Commented Nov 29, 2022 at 10:11

1 Answer 1

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The straight-forward manipulations of the equations of motion does not necessarily give the correct expressions for the Green-functions. It can be proven (e.g. in the book from Altland, nowhere explicitly but with the framework developed there one can prove it) that the equation of motion technique gives the correct answer for the retarded Green-function. But not for the time-ordered. The third equation in the question that I posted is simply wrong; When one performs the Fourier-transformations one has to be particularly careful about the boundary-conditions.

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