Why is the Matsubara Green function $\mathscr{G}(i\omega_n)$ equal to the retarded Green function (also the linear response susceptibility) $\chi(\omega+i\epsilon)$ under the substitution $i\omega_n \mapsto \omega+i\epsilon$.

I understand that you can compute the spectral representation and find that the 2 are equal. However, this seems rather unsatisfying, since there should be some motivation (hopefully) of defining the Masturbara Green function as $$ \mathscr{G}(\tau)=-\langle TA(\tau)A\rangle $$ which ultimately results in the fact that it is equal to $$ \chi(t)=-i\theta(t)\langle [A(t),A]\rangle $$ after Fourier transform.

Of course, it is also possible that the thought process is actually done in reverse, i.e., we first calculate $\chi(z)$ where $z \in \mathbb{C}^+$ is in the upper half-plane, and think: you know what? What if we tried computing $\chi(i\omega_n)$ and see what happens when we apply the inverse Fourier transform to get $\chi(\tau)$. What do you know, it happens to be a desired form of time-ordering correlation function in imaginary time.


1 Answer 1


Green's functions are mathematical objects, so there is no really satisfying physical interpretation of this fact.

Another way to look at it mathematically is in terms of the Lehmann representation - different Green's functions (retarted/advanced/time-ordered) are obtained depending on how you choose the integration contour in the frequency plane.

Yet another way is to discuss it in terms of Keldysh-Kadanoff-Baym choice of the time contour, which is a generalization of Matsubara approach to non-equilibrium situations (see, e.g., the introductory section of the review by Rammer and Smith).


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