# Matsubara Green Function vs Real Green Function

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

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).