Question about the retarded Green's function and the analytic continuation of the Matsubara Green's function

This question is a little bit mathematical. It is about the relation between correlation function in the Matsubara frequency and the retarded correlation function in the real frequency. The following question is based on the contents in the book by Flensberg and Bruus

The retarded correlation function $$G^R (\omega)$$ can be obtained by taking the analytic contiuation of the correlation function $$G(\omega)$$ in the Matsubara frequency, i.e. $$G^R (\omega) = G (i\omega_n \rightarrow \omega + i\eta)$$ with $$\omega_n$$ being the Matsubara frequency, $$\omega$$ being the real frequency and $$\eta \rightarrow 0$$ being a infinitesimal positive number. This theorem can be proved by using the Lehmann representation.

However, I found that, while reading the textbook by Flensberg and Bruus, the example of the density-density correlation function seems to violated the above theorem.

The density-denisty correlation function of non-interacting Fermi gas $$\chi$$ in the imaginary time domain is given by $$\begin{equation} \chi (q, \tau )=-\frac{1}{V}\langle T_\tau \rho (q,\tau)\rho(-q,0) \rangle, \end{equation}$$ with $$\tau$$ being the imaginary time. Here, $$V$$ is the system volume and $$\rho (q) = \sum_{k,\sigma = \pm} c^\dagger_{k\sigma}c_{k+q,\sigma}$$ is the particle density in the momentum space. While taking the analytic continuation of the Matsubara density-density correlation function, we obtain $$\begin{equation} \chi (q, i\omega_n \rightarrow \omega + i\eta )=-\frac{1}{V} \langle \rho _{q=0}\rangle \langle \rho_{q=0}\rangle+ \frac{1}{V} \sum_{k, \sigma} \frac{n_F (\epsilon_k)-n_F (\epsilon_{k+q})}{\omega + \epsilon_k-\epsilon_{k+q}+i\eta}. \end{equation}$$

On the other hand, the retarded density-density correlation $$\chi^R$$ is defined as $$\begin{equation} \chi^R (q, t-t')=-i\theta (t-t')\frac{1}{V} \langle [\rho(q,t), \,\,\rho (-q,t')] \rangle. \end{equation}$$ with $$t$$ being the real time. By Wick's contraction, $$\chi^R$$ in the frequency space takes the form $$\begin{equation} \chi^R (q, \omega )=\frac{1}{V} \sum_{k, \sigma} \frac{n_F (\epsilon_k)-n_F (\epsilon_{k+q})}{\omega + \epsilon_k-\epsilon_{k+q}+i\eta}. \end{equation}$$.

As compared $$\chi (q, i\omega_n \rightarrow \omega + i\eta )$$ with $$\chi^R (q, \omega )$$, there is an extra term $$-\frac{1}{V} \langle \rho _{q=0}\rangle \langle \rho_{q=0}\rangle$$ in $$\chi (q, i\omega_n \rightarrow \omega + i\eta )$$.

Does it mean that the theorem relevant to the analytic continuation mentioned above is violated, although the extra term is harmless?

Thanks!

Here the problem is that when we do the analytical continuation, we should start from a positive $$\omega_n$$ so that $$i\omega_n$$ is in the upper complex plane (so we continue to some $$\omega + i\eta$$ which is still in the upper half plane!). As a result, the disconnected part in your expression (the $$\langle \rho \rangle \langle \rho \rangle$$ part) would give rise to $$\int d\tau \ e^{i\omega_n \tau} \langle \rho \rangle \langle \rho \rangle = 0$$ for $$\omega_n = \frac{2 n \pi}{\beta},\ n \in Z_+$$.
• Hi, Chuan Chen, thank you for answering this question. The analytica continuation in the derivation for the retarded Green's function $i\omega_n\rightarrow \omega +i\eta$ is only restritcted to the upper half plane, namely it is not properly defined at $\omega_n = 0$. So we don't have to consider the contribution at $\omega_n = 0$ as we evaluate the retarded Green's function. Is it correct? – Chang Mar 18 at 3:50
• @Chang Yes, that's how I understand this, if we generalise the Matsubara Green function to the whole complex plane, i.e., $i\omega_n \rightarrow z$, we will see that it has a branch cut on the real axis (see Mahan or Bruus), so it's better to start from either the upper or lower complex plane. For the advanced Green function, the analytical continuation is done at the lower complex plane, i.e., $i\omega_n \rightarrow \omega -i\eta$, in this case, we consider a negative $\omega_n$. – Chuan Chen Mar 18 at 8:48