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}.

Comparing $\chi (q, i\omega_n \rightarrow \omega + i\eta )$ 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?



1 Answer 1


The relationship between retarded and imaginary time Green function (the analytical continuation here) can be shown explicitly by using the Lehmann representation, which is presented in many textbooks, e.g., Mahan.

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 $\omega_n = \frac{2 n \pi}{\beta},\ n \in Z_+$.

So the two should be the same.

  • $\begingroup$ 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? $\endgroup$
    – Chang
    Mar 18, 2019 at 3:50
  • 1
    $\begingroup$ @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 $\omega_n$. $\endgroup$
    – Ogawa Chen
    Mar 18, 2019 at 8:48
  • $\begingroup$ thanks! So for the correlation function of fermionic operators, there is no such problem. Firstly, there are no disconnected terms since fermion does not get condensed. Secondly, the values of the fermionic Matsubara frequency are all nonzero. $\endgroup$
    – Chang
    Mar 19, 2019 at 9:04
  • $\begingroup$ +1 The argument in Bruus & Flensberg is wrong. I remember them saying that there will be some background ionic cancellation for $q=0$, although it's clear from the analytic continuation procedure that one cannot analytically continue from a zero Matsubara frequency and the $q=0$ terms drop out, as one can also see directly from calculation of the retarded density-density correlator for the free electron gas! $\endgroup$
    – Vivek
    Oct 20, 2019 at 20:14

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