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There are many non-Fermi liquids where the imaginary part of the self-energy $\Sigma(\omega_n)$ is an odd function of the Matsubara frequency $\omega_n$.

For example, in Eq. 28 of PRB 69, 035111 (2004), the self energy depends on the energy of the excitation as $\text{Im}[\Sigma(\epsilon)] \sim\text{sign}(\epsilon) \epsilon^2 \log(1/|\epsilon|)$. Clearly, the imaginary part of the self-energy has opposite signs for opposite signs of $\epsilon$.

Another example is Eq. 30 of PRX 8, 031024 (2018), where $\text{Im}\Sigma(\epsilon) \sim - \epsilon \log(1/|\epsilon|)$.

The imaginary part of the self-energy stands for the inverse lifetime. If the lifetime becomes negative, it would signify an exponential growth (rather than a decay) of the quasiparticle excitations. In other words, the presumed ground state would be unstable towards formation of quasiparticles.

Is this conclusion true? In that case, this extremely unstable state would never be physically realized as the quasiparticles will spontaneously proliferate, and these non-Fermi liquid states are not true ground states (hence, not so interesting).

If that conclusion is wrong, then how does the system remain stable, with a positive, finite lifetime of quasiparticles? Am I missing something simple about the conversion from Matsubara self-energy to the lifetime of the quasiparticle?

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  • $\begingroup$ physics.stackexchange.com/q/688605/49107 $\endgroup$
    – Anyon
    Commented Nov 12 at 1:01
  • $\begingroup$ @Anyon I have read that post, but thanks for sharing. It does not answer my question. $\endgroup$ Commented Nov 12 at 1:03
  • $\begingroup$ @ArchismanPanigrahi, Anyon was not just sharing for the answer. The comments to that answer directly answers your question. $\endgroup$ Commented Nov 12 at 2:05
  • $\begingroup$ @naturallyInconsistent I found the answer and posted it below. It is something technical (although quite simple). Those comments do not answer the question either. $\endgroup$ Commented Nov 12 at 2:27
  • $\begingroup$ Yes, I noticed that your own answer stressed the important point that there is a distinction between retarded self-energy and Matsubara self-energy. Thanks. $\endgroup$ Commented Nov 12 at 4:31

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The imaginary part of the self-energy stands for the inverse lifetime

It is the imaginary part of the retarded self-energy, which can be obtained from the Matsubara self-energy using analytical continuation $i \omega_n \rightarrow \omega + i 0^+$.

In this analytical continuation, the sign of $\omega_n$ needs to be interpreted as,

$$\text{sign}(\omega_n) = \text{sign}(\text{Im}[i \omega_n]) \rightarrow \text{sign}(\text{Im}[\omega + i 0^+]) = 1,$$ which is an even function, i.e., it does not change sign when $\omega \rightarrow -\omega$.

This argument suggests that the imaginary part of the Matsubara self-energy needs to be odd in $\omega_n$ so that the imaginary part of the retarded self-energy is an even function of quasiparticle energy, which is necessary so that neither quasiparticles nor quasiholes don't spontaneously proliferate and the system remains stable.

Reference: https://johnof.folk.ntnu.no/green-2013.pdf. See Eq. 148-155 and the text between them.

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