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I'm reading Fetter Walecka on many body theory (Chapter 7). We have established the from of the many body green function \begin{equation} G(xt,x't')=\langle T \psi(xt)\, \psi^{\dagger}(x't')\rangle \end{equation} in the Lehmann representation \begin{equation} G(k,\omega)=\int_0^{\infty}dE \, \frac{A(k,E)}{\omega -\mu-E+i0}+\frac{B(k,E)}{\omega -\mu+E-i0} \end{equation}

$A$ and $B$ are positive semidefinite spectral functions. In the section on the physical interpretation, F&W make "an elementary model of the interacting assembly and assume that $G^R(\omega,k)$ has a simple pole close the the real axis at $\omega = \epsilon_k - i \gamma_k$ with residue $a$" (note that $\gamma$ is finite).

A short calculation yields

\begin{equation} G(k,t) = -i a e^{-i \epsilon_k t}\, e^{-\gamma_k t}, \end{equation} which describes the propagation of a quasiparticle with energy $\epsilon_k$ and lifetime $1/\gamma_k$.

Question: What is the microscopical justification for the simple pole at $\omega = \epsilon_k - i \gamma_k$? It seems to contradict the Lehmann representation, because the latter shows that all poles are infinitesimally close to the real axis.

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  • $\begingroup$ physics.stackexchange.com/questions/10921/… This post might help answer your question. $\endgroup$ Commented Aug 25, 2020 at 6:22
  • $\begingroup$ physics.stackexchange.com/questions/429301/… This one also may help! Essentially this says that even though the integrand of the spectral representation that you have written has poles at $\mu + E - i0$ and $\mu + E + i0$, the integral(i.e $G(k,\omega)$) can have poles at energies that have a finite imaginary part. $\endgroup$ Commented Aug 25, 2020 at 9:36
  • $\begingroup$ Interesting. This helps indeed. $\endgroup$
    – curio
    Commented Aug 27, 2020 at 14:34

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