5
$\begingroup$

Consider a one-particle retarded Green's function $$G^r(\alpha)=[\omega+i\eta-\varepsilon(\alpha)-\Sigma^r(\alpha)]^{-1}$$ with self-energy $\Sigma^r(\alpha)$ for some quantum number $\alpha$. It is argued that $-\mathrm{Im}\Sigma^r>0$ always holds as it signifies the quasiparticle lifetime. Is this true always?

$\endgroup$

1 Answer 1

0
$\begingroup$

Let us consider the following:

  • $\Sigma^r$ is a retarded self-energy, i.e., it is given by expressions like $$\Sigma^r(\omega)=\sum_k |V_k|^2G^r(\omega).$$ Here $V_k$ might be replace by vortex parts with more complex structure, but the rpesence of the retarged Green's function, $\Im \left[G^r\right]<0$ suggests that $\Im [\Sigma^r]<0$.
  • $G^r$ is a retarded Green's function - it would not be the case, if we had $\Im \left[\Sigma^r\right]>0$.

I cannot say, if there is a rigorous mathematical proof that $\Im[\Sigma^r]\leq 0$, for any type of dispersions laws and interactions. However, if this relation breaks, we are facing the breakdown of the causality in our calculations, which means that some of the premises of our derivations are wrong - the initial Hamiltonian requires revision (keeping in mind that sometimes the breakdowns of theory are meaningful - e.g., many quantities diverge near phase transitions).

$\endgroup$
4
  • 2
    $\begingroup$ to add to the answer: the point is that the single-particle retarded GF must be analytical in the upper half of the complex plane, because otherwise it means that for some $z=\omega + i\Gamma$ we have $\int_0^{\infty} dt e^{i\omega t -\Gamma t} \langle \{\psi^{\dagger}(0),\psi(t)\}\rangle = \infty$, which is unphysical. This means that the retarded self energy has to have a negative imaginary component $\endgroup$
    – user275556
    Commented Jan 14, 2022 at 9:51
  • $\begingroup$ @yyy I agree with that. $\endgroup$
    – Roger V.
    Commented Jan 14, 2022 at 9:53
  • $\begingroup$ @yyy Why is that divergence unphysical? Could you explain its physical meaning? $\endgroup$
    – xiaohuamao
    Commented Jan 14, 2022 at 10:44
  • 1
    $\begingroup$ we're doing a physical perturbation to the system, by adding one particle at time $0$, and then propagating the system in time, and looking at finite times responses, as the exponential suppression by $\Gamma$ ensures that. In essence, we are looking at ${\rm Tr}\langle e^{-\beta H} e^{iH t}\psi e^{-i H t} \psi^{\dagger} \rangle$ and integrating it with an exponentially decaying envelope in time. A physical system shouldn't have an infinite response to that $\endgroup$
    – user275556
    Commented Jan 14, 2022 at 11:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.