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In the Matsubara formalism, there is a commonly-made statement that an imaginary-time correlation function is related to a retarded correlation function via the replacement $i\omega_n \rightarrow \omega + i0^+$, also called analytic continuation.

In a mathematical sense, on the other hand, analytic continuation means to find an analytic function whose values match the "input data", which in this case are the values at the discrete Matsubara frequencies $i\omega_n$.

For analytic continuation to retarded correlation functions, I have two questions:

  1. What is the domain of analytic continuation and why? That is to say, do we only care about analytically continuing in the upper half plane, excluding all frequencies $\omega_n \leq 0$ as part of our input data to analytic continuation? Do we also exclude the real line, so that we ignore the zeroth bosonic frequency? And why is that the "physical" choice?
  2. In what sense is analytic continuation unique? Does one require additional "physical" conditions other than analyticity? There is a strong statement on the uniqueness of complex-analytic functions via the identity theorem; does the infinite Matsubara frequency count as an accumulation point and thus be sufficient for uniqueness?
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  • $\begingroup$ A relevant reference on this question: aip.scitation.org/doi/10.1063/1.1703704 $\endgroup$ Commented Apr 24, 2020 at 18:53
  • $\begingroup$ The case you're referring to is when $G^R(\omega)$ is analytically extended to the upper half plane. It is known from the imaginary time formalism that if you do the said replacement to the imaginary time GF in rational form , you'd get $G^R(\omega)$. The other option is to extend $G^A(\omega)$ to the lower half plane. Then you'd do the said replacement with the sign of the (imaginary) infinitesimal reversed on the imaginary time GFs obtained in the lower half plane to get $G^A$. You do exclude the real line - that's the branch cut, if you will, of the resolvent defined on the complex plane. $\endgroup$
    – Vivek
    Commented Jun 22, 2020 at 11:47
  • $\begingroup$ The paper that directly addresses your question head on on the analytic continuation part is: aip.scitation.org/doi/pdf/10.1063/1.1703704 $\endgroup$
    – Vivek
    Commented Jun 22, 2020 at 11:54

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I hope am not confusing something, the issue is that the retarted propagator $G^R$ is regular in the upper half plane, so by the second statement from your question, two functions equal on the set, with an accumulation point at infinity are equal at every point in the domain of analycity. The same statement also holds in lower plane, with $G^A$ instead of the retarded and $\omega - i 0$.

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  • $\begingroup$ Isn't this just a rephrasing of my question? $\endgroup$
    – Aaron
    Commented Apr 26, 2020 at 21:01

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