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I am computing the Matsubara sum $$S(i\omega_n) = \frac{1}{\beta}\sum_m \frac{1}{(i\omega_n - i\Omega_m +a)(\Omega_m^2 + b^2)}$$ where $\Omega_m$ is a bosonic Matsubara frequency. And, $\omega_n$ is a fermionic Matsubara frequency, not summed over.

From this Matsubara sum, I find a factor of $$\coth(\frac{\beta}{2}(i\omega_n + a)) \tag{1}$$ where $\omega_n = i\pi(2n+1)/\beta$ is the fermionic Matsubara frequency that is not summed over. By properties of hyperbolic functions, we have $$\coth(\frac{\beta}{2}(i\omega_n + a)) = \coth(i\pi n + i\frac{\pi}{2}+\frac{\beta}{2}a) = -\tanh(\frac{\beta}{2}a),$$ which is independent of the Matsubara frequency. This is trivially analytically continued. However, if I directly analytically continue the first expression $(1)$, I get $$\coth(\frac{\beta}{2}(\omega + a)),$$ which is very different. It depends on $\omega$, for one. Which procedure is correct and why? Note that the $\tanh$ expression seems to be used in the literature (as well as Wikipedia).

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  • $\begingroup$ Is this from a reference? Which page? $\endgroup$ Commented Jul 16 at 18:32
  • $\begingroup$ @Qmechanic It is implied by the expression written under "Fermion self energy" on the Matsubara frequency wikipedia page (en.wikipedia.org/wiki/Matsubara_frequency). I will add the actual Matsubara sum I am computing for clarity. $\endgroup$ Commented Jul 16 at 18:38
  • $\begingroup$ @Qmechanic A very similar sum (modulo some relative minus signs) is computed in G. D. Mahan Many-Particle Physics 3ed p 136-137. The manipulation from $\coth$ to $\tanh$ is done without any comment. More precisely, the manipulation from a Bose to Fermi distribution is done without any comment. $\endgroup$ Commented Jul 16 at 21:39
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    $\begingroup$ Could you add more details of how you arrive at the two results? As it is now, it is not even clear in what respect your first approach differs from the second. $\endgroup$ Commented Jul 17 at 8:48

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The problem is that knowing the value $f(z)$ on only certain discrete set of points ($(2n+1/2)\pi/\beta$ here) can not uniquely determine the $f(z)$. Since $\coth(i\beta\omega_n/2)=-1$ for any $\omega_n$, you can obtain infinite many functions $g(-\coth(\frac{\beta}{2}z))f(z)$ which is also "analytical continuation" of $f(i\omega_n)$ as long as $g(1)=1$, e.g., $g(z)=z,(1+z)/2,z^n$.

Whereas, the physically relevant "analytical continuation" is uniquely defined because it will be restricted by some other properties, e.g., the asymptotic behavior at $|z|\to \infty$. For example, in general, self-energy and any $n$-point correlation functions should behave analytically at $|z|\to \infty$, which can be seen from the Lehmann representations. $\langle AB\rangle_\omega\propto\int dt e^{i\omega t} \langle A e^{iHt}Be^{-iHt}\rangle\propto \int dt e^{i(\omega+E_m-E_0) t} \langle 0| A|m\rangle\langle m| B|0\rangle \propto \frac{\langle 0| A|m\rangle\langle m| B|0\rangle}{\omega+E_m-E_0} $ (Similar expression can be written down for $\langle BA\rangle_\omega,\langle [A,B]_{\pm}\rangle_\omega$ and so on.)

Thus at least we know we should use $−\tanh(\frac{\beta}{2}a)$ instead of $\coth(\frac{\beta}{2}(\omega+a))$ since the latter one has essential singularity at infinity. I don't know how to prove the uniqueness of the analytical continuation given the physical constraint, but in practice, this is enough to help us identify the true result.

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