I want to analytically continue an expression for a correlation function I have in terms of Matsubara frequencies in order to find the retarded correlation function. The expression involves a logarithm. To give an example, the expression looks something like this \begin{equation} C(i\omega_n)=f(i\omega_n)\ln\left[\frac{i\omega_n-E_1}{-i\omega_n-E_2}\right] \end{equation} where $f$ is a rational function. After analytical continuation by $i\omega_n\rightarrow\omega+i0^+$, I want to find the real and imaginary part of the retarded correlation function. In order to do that I have to specify where the branch cut of the logarithm is located. If I take it on the negative real axis, the imaginary part of the logarithm will be \begin{equation} \textrm{Im}\left\{\ln\left[\frac{i\omega_n-E_1}{i\omega_n-E_2}\right]\right\}=\pi\Theta(-\omega+E_1)+\pi\Theta(+\omega+E_2) \end{equation} whereas for a cut on another axis the above will be unaffected by the discontinuity and will thus obviously give a different result. Is there any physical/mathematical argument that recommends one particular branch cut as the right one?

I also came across the case where somebody put the branch cut on the negative imaginary axis and defined the zero of the complex argument to coincide with the positive imaginary axis. Then the above would be \begin{equation} \textrm{Im}\left\{\ln\left[\frac{i\omega_n-E_1}{i\omega_n-E_2}\right]\right\}=\frac{\pi}{2}\textrm{sgn}(-\omega+E_1)+\frac{\pi}{2}\textrm{sgn}(\omega-E_2) \end{equation}

I am confused by the amount of choice the branch cuts seem to offer and what could bring the results back together/determines the branch cut uniquely.

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    $\begingroup$ This might be too naive, but don't you have already choose the position of the branch cut when you get your first equation ? Then the subsequent calculation should be done with the same branch cut (at the end of the day, as long as you are consistent, the result should be independent of this choice). $\endgroup$ – Adam Jan 12 '17 at 7:57
  • $\begingroup$ The logarithm stems from an integral over a rational function. This rational function is real (i.e. $i\omega$ occurs only squared). Does this integration affect the choice of branch cut? I don't know too much complex analysis.. $\endgroup$ – loewe Jan 12 '17 at 19:33
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    $\begingroup$ Maybe you could give the integral (that could help). But I would say that in that case, the end result should be independent of the choice of the cut (that is, maybe you are just discussing an intermediate result, where the choice of the cut is important, but not in the long run). $\endgroup$ – Adam Jan 12 '17 at 19:50
  • $\begingroup$ The integral is $\int_a^b \frac{l\ \textrm{d}l}{4l^2-(i\omega)^2}=\frac{1}{8}\ln(\frac{(i\omega)^2-4b^2}{(i\omega)^2-4a^2})$. I used a simplified expression in the above to make the questions less confusing. I expected the end result to be independent of the choice of branch cut too, but does not seem to be, unless there is a mistake in my original question. $\endgroup$ – loewe Jan 12 '17 at 20:24
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    $\begingroup$ in that case, you should start by the analytical continuation, and then do the integral, thats much simpler. $\endgroup$ – Adam Jan 12 '17 at 20:29

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