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Aug 29, 2017 at 12:20 comment added Janosh @EverettYou Could you add the command you ran the obtain the result in your answer? Also, in the options inspector I set CellContext to "Notebook". Now Mathematica prefixes all global variables, e.g. running MatsubaraSum[1/(z - a)^2, z] gives -Global`\[Eta] \!\( \(\*SubsuperscriptBox[\(n\), \(Global`\[Eta]\), \(\[Prime]\)]\)[a]\). Is there some way to avoid this?
Mar 28, 2016 at 0:00 comment added Everett You @leongz The key point is to find poles and calculate residuals. Mathematica is good at these stuff.
Mar 27, 2016 at 8:58 comment added leongz @EverettYou, I just looked through your Mathematica package, and I am surprised how simple it seems. Could you briefly explain how it works? Typically calculating Matsubara sums by hand is quite a messy endeavor in contour integration for me.
Feb 17, 2016 at 17:35 comment added Everett You @MedullaOblongata Nice suggestion, I will consider it. For now, you can just set $n_\eta(x)\to(\Theta(x)-\eta)/2$ to get the zero temperature result.
Feb 16, 2016 at 11:30 comment added Medulla Oblongata Thanks. Just a suggestion, you could also include a zero temperature calculation option in the Mathematica package.
Feb 14, 2016 at 8:04 vote accept Medulla Oblongata
Feb 14, 2016 at 7:03 history edited Everett You CC BY-SA 3.0
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Feb 12, 2016 at 17:25 comment added Everett You @MedullaOblongata No, the result is invariant under the transformation $\xi_3\to-\xi_3$. The first term and the third term simply interchange, and the result remains the same. I do not see why +/- matters in the result.
Feb 12, 2016 at 10:27 comment added Medulla Oblongata If I set $\xi_3$ to be complex-valued, it doesn't matter if it's $\xi_3=\pm ia$ in $S_\eta$. However the +/- does matter in the result of the summation.
Feb 12, 2016 at 8:45 vote accept Medulla Oblongata
Feb 12, 2016 at 10:29
Feb 12, 2016 at 8:41 comment added Everett You @MedullaOblongata Yes, they can be imaginary. You can flip the sign back if you assume them to be imaginary. The result holds for generic complex $\xi$'s. The reason I change you sign is just to avoid a lot of imaginary units in the result.
Feb 12, 2016 at 8:30 comment added Medulla Oblongata Just to check - I know you changed the signs on my original expression e.g. $+\xi_3^2 \to -\xi_3^2$ and $+\xi_5^2 \to -\xi_5^2$. Do the $\xi$ constants have to be real or can they be imaginary?
Feb 12, 2016 at 8:15 history answered Everett You CC BY-SA 3.0