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By properties such as linearity, shifting, commutativity, etc. I was hoping to evaluate something like,

$$S_\eta = \dfrac{1}{\beta}\displaystyle\sum_{i\omega} \dfrac{i\omega-\xi_1}{[(i\omega-\xi_2)^2-\xi_3^2][(i\omega-\xi_4)^2-\xi_5^2]}$$

by using the results in this table. If not, would it be best to use partial fraction decomposition or is there another method?

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  • $\begingroup$ What do you mean "evaluate"? The thing there is just a fraction... there's no integral or sum anywhere in sight. Anyway, I would guess that yes, partial fractions is going to help you a lot. $\endgroup$
    – DanielSank
    Commented Feb 12, 2016 at 7:00

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The result of your summation is $$\frac{1}{2} \eta \left(-\frac{\left(\xi _1-\xi _2+\xi _3\right) n_{\eta }\left(\xi _2-\xi _3\right)}{\xi _3 \left(\xi _2^2-2 \left(\xi _3+\xi _4\right) \xi _2+\xi _3^2+\xi _4^2-\xi _5^2+2 \xi _3 \xi _4\right)}-\frac{\left(\xi _1-\xi _4+\xi _5\right) n_{\eta }\left(\xi _4-\xi _5\right)}{\left(\xi _2^2-2 \left(\xi _4-\xi _5\right) \xi _2-\xi _3^2+\left(\xi _4-\xi _5\right)^2\right) \xi _5}-\frac{\left(-\xi _1+\xi _2+\xi _3\right) n_{\eta }\left(\xi _2+\xi _3\right)}{\xi _3 \left(\xi _2^2+2 \left(\xi _3-\xi _4\right) \xi _2+\xi _3^2+\xi _4^2-\xi _5^2-2 \xi _3 \xi _4\right)}-\frac{\left(-\xi _1+\xi _4+\xi _5\right) n_{\eta }\left(\xi _4+\xi _5\right)}{\xi _5 \left(\xi _2^2-2 \left(\xi _4+\xi _5\right) \xi _2-\xi _3^2+\left(\xi _4+\xi _5\right)^2\right)}\right)$$

The Matsubara frequency summation can be calculated analytically using the Mathematica package MatsubaraSum, which can be found at this Github repository: https://github.com/EverettYou/MatsubaraSum.

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  • $\begingroup$ 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? $\endgroup$
    – Medulla Oblongata
    Commented Feb 12, 2016 at 8:30
  • $\begingroup$ @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. $\endgroup$
    – Everett You
    Commented Feb 12, 2016 at 8:41
  • $\begingroup$ 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. $\endgroup$
    – Medulla Oblongata
    Commented Feb 12, 2016 at 10:27
  • $\begingroup$ @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. $\endgroup$
    – Everett You
    Commented Feb 12, 2016 at 17:25
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    $\begingroup$ @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. $\endgroup$
    – Everett You
    Commented Feb 17, 2016 at 17:35

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