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How do I compute the Matsubara sum
$$\sum_n \log\left(-i\omega_n +\frac{k^2}{2m}+\mu\right)?$$
If I have sums like $\sum_n \frac{1}{i\omega_n -m}$, I can sum it up by calculating the sum of residues of the function $\frac{1}{z-m}g(z)$ at the poles where $g(z)=\begin{cases} \frac{\beta}{\exp (\beta z)+1} \text{ for Fermions}\\ \frac{\beta}{\exp (\beta z)-1} \text{ for Bosons} \end{cases}$
But how do I do I compute in this case where there is a $\log$ term and there are no poles.
For this kind of stuff you usually integrate by parts.
First change your sum to an integral:
$$\sum_n \log\left(-i\omega_n +\frac{k^2}{2m}+\mu\right) \Rightarrow \int \mathrm{d}\omega \, \log(f - \mathrm{i}\omega), $$ where $f$ here is $k^2/2m+\mu$ which I am assuming are not functions of $\omega$.
Then integrate by parts: $$\int \mathrm{d}\omega \, \underbrace{1}_{u'} \cdot\underbrace{\log(f - \mathrm{i}\omega)}_v = \underbrace{\omega}_{u}\underbrace{\log(f - \mathrm{i}\omega)}_{v}\bigg\vert_0^{\omega_{\text{max}}} - \int\mathrm{d}\omega\, \underbrace{\omega}_u\cdot\underbrace{\frac{\mathrm{-i}}{f-\mathrm{i}\omega}}_{v'}$$ $$\Rightarrow \omega_{\text{max}}\log(f - \mathrm{i}\omega_{\text{max}})+ \int_0^{\omega_{\text{max}}} \mathrm{d}\omega\,\omega\cdot\frac{\mathrm{i}}{f-\mathrm{i}\omega}.$$
The first term is a constant energy offset. Usually it cancels out when you consider differences in energy.
S=$\sum_n\ln(-i\omega_n+\epsilon)=\sum_n\ln(i\omega_n-\epsilon)+C$
(usually this is an action and the constant is irrelevant. This transform is unneccessary just for convenient)
$=\mathrm{Res}\left\{\ln(z-\epsilon)g(z)\right\}$
when $g(z)$ is what you've mentioned. Then the problem is to evaluate this integral. We could select the branch cut as $(-\infty,\epsilon)$ and the contour $-\infty+i\delta\to\epsilon\to(small\;circle\;around\;\epsilon)\to\epsilon\to-\infty-i\delta\to(circle\;in\;infinte\;distant)$ and $$S=\frac{1}{2\pi i}\int_C\ln(z-\epsilon)g(z)=\frac{1}{2\pi i}\int_{-\infty}^{\infty}(\ln(z^+-\epsilon)-\ln(z^--\epsilon))g(z)$$ using $g(z)=\xi\partial_z\ln(1-\xi e^{-\beta z})$ and integrate by part $$S=-\frac{\xi}{2\pi i}\int_{-\infty}^{\infty}\ln(1-\xi e^{-\beta\epsilon})\left(\frac{1}{z+i\delta-\epsilon}-\frac{1}{z-i\delta-\epsilon}\right)$$ and the relation $\lim_{\delta\to0}\frac{1}{x+i\delta}=\frac{1}{x}-i\pi\delta(x)$ we could get the result.
Reference: Altland,Simons Condense Matter Field Theory
One possible approach is writing the sum of logs as a log of product and using the formulas for infinite products in Gradshtein and Ryzhik.
