# Matsubara sum with log term [closed]

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.

• What would $\omega_{\text{max}}$ be there? I thought the Matsubara frequencies have no upper bound. – user824530 Nov 10 '20 at 0:43
• Whatever your upper integration limit is. If it's infinity, then it will be an UV divergence. Which is usually eliminated by looking at differences of energies so that both contribute the same term and it cancels. – SuperCiocia Nov 10 '20 at 0:44

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

• What is $\xi$ in your expression for $g(z)$ also how are these different $g(z)$ chosen. I saw two different functions in Altland and Simon. The one you have is a different one. – user824530 Nov 10 '20 at 14:16

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.