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---
title: "Matsubara sum"
firstLetter: "M"
publishDate: 2021-11-13
categories:
- Physics
- Quantum mechanics
date: 2021-11-05T15:19:38+01:00
draft: false
markup: pandoc
---
# Matsubara sum
A **Matsubara sum** is a summation of the following form,
which notably appears as the inverse
[Fourier transform](/know/concept/fourier-transform/) of the
[Matsubara Green's function](/know/concept/matsubara-greens-function/):
$$\begin{aligned}
S_{B,F}
= \frac{1}{\hbar \beta} \sum_{i \omega_n} g(i \omega_n) \: e^{i \omega_n \tau}
\end{aligned}$$
Where $i \omega_n$ are the Matsubara frequencies
for bosons ($B$) or fermions ($F$),
and $g(z)$ is a function on the complex plane
that is [holomorphic](/know/concept/holomorphic-function/)
except for a known set of simple poles,
and $\tau$ is a real parameter
(e.g. the [imaginary time](/know/concept/imaginary-time/))
satisfying $-\hbar \beta < \tau < \hbar \beta$.
Now, consider the following integral
over a (for now) unspecified counter-clockwise contour $C$,
with a (for now) unspecified weighting function $h(z)$:
$$\begin{aligned}
\oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z}
= \sum_{z_p} e^{z_p \tau} \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) h(z) \big)
\end{aligned}$$
Where we have applied the residue theorem
to get a sum over all simple poles $z_p$
of either $g$ or $h$ (but not both) enclosed by $C$.
Clearly, we could make this look like a Matsubara sum,
if we choose $h$ such that it has poles at $i \omega_n$.
Therefore, we choose the weighting function $h(z)$ as follows,
where $n_B(z)$ is the [Bose-Einstein distribution](/know/concept/bose-einstein-distribution/),
and $n_F(z)$ is the [Fermi-Dirac distribution](/know/concept/fermi-dirac-distribution/):
$$\begin{aligned}
h(z)
=
\begin{cases}
n_{B,F}(z) & \mathrm{if}\; \tau \ge 0
\\
-n_{B,F}(-z) & \mathrm{if}\; \tau \le 0
\end{cases}
\qquad \qquad
n_{B,F}(z)
= \frac{1}{e^{\hbar \beta z} \mp 1}
\end{aligned}$$
The distinction between the signs of $\tau$ is needed
to ensure that the integrand $h(z) e^{z \tau}$ decays for $|z| \to \infty$,
both for $\Re(z) > 0$ and $\Re(z) < 0$.
This choice of $h$ indeed has poles at the respective
Matsubara frequencies $i \omega_n$ of bosons and fermions,
and the residues are:
$$\begin{aligned}
\underset{z \to i \omega_n}{\mathrm{Res}}\!\big( n_B(z) \big)
&= \lim_{z \to i \omega_n}\!\bigg( \frac{z - i \omega_n}{e^{\hbar \beta z} - 1} \bigg)
= \lim_{\eta \to 0}\!\bigg( \frac{i \omega_n + \eta - i \omega_n}{e^{i \hbar \beta \omega_n} e^{\hbar \beta \eta} - 1} \bigg)
\\
&= \lim_{\eta \to 0}\!\bigg( \frac{\eta}{e^{\hbar \beta \eta} - 1} \bigg)
= \lim_{\eta \to 0}\!\bigg( \frac{\eta}{1 + \hbar \beta \eta - 1} \bigg)
= \frac{1}{\hbar \beta}
\\
\underset{z \to i \omega_n}{\mathrm{Res}}\!\big( n_F(z) \big)
&= \lim_{z \to i \omega_n}\!\bigg( \frac{z - i \omega_n}{e^{\hbar \beta z} + 1} \bigg)
= \lim_{\eta \to 0}\!\bigg( \frac{i \omega_n + \eta - i \omega_n}{e^{i \hbar \beta \omega_n} e^{\hbar \beta \eta} + 1} \bigg)
\\
&= \lim_{\eta \to 0}\!\bigg( \frac{\eta}{e^{\hbar \beta \eta} + 1} \bigg)
= \lim_{\eta \to 0}\!\bigg( \frac{\eta}{- 1 - \hbar \beta \eta + 1} \bigg)
= - \frac{1}{\hbar \beta}
\end{aligned}$$
In the definition of $h$, the sign flip for $\tau \le 0$
is introduced because negating the argument also negates the residues,
i.e. $\mathrm{Res}\big( n_F(-z) \big) = -\mathrm{Res}\big( n_F(z) \big)$.
With this $h$, our contour integral can be rewritten as follows:
$$\begin{aligned}
\oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z}
&= \sum_{z_p} e^{z_p \tau} n_{B,F}(z_p) \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) \big)
+ \sum_{i \omega_n} e^{i \omega_n \tau} g(i \omega_n) \: \underset{z \to i \omega_n}{\mathrm{Res}}\!\big( n_{B,F}(z) \big)
\\
&= \sum_{z_p} e^{z_p \tau} n_{B,F}(z_p) \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) \big)
\pm \frac{1}{\hbar \beta} \sum_{i \omega_n} g(i \omega_n) \: e^{i \omega_n \tau}
\end{aligned}$$
Where $+$ is for bosons, and $-$ for fermions.
Here, we recognize the last term as the Matsubara sum $S_{F,B}$,
for which we isolate, yielding:
$$\begin{aligned}
S_{B,F}
= \mp \sum_{z_p} e^{z_p \tau} n_{B,F}(z_p) \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) \big)
\pm \oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z}
\end{aligned}$$
Now we must choose $C$. Assuming $g(z)$ does not interfere,
we know that $h(z) e^{z \tau}$ decays to zero
for $|z| \to \infty$, so a useful choice would be a circle of radius $R$.
If we then let $R \to \infty$, the contour encloses
the whole complex plane, including all of the integrand's poles.
However, thanks to the integrand's decay,
the resulting contour integral must vanish:
$$\begin{aligned}
C
= R e^{i \theta}
\quad \implies \quad
\lim_{R \to \infty}
\oint_C g(z) \: h(z) \: e^{z \tau} \dd{z}
= 0
\end{aligned}$$
We thus arrive at the following results
for bosonic and fermionic Matsubara sums $S_{B,F}$:
$$\begin{aligned}
\boxed{
S_{B,F}
= \mp \sum_{z_p} e^{z_p \tau} n_{B,F}(z_p) \: \underset{{z \to z_p}}{\mathrm{Res}}\big(g(z)\big)
}
\end{aligned}$$
## References
1. H. Bruus, K. Flensberg,
*Many-body quantum theory in condensed matter physics*,
2016, Oxford.
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