---
title: "Matsubara sum"
sort_title: "Matsubara sum"
date: 2021-11-13
categories:
- Physics
- Quantum mechanics
layout: "concept"
---

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/):

$$\begin{aligned}
    \boxed{
        S_{B,F}
        \equiv \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty g(i \omega_n) \: e^{i \omega_n \tau}
    }
\end{aligned}$$

$$g(z)$$ is a *meromorphic* function on the complex frequency plane,
i.e. it is [holomorphic](/know/concept/holomorphic-function/)
except for a known set of simple poles,
and $$\tau \in [-\hbar \beta, \hbar \beta]$$ is a real parameter.
The Matsubara frequencies $$i \omega_n$$ are defined as follows
for bosons (subscript $$B$$) or fermions (subscript $$F$$):

$$\begin{aligned}
    \omega_n \equiv
    \begin{cases}
        \displaystyle\frac{2 n \pi}{\hbar \beta}
        & \mathrm{bosons}
        \\
        \displaystyle\frac{(2 n + 1) \pi}{\hbar \beta}
        & \mathrm{fermions}
    \end{cases}
\end{aligned}$$

How do we evaluate Matsubara sums?
Given a counter-clockwise closed contour $$C$$,
recall that the [residue theorem](/know/concept/residue-theorem/)
turns an integral over $$C$$ into a sum of the residues
of all the integrand's simple poles $$p_g$$ that are enclosed by $$C$$:

$$\begin{aligned}
    \oint_C \frac{g(z) \: e^{z \tau}}{i 2 \pi} \dd{z}
    = \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \: e^{z \tau} \Big\}
    = \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \Big\} \: e^{p_g \tau}
\end{aligned}$$

Now, the trick is to manipulate this relation
until a Matsubara sum appears on the right.

Let us introduce a (for now) unspecified weight function $$h(z)$$,
which crucially does not share any simple poles with $$g(z)$$,
so $$\{p_g\} \cap \{p_h\} = \emptyset$$.
This constraint allows us to split the sum:

$$\begin{aligned}
    \oint_C \frac{g(z) \: h(z) \: e^{z \tau}}{i 2 \pi} \dd{z}
    &= \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \: h(z) \: e^{z \tau} \Big\}
    + \sum_{p_h} \underset{z \to p_h}{\mathrm{Res}}\Big\{ g(z) \: h(z) \: e^{z \tau} \Big\}
    \\
    &= \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \Big\} \: h(p_g) \: e^{p_g \tau}
    + \sum_{p_h} g(p_h) \: \underset{z \to p_h}{\mathrm{Res}}\Big\{ h(z) \Big\} \: e^{p_h \tau}
\end{aligned}$$

Here, we could make the rightmost term look like a Matsubara sum
if we choose $$h$$ such that it has poles at $$i \omega_n$$.
We make the following choice,
where $$n_B(z)$$ is the [Bose-Einstein distribution](/know/concept/bose-einstein-distribution/) for bosons,
and $$n_F(z)$$ is the [Fermi-Dirac distribution](/know/concept/fermi-dirac-distribution/) for fermions:

$$\begin{aligned}
    h(z)
    \equiv
    \begin{cases}
        n_{B,F}(z) & \mathrm{if}\; \tau \ge 0
        \\
        -n_{B,F}(-z) & \mathrm{if}\; \tau \le 0
    \end{cases}
\end{aligned}$$

The distinction between the signs of $$\tau$$ is necessary
to ensure that $$h(z) \: e^{z \tau} \to 0$$ for all $$z$$ when $$|z| \to \infty$$
(take a moment to convince yourself of this).
The sign flip for $$\tau \le 0$$ is also needed,
as negating the argument negates the residues
$$\mathrm{Res}\{ n_{B,F}(-i \omega_n) \} = -\mathrm{Res}\{ n_{B,F}(i \omega_n) \}$$.

Indeed, this choice of $$h$$ has poles at the respective
Matsubara frequencies $$i \omega_n$$ of bosons and fermions,
and the residues are given by:

$$\begin{aligned}
    \underset{z \to i \omega_n}{\mathrm{Res}}\!\Big\{ n_B(z) \Big\}
    &= \lim_{z \to i \omega_n}\!\bigg( \frac{z - i \omega_n}{e^{\hbar \beta z} - 1} \bigg)
    = \lim_{\eta \to 0}\!\bigg( \frac{i \omega_n + \eta - i \omega_n}{e^{i \hbar \beta \omega_n} e^{\hbar \beta \eta} - 1} \bigg)
    \\
    &= \lim_{\eta \to 0}\!\bigg( \frac{\eta}{e^{\hbar \beta \eta} - 1} \bigg)
    = \lim_{\eta \to 0}\!\bigg( \frac{\eta}{1 + \hbar \beta \eta - 1} \bigg)
    = \frac{1}{\hbar \beta}
    \\
    \underset{z \to i \omega_n}{\mathrm{Res}}\!\Big\{ n_F(z) \Big\}
    &= \lim_{z \to i \omega_n}\!\bigg( \frac{z - i \omega_n}{e^{\hbar \beta z} + 1} \bigg)
    = \lim_{\eta \to 0}\!\bigg( \frac{i \omega_n + \eta - i \omega_n}{e^{i \hbar \beta \omega_n} e^{\hbar \beta \eta} + 1} \bigg)
    \\
    &= \lim_{\eta \to 0}\!\bigg( \frac{\eta}{e^{\hbar \beta \eta} + 1} \bigg)
    = \lim_{\eta \to 0}\!\bigg( \frac{\eta}{- 1 - \hbar \beta \eta + 1} \bigg)
    = - \frac{1}{\hbar \beta}
\end{aligned}$$

With this, our contour integral can now be rewritten as follows:

$$\begin{aligned}
    \oint_C \frac{g(z) \: h(z) \: e^{z \tau}}{i 2 \pi} \dd{z}
    &= \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \Big\} \: n_{B,F}(p_g) \: e^{p_g \tau}
    + \sum_{i \omega_n} g(i \omega_n) \underset{z \to i \omega_n}{\mathrm{Res}}\!\Big\{ n_{B,F}(z) \Big\} \: e^{i \omega_n \tau}
    \\
    &= \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \Big\} \: n_{B,F}(p_g) \: e^{p_g \tau}
    \pm \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty g(i \omega_n) \: e^{i \omega_n \tau}
\end{aligned}$$

Where the top sign ($$+$$) is for bosons,
and the bottom sign ($$-$$) is for fermions.
Here, we recognize the last term as the Matsubara sum $$S_{F,B}$$.
Isolating for that yields:

$$\begin{aligned}
    S_{B,F}
    = \mp \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \Big\} \: n_{B,F}(p_g) \: e^{p_g \tau}
    \pm \oint_C \frac{g(z) \: h(z) \: e^{z \tau}}{i 2 \pi} \dd{z}
\end{aligned}$$

Now we must choose $$C$$.
Earlier, we took care that $$h(z) \: e^{z \tau} \to 0$$ for $$|z| \to \infty$$,
so a good choice would be a circle of radius $$R$$.
If $$R \to \infty$$, then $$C$$ encloses the whole complex plane,
including all of the integrand's poles.
However, because the integrand decays for $$|z| \to \infty$$,
we conclude that the contour integral must vanish
(also for other choices of $$C$$):

$$\begin{aligned}
    C
    = R e^{i \theta}
    \quad \implies \quad
    \lim_{R \to \infty}
    \oint_C g(z) \: h(z) \: e^{z \tau} \dd{z}
    = 0
\end{aligned}$$

We thus arrive at the following results
for bosonic and fermionic Matsubara sums $$S_{B,F}$$:

$$\begin{aligned}
    \boxed{
        S_{B,F}
        = \mp \sum_{p_g} \underset{ {z \to p_g}}{\mathrm{Res}}\Big\{ g(z) \Big\} \: n_{B,F}(p_g) \: e^{p_g \tau}
    }
\end{aligned}$$



## References
1.  H. Bruus, K. Flensberg,
    *Many-body quantum theory in condensed matter physics*,
    2016, Oxford.