---
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/):

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

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 $$\Real(z) > 0$$ and $$\Real(z) < 0$$.
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.