---

Categories: Physics, Quantum mechanics.

Matsubara sum

A Matsubara sum is a summation of the following form, which notably appears as the inverse Fourier transform of the Matsubara Green’s 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 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 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 for bosons, and $n_F(z)$ is the 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.

---