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} 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 except for a known set of simple poles, and $$\tau$$ is a real parameter (e.g. the 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, and $$n_F(z)$$ is the 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}

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

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.