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 \(S_{B,F}\):

\[\begin{aligned} \boxed{ S_{B,F} = \mp \sum_{z_p} e^{z_p \tau} n_{B,F}(z_p) \: \underset{{z \to z_p}}{\mathrm{Res}}\big(g(z)\big) } \end{aligned}\]

References

  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.
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