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+---
+title: "Matsubara sum"
+firstLetter: "M"
+publishDate: 2021-11-13
+categories:
+- Physics
+- Quantum mechanics
+
+date: 2021-11-05T15:19:38+01:00
+draft: false
+markup: pandoc
+---
+
+# Matsubara sum
+
+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 $\Re(z) > 0$ 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.