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+---
+title: "Matsubara summation"
+sort_title: "Matsubara summation"
+date: 2021-11-13
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+**Matsubara summation** is a technique for evaluating sums 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/):
+
+$$\begin{aligned}
+    \boxed{
+        S_{B,F}
+        \equiv \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty g(i \omega_n) \: e^{i \omega_n \tau}
+    }
+\end{aligned}$$
+
+$$g(z)$$ is a *meromorphic* function on the complex frequency plane,
+i.e. it is [holomorphic](/know/concept/holomorphic-function/)
+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](/know/concept/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](/know/concept/bose-einstein-distribution/) for bosons,
+and $$n_F(z)$$ is the [Fermi-Dirac distribution](/know/concept/fermi-dirac-distribution/) for fermions:
+
+$$\begin{aligned}
+    h(z)
+    \equiv
+    \begin{cases}
+        n_{B,F}(z) & \mathrm{if}\; 0 \le \tau < \hbar \beta
+        \\
+        -n_{B,F}(-z) & \mathrm{if}\; \!-\!\hbar \beta < \tau \le 0
+    \end{cases}
+\end{aligned}$$
+
+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.