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----
-title: "Matsubara sum"
-sort_title: "Matsubara sum"
-date: 2021-11-13
-categories:
-- Physics
-- Quantum mechanics
-layout: "concept"
----
-
-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/):
-
-$$\begin{aligned}
-    \boxed{
-        S_{B,F}
-        \equiv \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty g(i \omega_n) \: e^{i \omega_n \tau}
-    }
-\end{aligned}$$
-
-$$g(z)$$ 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.