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:

SB,F1βn=g(iωn)eiωnτ\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)g(z) is a meromorphic function on the complex frequency plane, i.e. it is holomorphic except for a known set of simple poles, and τ[β,β]\tau \in [-\hbar \beta, \hbar \beta] is a real parameter. The Matsubara frequencies iωni \omega_n are defined as follows for bosons (subscript BB) or fermions (subscript FF):

ωn{2nπβbosons(2n+1)πβfermions\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 CC, recall that the residue theorem turns an integral over CC into a sum of the residues of all the integrand’s simple poles pgp_g that are enclosed by CC:

Cg(z)ezτi2πdz=pgReszpg{g(z)ezτ}=pgReszpg{g(z)}epgτ\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)h(z), which crucially does not share any simple poles with g(z)g(z), so {pg}{ph}=\{p_g\} \cap \{p_h\} = \emptyset. This constraint allows us to split the sum:

Cg(z)h(z)ezτi2πdz=pgReszpg{g(z)h(z)ezτ}+phReszph{g(z)h(z)ezτ}=pgReszpg{g(z)}h(pg)epgτ+phg(ph)Reszph{h(z)}ephτ\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 hh such that it has poles at iωni \omega_n. We make the following choice, where nB(z)n_B(z) is the Bose-Einstein distribution for bosons, and nF(z)n_F(z) is the Fermi-Dirac distribution for fermions:

h(z){nB,F(z)if  τ0nB,F(z)if  τ0\begin{aligned} h(z) \equiv \begin{cases} n_{B,F}(z) & \mathrm{if}\; \tau \ge 0 \\ -n_{B,F}(-z) & \mathrm{if}\; \tau \le 0 \end{cases} \end{aligned}

The distinction between the signs of τ\tau is necessary to ensure that h(z)ezτ0h(z) \: e^{z \tau} \to 0 for all zz when z|z| \to \infty (take a moment to convince yourself of this). The sign flip for τ0\tau \le 0 is also needed, as negating the argument negates the residues Res{nB,F(iωn)}=Res{nB,F(iωn)}\mathrm{Res}\{ n_{B,F}(-i \omega_n) \} = -\mathrm{Res}\{ n_{B,F}(i \omega_n) \}.

Indeed, this choice of hh has poles at the respective Matsubara frequencies iωni \omega_n of bosons and fermions, and the residues are given by:

Resziωn ⁣{nB(z)}=limziωn ⁣(ziωneβz1)=limη0 ⁣(iωn+ηiωneiβωneβη1)=limη0 ⁣(ηeβη1)=limη0 ⁣(η1+βη1)=1βResziωn ⁣{nF(z)}=limziωn ⁣(ziωneβz+1)=limη0 ⁣(iωn+ηiωneiβωneβη+1)=limη0 ⁣(ηeβη+1)=limη0 ⁣(η1βη+1)=1β\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:

Cg(z)h(z)ezτi2πdz=pgReszpg{g(z)}nB,F(pg)epgτ+iωng(iωn)Resziωn ⁣{nB,F(z)}eiωnτ=pgReszpg{g(z)}nB,F(pg)epgτ±1βn=g(iωn)eiωnτ\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 SF,BS_{F,B}. Isolating for that yields:

SB,F=pgReszpg{g(z)}nB,F(pg)epgτ±Cg(z)h(z)ezτi2πdz\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 CC. Earlier, we took care that h(z)ezτ0h(z) \: e^{z \tau} \to 0 for z|z| \to \infty, so a good choice would be a circle of radius RR. If RR \to \infty, then CC encloses the whole complex plane, including all of the integrand’s poles. However, because the integrand decays for z|z| \to \infty, we conclude that the contour integral must vanish (also for other choices of CC):

C=Reiθ    limRCg(z)h(z)ezτdz=0\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 SB,FS_{B,F}:

SB,F=pgReszpg{g(z)}nB,F(pg)epgτ\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}


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