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diff --git a/content/know/concept/matsubara-sum/index.pdc b/content/know/concept/matsubara-sum/index.pdc new file mode 100644 index 0000000..91183e6 --- /dev/null +++ b/content/know/concept/matsubara-sum/index.pdc @@ -0,0 +1,148 @@ +--- +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. |