From 7c412050570ef229dd78cbcffbf80f23728a630d Mon Sep 17 00:00:00 2001 From: Prefetch Date: Sat, 13 May 2023 15:42:47 +0200 Subject: Improve knowledge base --- source/know/concept/matsubara-sum/index.md | 137 +++++++++++++++++------------ 1 file changed, 82 insertions(+), 55 deletions(-) (limited to 'source/know/concept/matsubara-sum') diff --git a/source/know/concept/matsubara-sum/index.md b/source/know/concept/matsubara-sum/index.md index aef8379..0e04455 100644 --- a/source/know/concept/matsubara-sum/index.md +++ b/source/know/concept/matsubara-sum/index.md @@ -14,60 +14,88 @@ which notably appears as the inverse [Matsubara Green's function](/know/concept/matsubara-greens-function/): $$\begin{aligned} - S_{B,F} - = \frac{1}{\hbar \beta} \sum_{i \omega_n} g(i \omega_n) \: e^{i \omega_n \tau} + \boxed{ + S_{B,F} + \equiv \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty g(i \omega_n) \: e^{i \omega_n \tau} + } \end{aligned}$$ -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/) +$$g(z)$$ 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 $$\tau$$ is a real parameter -(e.g. the [imaginary time](/know/concept/imaginary-time/)) -satisfying $$-\hbar \beta < \tau < \hbar \beta$$. +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 $$F$$): + +$$\begin{aligned} + \omega_n \equiv + \begin{cases} + \displaystyle\frac{2 n \pi}{\hbar \beta} + & \mathrm{bosons} + \\ + \displaystyle\frac{(2 n + 1) \pi}{\hbar \beta} + & \mathrm{fermions} + \end{cases} +\end{aligned}$$ -Now, consider the following integral -over a (for now) unspecified counter-clockwise contour $$C$$, -with a (for now) unspecified weighting function $$h(z)$$: +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 $$C$$: $$\begin{aligned} - \oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z} - = \sum_{z_p} e^{z_p \tau} \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) h(z) \big) + \oint_C \frac{g(z) \: e^{z \tau}}{i 2 \pi} \dd{z} + = \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \: e^{z \tau} \Big\} + = \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \Big\} \: e^{p_g \tau} \end{aligned}$$ -Where we have applied the [residue theorem](/know/concept/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 $$i \omega_n$$. +Now, the trick is to manipulate this relation +until a Matsubara sum appears on the right. -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/): +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: $$\begin{aligned} h(z) - = + \equiv \begin{cases} n_{B,F}(z) & \mathrm{if}\; \tau \ge 0 \\ -n_{B,F}(-z) & \mathrm{if}\; \tau \le 0 \end{cases} - \qquad \qquad - n_{B,F}(z) - = \frac{1}{e^{\hbar \beta z} \mp 1} \end{aligned}$$ -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 $$\Real(z) > 0$$ and $$\Real(z) < 0$$. -This choice of $$h$$ indeed has poles at the respective +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: +and the residues are given by: $$\begin{aligned} - \underset{z \to i \omega_n}{\mathrm{Res}}\!\big( n_B(z) \big) + \underset{z \to i \omega_n}{\mathrm{Res}}\!\Big\{ n_B(z) \Big\} &= \lim_{z \to i \omega_n}\!\bigg( \frac{z - i \omega_n}{e^{\hbar \beta z} - 1} \bigg) = \lim_{\eta \to 0}\!\bigg( \frac{i \omega_n + \eta - i \omega_n}{e^{i \hbar \beta \omega_n} e^{\hbar \beta \eta} - 1} \bigg) \\ @@ -75,7 +103,7 @@ $$\begin{aligned} = \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) + \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) \\ @@ -84,37 +112,36 @@ $$\begin{aligned} = - \frac{1}{\hbar \beta} \end{aligned}$$ -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: +With this, our contour integral can now be rewritten as follows: $$\begin{aligned} - \oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z} - &= \sum_{z_p} e^{z_p \tau} n_{B,F}(z_p) \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) \big) - + \sum_{i \omega_n} e^{i \omega_n \tau} g(i \omega_n) \: \underset{z \to i \omega_n}{\mathrm{Res}}\!\big( n_{B,F}(z) \big) + \oint_C \frac{g(z) \: h(z) \: e^{z \tau}}{i 2 \pi} \dd{z} + &= \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \Big\} \: n_{B,F}(p_g) \: e^{p_g \tau} + + \sum_{i \omega_n} g(i \omega_n) \underset{z \to i \omega_n}{\mathrm{Res}}\!\Big\{ n_{B,F}(z) \Big\} \: e^{i \omega_n \tau} \\ - &= \sum_{z_p} e^{z_p \tau} n_{B,F}(z_p) \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) \big) - \pm \frac{1}{\hbar \beta} \sum_{i \omega_n} g(i \omega_n) \: e^{i \omega_n \tau} + &= \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \Big\} \: n_{B,F}(p_g) \: e^{p_g \tau} + \pm \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty g(i \omega_n) \: e^{i \omega_n \tau} \end{aligned}$$ -Where $$+$$ is for bosons, and $$-$$ for fermions. -Here, we recognize the last term as the Matsubara sum $$S_{F,B}$$, -for which we isolate, yielding: +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: $$\begin{aligned} S_{B,F} - = \mp \sum_{z_p} e^{z_p \tau} n_{B,F}(z_p) \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) \big) - \pm \oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z} + = \mp \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \Big\} \: n_{B,F}(p_g) \: e^{p_g \tau} + \pm \oint_C \frac{g(z) \: h(z) \: e^{z \tau}}{i 2 \pi} \dd{z} \end{aligned}$$ -Now we must choose $$C$$. 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: +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 $$C$$): $$\begin{aligned} C @@ -131,7 +158,7 @@ for bosonic and fermionic Matsubara sums $$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) + = \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}$$ -- cgit v1.2.3