diff options
Diffstat (limited to 'source/know/concept/matsubara-sum/index.md')
| -rw-r--r-- | source/know/concept/matsubara-sum/index.md | 56 |
1 files changed, 28 insertions, 28 deletions
diff --git a/source/know/concept/matsubara-sum/index.md b/source/know/concept/matsubara-sum/index.md index 45381ba..8b903d4 100644 --- a/source/know/concept/matsubara-sum/index.md +++ b/source/know/concept/matsubara-sum/index.md @@ -18,18 +18,18 @@ $$\begin{aligned} = \frac{1}{\hbar \beta} \sum_{i \omega_n} g(i \omega_n) \: e^{i \omega_n \tau} \end{aligned}$$ -Where $i \omega_n$ are the Matsubara frequencies -for bosons ($B$) or fermions ($F$), -and $g(z)$ is a function on the complex plane +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 +and $$\tau$$ is a real parameter (e.g. the [imaginary time](/know/concept/imaginary-time/)) -satisfying $-\hbar \beta < \tau < \hbar \beta$. +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 $h(z)$: +over a (for now) unspecified counter-clockwise contour $$C$$, +with a (for now) unspecified weighting function $$h(z)$$: $$\begin{aligned} \oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z} @@ -37,14 +37,14 @@ $$\begin{aligned} \end{aligned}$$ Where we have applied the residue theorem -to get a sum over all simple poles $z_p$ -of either $g$ or $h$ (but not both) enclosed by $C$. +to get a sum over all simple poles $$z_p$$ +of either $$g$$ or $$h$$ (but not both) enclosed by $$C$$. Clearly, we could make this look like a Matsubara sum, -if we choose $h$ such that it has poles at $i \omega_n$. +if we choose $$h$$ such that it has poles at $$i \omega_n$$. -Therefore, we choose the weighting function $h(z)$ as follows, -where $n_B(z)$ is the [Bose-Einstein distribution](/know/concept/bose-einstein-distribution/), -and $n_F(z)$ is the [Fermi-Dirac distribution](/know/concept/fermi-dirac-distribution/): +Therefore, we choose the weighting function $$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) @@ -59,11 +59,11 @@ $$\begin{aligned} = \frac{1}{e^{\hbar \beta z} \mp 1} \end{aligned}$$ -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 -Matsubara frequencies $i \omega_n$ of bosons and fermions, +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 +Matsubara frequencies $$i \omega_n$$ of bosons and fermions, and the residues are: $$\begin{aligned} @@ -84,10 +84,10 @@ $$\begin{aligned} = - \frac{1}{\hbar \beta} \end{aligned}$$ -In the definition of $h$, the sign flip for $\tau \le 0$ +In the definition of $$h$$, the sign flip for $$\tau \le 0$$ is introduced because negating the argument also negates the residues, -i.e. $\mathrm{Res}\big( n_F(-z) \big) = -\mathrm{Res}\big( n_F(z) \big)$. -With this $h$, our contour integral can be rewritten as follows: +i.e. $$\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} @@ -98,8 +98,8 @@ $$\begin{aligned} \pm \frac{1}{\hbar \beta} \sum_{i \omega_n} g(i \omega_n) \: e^{i \omega_n \tau} \end{aligned}$$ -Where $+$ is for bosons, and $-$ for fermions. -Here, we recognize the last term as the Matsubara sum $S_{F,B}$, +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: $$\begin{aligned} @@ -108,10 +108,10 @@ $$\begin{aligned} \pm \oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z} \end{aligned}$$ -Now we must choose $C$. Assuming $g(z)$ does not interfere, -we know that $h(z) e^{z \tau}$ decays to zero -for $|z| \to \infty$, so a useful choice would be a circle of radius $R$. -If we then let $R \to \infty$, the contour encloses +Now we must choose $$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$$, the contour encloses the whole complex plane, including all of the integrand's poles. However, thanks to the integrand's decay, the resulting contour integral must vanish: @@ -126,7 +126,7 @@ $$\begin{aligned} \end{aligned}$$ We thus arrive at the following results -for bosonic and fermionic Matsubara sums $S_{B,F}$: +for bosonic and fermionic Matsubara sums $$S_{B,F}$$: $$\begin{aligned} \boxed{ |