From 6e70f28ccbd5afc1506f71f013278a9d157ef03a Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 27 Oct 2022 20:40:09 +0200
Subject: Optimize last images, add proof template, improve CSS
---
source/know/concept/greens-functions/index.md | 22 ++++++++--------------
1 file changed, 8 insertions(+), 14 deletions(-)
(limited to 'source/know/concept/greens-functions')
diff --git a/source/know/concept/greens-functions/index.md b/source/know/concept/greens-functions/index.md
index ddba2cd..eda5671 100644
--- a/source/know/concept/greens-functions/index.md
+++ b/source/know/concept/greens-functions/index.md
@@ -21,6 +21,7 @@ but in general they are not the same,
except in a special case, see below.
+
## Single-particle functions
If the two operators are single-particle creation/annihilation operators,
@@ -146,11 +147,8 @@ $$\begin{gathered}
G_{\nu \nu'}^<(t, t') = G_{\nu \nu'}^<(t - t')
\end{gathered}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-time-delta" -%}
We will prove that the thermal expectation value
$$\expval{\hat{A}(t) \hat{B}(t')}$$ only depends on $$t - t'$$
for arbitrary $$\hat{A}$$ and $$\hat{B}$$,
@@ -189,8 +187,7 @@ because $$\hat{H}$$ is time-independent by assumption.
Note that thermodynamic equilibrium is crucial:
intuitively, if the system is not in equilibrium,
then it evolves in some transient time-dependent way.
-
-
+{% include proof/end.html id="proof-time-delta" %}
If the Hamiltonian is both time-independent and non-interacting,
then the time-dependence of $$\hat{c}_\nu$$
@@ -214,6 +211,7 @@ $$\begin{aligned}
\end{aligned}$$
+
## As fundamental solutions
In the absence of interactions,
@@ -237,11 +235,8 @@ $$\begin{aligned}
= \frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r})
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-commutator" -%}
In the second quantization,
the Hamiltonian $$\hat{H}_0$$ is written like so:
@@ -307,9 +302,8 @@ $$\begin{aligned}
&= \frac{\hbar^2}{2 m} \sum_{\nu'} \hat{c}_{\nu'} \nabla^2 \psi_{\nu'}(\vb{r})
= \frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r})
\end{aligned}$$
+{% include proof/end.html id="proof-commutator" %}
-
-
After substituting this into the equation of motion,
we recognize $$G^R(\vb{r}, t; \vb{r}', t')$$ itself:
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