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/lindhard-function/index.md | 23 +++++++++--------------
 1 file changed, 9 insertions(+), 14 deletions(-)

(limited to 'source/know/concept/lindhard-function/index.md')

diff --git a/source/know/concept/lindhard-function/index.md b/source/know/concept/lindhard-function/index.md
index 4033148..fd620df 100644
--- a/source/know/concept/lindhard-function/index.md
+++ b/source/know/concept/lindhard-function/index.md
@@ -149,11 +149,8 @@ $$\begin{aligned}
     = \hat{n}^\dagger(\vb{q})
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-density"/>
-<label for="proof-density">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-density">Proof.</label>
+
+{% include proof/start.html id="proof-density" -%}
 Starting from the general definition of $$\hat{n}$$,
 we write out the field operators $$\hat{\Psi}(\vb{r})$$,
 and insert the known non-interacting single-electron orbitals
@@ -210,8 +207,8 @@ $$\begin{aligned}
 
 The summation variable $$\vb{k}$$ has an associated spin $$\sigma$$,
 and $$\hat{n}$$ does not carry any spin.
-</div>
-</div>
+{% include proof/end.html id="proof-density" %}
+
 
 When neglecting interactions, it is tradition to rename $$\chi$$ to $$\chi_0$$.
 We insert $$\hat{n}$$, suppressing spin:
@@ -290,12 +287,10 @@ $$\begin{aligned}
     = \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}} - \hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k} + \vb{q}}
 \end{aligned}$$
 
-<div class="accordion">
-<input type="checkbox" id="proof-commutator"/>
-<label for="proof-commutator">Proof</label>
-<div class="hidden" markdown="1">
-<label for="proof-commutator">Proof.</label>
+
+{% include proof/start.html id="proof-commutator" -%}
 In general, for any single-particle states labeled by $$m$$, $$n$$, $$o$$ and $$p$$, we have:
+
 $$\begin{aligned}
     \comm{\hat{c}_m^\dagger \hat{c}_n}{\hat{c}_o^\dagger \hat{c}_p}
     &= \hat{c}_m^\dagger \hat{c}_n \hat{c}_o^\dagger \hat{c}_p - \hat{c}_o^\dagger \hat{c}_p \hat{c}_m^\dagger \hat{c}_n
@@ -319,8 +314,8 @@ $$\begin{aligned}
 
 In this case, $$m = p = \vb{k}$$ and $$n = o = \vb{k} \!+\! \vb{q}$$,
 so the Kronecker deltas are unnecessary.
-</div>
-</div>
+{% include proof/end.html id="proof-commutator" %}
+
 
 We substitute this result into $$\chi_0$$,
 and reintroduce the spin index $$\sigma$$ associated with $$\vb{k}$$:
-- 
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