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Diffstat (limited to 'source/know/concept/lindhard-function')
| -rw-r--r-- | source/know/concept/lindhard-function/index.md | 23 |
1 files changed, 9 insertions, 14 deletions
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}$$: |