X-Git-Url: https://hackdaworld.org/gitweb/?p=lectures%2Flatex.git;a=blobdiff_plain;f=physics_compact%2Fsolid.tex;h=50df24d29d03741e8a4efb3933ae5b98316dcbba;hp=20ecfef44e2132a8f2f8eb73c9dd44480b5cf773;hb=9fac15a186e73cf587fb7d9dfa1c737c778cd08a;hpb=a9f93985e52272cdecf902c4b559173a80f4b41d

diff --git a/physics_compact/solid.tex b/physics_compact/solid.tex
index 20ecfef..50df24d 100644
--- a/physics_compact/solid.tex
+++ b/physics_compact/solid.tex
@@ -27,21 +27,33 @@ Suppose two potentials $V_1$ and $V_2$ exist, which yield the same electron dens
 The corresponding Hamiltonians are denoted $H_1$ and $H_2$ with the respective ground-state wavefunctions $\Psi_1$ and $\Psi_2$ and eigenvalues $E_1$ and $E_2$.
 Then, due to the variational principle (see \ref{sec:var_meth}), one can write
 \begin{equation}
-E_1=\langle \Psi_1 | H_1 | \Psi_1 \rangle < \langle \Psi_2 | H_1 | \Psi_2 \rangle
+E_1=\langle \Psi_1 | H_1 | \Psi_1 \rangle <
+\langle \Psi_2 | H_1 | \Psi_2 \rangle \text{ .}
+\label{subsub:hk01}
 \end{equation}
-Expressing $H_1$ by $H_2+H_1-H_2$
+Expressing $H_1$ by $H_2+H_1-H_2$, the last part of \eqref{subsub:hk01} can be rewritten:
 \begin{equation}
 \langle \Psi_2 | H_1 | \Psi_2 \rangle =
 \langle \Psi_2 | H_2 | \Psi_2 \rangle +
 \langle \Psi_2 | H_1 -H_2 | \Psi_2 \rangle
 \end{equation}
-and the fact that the two Hamiltonians, which describe the same number of electrons, differ only in the potential
+The two Hamiltonians, which describe the same number of electrons, differ only in the potential
 \begin{equation}
 H_1-H_2=V_1(\vec{r})-V_2(\vec{r})
 \end{equation}
-one obtains
+and, thus
 \begin{equation}
 E_1<E2+\int n(\vec{r}) \left( V_1(\vec{r})-V_2(\vec{r}) \right) d\vec{r}
+\text{ .}
+\label{subsub:hk02}
 \end{equation}
-By switching the indices ...
+By switching the indices of \eqref{subsub:hk02} and adding the resulting equation to \eqref{subsub:hk02}, the contradiction
+\begin{equation}
+E_1 + E_2 < E_2 + E_1 +
+\underbrace{
+\int n(\vec{r}) \left( V_1(\vec{r})-V_2(\vec{r}) \right) d\vec{r} +
+\int n(\vec{r}) \left( V_2(\vec{r})-V_1(\vec{r}) \right) d\vec{r}
+}_{=0}
+\end{equation}
+is revealed.