more detailed

[lectures/latex.git] / physics_compact / solid.tex

diff --git a/physics_compact/solid.tex b/physics_compact/solid.tex
index 6c25ff1..30e078e 100644 (file)
--- a/physics_compact/solid.tex
+++ b/physics_compact/solid.tex
@@ -18,11 +18,58 @@
 
 \subsubsection{Hohenberg-Kohn theorem}
 
+The Hamiltonian of a many-electron problem has the form
+\begin{equation}
+H=T+V+U\text{ ,}
+\end{equation}
+where
+\begin{eqnarray}
+T & = & \langle\Psi|\sum_{i=1}^N\frac{-\hbar^2}{2m}\nabla_i^2|\Psi\rangle\\
+  & = & \frac{-\hbar^2}{2m} \sum_{i=1}^N \int d\vec{r} d\vec{r}' \,
+        \langle \Psi | \vec{r} \rangle \langle \vec{r} |
+        \nabla_i^2
+        | \vec{r}' \rangle \langle \vec{r}' | \Psi \rangle\\
+  & = & \frac{-\hbar^2}{2m} \sum_{i=1}^N \int d\vec{r} d\vec{r}' \,
+        \langle \Psi | \vec{r} \rangle \nabla_{\vec{r}_i}
+        \langle \vec{r} | \vec{r}' \rangle
+        \nabla_{\vec{r}'_i} \langle \vec{r}' | \Psi \rangle\\
+  & = & \frac{-\hbar^2}{2m} \sum_{i=1}^N \int d\vec{r} d\vec{r}' \,
+        \nabla_{\vec{r}_i} \langle \Psi | \vec{r} \rangle
+        \delta_{\vec{r}\vec{r}'}
+        \nabla_{\vec{r}'_i} \langle \vec{r}' | \Psi \rangle\\
+  & = & \frac{-\hbar^2}{2m} \sum_{i=1}^N \int d\vec{r} \,
+        \nabla_{\vec{r}_i} \Psi^*(\vec{r}) \nabla_{\vec{r}_i} \Psi(\vec{r})
+        \text{ ,} \\
+V & = & \int V(\vec{r})\Psi^*(\vec{r})\Psi(\vec{r})d\vec{r} \text{ ,} \\
+U & = & \frac{1}{2}\int\frac{1}{\left|\vec{r}-\vec{r}'\right|}
+        \Psi^*(\vec{r})\Psi^*(\vec{r}')\Psi(\vec{r}')\Psi(\vec{r})
+        d\vec{r}d\vec{r}'
+\end{eqnarray}
+represent the kinetic energy, the energy due to the external potential and the energy due to the mutual Coulomb repulsion.
+
+\begin{remark}
+As can be seen from the above, two many-electron systems can only differ in the external potential and the number of electrons.
+The number of electrons is determined by the electron density.
+\begin{equation}
+N=\int n(\vec{r})d\vec{r}
+\end{equation}
+Now, if the external potential is additionally determined by the electron density, the density completely determines the many-body problem.
+\end{remark}
+
 Considering a system with a nondegenerate ground state, there is obviously only one ground-state charge density $n_0(\vec{r})$ that corresponds to a given potential $V(\vec{r})$.
+\begin{equation}
+n_0(\vec{r})=\int \Psi_0^*(\vec{r},\vec{r}_2,\vec{r}_3,\ldots,\vec{r}_N)
+                  \Psi_0(\vec{r},\vec{r}_2,\vec{r}_3,\ldots,\vec{r}_N)
+             d\vec{r}_2d\vec{r}_3\ldots d\vec{r}_N
+\end{equation}
 In 1964, Hohenberg and Kohn showed the opposite and far less obvious result \cite{hohenberg64}.
+
+{\begin{theorem}
 For a nondegenerate ground state, the ground-state charge density uniquely determines the external potential in which the electrons reside.
-The proof presented by Hohenberg and Kohn proceeds by {\em reductio ad absurdum}.
+\end{theorem}
 
+\begin{proof}
+The proof presented by Hohenberg and Kohn proceeds by {\em reductio ad absurdum}.
 Suppose two potentials $V_1$ and $V_2$ exist, which yield the same electron density $n(\vec{r})$.
 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
@@ -55,5 +102,6 @@ E_1 + E_2 < E_2 + E_1 +
 \int n(\vec{r}) \left( V_2(\vec{r})-V_1(\vec{r}) \right) d\vec{r}
 }_{=0}
 \end{equation}
-is revealed, which proofs the Hohenberg Kohn theorem.
+is revealed, which proofs the Hohenberg Kohn theorem. \qed
+\end{proof}