where
\begin{eqnarray}
T & = & \langle\Psi|\sum_{i=1}^N\frac{-\hbar^2}{2m}\nabla_i^2|\Psi\rangle\\
- & = & \sum_{i=1}^N \int d\vec{r} d\vec{r}' \,
+ & = & \frac{-\hbar^2}{2m} \sum_{i=1}^N \int d\vec{r} d\vec{r}' \,
\langle \Psi | \vec{r} \rangle \langle \vec{r} |
- \frac{-\hbar^2}{2m}\nabla_i^2
+ \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_i \Psi^*(\vec{r}) \nabla_i \Psi(\vec{r})
+ \nabla_{\vec{r}_i} \Psi^*(\vec{r}) \nabla_{\vec{r}_i} \Psi(\vec{r})
\text{ ,} \\
-V & = & V(\vec{r})\Psi^*(\vec{r})\Psi(\vec{r})d\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{equation}
In 1964, Hohenberg and Kohn showed the opposite and far less obvious result \cite{hohenberg64}.
-{\begin{theorem}
+\begin{theorem}[Hohenberg / Kohn]
For a nondegenerate ground state, the ground-state charge density uniquely determines the external potential in which the electrons reside.
\end{theorem}
\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. \qed
+is revealed, which proofs the Hohenberg Kohn theorem.% \qed
\end{proof}