+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}