For classical ion, the DFT solution of the ground-state electronic system is given by $$ \left[-\frac{1}{2}\nabla^2 + V_H(\mathbf{r}) + V_{ei}(\mathbf{r}) + V_{xc}(\mathbf{r})\right]\psi(\mathbf{r}) = \varepsilon \psi(\mathbf{r}) $$ where $V_{ei}(\mathbf{r})$ is the Coulomb potential for electron-ion interaction, $V_{xc}(\mathbf{r})$ is the potential due to the exchange-correlation energy, and $V_H(\mathbf{r})$ is the classical Hartree potential defined as $$ V_H(\mathbf{r}) = \int\frac{n(\mathbf{r'})}{|\mathbf{r} - \mathbf{r'}|}\ \mathrm{d}^3\mathbf{r'}. $$
Based on many references that I read, I get an impression that the density in the Hartree potential is the total density from all orbitals, not the total density of electrons in other orbitals. That means even if there is one electron, the classical Hartree potential should be non-zero.
By rewriting the exact Schrodinger equation for one electron which is, $$ \left[-\frac{1}{2}\nabla^2 + V_{ei}(\mathbf{r})\right]\psi(\mathbf{r}) = \varepsilon \psi(\mathbf{r}), $$ does that mean for systems with only one electron, $V_{xc}(\mathbf{r}) = -V_H(\mathbf{r})$?
