# Exchange-correlation potential for one electron system

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, $$V_{xc}(\mathbf{r}) = -V_H(\mathbf{r})$$?

Since $$V_{xc}=-V_H$$ would be a poor choice for multi-electron systems, meta-exchange-correlation funtionals take advantage of the fact that one can approximately deduce from the Kohn-Sham kinetic energy density whether the charge density is locally composed mostly of a single orbital or of several ones. For the limit of single orbitals, these meta-exchange-correlation functionals are typically constructed such that the exchange energy cancels the Hartree energy of the hydrogen atom (see e.g. page 8 of Sun, Ruzsinszky, & Perdew "Strongly Constrained and Appropriately Normed Semilocal Density Functional" arXiv:1504.03028v3).