How can we say that the KS equation is describing a noninteracting many-electron system?

Question

Based on HK's two theorems, the density functional theory was built. Because one can't find the universal energy functional $F_{HK}[n(r)]$, Kohn and Sham further proposed the Kohn-Sham ansatz: mapping the interacting many-electron system to a noninteracting many-electron system (KS reference system) by keeping the ground-state electron density fixed. With that, the kinetic energy functional can be expressed with the help of the KS reference system. The variation of total energy functional respect to density $n(r)$ will result in the famous single-particle Schr$\ddot{\text{o}}$dinger-like equation (KS equation):

$$\tag{1}\left[ -\dfrac{1}{2}\nabla^2+V_{ext}+V_{hartree}+V_{xc} \right]\psi_i(\vec{r})=E_i\psi_i(\vec{r})$$

This equation is derived based on the KS reference system, which is a noninteracting one. Then what does the $V_{hartree}$ term describe? Does it still describe the electron-electron interaction of the original interacting many-electron system? If so, how can we say that the KS equation is describing a noninteracting many-electron system? Or more directly:

What's the difference for the Hartree term between the original interacting many-electron system and the noninteracting KS reference system?

Answer 1

As you note, the interacting electrons and the Kohn-Sham non-interacting electrons have the same density. How is this possible when the Hamiltonians for the two systems are so different?

The answer is that the Kohn-Sham potential—the potential felt by the non-interacting electrons—is constructed very carefully. Namely, if we want to remove electron-electron interactions but preserve the electron density, we must add $v_{Hartree}+v_{xc}$ to the original external potential: $$ v_{KS} = v_{ext}+v_{Hartree}+v_{xc} . $$ See the Kohn-Sham paper for the derivation.

Turning now to your main question:

What's the difference for the Hartree term between the original interacting many-electron system and the noninteracting KS reference system?

In the original interacting system, the Hartree term accounts for electron-electron interactions, and it is a prominent contribution to the total electron-electron interaction. It is completely separate from the external potential term.

In the non-interacting system, there are no electron-electron interactions, and the Hartree potential appears in the effective external potential. Moreover, it appears purely as a consequence of electron-electron interactions in the original system, and its purpose here is to keep the electron densities of the two systems identical. Why? So that we may ultimately obtain the non-interacting kinetic energy for that density.

Finally, one might ask: practically, how can $v_{KS}$ depend on the electron density? Answer: the Kohn-Sham equations must be solved self-consistently.

Answer 2

First of all, let me emphasize that it is more appropriate to speak of KS equations (plural), which you correctly denoted by an index $i$ in your post. This index goes over all KS orbitals (i.e. single-particle wavefunctions) of the system. Additionally, as you mentioned, these equations have the same form as the single-particle Schrödinger equation. And combined that is the reason why density-functional theory (DFT) is said to be a noninteracting theory.

Nevertheless, exchange and correlation (xc) effects are effectively taken into account through the energy functional that can be computed self-consistently in order to resemble the true quantum-mechanical state, \begin{equation}\tag{1} \Psi(\vec{r}_1\sigma_1,\vec{r}_2\sigma_2, ...,\vec{r}_N\sigma_N), \end{equation} that can be described by a many-particle wavefunction.

Concerning the form of $V_{hartree}$ and $V_{xc}$, assume we can determine the KS orbitals $\{\psi_i\}$ with eigenenergies $E_i$ below the Fermi energy, that allows us to express the electronic density as \begin{equation}\tag{2} n(\vec{r}) = \sum_{i=1}^N \psi^*_i(\vec{r})\psi_i(\vec{r}). \end{equation} Then, it is convenient to split the electron-electron interaction $U$ into the Hartree term $U_{hartree}$ and the remainder $U_{xc}$. That is because the Hartree potential $V_{hartree}$ explicitly depends on the electron density: \begin{equation}\tag{3} V_{hartree} = 2 \int \text{d} \vec{r}' \frac{n(\vec{r}')}{|\vec{r} - \vec{r}'|}. \end{equation} Mind, that also the kinetic energy is split into the kinetic energy of the noninteracting system $T_0$ and the remainder $T_{xc}$. The xc energy functional collects these terms: \begin{equation}\tag{4} E_{xc}[n]=T_{xc}[n] + U_{xc}[n]. \end{equation} Finally, the xc potential is obtained via variation w.r.t. the electron density: \begin{equation}\tag{5} V_{xc}(\vec{r})=\frac{\delta E_{xc}[n(\vec{r})]}{\delta n(\vec{r})} . \end{equation}

Units etc. as in J. Kübler, Theory of Itinerant Electron Magnetism (Oxford University Press, 2017)