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

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?

• +1. I think a lot of people here will be able to answer this, but it's a good questions nonetheless, so thanks for contributing it here! Apr 9, 2021 at 2:55
• @NikeDattani Learn a lot from your community. And it's a good place to record some even naive ideas. :)
– Jack
Apr 9, 2021 at 5:08

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.

• So the reference KS system provides two things: (1) the KS-orbitals to express the kinetic energy of the original system (2) the ground-state density to calculate the Hartree potential?
– Jack
Apr 14, 2021 at 10:35
• I think that's reasonable, although important to distinguish the kinetic energy of the original system from the kinetic energy of the Kohn-Sham system, which are close in magnitude but not identical. Their difference is subsumed into the exchange-correlation functional.
– wcw
Apr 14, 2021 at 22:25
• Also, the generalized Kohn-Sham scheme uses the orbitals in additional ways (e.g, exact exchange or some meta-GGAs).
– wcw
Apr 14, 2021 at 22:25

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)

• Thanks for your answer. I know that the exchange-correction potential contains the energy contribution described by (4) in your post, which of course includes many-body effects. But here I am only curious about the Hartree term, which is describing the interaction between two electrons, right? How can we talk about that for the noninteracting KS reference system?
– Jack
Apr 9, 2021 at 5:22
• Maybe there is a missunderstanding. The Hartree term describes the interaction of one electron with the electron density. There is no Hartree term in the original Hamiltonian, but an electron-electron Coulomb interaction $U= \sum_{i,j}^N \frac{1}{|\vec{r}_i -\vec{r}_j|}$, with $i\neq j$. Apr 9, 2021 at 6:12
• The Hartree term is not inherited from the electron-electron Coulomb interaction term? [The Hartree term describes the interaction of one electron with the electron density.] How can we say this for the KS reference system if it is a noninteracting one?
– Jack
Apr 9, 2021 at 6:22
• If you solve a particle in a box, you would not call this particle an interacting particle. Its state is described by a single-particle wavefunction. In the same way, DFT is a noninteracting theory even if there is a Hartree term, because the KS orbitals are single-particle wavefunctions. And yes, the Hartree term (and Fock term) are the first order correction if you expand the electron-electron interaction, so in that sense it is inherent. The difference to the KS system lies in the definition of the density, which must be expressed using the many-body wavefunction. Apr 12, 2021 at 13:55
• @Jack , mt.huebsch: Nice to see so much conversation here. If the system recommends you to move to chat, please use an existing room such as [this one[(chat.stackexchange.com/rooms/119831/…) rather than clicking on the link which creates a new room. We would like to avoid this problem: meta.stackexchange.com/q/353643/391772 Apr 12, 2021 at 17:30