Why Kohn-Sham equations are regarded as single-particle equations?

Question

The Kohn-Sham equations are given by:

$$ \left(-\frac{\hbar^2}{2m} \nabla_{i}^{2}+V_{s}\left(\hat{\boldsymbol{r}}_{i}\right)+V_{H}\left(\hat{\boldsymbol{r}}_{i}\right)+V_{X C}\left(\hat{\boldsymbol{r}}_{i}\right)\right) \psi_{i}(\boldsymbol{r})=\epsilon_{i} \psi_{i}(\boldsymbol{r}) $$

where $V_s$ is the electrostatic potential due to $N$ nuclei felt by the $i^{\text {th}}$ electron,

$$ V_{s}\left(\boldsymbol{r}_{i}\right)=-\frac{e^{2}}{4 \pi \epsilon_{0}} \sum_{I=1}^{N} \frac{Z_{I}}{\left|\boldsymbol{r}_{i}-\boldsymbol{R}_{I}\right|} $$

and the mean-field potential (also called the Hartree potential) felt by the $i^{\text {th}}$ electron is given by averaging the remaining $(n-1)$ electrons as a smooth distribution of negative charge with electron number density $\rho_{n}$,

$$ V_{H}\left(\boldsymbol{r}_{i}\right)=\frac{e^{2}}{4 \pi \epsilon_{0}} \int \frac{\rho_{n}\left(\boldsymbol{r}^{\prime}\right)}{\left|\boldsymbol{r}_{i}-\boldsymbol{r}^{\prime}\right|} d \boldsymbol{r}^{\prime} $$

and $V_{XC}({\boldsymbol{r}})$ is the exchange-correlation "potential" felt by the $i^{\text {th}}$ electron,

$$ V_{XC}(\boldsymbol{r}_i)=\frac{\delta E_{X C}\left[\rho_{n}\right]}{\delta \rho_{n}(\boldsymbol{r}_i)} $$

The exchange-corelation functional contains all the many-body effects. It includes all classical and quantum effects and corrections not already accounted for by $V_s$ and $V_H$. These include the exchange effects due to the Pauli exclusion principle, the short-range Coulomb correlations not accounted by the Hartree term, and the kinetic energy difference between the interacting and non-interacting electrons.

My question is how come the Kohn-Sham equations are considered single-particle equations when the $V_{XC}$ term is essentially a many-particle quantity?

I found a related question here How can we say that the KS equation is describing a noninteracting many-electron system? which is similar to my question; however, the question there is about $V_H$ not $V_{XC}$. I can perceive that $V_s$ and $V_H$ are external potentials coming from the nuclei and the mean-field approximation of the electrons, but how the presence of a many-body term is reconciled with this single-particle picture?

Answer 1

I would argue that because the exact potential, like the ground state density, is a property of the system. The Hohenberg-Kohn theorems show that there is a scalar single-particle potential $v({\bf r})$ that generates orbitals that minimize the energy, no? And the Kohn-Sham equation for the exact potential is a single-particle equation.

It does not matter that the potential depends implicitly on the orbitals, since it is still only a single-particle operator unlike the real molecular Hamiltonian that has terms that couple orbitals together.

Answer 2

You are exactly correct in saying the Kohn-Sham equations are single particle equations

The top equation is the classic eigenvalue equation with the spin orbitals being the eignfunctions and the associated eigenvalue being the energy of that spin orbital. Here the set of spin orbitals (psi in top equation) represent the single particle basis. Hence "single particle equations".

What is somewhat confusing is how standard DFT approximates the exchange-correlation energy. This is due to some approximate exchange-correlation functional being evaulated, which usually requires integration over a real space grid (the Vxc matrix is built in a similar fashion). Normally this doesn't describe a systems true exchange-correlation and is why DFT is approximate; however, once the approximation is made then yes you are just solving single particle equations (the top equation you posted).

HOWEVER, you can completely ignore this step and use wavefunction methods to generate the TRUE exchange-correlation energy. This is known as the Levy-Lieb density functional and very much shows how the ground state can be an entangled superposition state.

^Further info on Levy-Lieb. Lots of wavefunctions give a single electron density... the idea of Levy-Lieb density functional is given a particular electron denstiy find all wavefunctions that will give that particular electron density. Then evaluate the true Colomb and Exchange operators using wavefucntion methods for all these wavefunctions and pick the wavefunction that gives the lowest energy. This will be the ground state!