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?

  • 1
    $\begingroup$ +1 But I would recommend to ask a new question if you have further inquiries, rather than changing a question that already has two answers. $\endgroup$ Commented Aug 20, 2022 at 15:04
  • $\begingroup$ @NikeDattani thanks for the edit and recommendation, but I was trying to express the same question more clearly, based on the comments, not asking a new question. As for the present two answers I was definitely going to accept them if they were answering my question. $\endgroup$
    – Sha
    Commented Aug 23, 2022 at 6:59
  • $\begingroup$ Thanks Sha! Once answers are written, any changes to the question are strongly discouraged, so the recommendation is to spend a lot of time making sure the original question is as close to perfect as possible. You can first write your question as an answer to this: mattermodeling.meta.stackexchange.com/q/360/5, and get comments there before publishing a final version on the main site here :) $\endgroup$ Commented Aug 23, 2022 at 15:01
  • $\begingroup$ @NikeDattani Oh I did not know this "meta" feature. Thank you for all your hard work in this fantastic stackexchange community. $\endgroup$
    – Sha
    Commented Aug 23, 2022 at 15:17

2 Answers 2


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.

  • 1
    $\begingroup$ Dear @Susi Lehtola thank you for taking your time answering the question, but I am afraid this does not answer my question. Yes, by HK theorems the exact potential is a unique functional (up to a constant) of the ground-state density. How this is supposed to answer the question? $\endgroup$
    – Sha
    Commented Aug 11, 2022 at 13:23
  • 2
    $\begingroup$ @Sha according to Hohenberg-Kohn, the ground state density (or accordingly, the ground state potential) fully characterizes the ground state. Kohn-Sham is a variant of DFT where one avoids the need for a kinetic energy functional by evaluating the exact kinetic energy for the non-interacting system. In any case, the point is that v(r) is a one-particle operator. It does not couple determinants. $\endgroup$ Commented Aug 11, 2022 at 22:11
  • 1
    $\begingroup$ How come one can think of the Kohn-Sham effective potential as a one-particle operator while it contains the many-body term $V_{xc}$. This is my question. Thank you for your time but either I am not understanding your argument or you are not understanding my question. $\endgroup$
    – Sha
    Commented Aug 12, 2022 at 8:37
  • 1
    $\begingroup$ @Sha Perhaps it would help how you define a one-body operator!? $\endgroup$
    – Jakob
    Commented Aug 12, 2022 at 14:57
  • 2
    $\begingroup$ I think that the part that you don’t understand is a matter of semantics. Yes, the exchange correlation energy is there to take into account the many body interactions, but that doesn’t make the Hamiltonian a many body Hamiltonian. The comment about defining a one body operator would lead you to see that the he KS Hamiltonian is a single particle Hamiltonian of some auxiliary non-interacting system. There happens to be formal results that tell you that the KS and actual system have the same ground state energy and density, but the KS system is still a non-interacting system. $\endgroup$
    – AGS
    Commented Aug 13, 2022 at 7:06

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!


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .