The potential terms of the Hamiltonian in the Kohn-Sham equation are the external, Hartree, and exchange-correlation terms, which can be written as:

$V_{ext}(r) = - \sum_{i=1}^{M} \frac{Z_i} {||r - R_i||} \\ V_{H}(r) = \int \frac {\rho(r')} {||r - r'||} dr' \\ V_{xc}(r) = \int f(\rho(r)) dr$

The external potential is always negative because it contains the atomic number but has minus sign; the Hartree is always positive because the density $\rho$ is always positive.

Here, $V_{xc}$ can be written using a function $f$ by various approximation approaches (e.g., LDA), but is this $V_{xc}$ term always positive? Or can this $V_{xc}$ be negative depending on the type of approximation?

  • $\begingroup$ I think that $V_{ext}$ is not the external potential, it is the Coulomb potential between electrons and nuclei. The external potential, as the name, is a potential where your system can be, for example, an electric field. $\endgroup$
    – Camps
    Aug 10, 2021 at 11:33
  • $\begingroup$ In the 29 page of this slide (physics.gu.se/~tfkhj/lecture_VIII_DFT-3.pdf), the $v_{ext}$ is defined as I described. This equation is not correct? $\endgroup$
    – neco
    Aug 10, 2021 at 12:40
  • 1
    $\begingroup$ The choose of external as a label, looks for me as an unhappy choice. Following the description of the term in the slide, $V_{ext}$ is the potential from the nuclei. Of course is external to electrons, but the nuclei are part of the system! $\endgroup$
    – Camps
    Aug 10, 2021 at 12:54

1 Answer 1


I think this is true at least for the exchange energy $$ E = \int \epsilon_{x}({\bf r}) {\rm d}^3r $$ which is usually written in terms of an enhancement function $F$ times the local density approximation's exchange energy (which is always negative) $$ \epsilon_{x}({\bf r}) = - C n^{4/3}({\bf r}) F(\gamma, \tau) {\rm d}^3r $$ where $\gamma$ is the reduced gradient and $\tau$ is the local kinetic energy density. If the enhancement function is nonnegative, $\epsilon_x$ will also be nonnegative.

The correlation energy is a correction to the mean field, and it is much smaller than the exchange energy, thus on this argument alone it should be always negative. However, I would expect the correlation energy to also be always negative, because that is basically the definition of correlation energy: it's the energy gain you get by switching from Hartree-Fock to a many-electron description like MP2, CCSD or FCI.

  • $\begingroup$ No, Pauli repulsion is already included in the Kohn-Sham determinant, the same way it is also included in Hartree-Fock. You need to worry about the Pauli potential only in orbital-free density functional theory. $\endgroup$ Aug 16, 2021 at 17:22

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