Is the exchange-correlation term in Kohn-Sham equation always positive or negative?

Question

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?

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.