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

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?

• 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.
– Camps
Aug 10 '21 at 11:33
• 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?
– neco
Aug 10 '21 at 12:40
• 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!
– Camps
Aug 10 '21 at 12:54

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.