# Why does exchange lower energy with increasing electron density in Hartree-Fock

The Hartree-Fock equations include a term for the exchange interaction, which is usually explained as a repulsive force due to the Pauli exclusion principle. (It says so right in the description for the "exchange-interaction" tag.)

I expect that, in a system of repulsive particles, increasing the particle density would increase the energy of the system. (E.g. if you have a bunch of particles connected by springs, you would have to work to compress the system, so the energy of the system increases when compressed.)

However, for a free electron gas (and for the local density approximation), the exchange energy becomes more negative as the density is increased [Ashcroft and Mermin eq. (17.26)]:

$$U_{ex} \propto -\left(n\left(\vec{r}\right)\right)^{1/3},$$

where $n\left(\vec{r}\right)$ is the electron density. [See also Ashcroft and Mermin eq. (17.24) and the two sections on Hartree-Fock.] I interpret this to mean that increasing the electron density results in lower energies due to the exchange interaction.

Why does the exchange interaction -- a repulsive force -- lead to lower energy as the particle density is increased?