The Coulomb (Thomas-Fermi) functional

For the classical part of the potential, Thomas and Fermi employed the Coulomb potential energy functional

$$U[n(r)] = \frac{1}{2}\int\int \frac{n({\bf r}) n({\bf r}')}{\vert \mathbf{r}-\mathbf{r}' \vert} d^3r d^3r'$$ (141)

Again, $U[n]$ depends only on the charge density $n$ and does not depend on its gradient, Laplacian, or other higher-order derivatives. Therefore,

$$U_0 = 0; U_1 = 0; U_2({\bf r},{\bf r}') = \fra... ...{2}\frac{1}{\vert{\bf r} - {\bf r}'\vert}; U_{n>2} = 0;$$ (142)

and we find

$$\frac{\delta U[n]}{\delta n} = \int \frac{n(\mathbf{r}')}{\vert \mathbf{r}-\mathbf{r}' \vert} d^3r'$$ (143)

The second functional derivative of the Coulomb potential energy functional is

$$\frac{\delta^2 U[n]}{\delta n^2} = \frac{1}{\vert \mathbf{r}-\mathbf{r}' \vert}$$ (144)

We should make the following observation. Thsi potential contains a self-interaction between a particle and itself, which is unphysical. It does not take into account the Pauli principle, and therefore, this functional should be corrected to cancel the effects of the self-interaction and take exchange into account properly. Notice that this is equivalent to the Hartree potential in the Hartree-Fock approximation.