Coulomb potential in Kohn-Sham DFT

[molkee]

In Kohn-Sham DFT, the Coulomb potential, which is a component of the Kohn-Sham potential, is given by:

$v_H(\mathbf{r}) = \int \frac{\rho(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|}d\mathbf{r'}$

where $\rho(\mathbf{r'})$ is the electron density.

For molecular systems with exponential densities, this potential is known to be finite at any $\textbf{r}$.

How to prove it analytically just based on its definition?

Would a potential

$v(\mathbf{r}) = \int \frac{\rho(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|^n}d\mathbf{r'}$

(where $n$ is some nonnegative integer)

be also finite at any $\mathbf{r}$?

 

[azntoon]

Yes, the Coulomb potential (v_H) and any other potential of the form v(\mathbf{r}) = \int \frac{\rho(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|^n}d\mathbf{r'}, where n is some nonnegative integer, will be finite at any \mathbf{r}. This can be proven analytically by considering the integral:\int \frac{\rho(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|^n}d\mathbf{r'}If we assume that the electron density, \rho(\mathbf{r'}), is an exponential function, then this integral can be shown to be bounded. This is because the exponential function decays to 0 as |\mathbf{r'}| increases. As a result, the integral will always be finite for any value of \mathbf{r}. Therefore, the Coulomb potential and any other potential of the form v(\mathbf{r}) = \int \frac{\rho(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|^n}d\mathbf{r'} will be finite at any \mathbf{r}.