Note on electrostatic potential

In the PAW formalism, the electrostatic potential from the pseudo charge \(\tilde{\rho}(\mathbf{r})\) is obtained by solving a Poisson equation:

\[\nabla^2 \tilde{v}_H(\mathbf{r})=-4\pi\tilde{\rho}(\mathbf{r}).\]

To get the real all-electron electrostatic potential, we need the all-electron charge density:

\[\rho(\mathbf{r}) = \tilde{\rho}(\mathbf{r}) + \sum_a \Delta\tilde{\rho}^a(\mathbf{r} - \mathbf{R}^a),\]

where \(\Delta\tilde{\rho}^a\) is an atomic PAW correction to the pseudo charge density:

\[\Delta\tilde{\rho}^a(\mathbf{r}) = n_c^a(r) - \tilde{n}_c^a(r) - \mathbb{Z}^a\delta(\mathbf{r}) - \sum_{\ell=0}^{\ell_{\text{max}}} \sum_{m=-\ell}^\ell Q_{\ell m}^a \hat{g}_{\ell m}^a(\mathbf{r}) + \sum_{\sigma i_1 i_2} D_{\sigma i_1 i_2}^a (\phi_{i_1}^a(\mathbf{r})\phi_{i_2}^a(\mathbf{r}) - \tilde{\phi}_{i_1}^a(\mathbf{r})\tilde{\phi}_{i_2}^a(\mathbf{r})).\]

See here for details.

So, the all-electron potential is:

\[v_H(\mathbf{r}) = \tilde{v}_H(\mathbf{r}) + \sum_a \Delta\tilde{v}_H^a(\mathbf{r} - \mathbf{R}^a)\]


\[\Delta\tilde{v}_H^a(\mathbf{r}) = \int d\mathbf{r}' \frac{\Delta\tilde{\rho}^a(\mathbf{r}')} {|\mathbf{r}-\mathbf{r}'|}.\]

Notice that the \(Q_{\ell m}^a\) have been chosen so that all multipole moments of \(\Delta\tilde{\rho}^a\) are zero and therefore, the potential from these correction charges (\(\Delta\tilde{v}_H^a\)) will be non-zero only inside the atomic augmentation spheres.

The get_electrostatic_corrections() method will return an array of integrated corrections:

\[\int d\mathbf{r} \Delta\tilde{v}_H^a(\mathbf{r})\]

in units of eV Å3.