# 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:

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

and

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

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.