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.