If we know the eigenvector $P\left(r\right)$, then the eigenvalue $\varepsilon$ can be calculated by integration of the radial Kohn-Sham equation (see Task 2, formula (4)):
- $${\varepsilon} = 4\pi \int\limits_0^\infty {{P}\left( r \right){\bf{H}}{P}\left( r \right)dr} $$
where ${\bf{H}} = - {1 \over 2}{{{d^2}} \over {d{r^2}}} + {V^{KS}}\left( r \right)$ is the KS Hamiltonian of the system, whereas $V^{KS}\left(r\right)$ is the total KS potential. Then the eigenvalue can be calculated as
- $${\varepsilon} = 4\pi \int\limits_0^\infty {{P}\left( r \right)\left\{ { - {1 \over 2}{{{d^2}} \over {d{r^2}}}{P}\left( r \right) + {V^{KS}}\left( r \right){P}\left( r \right)} \right\}dr} $$
We calculate the integral in Eq. (2) on the region $\left[ {{r_0},{r_f}} \right]$. We suppose, that r0 is a sufficiently small number and rf is sufficiently big number so, that the integrals $\int\limits_0^{{r_0}} {dr} $ and $\int\limits_{{r_f}}^\infty {dr} $ can be neglected.
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