Electron-electron repulsion potential

The electron-electron repulsion potential has a form

  1. $${V^{ee}}\left( {\bf{r}} \right) = \int {{{\rho \left( {{\bf{r'}}} \right)} \over {\left| {{\bf{r}} - {\bf{r'}}} \right|}}} d{\bf{r'}}$$

The potential (1) satisfies the Poisson equation

  1. $$\Delta {V^{ee}}\left( {\bf{r}} \right) =  - 4\pi \rho \left( {\bf{r}} \right)$$

In the case of spherical symmetry we can write

$${1 \over r}{{{d^2}} \over {d{r^2}}}\left( {r{V^{ee}}\left( r \right)} \right) =  - 4\pi \rho \left( r \right)$$

or

  1. $${{{d^2}} \over {d{r^2}}}\left( {r{V^{ee}}\left( r \right)} \right) =  - {{4\pi } \over r}{P^2}\left( r \right)$$

where $P\left( r \right) = r\Psi \left( r \right)$ is the radial wave function.

We can solve differential equation (3) on a range $\left[ {{r_0},{r_f}} \right]$, where r0 is a very small number and rf is sufficiently big number.

For a very big value of r the potential is satisfying the equation

  1. $${\left. {{V^{ee}}\left( r \right)} \right|_{r \to  + \infty }} \sim  - {Q \over r}$$

where Q = -1 is the charge of the electron.

For the point r = 0 we can calculate the value of ${V^{ee}}\left( r \right)$ with the help of (1) and use analytical solution of Schrödinger equation for H-like atom

  1. $${V^{ee}}\left( 0 \right) = 4\pi \int\limits_0^{ + \infty } {{{{{\left( {{1 \over {\sqrt \pi  }}{Z^{{3 \over 2}}}{e^{ - Zr'}}} \right)}^2}} \over {r'}}} {r'^2}dr' = Z$$

Using (4) and (5) we can write the boundary conditions for function $r{V^{ee}}$

  1. $$\eqalign{
      & {\left. {r{V^{ee}}\left( r \right)} \right|_{r \to 0}} = 0  \cr
      & {\left. {r{V^{ee}}\left( r \right)} \right|_{r \to  + \infty }} = 1 \cr} $$

If we know the wave function $P\left( r \right)$, then we can calculate e-e potential by solving differential equation (3) with boundary conditions (6).

 

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