Angular Average, Shells, and Periodic Table

A further simplification of the Hartree equations, Eq. (IV.1.3), is achieved by replacing the Hartree potential by its angular average,

$$V_{\rm Hartree}({\bf r})\to \left<V_{\rm Hartree}\right>(r)\equiv \int \frac{d\Omega}{4\pi}V_{\rm Hartree}({\bf r}).$$ (1.5)

This still depends on all the wave functions $\psi_{\nu_i}$, but as the one-particle potential now is spherically symmetric, we can use the decomposition into spherical harmonics, radial wave functions, and spin,

$$\langle \xi \vert \nu_i\rangle = \psi_{\nu_i}(\xi)= R_{n_i,l_i}(r) Y_{l_i,m_i}(\theta,\phi)\vert\sigma_i\rangle,\quad \nu_i=(n_i,l_i,m_i,\sigma_i).$$ (1.6)

Here, the index $\nu_i=(n_i,l_i,m_i,\sigma_i)$ indicates that we are back to our usual quantum numbers $nlm\sigma$ that we know from the hydrogen atom. In contrast to the latter, the radial functions now depend on $n$ and $l$ because we do not have the simple $1/r$ Coulomb potential as one-particle potential.

An even cruder approximation to $V_{\rm Hartree}({\bf r})$ would be a parametrization of the form

$$V_{\rm Hartree}({\bf r})+\frac{e^2}{4\pi \varepsilon_0}\frac{Z}{r} \to V_{\rm eff}(r) \equiv \frac{e^2}{4\pi \varepsilon_0}\frac{Z(r)}{r}$$ (1.7)

$$Z(r\to 0) = Z,\quad Z(r\to \infty) = 1.$$ (1.8)

by which one loses the self-consistency and ends up with one single Schrödinger equation for a particle in the potential $V_{\rm eff}(r)$.

Exercise: Give a physical argument for the condition $Z(r\to 0) = Z,\quad Z(r\to \infty) = 1$ in the above equation.