Atomic terms and perturbation theory

Question

I have a question about the derivation of the possible atomic terms of an atom. Consider an atom with $N$ electrons in the central-field approximation, with hamiltonian

$$ H_{0} = \sum_{i=1}^{N} \left[ - \frac{1}{2} \nabla ^{2}_{i} + V _{c}(ri)\right] $$

We can write its states as the product of the states of one electron, which are characterized by the quantum numbers $n,\ l,\ m_{l}$ and $m_{s}$. We know that the configurations we get like this are degenerate: all configurations with same values of $n$ and $l$ but differents $m_{l}$ and $m_{s}$ have the same energy. These configurations also have well-defined values ​​of total orbital angular momentum and spin $L$ and $S$, and we can classify them in atomic terms $^{2S + 1}L$, that have the same energy. However, if we consider corrections for the central field approximation, the degeneracy of these terms is broken, and we can determine which ones have higher or lower energy by Hund's rules.

My doubt is the following: thinking from the perspective of perturbation theory, I am considering the various states of the same electronic configuration as unperturbed states:

$$ | \psi^{(0)} \rangle = |n_{1},l_{1},m_{l1},m_{s1}, ..., n_{N},l_{N},m_{lN},m_{sN} \rangle $$

and taking as perturbation a potential equal to $V_{ee} - V_{c}$, where $V_{ee}$ is the Coulomb interaction between the electrons. The use of degenerate perturbation theory generates $|\psi \rangle$ eigenstates that are also functions with well-defined $L,\ S,\ M_{L}$ and $M_{S}$, since the perturbed hamiltonian commutes with the spin and orbital angular momentum operators (I'm ignoring spin-orbit interaction for now). But why do the possible $L,\ S,\ M_{L},\ M_{S}$ values ​​of the disturbed states have to be the same as the values ​​of the undisturbed states? Couldn't the perturbation generate states with well-defined angular moments but with different eigenvalues?

Answer 1

As you know very well, this angular degeneracy is due to the fact that the potential $V_c=V_c(r)$ is purely radial and the angular part do not affect the energy. The angular solution are degenerate if we do not include interactions like spin-orbit interaction.

The interaction $V_{ee} \propto \frac{e^2}{|\vec{r}'-\vec{r}|}=V_{ee}(r,\phi)$ has an angular dependency. If you solve the equation $H|\psi \rangle = E |\psi \rangle $ with $H = H_0 + V_{ee}$ the angular degeneracy is broken.

If you consider $V_{ee}$ as a perturbation, acting on the state $|\psi^0\rangle$, the correction $\langle \psi^0 |V_{ee}|\psi^0\rangle$ will not break the degeneracy, the eigenvalues will be only shifted compared to the previous equation.