Are $|n,l,s,j,m_j \rangle$ states exact energy eigenstates for one electron spin orbit coupling or just zero order approximations?

Question

In my atomic lecture notes calculating the changes with spin-orbit coupling for a one electron atom they describe the $|n,l,s,j,m_j \rangle$ states as being used to 'diagonalise the spin-orbit perturbation' since the previous $|n,l,s,m_l,m_s \rangle$ basis was degenerate.

However if the perturbation $\Delta H \propto \vec{s} \cdot \vec{l} \propto \frac{1}{2} (j(j+1) -l(l+1) - s(s+1))$ then don't the states $|n,l,s,m_l,m_s \rangle$ completely diagonalise both the perturbation and the initial $H_{atom}$ Hamiltonian i.e. aren't they just the exact energy eigenstates, rather than just the $0^{th}$ order approximation?

I should note that the initial $H_{atom} = \frac{p^2}{2m} + U(r)$ where $U(r)$ is some central potential, but we have not included other electron-electron repulsion effects yet.

Is something similar also true for multi-electron atoms with spin-orbit coupling?

Answer 1

Yes, I would agree with you. The central potential does not break the symmetry under rotation of the Hamiltonian so the stationary states are eigenstates of $L_i^2$, $L_i^z$, $S_i^2$ and $S_i^z$ for each electron (without spin-orbit). The eigenstates of $H_0$ are tensor products $\bigotimes_i |n_i,l_i,m_i,s_i,{m_s}_i\rangle$. When considering the spin-orbit coupling $\sum_i \vec L_i.\vec S_i$, each term $\vec L_i.\vec S_i$ acts on a single electron and can be diagonalized independently of the others. The exact eigenstates are finally a tensor product $\bigotimes_i |n_i,l_i,j_i,{m_j}_i\rangle$. All this breaks down when electron-electron repulsion cannot be neglected.

To take into account the indiscernability of the electrons, first diagonalize the one-electron Hamiltonian, which leads here to $|n_i,l_i,j_i,{m_j}_i\rangle$, and then anti-symmetrize the tensor products $\bigotimes_i |n_i,l_i,j_i,{m_j}_i\rangle$. The result can be put under the form of a Slater determinant.