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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?

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  • $\begingroup$ Do you neglect electron-electron repulsion or not? If not, $|n,l,s,j,m_j\rangle$ are certainly not the exact eigenstates of the full Hamiltonian. $\endgroup$
    – Christophe
    Feb 6, 2021 at 13:22
  • $\begingroup$ @Christophe I agree definitely if you are including electron-electron repulsion then they would not be exact eigenstates. I think the situation in my lecture notes here are that we have included $H_{atom}$ which is the KE + some central field radial potential function, and we have only included the above $\Delta H_{so}$ but ignored other electron-electron repulsion effects (I think because for the alkali case here the spin-orbit effect is a larger perturbation). Would you agree in this case of H = KE + central potential + spin-orbit coupling that the above energy eigenstates are exact? $\endgroup$
    – Alex Gower
    Feb 6, 2021 at 13:25

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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.

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  • $\begingroup$ Ah thanks! I hadn't thought of the multi-electron case, but I guess those tensor products would work so long as you used a slater determinant right? $\endgroup$
    – Alex Gower
    Feb 6, 2021 at 13:56
  • $\begingroup$ I have updated the answer. Happy to have been helpful! $\endgroup$
    – Christophe
    Feb 6, 2021 at 14:10
  • $\begingroup$ Thank you very much, if you still have time, this question links to my other question here about precession in spin-orbit coupling. physics.stackexchange.com/questions/612676/… $\endgroup$
    – Alex Gower
    Feb 6, 2021 at 14:12

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