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In multi-electron atoms, the electronic state of the optically active "subshell" is often expressed in "term symbols" notation. I.e. $^{2S+1}L_J$. This presumes that the system of electrons has definite $L^2$, $S^2$ and $\mathbf{J}$ eigenstates.

In order to determine all the possible electronic states compatible with the Pauli's exclusion principle, the technique is to assign to each electron its $m_s$ and $m_l$ and to see what states can be found in terms of eigenvalues of $L^2$, $S^2$, $J^2$ and $J_z$.

What justifies this procedure? In other words, the assumption so far was:

  • heavy nucleus -> the Hamiltonian is spherically symmetric -> total $\mathbf{J}$ is conserved
  • spin-orbit is negligible to some extent -> total $S^2$ and $L^2$ are conserved

What justifies, apart from intuitiveness, returning to the single electron picture (assigning to every electron its spin and angular momentum) to find what states are allowed or not according to the requirement that the total wavefunction should be antisymmetric? And could be there exceptions?

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  • $\begingroup$ Term symbol $^{2S+1}L_{J}$ is a collection of approximated good quantum numbers. The assumptions may be rephrased as approximations. As you mentinoed, for Dirac equation, $L^2$ nor $S^2$ does not commute with Hamiltonian even in hydrogen atom (ref. Advanced Quantum Mechanics by Sakurai). Other approximation like single-electron picture does not appear, since $[H,L^2]=0$ anyway. You may image the coupled angular momentum is a generator for rotating all electrons together, which commutes with the Hamiltonian. Cohen-Tannoudji's QM vol II, X.2 has a excellent discussion for this point. $\endgroup$ – user26143 Feb 7 '14 at 6:54
  • $\begingroup$ Here the single-electron picture I mean using a single Slater determinant to approximate the many-electron wave function. $\endgroup$ – user26143 Feb 7 '14 at 7:00

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