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Since the mutual repulsion term between electrons orbiting the same nucleus does not commute with either electron's angular momentum operator (but only with their sum), I'd assume that the electrons don't really have a well-defined angular momentum (i.e., they do not occupy a pure $\left|lm\right>$ state). I would assume that the actual wavefunction is dominated by one such state compared to others, so it is approximately pure, but is there really a points in enumerating electrons according to their momenta, like the s, p, d, f and so on sub-shells?

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up vote 6 down vote accepted

It's not often that dmckee and I differ (mainly because he's usually right :-) but we differ on this on. Or at least we differ if I've correctly understood what you're asking.

In a hydrogen atom the 1s, 2s, etc wavefunctions are (subject to various approximations) good descriptions of the single electron and have well defined angular momentums. In multielectron atoms it's convenient to think of electrons populating successive 1s, 2s, etc levels, but this is only a conceptual model and not an accurate representation. You're quite correct that while there is a well defined angular momentum for the whole atom, you cannot define the angular momentum of individual electrons.

In the old days (maybe it's still done) we'd calculate atomic structure using a Hartree-Fock method with individual electron wavefunctions as the basis, and as dmckee points out, atoms have spectral lines that can often be approximately thought of as exciting a specific electron between individual electron wavefunctions. However what you're really doing is labelling the whole atom as an $l,m$ state and not an individual electron.

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Hartree-Fock assumes that the multi-electron wavefunction is a single Slater determinant (in the 1-electron wavefn basis), rather than the sum of multiple Slater determinants. Your answer raises the question: To what extent is that true for light atoms and for heavy atoms? I don't know.. Another thing that OP may have been thinking of was the prescription for calculating a ground-state term symbol -- -- I always figured the prescription was more-or-less a mnemonic, not a literal description of the electrons. But I don't know. – Steve B Jan 20 '13 at 13:49
Do people still use term symbols? That's a genuine question, it was several decades ago as part of an undergrad project I was doing this work. Also, though my memory of the period is vague, I'm fairly certain most of our code was Hartree-Fock + tweaks rather than a pure HF approach. – John Rennie Jan 20 '13 at 14:42
I agree entirely that the states are not simply those of the hydrogen atom, I only refer to the experimental fact that they behave very much as if $l$ and $m$ are good quantum numbers for the states. – dmckee Jan 20 '13 at 18:09

Boy. I think the professor of my graduate spectroscopy lab1 would differ.

I mean you can measure the fine and hyperfine splitting of a variety of atoms in situ and it agrees very well with computations based on the assumption that these are good quantum numbers. I think that we can conclude that any mixing is at a very small level.

1 A crusty old emeritus who only taught a couple of advanced labs and seemed to have a grand time doing it.

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It might indeed be small, but why is it small? One naively expects total e-e repulsion to be comparable to e-nucleus attraction, right? (I expect that my question has one answer for lithium and a different answer for uranium.) – Steve B Jan 20 '13 at 13:55

The problem from your question is really a problem, but it is a small problem if you compare with a problem of quite strong interaction between these electrons which may not be treated quasiclassically as a interaction of two particles in defined quantum states. If you assume (as people usually do) that the atom with many electrons may be treated in one-electron approximation (we explicitly consider that we may pick one electron and describe its state, and not the state of the whole system), which is very rude approximation, then it is not a big deal to think that this electron has definite angular momentum.

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