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Let's make the question easier by considering two-level atoms(with spin states, i.e. spin up $|\uparrow\rangle$ and spin down $|\downarrow\rangle$). An article I recently read claims that atoms do not have dipole moments when they are in energy eigenstates (i.e. when you put it into a magnetic field in z direction).

I was thinking if it's an eigenstate, since the spin cannot be pointing in z direction (remember it's actually pointing in a cone area), how could it not have a dipole moment?

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Electric dipole moment or magnetic? I believe the currently measured value of the EDM of the electron is experimentally consistent with zero. The magnetic moment, however, is definitely not. – Jerry Schirmer Jul 26 '13 at 6:39
Also an article claims that the moon is made of green cheese. To which article are you referring? :) – Michael Brown Jul 26 '13 at 6:42

The claim you read is probably

Electronic eigenstates of an atom, or of an inversion-symmetric molecule, always have vanishing electronic dipole moment about the atomic nucleus or the molecular inversion centre.

This is because if you invert the system you must be left with exactly the same eigenstates, but the electric dipole, being a vector quantity, must switch sign.

Magnetic dipole moments, on the other hand, can indeed be nonzero. This is because magnetic dipoles are pseudovector quantities and do not change sign upon inversion. Thus an atom like boron, which has five electrons and therefore a single 2p electron and orbital angular momentum $l=1$, can have a nonvanishing magnetic dipole moment (which will be proportional to the angular momentum of the p electron, so it obeys the standard QM restrictions: e.g. it can only have well-defined components along only one axis at a time).

Electron spin is also a pseudovector quantity, and so atoms and molecules can have spin-induced magnetic dipole moments without violating inversion symmetry.

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This argument doesn't seem to use any facts about the electromagnetic interaction, except perhaps, implicitly, the fact that they conserve parity. Since nuclear structure is determined by the electromagnetic and strong forces, and both of these conserve parity, it should apply to nuclear physics as well. Also, it would appear to apply to any odd electric multipole, so it would seem to rule out static electric octupole moments. And yet some nuclei are believed to have static electric octupole moments: Gaffney et al., "Studies of pear-shaped nuclei using accelerated radioactive beams." – Ben Crowell Sep 5 '13 at 2:08
@BenCrowell You're right, I'm not sure why that happens, and I've asked the corresponding question. It appears the weak interaction (which does break parity) is to blame. – Emilio Pisanty Sep 5 '13 at 15:54

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