In a recent question, Ben Crowell raised an observation which really puzzled me. I obtained a partial answer by looking in the literature, but I would like to know if it's on the right track, and a fuller explanation for it.
It is a well-known fact in atomic and molecular physics that electronic eigenstates of inversion-symmetric molecules never have electronic dipole moments. This is because the electromagnetic hamiltonian that governs molecular physics is parity invariant: under a reflection the eigenstates must map to themselves, but nonzero vector quantities - like dipole moments - must switch signs.
However, it was a fairly big news item earlier this year (see e.g. the University of York press release or the piece in Nature News, 8 May 2013) that atomic nuclei can be 'pear-shaped'. This was predicted in the fifties, such as e.g.
Stability of Pear-Shaped Nuclear Deformations. K. Lee and D. R. Inglis. Phys. Rev. 108 no. 3, pp. 774-778 (1957)
and was experimentally confirmed this year in
A pear-shaped nucleus is one that has a nonzero electric octupolar moment. The pear shape arises from the added contributions of quadrupole and octupole perturbations on a spherical shape, winding up with something like this:
However, this poses an immense problem, because octupole moments have odd parity. If you reflect a pear-shaped nucleus (as opposed to a rugby-ball-shaped quadrupolar one), you get a pear pointing the other way. Having such a nucleus requires a mixing of parity-even and -odd contributions to an energy eigenstate, and this is not allowed for eigenstates of the parity-conserving electromagnetic and strong interactions that (presumably) shape atomic nuclei.
To put this another way, having a pear-shaped nucleus requires a way to tell which way the pear will point. The nuclear angular momentum can break isotropy and provide a special axis, but the 'pear' is a vector (pointing from the base to the stem) and one needs parity-violating machinery to turn a pseudovector angular momentum into a vector quantity.
Another way to phrase this is by saying that if such an eigenstate were possible for a parity-conserving hamiltonian, then the reflected version should also be a degenerate, inseparable eigenstate. Having a unique such ground state means having a way to lift that degeneracy.
I can then pose my question: why are pear-shaped nuclei possible? Is my reasoning incorrect? That is, can parity-conserving hamiltonians lead to such parity-mixed eigenstates? Or are there in fact parity-violating interactions that decisively lift the degeneracies and shape these nuclei? If so, what are they?