Why are pear-shaped nuclei possible? 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

Studies of pear-shaped nuclei using accelerated radioactive beams. L. P. Gaffney, P. A. Butler et al. Nature 497, 199–204 (09 May 2013). E-print on L.P. Gaffney's L.U. page.

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?
 A: I have a partial answer to my question, which I'm posting down here because the question was getting too long.
After a good look online at a bunch of confusing (to me) papers from the nuclear physics literature, I came upon this review:

Intrinsic reflection asymmetry in atomic nuclei. P. A. Butler and W. Nazarewicz. Rev. Mod. Phys. 68 no. 2, pp. 349-421 (1996).

Here they draw an analogy to the Jahn-Teller effect, which is originally a molecular physics principle that states that spatially degenerate ground states are in general not possible: there will always be some deformation of the molecule - or some other interaction, however small - that breaks the symmetry of the system and therefore necessarily lowers the energy of at least one of the degenerate states. Thus they explicitly state that

Stable reﬂection-asymmetric deformations in the body-ﬁxed frame can be attributed to a parity-breaking odd-multipolarity interaction which couples intrinsic states of opposite parity.

Later on, they pinpoint the cause of this parity breaking to the weak interaction:

Parity violation (in the laboratory frame) is caused by the parity-nonconserving component, $V^\text{PNC}$, of the weak interaction. The magnitude of this effect is of the order of $\alpha_p=G_Fm_\pi^2 /G_S\sim10^{-7}$, where $G_F$ is the Fermi constant and $G_S$ is the strong coupling constant.

If this is the case, my natural assumption is that a nucleus whose ground state is pear-shaped must be part of a nearly-degenerate doublet (which comes from the original degenerate ground state of the strong and electromagnetic interactions) separated by about $10^{-7}$ of the gap to the next excited state. The first, 'weakly' excited state would then also be pear-shaped, and would have the projection $$\text{(pear shape vector)}\cdot\text{(angular momentum)}$$ in the opposite direction to the ground state. Is this intuition correct?
