The even-even nucleus has ground-state spin and parity

PHYSICS · Subjective · Fermi University Physics · Quantum Mechanics · Nuclear Physics · Nuclear constituents, size and binding

The Nucleus Resolved, Ch Spin, parity and the electromagnetic moments of a nucleus

Problem

The even-even nucleus 154Sm^{154}\mathrm{Sm} has ground-state spin and parity Jπ=0+J^\pi = 0^+. An independent electromagnetic measurement, exciting the nucleus with the electric field of a passing charged particle, shows that its charge distribution in its own body-fixed frame is strongly elongated, with an intrinsic quadrupole moment Q0≈+6.6 e bQ_0 \approx +6.6\,e\,\mathrm{b}. Atomic-beam magnetic resonance on the ground state, however, measures a spectroscopic quadrupole moment Qsp=0Q_{\rm sp} = 0. A colleague argues: "Qsp=0Q_{\rm sp}=0 is a direct measurement of the charge distribution, so 154Sm^{154}\mathrm{Sm} must in fact be spherical; the Coulomb-excitation result must be in error." Is this argument correct? Justify your answer using the triangle condition on the multipole order and the meaning of QspQ_{\rm sp} as an expectation value in the state m=Jm=J.

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