Spectroscopic nuclear quadrupole moment ($Q_s$)

The laboratory expectation value of the axial electric quadrupole moment in the stretched nuclear quantum state $|I,M=I⟩$; $M\hslash$ is the laboratory $z$ component of total angular momentum. It differs from the intrinsic nuclear quadrupole moment $Q_0$. In an axial nuclear rotational band with $K=0$, projection onto definite angular momentum givesHere $K\hslash$ denotes the component of total angular momentum along the intrinsic symmetry axis. Consequently a positive $Q_0$ for a prolate spheroid corresponds to a negative $Q_s(2)$ in this model. The sign of the laboratory moment alone must be interpreted together with the band assignment.

The $I=0$ result has a short symmetry explanation. A spin-zero state has no preferred spatial direction. Its second charge moments satisfy $⟨x^2⟩=⟨y^2⟩=⟨z^2⟩$, so $⟨2z^2-x^2-y^2⟩=0$. Equivalently, averaging an intrinsic axis uniformly over directions gives $⟨P_2(\cos \theta )⟩=0$, where $P_2(u)=(3u^2-1)/2$ is a Legendre polynomial. This is a statement about a quantum expectation value, not a classical picture of a spinning object sweeping through orientations.

For an independent check of the $I=2$ sign, the $K=0$, $M=2$ rotor orientation probability is proportional to $|Y_2^2{|}^2\propto {\sin }^4\theta$, using a spherical harmonic. With $u=\cos \theta$,A zero static laboratory moment does not make every quadrupole observable zero. With electric multipole operator ${\mathcal{M}}_{2\mu }=\int {\rho }_{\mathrm{ch}}{r}^2{Y}_2^{\mu }{d}^{3}r$, define the reduced transition strength using a reduced matrix element:In the axial $K=0$ rotor, angular projection gives $B(E2;0^{+}\to 2^{+})=5Q_0^2/(16\pi )$ and $B(E2;2^{+}\to 0^{+})=Q_0^2/(16\pi )$. The factor of five is the ratio of the numbers of magnetic substates. Thus an electric quadrupole transition can reveal collective deformation even when $Q_s(0)=0$. These strengths determine $|Q_0|$ within the model; their squares alone cannot distinguish its sign.