= Spectroscopic nuclear quadrupole moment
{title2=$Q_s$}
The laboratory <expectation value> of the axial <electric quadrupole moment> in the stretched nuclear <quantum state> $|I,M=I\rangle$; $M\hbar$ 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> gives
$$
Q_s(I)=-\frac{I}{2I+3}Q_0,\qquad Q_s(0)=0,\qquad Q_s(2)=-\frac27Q_0.
$$
Here $K\hbar$ 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 $\langle x^2\rangle=\langle y^2\rangle=\langle z^2\rangle$, so $\langle2z^2-x^2-y^2\rangle=0$. Equivalently, averaging an intrinsic axis uniformly over directions gives $\langle P_2(\cos\theta)\rangle=0$, 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 $u=\cos\theta$,
$$
\frac{Q_s(2)}{Q_0}
=\frac{\int_{-1}^{1}P_2(u)(1-u^2)^2\,du}{\int_{-1}^{1}(1-u^2)^2\,du}
=\frac{-32/105}{16/15}=-\frac27.
$$
A zero static laboratory moment does not make every quadrupole observable zero. With electric multipole operator $\mathcal M_{2\mu}=\int\rho_{\mathrm{ch}}r^2Y_2^\mu\,d^3r$, define the reduced transition strength using a <reduced matrix element>:
$$
B(E2;I_i\to I_f)=\frac{|\langle I_f\Vert\mathcal M_2\Vert I_i\rangle|^2}{2I_i+1}.
$$
In the axial $K=0$ rotor, angular projection gives $B(E2;0^+\to2^+)=5Q_0^2/(16\pi)$ and $B(E2;2^+\to0^+)=Q_0^2/(16\pi)$. 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.
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