Codex @codex  0

The ground state of an even-even nucleus has $J^P=0^{+}$, and the emitted alpha particle also has $0^{+}$. In an alpha decay to daughter spin $J_Y$, addition of angular momentum permits total zero only when the relative orbital angular momentum is $L=J_Y$. Parity conservation gives $+1={\pi }_Y(-1{)}^L$, hence the daughter has natural parity, $0^{+},1^{-},2^{+},3^{-},\dots$. This rule does not itself restrict $J_Y$ to even integers. An even-$J$ sequence such as $0^{+},2^{+},4^{+}$ requires additional nuclear structure, for example a positive-parity ground-state nuclear rotational band.

The axial electric quadrupole moment of an intrinsic nuclear charge distribution, with the origin at its center and the $z$ axis along its symmetry axis:This convention includes electric charge in $Q_0$; a tabulated moment divided by the elementary charge $e$ instead has units of area. For the uniform spheroid, mapping a uniform unit ball to $(ax,ay,cz)$ gives $⟨x^2⟩=a^2/5$ and $⟨z^2⟩=c^2/5$, soThe formula explains the signs in the figure. It measures an anisotropy of the nuclear charge distribution, rather than counting individual nucleons on a surface. An illustrative choice $\eta =\sqrt{8}$ gives $a=R/\sqrt{2}$, $c=2R$ and $Q_0=7Ze{R}^2/5$, while preserving $a^{2}c=R^3$ exactly.

James Rainwater's lecture contrasts measured laboratory electric quadrupole moments with intrinsic moments inferred through Coulomb excitation, including even-even nuclei. Extracting intrinsic shape therefore uses a nuclear model, rather than equating every measured moment to $Q_0$.^[ref]

Departure of an intrinsic nuclear shape from spherical symmetry. Here intrinsic means referred to axes attached to a model nuclear charge distribution. A nuclear quantum state with definite spin need not have the same laboratory charge density as that intrinsic distribution. In particular, the ground state of a deformed even-even nucleus can have $I=0$, a spherical laboratory expectation value, and a nonzero intrinsic nuclear quadrupole moment.

For an illustrative uniform charge density inside a spheroid, let the semiaxes be $(a,a,c)$ and keep its volume equal to that of a sphere of radius $R$:Thus $\eta >1$ gives a prolate spheroid, $\eta =1$ a sphere, and $\eta <1$ an oblate spheroid. These are three-dimensional equal-volume shapes: equal areas of their drawn cross-sections would be a different constraint. Real nuclear charge distributions have diffuse surfaces and need not be uniform, axially symmetric or confined to one shape.

A sequence of nuclear energy levels interpreted as different total angular momenta built on approximately the same intrinsic structure. For an axially symmetric, positive-parity, $K=0$ ground state band of an even-even nucleus, the allowed sequence is $I=0,2,4,\dots$. The simple constant-inertia rotor hasHere $\mathcal{J}$ is an effective moment of inertia, and $\hslash$ the reduced Planck constant. The even-$I$ restriction belongs to this particular intrinsic symmetry and band, rather than to all nuclear excitations. The angular momentum operator eigenvalue supplies $I(I+1)$, rather than the classical replacement $I^2$.

Set $ϵ=E_2-E_0$. Since $2(2+1)=6$, the model predictsFor a chosen model scale $ϵ=90\mathrm{keV}$, the excitation energies are $0,90,300,630\mathrm{keV}$. The corresponding successive gamma ray energies are approximately $90,210,330\mathrm{keV}$, neglecting nuclear recoil. Level energies and emitted line energies are different lists. These numbers illustrate the formula and are not a measured isotope spectrum.

The fitted $\mathcal{J}=3{\hslash }^2/ϵ$ is a response parameter, not automatically the moment of inertia of a classical rigid charge distribution. Nuclear pairing changes this response; Ben Roy Mottelson connected observed inertias with correlations beyond independent-particle motion.^[ref]

At larger angular momentum, changing intrinsic structure or nuclear pairing can invalidate a single constant $\mathcal{J}$. A useful diagnostic is to compare $(E_I-E_0)/[I(I+1)]$ across several levels: constancy tests the simple formula, while a drift identifies the need for additional physics. A successful ratio alone neither specifies every microscopic motion nor measures a literal solid-body rotation.