Paired nucleons in a rotating nucleus

Turn rotational kinetic energy into a spectrum

Classical rotational energy is $L^2/(2\mathcal I)$. A larger moment of inertia $\mathcal I$ carries the same angular momentum at lower energy. For an axially deformed nucleus, take rotation perpendicular to the symmetry axis and replace squared angular momentum with its quantum eigenvalue.

The axially symmetric even-even ground band has intrinsic projection $K=0$ and allows $J=0,2,4,\ldots$. Its members share an intrinsic configuration. The drawn shape belongs to a body-fixed frame; a zero-spin laboratory ground state has no fixed spatial orientation.

Source: R. F. Casten

Worked example: one calibration predicts three levels

The March 2008 ENSDF erbium-166 ground band starts at $0,80.5776,264.990,545.454,911.208\,\mathrm{keV}$, with spins increasing by two. The first excited level sets $A=80.5776/6=13.4296\,\mathrm{keV}$.

For the three higher spins the rotor gives $268.592,564.0432,966.9312\,\mathrm{keV}$. Subtracting the original evaluated values gives residuals $+3.602,+18.5892,+55.7232\,\mathrm{keV}$. Increasing overprediction at higher spin suggests that the effective inertia changes as the nucleus is excited.

A quick pattern check is $E_4/E_2$. The rigid rotor gives $10/3$ and the evaluated ratio is about $3.29$. Yet the quoted energy errors range from $0.0020$ to $0.006\,\mathrm{keV}$, much smaller than the model discrepancy. The rotor explains the coarse pattern; precision spectroscopy needs finer dynamics.

Source: NNDC

What would a vibration ladder look like?

Consider a different starting point: a small quadrupole distortion about a spherical equilibrium. One quantum of this vibration is a phonon and carries angular momentum two. In the harmonic limit each additional phonon costs the same energy.

One phonon gives $2^+$. Two identical quadrupole phonons combine into degenerate $0^+,2^+,4^+$ states. Following only the maximum-spin member at each phonon number gives $E_4/E_2=2$, well below the evaluated erbium ratio. The workbench lists the other two-phonon states alongside this stretched sequence.

Anharmonicity and configuration mixing split real multiplets. Quadrupole transition strengths and quadrupole moments provide further tests. Energy ratios identify a promising explanation. Reconstructing the shape then requires the spectrum together with these electromagnetic observables.

Source: R. F. Casten

Where does the cost of breaking a pair come from?

Within a fixed $j$ shell, magnetic projections $m_j$ and $-m_j$ are time-reversed partners. An attractive zero-angular-momentum pairing interaction mixes configurations of these occupied pairs and lowers their energy. Pairing concerns nucleon correlations; rotation concerns collective orientation. Both can occur in a deformed nucleus.

A minimal exact example has $2\Omega=2j+1$ magnetic substates and $\Omega$ available pair positions. Put only one pair in the shell. Suppose every matrix element between pair positions is $-G$. An equal-amplitude superposition receives $\Omega$ equal contributions in each row, so its energy is $-G\Omega$. Orthogonal combinations have zero energy.

Take $j=5/2$ and $G=0.30\,\mathrm{MeV}$. Three pair positions give correlation energy $-0.90\,\mathrm{MeV}$. In this restricted model a broken pair loses that attraction, at an energy cost of $0.90\,\mathrm{MeV}$. As $G\rightarrow0$, the cost vanishes. These illustrative parameters are not fitted to erbium masses or spectra.

For more particles, seniority $v$ counts particles outside these zero-angular-momentum pairs. The exact degenerate-shell result below extends the one-pair calculation. Real shells have level spacings and additional correlations that couple pairing to deformation and rotation.

Source: J. Dobaczewski et al.