---
title: "Paired nucleons in a rotating nucleus"
description: "Derive the erbium-166 rotational pattern, compare harmonic vibration and solve a small pairing example that connects single-particle and collective motion."
canonical_url: "https://sandboxphysics.com/stories/nuclear-bridge/pairing-and-collectivity/"
language: "en"
page_type: "story"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/stories/nuclear-bridge/pairing-and-collectivity/"
last_modified: "2026-10-27"
---

[Nuclear physics](https://sandboxphysics.com/stories/nuclear-bridge/)

L3 · Chapter 4

# Paired nucleons in a rotating nucleus

Sandbox Physics · 2026-10-27

The lowest erbium-166 levels spread farther apart as their angular momentum grows. A shared deformed configuration explains this pattern through collective rotation. Meanwhile, nucleons can correlate in zero-angular-momentum pairs. Separating rotation from the cost of breaking a pair shows why both pictures belong in nuclear structure.

## Before you begin

Total angular momentum, spin-orbit partners and E2 selection from the preceding chapter. Rotational kinetic energy is assumed; phonons and time-reversed pairing are introduced below.

[Review the preceding prerequisite](https://sandboxphysics.com/stories/nuclear-bridge/spectra-and-selection/)

- Predict a rotational band from its first excited energy and test constant inertia against assessed levels.
- Compare rotational, vibrational and pair-breaking energy scales.

## 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.

$$
E_J=A J(J+1),\qquad A=\frac{\hbar^2}{2\mathcal I}
$$

These are excitation energies relative to the ground state. Constant inertia is the approximation being tested.

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.

$$
E_n=n\hbar\omega,\qquad\left.\frac{E_4}{E_2}\right|_{\rm stretched}=2
$$

The ground-state zero-point energy is subtracted. The energy subscripts label spin, whereas $n$ counts phonons.

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.

$$
E(n,v)=-\frac G4(n-v)(2\Omega-n-v+2)
$$

Here $n$ counts particles, $\Omega$ counts available pair positions and $G$ is a positive attraction. Allowed seniority has the same parity as particle number and must fit within both particle and hole spaces.

Source: J. Dobaczewski et al.

## Try the reasoning in an experiment

### [Predict the change before increasing inertia](https://sandboxphysics.com/labs/nuclear-spectroscopy/)

1. Choose An erbium rotational band. Calculate the next three energies from the first excitation, then compare model values and residuals.
2. Predict the effect of doubling inertia, then set Moment of inertia relative to reference to two. Explain why all energies halve but the ratios stay fixed.
3. Choose Test a vibration ladder. Identify the shared calibration energy, explain the diverging higher levels and list the two-phonon spins omitted from the stretched sequence.

## Exercises and explanations

1. With $E_2=120\,\mathrm{keV}$, predict $E_4$ and $E_6$ for the rotor and stretched harmonic sequence.

   Read the explanation

   The rotor multiplies by $20/6$ and $42/6$, giving $400$ and $840\,\mathrm{keV}$. Two and three phonons give $240$ and $360\,\mathrm{keV}$. An identical first excitation therefore leads to clearly different higher spacings.

2. Place two particles in a degenerate $j=7/2$ shell with $G=0.20\,\mathrm{MeV}$. Find the paired ground energy and breaking cost. Does a filled shell have zero interaction energy?

   Read the explanation

   There are $\Omega=4$ pair positions, giving $-G\Omega=-0.80\,\mathrm{MeV}$ and breaking cost $0.80\,\mathrm{MeV}$. With $n=8,v=0$, the filled-shell result is still $-0.80\,\mathrm{MeV}$. The Hamiltonian includes diagonal attraction on occupied pair positions; its energy convention must stay fixed.

[Previous: Read a nuclear spectrum one neutron at a time](https://sandboxphysics.com/stories/nuclear-bridge/spectra-and-selection/) [Next: Where the wave goes after a neutron meets a nucleus](https://sandboxphysics.com/stories/nuclear-bridge/complex-potential-scattering/)

## Sources and reproduced scope

Evaluated level energies and published reaction measurements retain their source versions and uncertainties. Model curves, schematic figures and synthetic scattering records are identified separately. The offline benchmark files state which calculations have been reproduced.

1. [NNDC, ENSDF adopted levels and gammas for erbium-166](https://www.nndc.bnl.gov/ensnds/166/Er/adopted.pdf)

   Selected ground-band levels in the frozen adopted evaluation; the dataset records the evaluation date and errors. Only selected numerical facts are redistributed; model curves and artwork are original.

2. [J. Dobaczewski et al., Odd-even staggering, degenerate-shell seniority model](https://users.fuw.edu.pl/~dobaczew/oes-25w/node7.html)

   Exactly solvable degenerate-shell pairing example. Its fixed interaction and restricted shell are a teaching limit, not an isotope mass evaluation.

3. [R. F. Casten, Lectures on Nuclear Structure, NNPSS 2013](https://archive.int.washington.edu/NNPSS/2013/lectures/Casten1.pdf)

   Collective rotational sequences and intrinsic deformation. The rigid rotor and harmonic vibrator are limiting models; the evaluated band residuals test their range.