advanced30 min studyLesson 576 in Physics

Spin-orbit force and shell ordering

Derive the sign, magnitude, and degeneracy balance of a nuclear spin-orbit doublet in a declared effective Hamiltonian.

Before you begin

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The question for today

What you’ll understand

Evaluate angular-momentum scalar products for both spin-orbit partners.
Calculate splitting and the degeneracy-weighted doublet centroid.
Explain how a strong nuclear spin-orbit term reorganizes shell gaps.

A spherical central potential alone misses important nuclear shell closures. A strong effective spin-orbit term separates states with the same orbital motion but different coupling between orbital and intrinsic spin. The observed nuclear ordering favors the higher-j partner for the common sign convention. Deriving that sign explicitly is safer than importing an atomic fine-structure mnemonic. The resulting splitting also depends on radial dynamics, so angular algebra supplies a pattern rather than a complete universal level scheme.

Derive the two scalar-product eigenvalues

Write total angular momentum as orbital plus spin. Squaring this relation isolates the scalar product of orbital and spin operators. For spin one half, substituting the two allowed j values gives a positive orbital-spin scalar product for the higher-j partner and a negative one for the lower-j partner. The result uses quantum operator eigenvalues, not the cosine between two classical arrows of fixed lengths.

Choose the illustrative effective interaction to be minus kappa times this scalar product divided by the reduced Planck constant squared, with positive kappa. The higher-j state is then lowered and the lower-j state raised. Other authors may absorb the minus sign or the Planck factors into their coupling coefficient. Comparing formulas therefore requires checking the Hamiltonian definition before comparing quoted positive or negative constants.

\frac{\langle\mathbf l\cdot\mathbf s\rangle}{\hbar^2}=\frac{j(j+1)-l(l+1)-3/4}{2}=\begin{cases}l/2,&j=l+1/2,\\-(l+1)/2,&j=l-1/2.\end{cases}

The orbital quantum number l is a nonnegative integer. The lower-j branch exists only for l at least one; the scalar product has angular-momentum-squared units.↔ Scroll wide equations horizontally.

Keep the centroid and the splitting distinct

Within one radial orbital, approximate the radial spin-orbit matrix element by a single kappa shared by both partners. Their energy shifts then differ by kappa times twice l plus one divided by two. Higher orbital angular momentum produces a larger angular factor at fixed kappa, but the radial overlap need not stay fixed across different orbitals. The l-zero orbital has no splitting because it has no lower partner and its scalar product vanishes.

The two partners have different magnetic degeneracies: twice l plus two for the higher-j state and twice l for the lower-j state. Their degeneracy-weighted shift sums to zero in this simple approximation. Thus spin-orbit coupling redistributes one unsplit orbital's substates around its centroid, rather than moving every state downward. The arithmetic mean of the two energy shifts is not the correct centroid because it ignores the unequal numbers of magnetic states.

H_{ls}=-\kappa\frac{\mathbf l\cdot\mathbf s}{\hbar^2},\qquad \Delta E_{+}=-\frac{\kappa l}{2},\quad\Delta E_{-}=\frac{\kappa(l+1)}{2},\quad E_{-}-E_{+}=\frac{\kappa(2l+1)}2

Kappa and the energy shifts have energy units. A common radial matrix element and a perturbative doublet picture are assumptions, not exact identities for self-consistent nuclear orbitals.↔ Scroll wide equations horizontally.

Connect a local splitting to shell rearrangement

A sufficiently lowered high-j orbital can join a lower group of levels. For example, an f orbital's higher partner has j seven halves and capacity eight, so lowering it beyond a closure at twenty can help form a new gap at twenty-eight. This counting explains a mechanism without predicting the complete spectrum: central-field differences, radial nodes, residual interactions, and other spin-orbit partners also determine where a gap appears.

Both partners retain parity determined by the same orbital angular momentum. A spin-orbit splitting therefore changes j and energy without changing the pair's parity. Across isotopic chains the effective monopole interaction, tensor components, deformation, and continuum coupling can alter ordering. A level crossing is not evidence that angular-momentum addition failed. It indicates that the radial and many-body contributions in the effective Hamiltonian have changed, requiring comparison with the appropriate states and observables.

(2l+2)\Delta E_+ +(2l)\Delta E_-=0,\qquad \pi_+=\pi_-=(-1)^l

The centroid identity holds for the displayed shared-kappa model, with one radial parent orbital. The parity identity follows from spatial inversion and is independent of the splitting strength.↔ Scroll wide equations horizontally.

Work it through

Split a toy f doublet

Use orbital angular momentum $l=3$ and the defined interaction with $\kappa=2\,\mathrm{MeV}$ . Find both j values, shifts, capacities, parity, splitting, and the weighted centroid shift.

The partners are $j=7/2$ and $j=5/2$ , with scalar-product eigenvalues $3/2$ and $-2$ in units of hbar squared. The declared negative interaction sign gives shifts minus three and plus four MeV.
Their capacities are eight and six, respectively. Both have parity $(-1)^3=-1$ . The energy separation is $4-(-3)=7\,\mathrm{MeV}$ , agreeing with $2(2\times3+1)/2$ .
Weighting by the capacities gives total shift $8(-3)+6(4)=0\,\mathrm{MeV}$ . Dividing by fourteen leaves zero centroid shift. The simple arithmetic average would instead give plus half a MeV and is not the orbital centroid.

The result

The toy seven-halves state lies 3 MeV below its centroid and the five-halves state 4 MeV above it, with a 7 MeV splitting and common negative parity.

The living notebook

Explore the model

Interactive lab

Interactive lab · vectors

A scalar-projection sign analogy

For a classical sign analogy only, Reset and set angle $0^\circ$ , magnitude one; predict x projection plus one. Compare a fresh Reset run at $180^\circ$ , magnitude one; predict minus one. Multiplying these projections by a negative coefficient reverses the energy sign. This planar-vector lab does not calculate quantum orbital-spin eigenvalues, j capacities, or nuclear shell ordering.

Vectorx / y projectionsTrace marker

x component: 1.393
y component: 0.975
Magnitude: 1.7

Angle35°Magnitude1.7

t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA sign convention must accompany any coupling interpretation; classical projections do not replace angular-momentum algebra.

Pause & reason

Check your understanding

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Read the checkpoint explanation

The higher-j partner has positive orbital-spin scalar product l over two in hbar-squared units. The declared negative Hamiltonian coefficient therefore lowers its energy.

Your turn

Try it yourself

For a toy d orbital with $l=2$ and $\kappa=1.2\,\mathrm{MeV}$ , calculate shifts, capacities, splitting, and parity. Check the weighted centroid and the l-zero limiting case.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Substitute l into both branches, then weight by two j plus one rather than averaging the two shifts equally.

Compare with the solution

The $j=5/2$ partner shifts by $-1.2(2)/2=-1.2\,\mathrm{MeV}$ and has capacity six. The $j=3/2$ partner shifts by $1.2(3)/2=1.8\,\mathrm{MeV}$ and has capacity four. Their splitting is three MeV and both parities are positive. The weighted numerator is $6(-1.2)+4(1.8)=0$ . For l zero the sole j-one-half orbital has zero scalar product and no spin-orbit doublet splitting in this model.

Keep the idea

Take it with you

The sign of a spin-orbit shift follows from an explicitly defined Hamiltonian.
Doublet centroids require magnetic-degeneracy weights.
Shell ordering combines angular splitting with radial and many-body effects.

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