University Physics V · Nuclear Physics · 14.8

The Shell Model & Spin–Orbit Splitting

Fill a Woods–Saxon well and the closures come out 2, 8, 20, then 40, 70, 112 — and experiment reports 28, 50, 82, 126. This is the lesson that adds the one operator which repairs that, l⋅s with the sign reversed from an atom's, and shows why its growth as 2l + 1 lets only the top orbital of each shell fall through the gap.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The shell model is the claim that a nucleon moves in an average potential made by the other A − 1, so a hopeless A-body problem becomes one eigenvalue problem repeated A times. Take that potential to be Woods–Saxon — a flat-bottomed well about 50 MeV deep, of radius 1.25 A¹⁄³ fm and surface thickness 0.55 fm — and diagonalise p²/2m_N + V_WS(r) in a truncated harmonic-oscillator basis. You get 2, 8 and 20 exactly, and then nothing: 40, 70 and 112 where experiment reports 28, 50, 82 and 126.

Mayer and Jensen's repair in 1949 was one more one-body operator, Vₗₛ(r) l⋅s, with Vₗₛ negative and confined to the surface by the Thomas shape (1/r) dV_WS/dr. In the coupled basis |n l j mⱼ⟩ that operator is just a number, (ħ²/2)[j(j+1) − l(l+1) − 3/4], so the Hamiltonian stays block diagonal in (l, j) and the gap between the two members of an l multiplet is (2l+1)ħ²|Vₗₛ|/2 — linear in l. That linearity is the physics: only the largest-l orbital of an oscillator shell splits enough for its j = l + ½ member to fall a whole ħω, and when 1f7/2, 1g9/2, 1h11/2 and 1i13/2 do, their 8, 10, 12 and 14 places land on 20, 40, 70 and 112 and give exactly the observed gaps.

The cost is that this is still a mean field. It has no pairing, so even-even 0⁺ ground states must be put in by hand; it has no deformation, so ¹⁹F comes out 5/2⁺ against a measured 1/2⁺; and it gives energies, never rates.

Simple definition: The nuclear spin–orbit term is a one-body operator Vₗₛ(r) l⋅s with Vₗₛ negative, which splits every l > 0 orbital into j = l ± ½ and, because the splitting grows as 2l + 1, drives the largest-l j = l + ½ orbital out of its own oscillator shell.
Example: For 1g (l = 4) with |Vₗₛ| = 1.4 MeV the splitting is (2l+1)|Vₗₛ|/2 = 9 × 1.4/2 = 6.3 MeV, and the barycentre rule puts 1g9/2 four-ninths of that, 2.8 MeV, below the multiplet centre while 1g7/2 sits five-ninths, 3.5 MeV, above it.

Single-particle shell-model HamiltonianĤ = p̂²/2m_N + V_WS(r) + Vₗₛ(r) l̂⋅ŝ/ħ²

One body, so A nucleons become A copies of one eigenproblem; u(0) = 0 and u(r→∞) = 0 come free with a harmonic-oscillator basis.

V_WS = −V₀/[1 + e^((r−R)/a)], V₀ ≈ 50 MeV, R = 1.25 A¹⁄³ fm, a ≈ 0.55 fm; ħ²/2m_N = 20.7 MeV fm²

l⋅s in the coupled basis⟨l⋅s⟩ = (ħ²/2)[j(j+1) − l(l+1) − 3/4]

Diagonal in |n l j mⱼ⟩, so Ĥ is block diagonal in (l, j) — no Clebsch–Gordan coefficients, just one small radial matrix per block.

+lħ²/2 for j = l + ½ and −(l+1)ħ²/2 for j = l − ½; 1f7/2 gives +1.5ħ², 1f5/2 gives −2ħ²

Splitting and the barycentre ruleΔE = (2l+1)ħ²|Vₗₛ|/2shifts −l/(2l+1) and +(l+1)/(2l+1) of ΔE

Growth linear in l is why only the top-l orbital of a shell can fall a whole ħω, and why the fatter level is the one that moves less.

ΔE in MeV; the (2j+1)-weighted mean is unmoved, since (2l+2)l − 2l(l+1) = 0

Thomas radial shape of the spin–orbit termVₗₛ(r) = −κ (1/r) dV_WS/dr, peaked at r = R

Confines the force to a surface skin of width a, so the high-l orbitals that the centrifugal barrier pushes outward are the ones that feel it.

κ ≈ 0.5 fm²; |dV_WS/dr| peaks at V₀/4a ≈ 23 MeV fm⁻¹ for V₀ = 50 MeV and a = 0.55 fm

The oscillator gap an intruder must crossħω ≈ 41 A^(−1/3) MeVoscillator closures 2, 8, 20, 40, 70, 112

The benchmark for every intruder claim: an orbital only changes shells once l⋅s plus the ℓ² depression carry it past this gap.

Shell N = 2(n−1) + l holds (N+1)(N+2) nucleons with spin; A = 90 gives ħω = 9.1 MeV

Odd-A ground-state assignmentJ^π = j of the last odd nucleon, π = (−1)^l of that orbital

⁴¹Ca → 1f7/2 → 7/2⁻ and ⁹¹Zr → 2d5/2 → 5/2⁺, while ¹⁹F predicts 5/2⁺ and measures 1/2⁺ — deformation, not a bad orbital.

Even-even is 0⁺ from pairing; odd-odd needs Nordheim's rules; a hole carries the particle's j and π

Name the operator, the basis, and the boundary condition

The shell model trades the A-body Hamiltonian for a one-body one, Ĥ = p̂²/2m_N + V_WS(r) + Vₗₛ(r) l̂⋅ŝ/ħ², with V_WS(r) = −V₀/[1 + exp((r − R)/a)], V₀ ≈ 50 MeV, R = 1.25 A¹⁄³ fm and a ≈ 0.55 fm. Because Ĥ is one-body its eigenkets are single-nucleon orbitals, and the nucleus is built by filling them. Work in the coupled basis |n l j mⱼ⟩ = |n l⟩ ⊗ [Yₗ ⊗ χ_½]^j, and expand the radial factor in three-dimensional harmonic-oscillator functions Rₙₗ(r; b) with b = √(ħ/m_N ω). That basis choice is also the boundary condition: every Rₙₗ vanishes as r^l at the origin and dies as a Gaussian at infinity, so u(0) = 0 and u(r→∞) = 0 are imposed by construction rather than searched for by shooting. With ħ²/2m_N = 20.7 MeV fm², about fifteen radial functions per block converge the bound spectrum to tens of keV — and the truncation, not the physics, is the first thing to check.

l⋅s is a number in this basis, and that is the whole trick

Square ĵ = l̂ + ŝ: ĵ² = l̂² + ŝ² + 2 l̂⋅ŝ, so l̂⋅ŝ = ½(ĵ² − l̂² − ŝ²), and in |n l j mⱼ⟩ it is diagonal with eigenvalue (ħ²/2)[j(j+1) − l(l+1) − 3/4]. For j = l + ½ that is +lħ²/2; for j = l − ½ it is −(l+1)ħ²/2. Nothing else in Ĥ touches spin, so the matrix never connects different (l, j): it is block diagonal, one real symmetric radial block per (l, j), each perhaps 15 × 15, and numpy.linalg.eigh returns the lot in milliseconds. No Clebsch–Gordan coefficient is ever needed, because the coupled basis has already done the coupling, and the 2j + 1 values of mⱼ come out degenerate for free since Ĥ commutes with ĵ. Concretely 1f7/2 carries ⟨l⋅s⟩ = +1.5ħ² and 1f5/2 carries −2ħ²; those two numbers are the same in every nucleus, and only the coefficient Vₗₛ in front of them moves with A.

Splitting grows as 2l + 1; the centre of gravity does not move

Subtract the two eigenvalues: ΔE = E(j = l − ½) − E(j = l + ½) = (2l+1)ħ²|Vₗₛ|/2. The gap is linear in l, so a p orbital splits by 3|Vₗₛ|/2 while an i orbital splits by 13|Vₗₛ|/2 — more than four times as much for the same coefficient. Where the two levels sit is then fixed by a sum rule: weight by 2j + 1 and note that (2l+2)(l) − (2l)(l+1) = 0, so l̂⋅ŝ leaves the multiplet's centre of gravity exactly where it was. The j = l + ½ level therefore falls by l/(2l+1) of ΔE and the j = l − ½ level rises by (l+1)/(2l+1) of it; the fatter level moves less. For 1g with |Vₗₛ| = 1.4 MeV, ΔE = 9 × 1.4/2 = 6.3 MeV, 1g9/2 drops 4/9 × 6.3 = 2.8 MeV, 1g7/2 climbs 5/9 × 6.3 = 3.5 MeV, and 10(−2.8) + 8(+3.5) = 0 as it must. Use that as an arithmetic check on every level scheme you draw.

The force lives in the surface, and its sign is not the atom's

The radial shape is Thomas-like: Vₗₛ(r) = −κ (1/r) dV_WS/dr. Woods–Saxon is flat inside, so its derivative is a bell of width a centred on r = R with peak V₀/4a = 50/(4 × 0.55) = 23 MeV fm⁻¹. At A = 90, R = 5.60 fm, and with κ ≈ 0.5 fm² the peak coefficient is 0.5 × 23/5.60 ≈ 2.0 MeV; averaging over a radial wavefunction that is not all in the skin brings it down to roughly 1.4 MeV. Two effects then point the same way — the 2l + 1 factor, and the centrifugal term l(l+1)ħ²/2m_N r² pushing high-l orbitals out into exactly the shell where the force acts. The sign is the part to memorise. Atomic fine structure has a positive coefficient and puts j = l + ½ higher; the nuclear term is negative and puts it lower. It is also twenty to thirty times too strong to be the electron's Thomas precession scaled up: it comes from the nucleon–nucleon interaction, and in a non-relativistic mean field both κ and its sign are fitted, not derived.

The intruder arithmetic that makes 28, 50, 82 and 126

Count the plain oscillator first. Shell N = 2(n − 1) + l holds (N + 1)(N + 2) nucleons once spin is included, so the closures are 2, 8, 20, 40, 70, 112: three right, the rest wrong. Flattening the well adds a depression roughly proportional to l(l+1), which spreads each N shell by l but leaves the shells intact. Spin–orbit is what regroups them, dragging the j = l + ½ member of the largest l in shell N down past the whole gap ħω ≈ 41 A^(−1/3) MeV and parking it on top of shell N − 1. After that it is only addition. Shell N = 3 loses 1f7/2, which holds 8: 20 + 8 = 28. The remaining 12 of N = 3 take you to 40, and 1g9/2's 10 give 50. The remaining 20 of N = 4 reach 70, and 1h11/2's 12 give 82. The remaining 30 of N = 5 reach 112, and 1i13/2's 14 give 126. Every gap above 20 is one intruder; 2, 8 and 20 need none, because the well alone makes them.

Reading J^π off the scheme, and the ways it fails

Even-even nuclei are 0⁺ in their ground state without exception, and the mean field does not explain that: it is the residual pairing interaction, left out of Ĥ by construction, coupling identical nucleons pairwise to J = 0. Grant it, and odd-A becomes a one-line prediction — J^π is the j of the last odd nucleon with parity (−1)^l for its orbital. ¹⁷O puts a neutron in 1d5/2 and is 5/2⁺; ⁴¹Ca puts one in 1f7/2 and is 7/2⁻; ⁹¹Zr's 51st neutron sits in 2d5/2 and gives 5/2⁺; ²⁰⁷Pb is a 3p1/2 hole in the N = 126 core and is 1/2⁻, a hole carrying the particle's j and parity. Then the failures. Odd-odd nuclei need Nordheim's rules to couple two odd nucleons. Deformed nuclei break spherical symmetry so j stops being a good quantum number: ¹⁹F should be 5/2⁺ and is 1/2⁺, ²³Na should be 5/2⁺ and is 3/2⁺, and both want Nilsson orbitals labelled Ω[N n_z Λ]. And a level scheme fixes which states exist, never how fast anything decays.

Change one variable at a time

Make the relationship visible.

Interactive model

Spin-orbit coefficient |Vₗₛ|1.40 MeV

l² depression coefficient D0.34 MeV

Mass number A92

Push the spin–orbit slider to its top with the ℓ² depression at its lowest: the closure still reads 40, so l⋅s alone cannot do it. Raise the depression as well and only 1g9/2 dives past the N = 3 shell top while 1g7/2 climbs; the moment it crosses, the count flips to 50.

1g SPLITTING ΔE6.30 MeV

1g9/2 BELOW BARYCENTRE9.60 MeV

OSCILLATOR GAP9.08 MeV

SHELL CLOSES AT50 nucleons

Live interpretation1g SPLITTING ΔE: 6.30 MeV. 1g9/2 BELOW BARYCENTRE: 9.60 MeV. OSCILLATOR GAP: 9.08 MeV. SHELL CLOSES AT: 50 nucleons

Catch the common trap

Explain before calculating.

Misconception check

The trap is importing atomic fine structure. In an atom the coefficient is positive, so j = l + ½ lies above j = l − ½; in a nucleus Vₗₛ is negative and j = l + ½ lies below, and the whole shell scheme hangs on that sign. Flip it and the intruders become 1f5/2, 1g7/2, 1h9/2 and 1i11/2, each holding two fewer nucleons, so the closures come out 26, 48, 80 and 124 — every one wrong — and ⁴¹Ca is predicted 5/2⁻ against a measured 7/2⁻. The second half of the error is calling the term relativistic. Scale the electron's Thomas precession up to nuclear fields and the splitting is twenty to thirty times too small, still with the atomic sign. The nuclear l⋅s is not electromagnetic: it belongs to the nucleon–nucleon interaction, and in relativistic mean-field language it comes from scalar and vector potentials of a few hundred MeV that cancel in the central well and add in the spin–orbit one.

A student writing a Woods–Saxon shell-model code enters the spin–orbit coefficient with the atomic sign, so Vₗₛ comes out positive. The well, the harmonic-oscillator basis and the ℓ² depression are all unchanged. What does the level scheme now predict for the magic numbers?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA neutron occupies the 1g orbital of a nucleus near A = 90, where the radially averaged spin–orbit coefficient is |Vₗₛ| = 1.4 MeV. Evaluate ⟨l⋅s⟩ for both members of the doublet, find the splitting, and place each level relative to the multiplet's centre of gravity.

l = 4, so j = l ± ½ gives 9/2 and 7/2, and l(l+1) = 20. For j = 9/2, j(j+1) = 99/4 = 24.75, so ⟨l⋅s⟩ = (ħ²/2)(24.75 − 20 − 0.75) = +2ħ².
For j = 7/2, j(j+1) = 63/4 = 15.75, so ⟨l⋅s⟩ = (ħ²/2)(15.75 − 20 − 0.75) = −2.5ħ². Check against the shortcut +l/2 = +2 and −(l+1)/2 = −2.5.
Splitting: ΔE = (2l+1)|Vₗₛ|/2 = 9 × 1.4/2 = 6.3 MeV. Vₗₛ is negative, so the j = l + ½ member, 1g9/2, is the lower one — the reverse of atomic fine structure.
Barycentre rule: 1g9/2 falls l/(2l+1) = 4/9 of 6.3 = 2.8 MeV and 1g7/2 rises 5/9 of 6.3 = 3.5 MeV. Weight by 2j + 1 to check: 10(−2.8) + 8(+3.5) = −28 + 28 = 0.

Answer⟨l⋅s⟩ = +2ħ² for 1g9/2 and −2.5ħ² for 1g7/2; ΔE = 6.3 MeV with 1g9/2 lower, sitting 2.8 MeV below the centre of gravity while 1g7/2 sits 3.5 MeV above it.

MediumStarting from the closure at 28, fill the orbitals in the standard order 2p3/2, 1f5/2, 2p1/2, 1g9/2 and show that the next gap falls at 50 rather than 40. Then assign J^π to ⁹¹Zr (Z = 40, N = 51) and to ⁸⁹Y (Z = 39, N = 50), saying in each case which nucleon does the work.

Capacities are 2j + 1: 2p3/2 holds 4, 1f5/2 holds 6, 2p1/2 holds 2. Adding them to 28 gives 28 + 4 + 6 + 2 = 40, which exhausts what is left of the N = 3 oscillator shell.
1g9/2 belongs to N = 4 but has been dragged below the rest of it, and it holds 2j + 1 = 10, so 40 + 10 = 50 with the next orbital, 2d5/2, across a gap. 40 is the plain oscillator's own closure and shows up only as a weak sub-shell; the strong gap is at 50, and it exists solely because of the intruder.
⁹¹Zr: Z = 40 is even and N = 50 is closed, so the odd neutron decides. It is the 51st, the first into 2d5/2, giving j = 5/2 and l = 2, so π = (−1)² = +1 and J^π = 5/2⁺.
⁸⁹Y: N = 50 is closed and even, so the odd proton decides. Protons fill to 32 in 2p3/2 and to 38 in 1f5/2, so the 39th is the first of the two in 2p1/2: j = 1/2, l = 1, π = (−1)¹ = −1, and J^π = 1/2⁻.
Both match the measured ground states, and both test the ordering rather than the spins alone: put 1g9/2 below 2p1/2 instead and the 39th proton would sit in 1g9/2, making ⁸⁹Y 9/2⁺.

AnswerThe next gap is at 40 + 10 = 50. ⁹¹Zr is 5/2⁺ from its 51st neutron in 2d5/2, and ⁸⁹Y is 1/2⁻ from its 39th proton in 2p1/2. Both agree with the measured ground states.

HardIn the lead region take ħω = 41 A^(−1/3) MeV, a radially averaged spin–orbit coefficient |Vₗₛ| = 0.80 MeV, and model the flat-bottomed well by an extra depression D l(l+1) with D = 0.16 MeV. For A = 208, find ħω, find the 1i splitting and how far 1i13/2 and 1i11/2 each sit below the l = 0 member of the N = 6 shell, decide which of them clears the gap, and assign J^π to ²⁰⁷Pb.

Oscillator gap: 208¹⁄³ = 5.925, so ħω = 41/5.925 = 6.92 MeV. That is the drop an orbital must manage to leave shell N = 6.
Splitting: l = 6, so ΔE = (2l+1)|Vₗₛ|/2 = 13 × 0.80/2 = 5.20 MeV. The barycentre rule puts 1i13/2 at 6/13 of that below the 1i centre, 2.40 MeV, and 1i11/2 at 7/13 above it, 2.80 MeV.
ℓ² depression: D l(l+1) = 0.16 × 42 = 6.72 MeV lowers the whole 1i multiplet relative to the l = 0 member of N = 6, and lowers both partners equally.
Totals below that l = 0 member: 1i13/2 sits 6.72 + 2.40 = 9.12 MeV down, 1i11/2 sits 6.72 − 2.80 = 3.92 MeV down.
Against ħω = 6.92 MeV only 1i13/2 clears the gap, since 9.12 > 6.92 while 3.92 < 6.92. Its 2j + 1 = 14 places land on the 112 already accumulated and close the shell at 126. Neither term does it alone: the spin–orbit contribution of 2.40 MeV is only 35% of ħω.
²⁰⁷Pb has Z = 82 closed and N = 125, one short of 126, so the ground state is a hole in the last orbital below the gap, 3p1/2. A hole carries the particle's j and parity: j = 1/2, l = 1, π = (−1)¹, so J^π = 1/2⁻, as measured.

Answerħω = 6.92 MeV and ΔE(1i) = 5.20 MeV. 1i13/2 lies 9.12 MeV below the N = 6 l = 0 member and clears the gap; 1i11/2, at 3.92 MeV, does not. The closure is 126, and ²⁰⁷Pb is 1/2⁻.