Nuclear force and the shell model

The deuteron is the whole bound two-nucleon spectrum. It binds by 2.224 MeV with J = 1⁺, and there is nothing else: no excited deuteron, no dineutron, no diproton. Model it as a square well of radius R = 2.1 fm and a first bound state needs V₀R² > π²ℏ²/8μ = 102 MeV fm², so V₀ > 23.2 MeV; the depth that actually reproduces 2.224 MeV is about 34 MeV. Binding is about 7% of the well depth — the force clears the bar and stops. Now flip the two spins antiparallel. The spin-singlet np state does not bind at all, and its scattering length aₛ = −23.7 fm, large and negative, says it misses by a hair; the triplet's is aₜ = +5.42 fm. So the force is strongly spin dependent. It is also not central: the deuteron carries a quadrupole moment of +0.286 fm², which a pure ³S₁ state cannot have, and the roughly 4% ³D₁ admixture from a tensor term is also why its magnetic moment is 0.857 μ_N rather than the 0.880 μ_N you get by adding the free proton and neutron moments.

If every nucleon attracted every other, binding would follow the pair count A(A−1)/2 and B/A would climb with A. Instead B/A sits near 8 MeV from A ≈ 20 to A ≈ 240, and R = 1.2 A¹⁄³ fm holds the mean density fixed near 0.14 nucleons per fm³ however many you add, with the flat interior near 0.17. That is saturation, and it forces a range of a fermi or two: a nucleon binds only to its neighbours. Short range alone would still let matter collapse to any density, so something must push back — the ¹S₀ nucleon-nucleon phase shift is attractive at low energy but changes sign near 250 MeV, which is a repulsive core of radius about 0.5 fm. Charge independence completes the picture. Strip the Coulomb term out of proton-proton scattering and the singlet scattering lengths come out close: aₚₚ ≈ −17.3 fm, aₙₙ ≈ −18.9 fm, aₙₚ ≈ −23.7 fm. A state that misses binding by a hair magnifies any change in well depth into its scattering length, so even that spread means strengths equal to within a few percent. To the strong force a proton and a neutron are one particle in two charge states, which is why mirror pairs such as ¹⁵N and ¹⁵O differ in binding by little more than their Coulomb energies.

How can independent orbits survive a force this strong? A nucleon crossing a nucleus ought to scatter within a fermi, and then no orbit would last long enough to quantise. The exclusion principle is the answer: a collision needs an empty final state for each partner, and deep in the Fermi sea every nearby state is occupied, so the scattering is blocked. The mean free path stretches beyond the nuclear radius and each nucleon moves smoothly in the average potential the others build. Take that potential to be Woods-Saxon, V(r) = −V₀/[1 + exp((r − R)/a)] with V₀ ≈ 50 MeV, R = 1.25 A¹⁄³ fm and a ≈ 0.52 fm: flat where the density is flat, falling across a 2-3 fm skin, and lying between the infinite square well and the harmonic oscillator, whose solvable orderings bracket it. Fill it and closed shells appear at 2, 8 and 20 — ⁴He, ¹⁶O and ⁴⁰Ca, all anomalously bound. Then it fails.

The oscillator's next closures are 40, 70 and 112, the square well's are 34, 58 and 92, and the data insist on 28, 50, 82 and 126 — read off high separation energies, small quadrupole moments, extra stable isotopes and low neutron-capture cross-sections. No choice of V₀, R or a repairs that, because the well has no term distinguishing the two ways l and s can couple. Mayer and Jensen (1949) added Vₗₛ(r) l⋅s. Its expectation follows from j = l + s: ⟨l⋅s⟩ = ½[j(j+1) − l(l+1) − 3/4]ℏ², which is +(ℏ²/2)l for j = l + 1/2 and −(ℏ²/2)(l+1) for j = l − 1/2. Choose the coefficient so that j = l + 1/2 is pushed down, the opposite of the atomic case, and the pair separates by ΔE = C(l + 1/2), linear in l. C runs from about 2 MeV in the sd shell to under 1 MeV in the heaviest nuclei, so the 1d pair in ¹⁷O is split by 5.08 MeV — comparable to the oscillator spacing ℏω ≈ 41 A^(−1/3) MeV rather than a correction to it.

Order the split levels and fill them. 1s1/2 closes at 2; 1p3/2 then 1p1/2 close at 8; 1d5/2, 2s1/2 and 1d3/2 close at 20. Now the new physics. Because ΔE grows as 2l + 1, the j = l + 1/2 member of the highest-l orbital in each shell is driven out of that shell and down beside the one below. 1f7/2 goes first, holding 8 and closing 28. 1g9/2 follows, holding 10 and closing 50 — and it carries even parity into an otherwise odd-parity neighbourhood, which is why ⁹³Nb has a 9/2⁺ ground state. 1h11/2 gives 82 and 1i13/2 gives 126. Two book-keeping checks keep this honest. The occupancy-weighted centre of gravity never moves, since 2l + 2 states drop by Cl/2 while 2l states rise by C(l+1)/2 and the two products are equal. And each partner keeps the parity of its l, so an intruder is always the odd one out in its new company.

Even-even nuclei are 0⁺ without exception, because pairing binds identical nucleons in time-reversed pairs coupled to zero. For odd A the single unpaired nucleon carries everything: J = j of its orbital and π = (−1)^l. ¹⁷O comes out 5/2⁺, ⁴¹Ca 7/2⁻, ²⁰⁷Pb a 3p1/2 hole at 1/2⁻, ²⁰⁹Bi a 1h9/2 proton at 9/2⁻ — all measured, all right. Then look at ¹⁹F, where the model says 5/2⁺ and the ground state is 1/2⁺, or ²³Na, where it says 5/2⁺ and the answer is 3/2⁺. Both sit mid-shell, where a spherical mean field is simply the wrong starting point: the nucleus is deformed, the last nucleon moves in a non-spherical well, and l and j stop being good labels. Two signatures give it away. Quadrupole moments in the rare earths and actinides run about ten times the single-particle estimate, and those nuclei show rotational bands with E proportional to J(J+1) — collective motion the independent-particle model has no coordinate for. Near closed shells the mean field is excellent; away from them it needs the deformed Nilsson basis.