Partial waves with spin: phase shifts for j = l ± ½

Units are SI, with nuclear numbers in MeV and fm (ħc = 197.327 MeV fm, 1 b = 10⁻²⁸ m²); everything here is non-relativistic, so no metric signature enters. Work in the centre-of-mass frame with reduced mass μ and ħk = √(2μE), and let V = V_c(r) + Vₛₒ(r) L⋅S act on a spin-½ projectile and a spinless target. L⋅S commutes with L², S² and J = L + S but not with L_z or S_z, so mₗ and mₛ are not conserved while l, j and mⱼ are. Parity finishes the job: the two candidates l = j ± ½ for a given j have opposite parity and cannot mix. The coupled states |l, ½, j, mⱼ⟩ therefore scatter independently, and on them L⋅S takes Quantum Mechanics II's values lħ²/2 for j = l + ½ and −(l + 1)ħ²/2 for j = l − ½ (quoted). Each l ≥ 1 carries two radial problems and two phase shifts, δₗ⁺ and δₗ⁻, found by QM-II's radial machinery unchanged. The s wave has only j = ½ and one phase shift.

The incident state is e^(ikz)χ = Σ iˡ(2l + 1) jₗ(kr) Pₗ(cos θ)χ, Rayleigh's expansion times a fixed spinor χ. Each term Pₗ(cos θ)χ has definite l but not definite j: it mixes j = l + ½ and j = l − ½ at the same mⱼ. Separate them with σ⋅L = 2L⋅S/ħ, whose eigenvalues are lħ and −(l + 1)ħ. The operators P₊ = (l + 1 + σ⋅L/ħ)/(2l + 1) and P₋ = (l − σ⋅L/ħ)/(2l + 1) equal 1 on one eigenvalue and 0 on the other and sum to 1; since σ⋅L is traceless, their traces over the 2(2l + 1) states of given l are 2(l + 1) and 2l, the 2j + 1 states of each channel. Far out, channel ± differs from the free wave only by e^(2iδₗ⁺) or e^(2iδₗ⁻) on its outgoing part, so the scattering operator is (1/k) Σ (2l + 1)[fₗ⁺P₊ + fₗ⁻P₋]Pₗ(cos θ), with fₗ = e^(iδₗ) sin δₗ. Substituting the projectors gives (1/k) Σ [(l + 1)fₗ⁺ + l fₗ⁻ + (fₗ⁺ − fₗ⁻)σ⋅L/ħ]Pₗ(cos θ), derived.

σ⋅L/ħ still acts on the angular function. For a function of θ alone, L = −iħ r × ∇ reduces to −iħ φ̂ ∂/∂θ, because r̂ × θ̂ = φ̂. Since dPₗ(cos θ)/dθ = −sin θ Pₗ′(cos θ) = −Pₗ¹(cos θ), this gives L Pₗ = iħ φ̂ Pₗ¹, so σ⋅L Pₗ/ħ = i(σ⋅φ̂)Pₗ¹. At the detector direction k′, φ̂ is exactly n = k × k′/|k × k′|, the normal to the scattering plane. Hence the amplitude is g(θ) + i h(θ)σ⋅n with g = (1/k) Σ [(l + 1)fₗ⁺ + l fₗ⁻]Pₗ and h = (1/k) Σ (fₗ⁺ − fₗ⁻)Pₗ¹, both derived: the form the previous topic built from rotation and parity, with its two functions now written in phase shifts. The sign of h rests on two conventions. With the Condon–Shortley Pₗ¹, which includes a factor −1, the same physics reads h = −(1/k) Σ (fₗ⁺ − fₗ⁻)Pₗ¹, and defining n as k′ × k flips it again. Mixing a formula from one source with a convention from another reverses every predicted polarisation while leaving |h|² and the cross-section untouched.

Set δₗ⁺ = δₗ⁻: h vanishes term by term and g becomes (1/k) Σ (2l + 1)fₗPₗ, QM-II's spinless amplitude. The s wave has only j = ½, so its weight in g is l + 1 = 1 with no l fₗ⁻ partner, and P₀¹ = 0: h needs p waves at least. Near threshold δₗ grows as k^(2l+1) (quoted from QM-II), so h ∝ k² while g → −a, and |h|²/|g|² falls as k⁴: spin effects on a spinless target die at low energy. Dimensions: fₗ is a number and 1/k a length, so g and h are lengths and |g|² + |h|², the unpolarised cross-section from the previous topic, is an area per steradian. Integrate it using ∫Pₗ² dΩ = 4π/(2l + 1) and ∫(Pₗ¹)² dΩ = 4πl(l + 1)/(2l + 1): the interference terms between fₗ⁺ and fₗ⁻ cancel exactly between |g|² and |h|², leaving σ = (4π/k²) Σ [(l + 1) sin²δₗ⁺ + l sin²δₗ⁻], derived. Because Pₗ(1) = 1, Pₗ¹(1) = 0 and Im fₗ = sin²δₗ, this equals (4π/k) Im g(0): the optical theorem survives the spin, and each channel is capped at (4π/k²)(j + ½).

⁴He has spin 0 and even parity, and below the ⁴He(n, d)³H threshold at 22 MeV n-⁴He scattering is purely elastic, so it is the clean case. Quoted data: the total cross-section peaks at 7.6 b at Eₙ = 1.15 MeV, and an R-matrix analysis places a p₃/₂ resonance, the unbound ⁵He ground state, at E_cm ≈ 0.77 MeV with width 0.64 MeV, and a p₁/₂ resonance near 2 MeV with width about 5 MeV. At 1.15 MeV, E_cm = 0.918 MeV; with μc² = 750.4 MeV this gives k = 0.188 fm⁻¹ and 4π/k² = 3.55 b. The s wave is repulsive, because the Pauli principle bars the neutron from the filled 0s shell, and acts like a hard sphere of radius near the scattering length a = 2.46 fm (quoted): δ₀ ≈ −ka ≈ −27°. If both p channels resonated together, as a spinless p wave would, σ would reach 3.55 b × (3 + 0.21) = 11.4 b. With only p₃/₂ at 90° and δ(p₁/₂) ≈ 8° (quoted), σ = 3.55 b × (2 + 0.21 + 0.02) = 7.9 b, 4% above the measurement. The weight j + ½ = 2 is what the experiment sees.

With s and p waves only, g = [f₀ + (2f₁⁺ + f₁⁻)cos θ]/k and h = (f₁⁺ − f₁⁻)sin θ/k. At 90° the p-wave part of g vanishes and h is at its largest, so k² dσ/dΩ(90°) = sin²δ₀ + sin²(δ₁⁺ − δ₁⁻), derived. Without a spin-orbit split the second term is zero and the unpolarised angular distribution dips deeply at 90°; with it, |h|² fills the dip. For n-⁴He at 1.15 MeV the bracket is 0.21 + 0.98 = 1.19 in units of 1/k² = 0.282 b, so dσ/dΩ(90°) ≈ 0.34 b sr⁻¹, nearly six times the s-wave floor of 0.058 b sr⁻¹. Wheeler and Barschall inferred a nuclear spin-orbit coupling from unpolarised neutron-helium angular distributions at 2.5 MeV in the 1940s, before the shell model needed one. The ordering, p₃/₂ below p₁/₂, says Vₛₒ is negative, attractive in j = l + ½, the reverse of atomic fine structure, and helium's large h is why it became the standard analyser of neutron polarisation, the subject of the next topic.