Identical spin-half fermions: ¼|f + f′|² + ¾|f − f′|²

Work in SI units in the centre-of-mass frame; nothing here is relativistic, so the course metric diag(1, −1, −1, −1) never enters. Two identical spin-half fermions interact through a central V(r) that ignores spin, so H commutes with S₁ and S₂ separately, the total spin S and its projection are conserved, and the singlet and the triplet scatter as separate problems. Exchange sends r = r₁ − r₂ to −r, that is (θ, φ) to (π − θ, φ + π), and multiplies the spin state by (−1)^(S+1): −1 for the singlet, +1 for the triplet. The whole state must change sign, so the spatial wave carries (−1)^S: incident e^(ikz) + (−1)^S e^(−ikz), scattered [f(θ) + (−1)^S f(π − θ)] e^(ikr)/r. Call these F₀ = f + f′ for the singlet and F₁ = f − f′ for the triplet, with f′ = f(π − θ). Both are lengths, so each |F_S|² is an area per steradian, counting either particle as in topic 2.05.

Random beam and target spins give the pair the density matrix ρ = I/4, with I the 4 × 4 identity. It looks the same in every basis, so the singlet and each triplet state are occupied with probability 1/4. The detector does not resolve spin, so the cross-section is a trace over final spins, and singlet and triplet waves, whose spin states are orthogonal, add with no cross term: dσ/dΩ = (1/4)|f + f′|² + (3/4)|f − f′|². The trace gives it in one line. Write the amplitude as a spin operator M = f − f′P_σ, with P_σ = (1 + σ₁⋅σ₂)/2 the spin-exchange operator, +1 on the triplet and −1 on the singlet; Tr P_σ = 2, so (1/4)Tr(M†M) = |f|² + |f′|² − Re(f f′*). For f = (3.0 + 1.0i) fm and f′ = (1.0 + 2.0i) fm the channels give 25 and 5 fm² sr⁻¹ and the weighted sum is 10 fm² sr⁻¹, against 15 for distinguishable particles.

For spin s a random pair sits in one of (2s + 1)² product states |m₁, m₂⟩. The force leaves each particle's spin alone, so a pair with m₁ ≠ m₂ is labelled: the particle reaching the detector with m₁ came by f(θ), the one with m₂ by f(π − θ), the two final spin states are orthogonal, and the rate is |f|² + |f′|² with no interference. A pair with m₁ = m₂ has a symmetric spin state and interferes fully, |f ± f′|², plus for bosons and minus for fermions. That happens with probability (2s + 1)/(2s + 1)² = 1/(2s + 1), so dσ/dΩ = |f|² + |f′|² + (−1)^(2s) 2 Re(f f′*)/(2s + 1). The trace agrees, because the spin-exchange operator has trace (s + 1)(2s + 1) − s(2s + 1) = 2s + 1. Spin zero gives topic 2.05's +2 Re(f f′*), spin half −Re(f f′*), deuterons +(2/3) Re(f f′*). With many spin states shared ones are rare, and the classical sum returns.

At θ = 90° the angle π − θ is θ itself, so f′ = f exactly, for any spin-independent central force at any energy. The triplet amplitude vanishes and the singlet's is 2f, so the unpolarised rate is (1/4)|2f|² = |f(90°)|²: half the distinguishable sum 2|f(90°)|² and a quarter of the spinless bosons' 4|f(90°)|². Polarising beam and target changes the singlet probability. With ρ = (1 + P₁⋅σ₁)(1 + P₂⋅σ₂)/4 and the singlet projector (1 − σ₁⋅σ₂)/4, the trace, using Tr σᵢσⱼ = 2δᵢⱼ for each particle, is (1 − P₁⋅P₂)/4, so dσ/dΩ(90°) = (1 − P₁⋅P₂)|f(90°)|². Fully parallel spins are pure triplet and nothing reaches the 90° counter. Antiparallel spins double the unpolarised 90° rate, and the interference term −(1 + P₁⋅P₂)Re(f f′*) cancels at every angle, because up-down pairs are labelled by spin. A beam polarised to 0.80 on a parallel target at 0.70 cuts the 90° rate to 0.44 of its unpolarised value.

Since Pₗ(−cos θ) = (−1)ˡ Pₗ(cos θ), f(π − θ) is f with every odd partial wave reversed. So f + f′ is twice the sum of the even waves and f − f′ twice the odd: l + S must be even, which allows ¹S₀, ³P_J and ¹D₂ for two like nucleons and forbids ³S₁. Every event puts one particle in each hemisphere, so integrating dσ/dΩ over 2π sr gives σ = (2π/k²) Σ (2l + 1) sin²δₗ over even l plus (6π/k²) Σ (2l + 1) sin²δₗ over odd l: even waves at half their distinguishable weight 4π/k², odd ones at 3/2 of it. At threshold sin δ₀/k → −a while σₗ falls as k^(4l), so σ → 2πa²; for two neutrons with aₙₙ ≈ −18.9 fm, quoted from π⁻d → nnγ, that is 22.4 b. Fermions in one spin state have no s-wave at all: in 1999 the JILA group measured collisions of spin-polarised ⁴⁰K atoms dying away on cooling as the p-wave threshold law requires, and cooled the first degenerate Fermi gas as a two-state mixture.

A force V_c(r) + V_σ(r)σ₁⋅σ₂ still conserves S, so topic 2.04's singlet and triplet amplitudes replace f: dσ/dΩ = (1/4)|fₛ + fₛ′|² + (3/4)|fₜ − fₜ′|², derived exactly as before. Only the even waves of fₛ and the odd waves of fₜ survive, so the triplet s-wave that binds the deuteron, aₜ = 5.4 fm in neutron-proton scattering, has no place in proton-proton scattering. For protons at a few MeV the amplitude is the Coulomb one, present in both channels and built in topics 2.07 and 2.08, plus a nuclear term that is s-wave and therefore singlet only. Phase-shift analyses of the measured data, quoted, put the ¹S₀ phase near 55° at 10 MeV laboratory energy while every ³P phase stays within a few degrees. An unpolarised experiment sees the nuclear force with weight 1/4; one with spins parallel sees none of it. The model fails where tensor and spin-orbit forces, which split the ³P phases by J as no σ₁⋅σ₂ term can, make the triplet amplitude a matrix in M_S.