Singlet and triplet weights in neutron-proton scattering

Units are SI, with nuclear lengths in femtometres (1 fm = 10⁻¹⁵ m) and cross-sections in barns (1 b = 10⁻²⁸ m² = 100 fm²); everything is non-relativistic, so no metric signature enters. Take a central force that feels spin only through S₁⋅S₂: V(r) = V_c(r) + V_σ(r) S₁⋅S₂/ħ². Since S₁⋅S₂ = (S² − S₁² − S₂²)/2, V commutes with S² and with S itself, so scattering can neither change the total spin nor tell one triplet m_S from another. The scattering operator is one function fₛ(θ) on the singlet and one fₜ(θ) times the identity on the triplet. With S₁⋅S₂ = [S(S + 1) − 3/2]ħ²/2, equal to −3ħ²/4 for S = 0 and +ħ²/4 for S = 1, the combinations P₀ = 1/4 − S₁⋅S₂/ħ² and P₁ = 3/4 + S₁⋅S₂/ħ² are 1 on their own space and 0 on the other, so f = fₛP₀ + fₜP₁, derived. One-channel check: if fₛ = fₜ = f₀, the operator is f₀ times the identity, the spinless amplitude of Quantum Mechanics II.

An unpolarised beam on an unpolarised target fills the four product states equally, ρ = 1/4, and the measured cross-section sums over final spins: dσ/dΩ = Tr(ρ f†f). P₀ and P₁ are orthogonal projectors, so f†f = |fₛ|²P₀ + |fₜ|²P₁ with no cross term, and Tr P₀ = 1, Tr P₁ = 3 give (1/4)|fₛ|² + (3/4)|fₜ|², derived. Nothing dynamical sets 1/4 and 3/4: they are the singlet's one state and the triplet's three. Polarise both partners and the count changes. For a product state ⟨S₁⋅S₂⟩ = (ħ²/4)Pₙ⋅Pₚ, so the singlet weight becomes (1 − Pₙ⋅Pₚ)/4 and the triplet weight (3 + Pₙ⋅Pₚ)/4. Fully parallel spins are pure triplet, m_S = +1. Fully antiparallel spins give 1/2 and 1/2, not pure singlet, because |↑↓⟩ = (|1, 0⟩ + |0, 0⟩)/√2. With the zero-energy lengths of the next section, that is 3.66 b parallel against 37.1 b antiparallel, a factor of ten that polarised-target experiments exploit.

At zero energy each channel's amplitude tends to minus its scattering length, f → −a, quoted from Quantum Mechanics II, and is isotropic. The unpolarised result is σ₀ = 4π[(1/4)aₛ² + (3/4)aₜ²] = π(aₛ² + 3aₜ²). With the quoted aₛ = −23.7 fm and aₜ = 5.4 fm, the singlet supplies π × 561.7 fm² = 17.65 b and the triplet 3π × 29.16 fm² = 2.75 b: 20.4 b in all, the measured cross-section for neutrons between about 1 eV and 1 keV, fast enough that the proton recoils as if free and slow enough that k|aₛ| stays small. Checks: a length squared is an area, and a spin-blind force, aₛ = aₜ = a, returns the spinless 4πa². Neither length alone works. The deuteron fixes the triplet, and 4πaₜ² = 3.7 b falls short by a factor of 5.6; the singlet alone, 4πaₛ² = 70.6 b, overshoots by 3.5. The singlet is occupied a quarter of the time and still gives 87% of the scattering, because its length is 4.4 times larger and enters squared.

A large positive scattering length signals a shallow bound state at E ≈ −ħ²/(2μa²), and a large negative one a virtual level just above threshold; both are quoted from Quantum Mechanics II, with μ = m_N/2 for neutron and proton. The deuteron, bound by B = 2.224 MeV, sets κ = √(m_N B)/ħ = 0.232 fm⁻¹, and the zero-range estimate aₜ ≈ 1/κ = 4.32 fm has the right sign and nearly the right size; the effective range of 1.75 fm, quoted, supplies the rest of 5.4 fm. In the singlet, |aₛ| = 23.7 fm is about twelve times the 2 fm range of the force, the mark of a level almost at zero energy, and the negative sign puts it on the unbound side, near ħ²/(m_N aₛ²) ≈ 0.07 MeV. The singlet misses binding by less than 0.1 MeV. The trouble is that σ₀ contains only aₛ² and aₜ², so a bound singlet with aₛ = +23.7 fm would give exactly the same 20.4 b. Free-proton scattering cannot see the sign; another experiment has to.

With sₙ the neutron's spin and s₁, s₂ the protons', each proton's zero-energy amplitude is −[a_coh + (aₜ − aₛ) sₙ⋅sᵢ/ħ²], where a_coh = (aₛ + 3aₜ)/4. A neutron whose wavelength dwarfs the 74 pm proton separation, kd → 0, adds the two amplitudes before squaring, and sₙ⋅s₁ + sₙ⋅s₂ = sₙ⋅S depends only on the total proton spin S. In para-hydrogen S = 0, the spin term vanishes, and σ(para) = 4π(2a_coh)² = π(aₛ + 3aₜ)², derived. In ortho-hydrogen S = 1 and ⟨(sₙ⋅S)²⟩ = ħ⁴/2, adding 2π(aₜ − aₛ)², also derived. Opposite signs make a_coh small: 1.77 b for para against 55.0 b for ortho, where like signs would give 50.0 b and 71.1 b. Schwinger and Teller proposed the comparison in 1937; later measurements, quoted, found para-hydrogen far more transparent, so the signs are opposite. The molecule's mass multiplies both by (4/3)², which cancels in the ratio, and finite kd adds form factors, quoted; neither turns 31 into 1.4. Neutron refraction now measures the same combination: hydrogen's bound coherent length, −3.74 fm, is 2a_coh to within 0.3%.

Two assumptions end it. The first is zero energy. In zero-range form each channel is σ = 4πa²/(1 + k²a²), quoted from Quantum Mechanics II with the effective range dropped. The singlet halves when k|aₛ| = 1, at a laboratory energy of 0.15 MeV; the triplet only at 2.8 MeV. The singlet's 87% share therefore erodes first, and the plateau ends well before the deuteron's own energy scale. The second assumption is a purely central force. The deuteron's measured electric quadrupole moment, 0.286 fm², is the signature of a tensor force, S₁₂ = 3(σ₁⋅r̂)(σ₂⋅r̂) − σ₁⋅σ₂, which mixes the ³S₁ and ³D₁ waves; it is named here, not treated. It annihilates the singlet and conserves S, so the singlet-triplet split survives, but the triplet amplitude becomes a 3 × 3 matrix in m_S that depends on the spins' orientation relative to the momentum. At zero energy only the s-wave survives, so σ₀ = π(aₛ² + 3aₜ²) is untouched; the tensor force shows itself in polarisations and angular distributions at MeV energies.