The spin-dependent amplitude g(θ) + i h(θ) σ⋅n

Quantum Mechanics II's asymptotic form carries over with one change: the projectile brings a spinor χ, and far from the target ψ → e^(ikz)χ + Mχ e^(ikr)/r, with M(k′, k) a 2 × 2 matrix acting on spin. The target has spin zero and stays in its ground state, so it is structureless and adds no labels. For an incident spinor χ the scattered flux gives dσ/dΩ = |Mχ|² = χ†M†Mχ, and for an unpolarised beam, whose density matrix is half the identity, dσ/dΩ = ½ Tr(M†M). Any 2 × 2 matrix is a combination of the identity and the three Pauli matrices, so M = a + b⋅σ: four complex functions of the outgoing direction at fixed energy, where the spinless problem had one. How many are genuinely free is settled by symmetry before any force is written down. Units are SI, with amplitudes in metres, usually quoted in fm, and 1 fm² sr⁻¹ = 10 mb sr⁻¹. Everything here is non-relativistic, so no metric signature enters.

If H commutes with J = L + S, rotating the whole experiment changes nothing: U(R) M(k′, k) U(R)† = M(Rk′, Rk), with U(R) Quantum Mechanics II's spin-half rotation. Because U(R) σ U(R)† turns σ like a vector, the parts of M must obey a(Rk′, Rk) = a(k′, k) and b(Rk′, Rk) = R b(k′, k). So a is a rotation scalar, and the only scalars on hand are k², k′² and k⋅k′; elastic scattering fixes |k′| = |k|, so a depends on θ alone. The vector b has to be assembled from the vectors on hand. Away from θ = 0 and π the three vectors k, k′ and k × k′ are independent, so b = β₁k + β₂k′ + β₃ k × k′, each β a scalar and so a function of θ. Rotations have turned four functions of direction into four functions of one angle, but removed none: σ⋅k and σ⋅k′ are perfectly good rotation scalars.

Parity sends k → −k and k′ → −k′ but leaves σ alone: spin, like L = r × p, is an axial vector, as Unit 1 sorted it. If H commutes with P, then M(−k′, −k) = M(k′, k). The scalar a survives, k⋅k′ being even, but β₁k and β₂k′ change sign while (−k) × (−k′) = k × k′ does not, so β₁ = β₂ = 0. Renaming a as g(θ) and writing β₃ k² sinθ ≡ i h(θ) gives M = g(θ) + i h(θ) σ⋅n with n = k × k′/|k × k′|, two functions of angle, derived from rotation and parity alone. Time reversal removes nothing more: with Unit 1's Θ = −iσ_y K, reciprocity requires M(k′, k) = σ_y M(−k, −k′)ᵀ σ_y, and since n(−k, −k′) = −n while σ_y σᵢᵀ σ_y = −σᵢ, the form obeys it already. The i is a convention chosen so that h is real in first Born approximation. For k along z and k′ at polar angle θ and azimuth φ, n = (−sinφ, cosφ, 0), so scattering towards +x has n along +y.

Quantise the spin along k. Then σ⋅n = −sinφ σₓ + cosφ σ_y, and M has g in both diagonal places, h e^(−iφ) above and −h e^(iφ) below: g keeps the spin component along the beam and h reverses it, so a neutron arriving spin-up leaves as g|↑⟩ − h e^(iφ)|↓⟩. The unpolarised cross-section needs only Tr σᵢ = 0 and Tr σᵢσⱼ = 2δᵢⱼ. Expanding, M†M = |g|² + |h|² − 2 Im(g*h) σ⋅n, the traceless last term drops out of ½ Tr(M†M), and dσ/dΩ = |g|² + |h|², derived: flip and non-flip end in orthogonal spin states and add as probabilities. The spin's direction responds to h at first order, the rate only at second. When g and h are real, as in first Born approximation, M = (g/cos(β/2)) exp(i(β/2)σ⋅n) with tan(β/2) = h/g, a pure rotation of the spin about n through −β that polarises nothing. Polarisation needs a phase between g and h, and lives in the discarded term: that is topic 2.3.

At θ = 0 the outgoing momentum equals the incident one, and at θ = π it is reversed. Either way only one vector, k, is on hand, so M can hold only the identity and σ⋅k, and parity forbids σ⋅k: at both ends M is a multiple of the identity. Yet n is undefined there, pointing a different way for each azimuth from which θ = 0 is approached, so M has a limit only if h(0) = h(π) = 0, derived. For a short-range force M is smooth in k′ and h falls as sinθ at both ends. Two consequences follow. The forward amplitude is spin-blind, so Quantum Mechanics II's optical theorem, applied to each incident spinor, gives σₜₒₜ = (4π/k) Im g(0) whatever the beam's polarisation. And a beam's helicity can change σₜₒₜ on a spin-zero target only through a parity-violating σ⋅k term. The weak interaction supplies one, and in proton scattering on helium-4 the helicity dependence is measured at a few parts in 10⁷. A Coulomb-range tail denies the forward amplitude, as the hard example shows.

If V depends on r alone and ignores spin, H commutes with σ, every order of perturbation theory returns a multiple of the identity, and h = 0 exactly. Add a spin-orbit term, H = p²/(2 m) + V(r) + Vₛₒ(r) L⋅S/ħ², with Vₛₒ an energy and m the reduced mass, and take Quantum Mechanics II's first Born amplitude −(m/2πħ²)⟨k′|V|k⟩, quoted. The new step is L e^(ik⋅r) = ħ(r × k) e^(ik⋅r). With q = k′ − k, ∫ e^(−iq⋅r) Vₛₒ(r) r d³r = i q̂ dṼₛₒ/dq and q × k = −k × k′, so h_B = (m k² sinθ/(4πħ²q)) dṼₛₒ/dq, real, derived; its units are kg m⁻² × J m⁴/(J² s² m⁻¹) = m. For V = −U e^(−r²/b²) and Vₛₒ = −W e^(−r²/b²) this gives h/g = (W/U)(k²b²/4) sinθ: zero at both ends, zero as k → 0, where only s waves scatter, and growing as k². In nuclei W/U is of order 0.1: the same coupling splits ¹⁷O's 0d₅/₂ and 0d₃/₂ neutron levels by a measured 5.08 MeV, which is (5/2)|⟨Vₛₒ⟩|, so |⟨Vₛₒ⟩| ≈ 2.0 MeV against a central well some tens of MeV deep.