Polarisation by scattering and the analysing power

Work in SI units in the centre-of-mass frame; nothing derived here is relativistic, so the course metric diag(1, −1, −1, −1) never enters. Amplitudes are lengths, conveniently in fm, with 1 fm² sr⁻¹ = 10 mb sr⁻¹. Quantum Mechanics II's beam density matrix ρ = (1 + P_b⋅σ)/2 scatters into M ρ M†, whose trace is dσ/dΩ and whose normalised trace with σ is the outgoing polarisation. With M = g + i h σ⋅n, n = k × k′/|k × k′| and (σ⋅n)² = 1, multiplying out gives M M† = |g|² + |h|² + 2 Im(g h*) σ⋅n. For an unpolarised beam ρ is half the unit matrix: the constant term gives dσ/dΩ = |g|² + |h|², and the σ⋅n term gives P = 2 Im(g h*)/(|g|² + |h|²), directed along n, derived. Checks: P is a ratio of areas, so dimensionless; |P| ≤ 1 because 2|g||h| ≤ |g|² + |h|²; and P = 0 at 0 and 180°, where h, proportional to sin θ, vanishes. Particles leaving to the left and to the right at the same θ have opposite n, so they carry opposite polarisations in space.

The i in i h σ⋅n is the whole story. In the eigenbasis of σ⋅n the two spin states scatter with g + ih and g − ih, and |g ± ih|² = |g|² + |h|² ± 2 Im(g h*). Topic 2.01 found the first Born g and h of V_c + Vₛₒ L⋅S both real, quoted from there, so g and ih are a quarter-cycle apart, add in quadrature, and both spin states scatter equally: P = 0 at every angle. In partial waves each channel brings e^(iδ) sin δ, and every term of Im(g h*) has the form sin δ₁ sin δ₂ sin(δ₁ − δ₂): two channels with different phases. A central force makes δₗ⁺ = δₗ⁻ and h = 0. With s and p waves only, s–p interference sets P(90°), while the p₃/₂–p₁/₂ term carries cos θ and is odd about 90°. Scale every phase by λ → 0 and Im(g h*) falls as λ³ while |g|² + |h|² falls as λ², so P is first order in the potential, yet a first-order amplitude cannot show it: it comes from the second-order imaginary parts of g and h. It is largest where the cross-section is smallest, near an angle where the first-order g passes through zero.

Give the beam a polarisation P_b. The cross-section is Tr(M ρ M†) = ½Tr(M M†) + ½Tr(M† M σ⋅P_b), using cyclicity on the second term. Here M depends on spin only through σ⋅n, so M† M = M M†, and dσ/dΩ = (|g|² + |h|²)(1 + A P_b⋅n) with A = 2 Im(g h*)/(|g|² + |h|²): the analysing power equals the polarisation, derived from the form of the amplitude. The spin-resolved picture says why in one line: both experiments compare the same two intensities, |g + ih|² and |g − ih|². Use the Madison convention, n pointing up for scattering to the left of the beam seen from behind, and polarise the beam up. To the left P_b⋅n = +P_b; to the right n reverses. Equal-angle counts then give ε = (N_L − N_R)/(N_L + N_R) = A P_b. Spin components in the scattering plane drop out of P_b⋅n, so a longitudinally polarised beam gives no asymmetry until its spin is turned transverse.

Cyclicity alone does not give the equality: it turns Tr(M σ⋅n M†) into Tr(M† M σ⋅n), but P needs Tr(M M† σ⋅n), and M M† ≠ M† M for a general amplitude. Unit 1's reciprocity closes the gap. With Θ = −iσ_y K it reads M(k′, k) = σ_y Mᵀ(−k, −k′) σ_y, and since σ_y σ σ_y = −σᵀ and a trace survives transposition, Tr(M M† σ⋅n) = −Tr(M̃ σ⋅n M̃†), where M̃ is the amplitude for −k′ → −k. A rotation by π about k − k′ carries −k′ to k and −k to k′, so rotation invariance makes M̃ = U† M U, and that rotation reverses n, which is perpendicular to its axis. The two minus signs cancel: Tr(M M† σ⋅n) = Tr(M σ⋅n M†), so P = A, derived. Nothing used the form of M. The same lines, with σ_y ⊗ σ_y and traces over four spin states, hold for nucleon-nucleon scattering, whose five amplitudes (Wolfenstein's, quoted) would make the form argument a long calculation. For a reaction a + b → c + d, M̃ belongs to c + d → a + b, so P of c equals the analysing power of the inverse reaction.

Scatter an unpolarised beam at θ to the left: it leaves with polarisation P(θ) along n. Send it at the same energy to an identical target and count left and right at the same θ in the same plane. By the equality just proved, ε₂ = A(θ)P(θ) = P(θ)², derived, and no calibrated analyser is needed, which is why double scattering is how nucleon polarisation was first measured. Two cautions. The energies must match, so energy lost in the first target or to recoil must be negligible or corrected, as it nearly is for elastic scattering from a heavy nucleus. And ε₂ ≥ 0 for either sign: a beam polarised down meets A = P, also negative, and the product is again positive, so the second scattering always prefers to turn the same way as the first. The sign of P needs other input, such as a phase-shift analysis that puts p₃/₂ below p₁/₂, as the nuclear spin-orbit force does. Measuring ε₂ = 0.160 ± 0.004 gives |P| = 0.400 ± 0.005, since d(P²) = 2P dP.

Mott predicted in 1929 that electrons scattered twice from heavy nuclei would show this asymmetry; Shull, Chase and Myers observed it in 1943. The spin-orbit force here is the relativistic one Unit 6 derives from the Dirac equation, Thomas factor included, and for gold Zα = 79/137 = 0.58 is not small. First Born gives P = 0, so the effect lives in higher orders and only exact Dirac partial waves deliver it: Sherman's 1956 numerical values, quoted here, give the single-atom analysing power S(θ, E), about 0.4 in magnitude near 120° and 100 keV, and double scattering from gold, which measures S² directly, is how such values are checked. A Mott polarimeter counts left and right, ε = S P_⊥, so it sees only the component normal to its scattering plane: the longitudinal polarisation of a strained-GaAs photocathode source, up to about 0.9, must be turned transverse first. Plural scattering in a real foil dilutes S to an effective S_eff, found by extrapolating to zero thickness, and statistics are costly: δP ≈ 1/(|S_eff| √N), so |S_eff| = 0.25 and δP = 0.01 need 1.6 × 10⁵ counts.