Time reversal in scattering: reciprocity, detailed balance

Units are SI, with reaction energies in MeV and momenta in MeV/c; the one invariant used, s = (E₁ + E₂)² − |p₁ + p₂|²c², the squared centre-of-mass energy, carries the signature (+, −, −, −). Quote the Lippmann-Schwinger equation from Quantum Mechanics II: |ψ⁺(k)⟩ = |k⟩ + (E − H₀ + iε)⁻¹V|ψ⁺(k)⟩, a plane wave plus outgoing spherical waves. Assume ΘH₀Θ⁻¹ = H₀ and ΘVΘ⁻¹ = V, and apply Θ to the whole equation. It sends |k⟩ to |−k⟩, commutes with E − H₀, and, being antilinear, sends +iε to −iε, so Θ(E − H₀ + iε)⁻¹Θ⁻¹ = (E − H₀ − iε)⁻¹. Out comes the same equation with −iε and −k: Θ|ψ⁺(k)⟩ = |ψ⁻(−k)⟩, derived. The reversed film, with spherical waves converging on the target and a plane wave leaving along −k, is not an imagined experiment but a stationary state of the same Hamiltonian at the same energy. With spin, Θ also flips each projection: |k, m⟩ becomes (−1)^(j−m)|−k, −m⟩.

For δ-normalised plane waves the amplitude is f(k → k′) = −(4π²μ/ħ²)⟨k′|V|ψ⁺(k)⟩. Write the scattering states with the full resolvent, |ψ⁺(k)⟩ = |k⟩ + (E − H + iε)⁻¹V|k⟩ and its −iε partner (quoted from Quantum Mechanics II), and the same number is ⟨ψ⁻(k′)|V|k⟩: both equal ⟨k′|V + V(E − H + iε)⁻¹V|k⟩. Now apply topic 1.02's rule for an antiunitary operator, ⟨β|A|α⟩ = ⟨Θα|ΘA†Θ⁻¹|Θβ⟩, with A = V Hermitian and T-even: ⟨k′|V|ψ⁺(k)⟩ = ⟨ψ⁻(−k)|V|−k′⟩ = ⟨−k|V|ψ⁺(−k′)⟩. Hence f(k → k′) = f(−k′ → −k), derived; with spin the phases (−1)^(j−m) combine into (−1)^{(m − m′)}, invisible in |f|². Read it exactly: initial and final swap, and every momentum and spin reverses. For a spinless particle on a central potential f depends only on k⋅k′, so the statement is automatic; its content lives in spin labels and in targets fixed in space. Dimensions: ⟨k′|V|ψ⁺⟩ is in J m³ and μ/ħ² in J⁻¹ m⁻², so f is a length, as it must be.

An unpolarised experiment averages over m and sums over m′. Summing reciprocity over both labels and renaming −m as m gives Σ|f(k, m → k′, m′)|² = Σ|f(−k′, m′ → −k, m)|². Now add rotation invariance and rotate the reversed process by π about n = k × k′. Every vector in the scattering plane changes sign, so −k′ becomes k′ and −k becomes k, while a sum over complete sets of spin states does not care which axis they were quantised along. Result, derived: dσ/dΩ(k → k′) = dσ/dΩ(k′ → k) for unpolarised beams and targets. Keep the two assumptions apart. Θ produced the reversed process; rotation invariance converted it into the inverse. Scatter neutrons from a single crystal held in one orientation, or electrons from molecules frozen in a lattice, and only the reversed statement survives. Limiting case: at θ = 0 the two sides are the same process and the relation holds trivially, which is a consistency check rather than a prediction.

For a + b → c + d the golden rule (quoted from Quantum Mechanics II) gives a rate (2π/ħ)|M|²ρ_f, and dividing by the incident flux gives σ. In the centre-of-mass frame the final density of states per unit energy goes as p_f² dp_f/dE = p_f²/v_f, with v_f the relative speed, and the flux goes as vᵢ, so σ(i → f) ∝ |M|² p_f²/(vᵢ v_f), averaged over gᵢ initial and summed over g_f final spin states. The reverse reaction at the same √s shares the spin-summed |M|² and the product vᵢ v_f, so the ratio holds no dynamics at all: gᵢ pᵢ² σ(i → f) = g_f p_f² σ(f → i), derived, and relativistic flux and phase space leave the cancellation intact. Count g = (2s₁ + 1)(2s₂ + 1), except g = 2 for a photon, whose two helicities replace three. Identical particles in a final state are double-counted over 4π, so that σ carries a 1/2. Limiting case: near threshold the endothermic σ vanishes as p_f and the exothermic one grows as 1/vᵢ, the law thermal capture obeys, measured at 0.332 b for n + p → d + γ at 2200 m s⁻¹.

Apply the counting to p + p ⇌ π⁺ + d. Two protons give g(pp) = 4; the deuteron has spin 1, so g(π⁺d) = 3(2s_π + 1); and the two protons leaving π⁺d → pp are identical, which halves that cross-section and puts a 2 into the ratio: σ(pp → π⁺d)/σ(π⁺d → pp) = (3/2)(2s_π + 1)(p_π/pₚ)², derived. Take a 400 MeV proton beam on hydrogen: √s = 2066.9 MeV, pₚ = 433.2 MeV/c and p_π = 124.7 MeV/c, so (p_π/pₚ)² = 0.0828 and the ratio is 0.124 for spin 0 against 0.373 for spin 1. The matching inverse run uses 56.3 MeV pions on deuterium at rest, not 400 MeV ones: the comparison is made at equal √s, never at equal beam energy. Measured in 1951, the pair agreed with s_π = 0, the first determination of the charged pion's spin (quoted). Checks: g and the momentum ratio are pure numbers, as a ratio of cross-sections must be, and at threshold p_π → 0, so production switches off while the exothermic absorption does not.

Detailed balance follows from T, but seeing it hold is weak evidence for T. To first order the amplitude is ⟨f|V|i⟩, and Hermiticity alone gives |⟨f|V|i⟩|² = |⟨i|V|f⟩|², derived: every Born or golden-rule rate balances whatever V does under Θ. Beyond first order the transition operator V + V(E − H + iε)⁻¹V is not Hermitian, the iε tells in from out, and a T-odd part of V can show through its interference with a T-even amplitude of different phase. Unitarity also gets in first: with two open channels, S S† = S†S = 1 gives |S₁₁|² + |S₁₂|² = 1 = |S₁₁|² + |S₂₁|², so |S₁₂| = |S₂₁| with no symmetry assumed, derived. A test needs three or more channels and amplitudes beyond Born; compound-nucleus comparisons such as ²⁷Al + p ⇌ ²⁴Mg + α meet both conditions and bound the ratio of T-odd to T-even amplitudes at a few parts in 10⁴ (quoted). And when an applied field B breaks T, the derivation still runs with Θ reversing B too: f(k → k′; B) = f(−k′ → −k; −B).