Identical bosons: the symmetrised amplitude f(θ) + f(π − θ)

Work in SI units in the centre-of-mass frame; the collision is non-relativistic, so no metric signature enters. Two particles at r₁ and r₂ have centre of mass (r₁ + r₂)/2, which a swap leaves alone, and relative position r = r₁ − r₂, which it reverses. Exchange is therefore r → −r: the same distance, with θ → π − θ and φ → φ + π. Far away, the distinguishable scattering state is ψ(r) ≈ e^(ikz) + f(θ) e^(ikr)/r. For spin-zero bosons the symmetrisation postulate, quoted from Quantum Mechanics II, admits only ψ(r) + ψ(−r) ≈ e^(ikz) + e^(−ikz) + [f(θ) + f(π − θ)] e^(ikr)/r, derived. The incoming part holds no surprise: e^(ikz) has particle 1 moving up the axis and particle 2 down it, e^(−ikz) the reverse, and for identical particles these are one situation. The outgoing part is the result: one spherical wave whose amplitude adds the path in which the projectile goes to θ to the path in which it goes to π − θ and the target arrives at θ.

Quote Quantum Mechanics II's expansion f(θ) = (1/k) Σ (2l + 1) e^(iδₗ) sin δₗ Pₗ(cos θ). The reflected angle needs only cos(π − θ) = −cos θ and the parity of the Legendre polynomials, Pₗ(−x) = (−1)ˡ Pₗ(x). Each term of f(π − θ) is the matching term of f(θ) times (−1)ˡ, so in the sum the s, d, g, … waves appear twice and the p, f, h, … waves cancel term by term, derived: f(θ) + f(π − θ) = (2/k) Σ over even l of (2l + 1) e^(iδₗ) sin δₗ Pₗ(cos θ). The phase shifts themselves are untouched, since δₗ is fixed by the potential in the relative coordinate whether or not the particles can be told apart. What changes is which waves the pair may occupy. A relative-motion state of angular momentum l has exchange eigenvalue (−1)ˡ, equal to its parity, so a spinless boson pair lives only in l = 0, 2, 4, …, with total J = l and parity +. A p-wave resonance in the potential, however sharp, cannot be seen.

A counter at angle θ fires for either particle and cannot say which, so it records dσ/dΩ = |f(θ) + f(π − θ)|²; the 1/√2 that normalises the symmetric state halves the incident flux and the outgoing current alike, so it cancels, derived. The fair comparison is not |f(θ)|² but the classical sum |f(θ)|² + |f(π − θ)|², the projectiles plus the recoils that would reach the same counter if they could be told apart. The two differ by the interference term 2 Re[f(θ) f*(π − θ)]. Writing A and B for the even and odd parts [f(θ) ± f(π − θ)]/2, the boson rate is 4|A|² and the classical sum 2|A|² + 2|B|², so their ratio 2|A|²/(|A|² + |B|²) lies between 0 and 2. At 90° every odd Legendre polynomial vanishes, B = 0, and the boson rate is exactly twice the classical sum, at any energy and for any short-range central force. Away from 90° it can be lower: a strong p wave lifts the classical sum forward and backward and leaves the boson rate where it was.

The two particles leave back to back in the centre-of-mass frame, so each collision puts one into each hemisphere. Integrating |f(θ) + f(π − θ)|² over all 4π sr counts every collision twice; the cross-section, collisions per unit incident flux, is the integral over one hemisphere. With the even-wave expansion and the orthogonality of the Pₗ it is σ = (8π/k²) Σ over even l of (2l + 1) sin² δₗ, derived, against (4π/k²) Σ over all l for distinguishable particles. Three checks hold. Dimensions: 1/k² is an area. Unitarity: f_B(0) = (2/k) Σ over even l of (2l + 1) e^(iδₗ) sin δₗ, so (4π/k) Im f_B(0) equals this σ, and the optical theorem keeps its form with the hemisphere count and with no other. The classical limit: when f is sharply forward-peaked, as at high energy, f(π − θ) is negligible wherever f(θ) is not, the interference term dies, and the boson rate becomes the classical sum, whose hemisphere integral is exactly the distinguishable σ.

When kR ≪ 1 for a force of range R, only the s wave scatters and f → −a, with a the scattering length, quoted from Quantum Mechanics II. Then f(θ) + f(π − θ) → −2a, dσ/dΩ → 4a² at every angle, and the hemisphere gives σ = 2π × 4a² = 8πa², derived: twice the distinguishable 4πa², not four times. ⁸⁷Rb atoms in the |F = 1, m_F = −1⟩ state share one internal state and so scatter as spin-zero bosons; their a = 100.4a₀ = 5.31 nm is quoted from cold-collision spectroscopy, so σ = 7.09 × 10⁻¹⁶ m², the value evaporative-cooling models of Bose–Einstein condensates use. There is a subtler gain. Quantum Mechanics II's threshold law, tan δₗ ∝ (kR)^(2l+1), quoted, makes the first angular structure in distinguishable scattering the s–p interference, of relative size (kR)². Identical bosons have no p wave, so their first structure is s–d interference, of relative size (kR)⁴. Halo images of colliding ⁸⁷Rb clouds (2004, quoted) show s and d waves interfering and no p wave.

The α particle is a spin-zero boson, so α–α scattering is the cleanest nuclear case, and the selection rule of the second section becomes a statement about a nucleus. A resonance in α + α is a level of ⁸Be through which the pair passes, and the pair exists only with l = 0, 2, 4, …, J = l and parity +. So only 0⁺, 2⁺, 4⁺, … levels can appear as α–α resonances, derived. A 1⁻ or 3⁻ level, open to distinguishable spinless particles, is excluded by symmetrisation; a 1⁺ is excluded even before it, because a spinless pair has parity (−1)ᴶ. The measured spectrum, quoted, fits: the 0⁺ ground state lies 91.84 keV above the α + α threshold with width 5.57 eV, the 2⁺ sits at 3.03 MeV excitation with width 1.51 MeV, and the 4⁺ at 11.35 MeV with width about 3.5 MeV. The 1⁺ level at 17.64 MeV, reached in p + ⁷Li, never shows as an α–α resonance. With helium at rest, E_lab = 2E_cm, so the 2⁺ appears under a 6.24 MeV beam. The α–α Coulomb amplitude must be symmetrised too, Topic 02.08's task; being central, it changes no selection rule.