The exact Coulomb amplitude in parabolic coordinates

Units are SI, with α = e²/(4πε₀ħc) ≈ 1/137.036; nothing here is relativistic, so no metric signature enters. Quantum Mechanics II's partial waves assume V falls faster than 1/r, so that far out each wave is free and carries a fixed phase shift. Coulomb's tail breaks that. Energy conservation gives the local wavenumber k(r) = √(k² − 2ηk/r) ≈ k − η/r, with η = Z₁Z₂e²/(4πε₀ħv), so the accumulated phase ∫k(r) dr = kr − η ln r + constant drifts at every distance (derived). No Coulomb wave ever becomes free, no phase shift can be measured against a free one, and the problem has to be solved exactly. η also says how classical the collision is: η/k = Z₁Z₂e²/(4πε₀μv²) is half the head-on turning distance, so η counts reduced wavelengths across it. A 5.0 MeV α on gold has η = 22.3 and η/k = 23.2 fm, a classical orbit; a 3.0 MeV proton on a proton at rest has η = 0.091, nearly Born.

Write the Schrödinger equation as (∇² + k² − 2ηk/r)ψ = 0 with k = μv/ħ. The beam picks out z, and the scattered wave depends on r − z = 2r sin²(θ/2), so try ψ = e^(ikz) w(u) with u = r − z. Three facts do the work: |∇u|² = 2u/r, ∇²u = 2/r and ∂u/∂z = −u/r. Then ∇²ψ = e^(ikz)[(2u/r)w″ + (2/r)w′ − (2iku/r)w′ − k²w], the k² cancels, and multiplying by r/2 leaves u w″ + (1 − iku)w′ − ηk w = 0. With x = iku this is Kummer's equation x w″ + (b − x)w′ − a w = 0 with a = −iη, b = 1. One solution is M(−iη, 1, x) = Σ (−iη)ₙ xⁿ/(n!)², finite at x = 0; the other carries ln x. Since u = 0 along the whole forward half-axis, not just at the origin, ln x would make ψ singular on a line, so only M survives. All of this is derived; that the equation separates at all reflects the Coulomb problem's extra symmetry, the one that conserves the Runge–Lenz vector.

Set ψ = N e^(ikz) M(−iη, 1, ik(r − z)). The large-|x| form of M is quoted from hydrogen's radial problem: M(a, b, x) ≈ Γ(b)(−x)^(−a)/Γ(b − a) + Γ(b)eˣ x^{(a − b)}/Γ(a). With x = iku and principal branches, the first term is e^(πη/2) e^{(iη ln ku)}/Γ(1 + iη) and the second is −i e^(πη/2) e^(iku) e^{(−iη ln ku)}/[ku Γ(−iη)]. Choosing N = Γ(1 + iη) e^(−πη/2) gives the incident wave unit amplitude: exp[i(kz + η ln k(r − z))]. Because Γ(1 − iη) = −iη Γ(−iη), the ratio −iΓ(1 + iη)/Γ(−iη) collapses to −η Γ(1 + iη)/Γ(1 − iη) = −η e^(2iσ₀). With e^(ikz) e^(iku) = e^(ikr) and k(r − z) = 2kr sin²(θ/2), the second term becomes f_C(θ) exp[i(kr − η ln 2kr)]/r, and f_C is the card's formula: derived here, from asymptotics that are quoted.

|f_C|² = (η/2k)²/sin⁴(θ/2) = (Z₁Z₂e²/(16πε₀E))²/sin⁴(θ/2). ħ has cancelled, so the exact quantum answer equals the classical orbit calculation and Quantum Mechanics II's first Born result at every energy. Dimensions check: η/k is a length, so dσ/dΩ is in m² sr⁻¹. The coincidence belongs to 1/r alone; a screened potential's Born cross-section depends on ħ through k times the screening length. The amplitude is another matter: f_C = f_B e^(iΦ), with f_B = −η/(2k sin²(θ/2)) (quoted from Quantum Mechanics II) and Φ = 2σ₀ − η ln sin²(θ/2). As η → 0, σ₀ ≈ −γη with γ = 0.5772 (quoted), so Φ → 0 and Born returns: the weak-coupling limit. For 5.0 MeV α on gold, |f_C(90°)| = 23.2 fm and dσ/dΩ = 5.39 b sr⁻¹. Geiger and Marsden's 1913 counts on gold followed 1/sin⁴(θ/2) from 15° to 150°, over which that factor falls 3000-fold: a measured check of the modulus, and blind to the phase.

Far away, ψ = exp[i(kz + η ln k(r − z))] + f_C exp[i(kr − η ln 2kr)]/r, and neither piece is ever free. The log phases leave the counts alone: their gradient is η/r, so the radial current (ħ/μ)|f_C|²(k − η/r)/r² differs from the free value by η/(kr), 2 × 10⁻¹³ for the 5.0 MeV α on gold at 10 cm. They do bound the result. The split needs k(r − z) ≫ 1 + η², which fails in a narrowing cone around the forward direction at any r; there f_C diverges, the total cross-section is infinite, and pure Coulomb scattering has no optical theorem. And the phase becomes physics once f_C is added to another amplitude. For identical nuclei f_C(θ) meets f_C(π − θ), and the two phases differ by η ln tan²(θ/2): Mott's oscillations, the next lesson. With a nuclear force the short-range wave arrives as (1/k) e^(2iσ₀) e^(iδ₀) sin δ₀ (quoted); σ₀ cancels in the interference but −η ln sin²(θ/2) does not, which the figure shows.

Atomic electrons screen the nucleus beyond about the Thomas–Fermi radius a_TF ≈ 0.885a₀Z^(−1/3) (quoted), 1.1 × 10⁴ fm for gold. Outside it the potential falls faster than 1/r, the log phase stops growing and the forward amplitude turns finite; classically only impact parameters (η/k)cot(θ/2) beyond a_TF are touched, angles below 0.24° for 5.0 MeV α on gold. Nuclear forces need contact. The classical closest approach at angle θ is (η/k)[1 + 1/sin(θ/2)], 46 fm at 180° for that α against about 9 fm for touching nuclei. The exact solution says how rarely the charges meet: N gives |ψ(0)|² = |Γ(1 + iη)|² e^(−πη) = 2πη/(e^(2πη) − 1), using |Γ(1 + iη)|² = πη/sinh(πη) (quoted). That is 2.5 × 10⁻⁵⁹ for η = 22.26, the unrounded value for this α. At large η it falls as 2πη e^(−2πη), whose exponential is the Gamow factor divided out of measured fusion cross-sections to define the astrophysical S-factor; as η → 0 it tends to 1. Spin and relativity, absent here, bring the Mott factor in Unit 5.