The Klein-Gordon radial equation in a Coulomb field

SI units throughout, with this unit's metric diag(1, −1, −1, −1). Minimal coupling, quoted from the gauge-invariance lesson, turns iħ∂ₜ into iħ∂ₜ − V, so a stationary state ψ = χ(r) e^(−iEt/ħ) obeys (E − V)²χ = (−ħ²c²∇² + m²c⁴)χ, with E the total energy, rest energy included. For charge −e and a point nucleus Ze, V = −Ze²/(4πε₀r) = −Zαħc/r. V only multiplies χ, and the Laplacian is Schrödinger's own, so χ = (u/r)Yₗₘ separates as in Quantum Mechanics I with the same integer l. Expanding (E − V)² = E² + 2EZαħc/r + (Zα)²ħ²c²/r² and dividing by (ħc)² gives, derived, u″ + [(E² − m²c⁴)/(ħc)² + 2EZα/(ħc r) − (l(l + 1) − (Zα)²)/r²]u = 0. Dimensions check: E²/(ħc)² is J² over J² m², so m⁻²; E/(ħc) and Zα/r are each m⁻¹; the last term is m⁻². Note that E now enters squared, so the eigenvalue problem is quadratic in E.

Set it beside Schrödinger's radial equation, u″ + [2mε/ħ² + 2mZαc/(ħr) − l(l + 1)/r²]u = 0 with ε = E − mc². First, the energy enters squared: (E² − m²c⁴)/(ħc)² = (2mε/ħ²)[1 + ε/(2mc²)], a change of −1.3 × 10⁻⁵ for hydrogen's ε = −13.6 eV. Second, the Coulomb term carries E/c² where Schrödinger's carries m: the particle couples to the field through its total energy, and a bound E lies below mc², so the attraction is weakened by B/mc² = 2.7 × 10⁻⁵ in hydrogen's 1s state. Third, l(l + 1) is lowered by (Zα)²: V² = (Zα)²ħ²c²/r² falls exactly as the centrifugal term does, and it sits on the energy side, so it attracts. For hydrogen (Zα)² = 5.3 × 10⁻⁵, the same order as the other two; at Z = 40 it is 0.085, already 4% of a p state's l(l + 1) = 2. The first two changes only rescale constants. The third changes a parameter that fixes the shape of the solution.

The three changes are one. With E = mc² + ε, (E − V)² − m²c⁴ = 2mc²(ε − V) + (ε − V)² exactly, so u″ + [2m(ε − V)/ħ² + (ε − V)²/(ħc)² − l(l + 1)/r²]u = 0. The first and last terms are Schrödinger's; everything relativistic is the middle term, whose ratio to the first is (ε − V)/(2mc²), the local kinetic energy over 2mc². Since ε − V is the classical p²/2m, the term is (p²/2m)²/(2mc²), the −p⁴/(8m³c²) of the non-relativistic-limit lesson. Drop it and Schrödinger's equation returns; with the reduced mass it binds hydrogen by 13.59829 eV against the measured 13.59843 eV, and the 0.15 meV left over is a tenth of α⁴mc² = 1.45 meV, the order at which this term acts. It rivals the first term only where |V| ≈ 2mc², at r ≈ Zαħ/(2mc) = 1.4 fm in hydrogen, close to the proton's measured 0.84 fm charge radius, where the point potential is itself no longer right.

Because V² has the centrifugal term's 1/r², define l′ by l′(l′ + 1) = l(l + 1) − (Zα)². The quadratic gives l′ = −1/2 ± √((l + 1/2)² − (Zα)²); take the + root, the one that returns l as Zα → 0. Expanding, l′ ≈ l − (Zα)²/(2l + 1), so the shift δ = l − l′ is largest for s states, which have no centrifugal term to share it with. For hydrogen's 1s, l′ = −5.33 × 10⁻⁵; for its 2p, l′ = 1 − 1.78 × 10⁻⁵. At Z = 40, l′ = −0.0940 for l = 0 and 0.9713 for l = 1. Two cautions. l′ belongs to the radial equation, not to the angular momentum: the angular equation never saw V, so the state is still Yₗₘ, L² is still l(l + 1)ħ², and the 2l + 1 values of m are still degenerate. And l′ is real only while Zα ≤ l + 1/2; beyond that the map fails, which is the subject of the lesson on fall to the centre.

A bound state has E below mc², so write (E² − m²c⁴)/(ħc)² = −κ² and set ρ = κr. The equation becomes d²u/dρ² = [1 − ρ₀/ρ + l′(l′ + 1)/ρ²]u with ρ₀ = 2EZα/(ħcκ): Quantum Mechanics II's hydrogen equation exactly, with l → l′ and its 2mcZα/(ħκ) → 2EZα/(ħcκ). Its solution is quoted, not redone: u = ρ^(l′+1) e^(−ρ) v(ρ), v = Σ cⱼρʲ, with cⱼ₊₁ = [2(j + l′ + 1) − ρ₀]cⱼ/[(j + 1)(j + 2l′ + 2)]. No step needs l to be an integer, only j + 2l′ + 2 ≠ 0, which holds for l′ above −1. Unless the series stops, v grows like e^(2ρ) and u blows up, so ρ₀ = 2(nᵣ + l′ + 1) for an integer nᵣ ≥ 0. That is one equation for E, solved in the next lesson. It already says ρ₀ must be positive, so a bound state has 0 < E < mc²: a negative-energy solution feels 2EZα/(ħcr) as a repulsion and cannot bind.

Near r = 0 the 1/r² term dominates and u ∝ rˢ with s(s − 1) = l′(l′ + 1), so s = l′ + 1 or s = −l′. For l ≥ 1 the second root is negative and u blows up: discard it, as in hydrogen. For l = 0 and 0 < Zα < 1/2 the roots are 1/2 ± √(1/4 − (Zα)²), both positive, so both solutions vanish at the origin. Two arguments keep s = l′ + 1. As Zα → 0 the other root tends to 0, giving R ∝ 1/r, which is no solution at all because ∇²(1/r) = −4πδ³(r). And a nucleus of radius b, inside which V is finite, matches onto an outside mixture whose second piece is suppressed by at least (b/a)^{(2√(1/4 − (Zα)²))}, with a = ħ/(Zαmc): at most of order 10⁻⁵ in hydrogen, and far less in fact, since there the regular inside solution, u ∝ r, already almost matches the kept u ∝ r^(l′+1). The kept solution still makes an s state's R ∝ r^(l′) diverge, but weakly: in hydrogen (r/a₀)^(l′) is only 1.0006 at the proton's 0.84 fm. As Zα → 0, l′ → 0 and the divergence goes.