Series termination and the exact Dirac–Coulomb energies

Units are SI with ħ and c kept. The course's metric, diag(1, −1, −1, −1), is not needed here, since everything happens in the rest frame of the nucleus. Topic 7.04's radial equations, for ψ = ((G/r)χ_κᵐ, i(F/r)χ_(−κ)ᵐ) with σ⋅r̂ χ_κᵐ = −χ_(−κ)ᵐ, read G′ = −(κ/r)G + (k₊ + Zα/r)F and F′ = (κ/r)F + (k₋ − Zα/r)G, with k± = (mc² ± E)/(ħc). Topic 7.05 supplied λ = √(k₊k₋) and γ = √(κ² − (Zα)²). Write G = e^(−λr) r^γ Σ gₛ rˢ and F = e^(−λr) r^γ Σ fₛ rˢ and substitute. Differentiation brings down (s + γ)gₛ rˢ⁻¹ − λgₛ rˢ, and matching the coefficient of rˢ⁻¹ on both sides gives (s + γ + κ)gₛ − Zα fₛ = λgₛ₋₁ + k₊fₛ₋₁ and (s + γ − κ)fₛ + Zα gₛ = λfₛ₋₁ + k₋gₛ₋₁, derived. Dimensions check order by order: λ and k± are inverse lengths, so λgₛ₋₁ carries the units of gₛ, and Zα is a pure number.

Read the pair as a matrix acting on (gₛ, fₛ). Its determinant is (s + γ)² − κ² + (Zα)² = s(s + 2γ), using γ² = κ² − (Zα)². At s = 0 nothing drives the step, so the determinant must vanish: that is topic 7.05's indicial equation, and it leaves f₀/g₀ = (γ + κ)/(Zα). For s ≥ 1 the determinant is non-zero, so g₀ fixes everything. Now the right-hand sides. Because λ² = k₊k₋, the second, λfₛ₋₁ + k₋gₛ₋₁, is exactly λ/k₊ times the first, λgₛ₋₁ + k₊fₛ₋₁. Every step is pushed along the fixed direction (1, λ/k₊) with one amplitude dₛ₋₁ = λgₛ₋₁ + k₊fₛ₋₁. Inverting the matrix and forming dₛ gives dₛ = dₛ₋₁ × 2[λ(s + γ) − ZαE/(ħc)]/[s(s + 2γ)], derived, the energy entering through k₊ − k₋ = 2E/(ħc). With N = ZαE/(ħcλ) the bracket is λ(s + γ − N). For hydrogen's 2S₁/₂ the recursion gives f₀/g₀ = −0.00365 and g₁/g₀ = −0.50001 Z/a₀, and d₁ comes out zero.

The ratio dₛ/dₛ₋₁ = 2λ(s + γ − N)/[s(s + 2γ)] tends to 2λ/s as s grows, the ratio of successive terms of e^(2λr). So Σ dₛ rˢ converges at every radius and nothing goes wrong at finite r; the failure is at infinity. Matching the recursion to Kummer's series identifies the sum as d₀ M(1 + γ − N, 2γ + 1, 2λr), the confluent hypergeometric function of Quantum Mechanics II's Schrödinger atom, now with the non-integer parameter 2γ + 1. Its large-argument form, quoted, is Γ(2γ + 1)/Γ(1 + γ − N) times e^(2λr)(2λr)^(−γ−N). Multiplied by the e^(−λr) factored off, G and F grow as e^(λr) r^(−N), and ∫(G² + F²) dr diverges. The prefactor 1/Γ(1 + γ − N) vanishes only when 1 + γ − N is zero or a negative integer, so the growing piece can be switched off only at N = γ + 1, γ + 2, …, or by making d₀ itself zero, the nᵣ = 0 case met two sections on. The next section reaches the same condition from the recursion alone.

To stop at degree nᵣ, the amplitude dₙᵣ must vanish while gₙᵣ and fₙᵣ do not; every later step then has nothing to drive it. For nᵣ ≥ 1 the scalar recursion allows this only through its factor (s + γ − N) at s = nᵣ, so ZαE/(ħcλ) = nᵣ + γ, derived; the nᵣ = 0 case, taken up below, gives the same condition with nᵣ = 0. Square it and use (ħcλ)² = m²c⁴ − E²: (Zα)²E² = (nᵣ + γ)²(m²c⁴ − E²), so E = mc²[1 + (Zα)²/(nᵣ + γ)²]^(−1/2), the positive root because N is positive. Dimensions: the bracket is a pure number and E carries mc². Limit: as Zα → 0, γ → |κ| and E → mc² − mc²(Zα)²/(2n²) with n = nᵣ + |κ|, Bohr's levels. Measurement: for hydrogen's 1S₁/₂, with α = 1/137.035999, mc² − E = 13.6058742 eV for an infinitely heavy point proton; the reduced-mass factor 0.99945568 brings it to 13.5984683 eV against the measured ionisation energy, 13.5984346 eV. The 33.7 μeV left over, 2.5 parts per million, is the 1S Lamb shift, which no Dirac–Coulomb level contains.

For nᵣ = 0 the series is a single term, so d₀ itself must vanish: f₀/g₀ = −λ/k₊ = −√((mc² − E)/(mc² + E)), the negative ratio the large-r equations demand. The origin has already fixed f₀/g₀ = (γ + κ)/(Zα). Since γ is less than |κ|, γ + κ has the sign of κ, so the two ratios can agree only for κ ≤ −1, where they agree at exactly one energy, E = mc²γ/|κ|, the nᵣ = 0 energy. For κ ≥ 1 the lowest degree is nᵣ = 1. Define n = nᵣ + |κ|. A shell then holds |κ| = 1 to n, with |κ| = n reached only through κ = −n: j = n − ½ appears once and every lower j twice, each with 2j + 1 = 2|κ| states, 2n² in all. For n = 2 that is 2S₁/₂ (κ = −1, nᵣ = 1), 2P₁/₂ (κ = +1, nᵣ = 1) and 2P₃/₂ (κ = −2, nᵣ = 0), 2 + 2 + 4 = 8 states. Because nᵣ + γ = n − |κ| + √(κ² − (Zα)²), E depends on n and |κ| = j + ½ alone, so 2S₁/₂ and 2P₁/₂ coincide to all orders in Zα. Fixing g₀ by ∫(G² + F²) dr = 1 normalises the state; its general closed form, a product of Γ functions, is not derived here, and no energy depends on it.