Solving the Coulomb radial equation

With u = rR the radial equation reads −(ħ²/2μ)u″ + [−e²/4πε₀r + ħ²l(l+1)/2μr²]u = Eu. For a bound state E < 0, so κ = √(−2μE)/ħ is real and has units of inverse length; measure r in units of 1/κ by setting ρ = κr. Dividing through by E and tidying gives d²u/dρ² = [1 − ρ₀/ρ + l(l+1)/ρ²]u, where ρ₀ = μe²/(2πε₀ħ²κ) = 2/(κa). Two physical constants have collapsed into one pure number. That is the entire point of the scaling: the energy no longer appears in two places at once, so the question "which E are allowed?" turns into "which ρ₀ are allowed?", and the answer will be a statement about integers.

As ρ → 0 the 1/ρ² term dominates and u″ ≈ l(l+1)u/ρ², whose solutions are ρ^(l+1) and ρ^(−l). Reject ρ^(−l): for l ≥ 1 the norm diverges at the origin. For l = 0 the rejected branch is u → constant, which is square-integrable, so the usual "it is not normalisable" excuse fails — it goes because R = u/r behaves as 1/r, and ∇²(1/r) = −4πδ³(r) puts a δ function at the nucleus that nothing in the equation can balance. As ρ → ∞ the bracket tends to 1 and u″ ≈ u, giving e^(±ρ), and e^(+ρ) is rejected outright. Now write u = ρ^(l+1)e^(−ρ)v(ρ). This is not an approximation but an exact change of unknown, valid at every ρ, with the whole solution carried by v.

Substituting gives ρv″ + 2(l+1−ρ)v′ + [ρ₀ − 2(l+1)]v = 0. Put v = Σ cⱼ ρ^j, collect equal powers of ρ, and each power returns the same relation: c_(j+1) = [2(j+l+1) − ρ₀] cⱼ / [(j+1)(j+2l+2)]. Two features matter. First, this fixes the entire function from c₀ alone — the second constant of a second-order equation was already spent rejecting ρ^(−l) at the origin, and c₀ is nothing but overall scale, so there is no third knob to turn. Second, ρ₀ sits in the numerator and not in the denominator, which is where the energy will make its move. Try l = 0 with ρ₀ = 6: c₁ = (2−6)c₀/(1⋅2) = −2c₀, then c₂ = (4−6)c₁/(2⋅3) = (2/3)c₀, then c₃ = (6−6)c₂/(3⋅4) = 0. The series has stopped of its own accord.

At large j the recursion tends to c_(j+1)/cⱼ ≈ 2j/j² = 2/j. Those are exactly the coefficients of e^(2ρ) = Σ (2ρ)^j/j!, whose consecutive ratio is 2/(j+1). So an unterminated v behaves as c₀e^(2ρ) at large ρ, and u = ρ^(l+1)e^(−ρ)v behaves as c₀ρ^(l+1)e^(+ρ). The branch discarded during the asymptotic step has walked back in through the series, and ∫|u|²dr diverges. This is the whole quantisation argument, and notice what it does not say: not that the growth is ugly, or unphysical, or hard to normalise. It says the object has no finite norm, so it is not a state, and the energy that produced it is not in the spectrum.

The only escape is that the numerator vanish: 2(jₘₐₓ + l + 1) = ρ₀ for some non-negative integer jₘₐₓ, after which c_{(jₘₐₓ+1)} and everything beyond it is zero and v is a polynomial of degree jₘₐₓ. Define n ≡ jₘₐₓ + l + 1, so ρ₀ = 2n with n a positive integer. Because jₘₐₓ ≥ 0, l ≤ n − 1 follows at once, and the number of radial nodes is jₘₐₓ = n − l − 1. Nothing was postulated anywhere in that sentence. Take n = 3, l = 0: jₘₐₓ = 2, and the recursion above gave v = c₀(1 − 2ρ + 2ρ²/3), whose roots are ρ = (3 ± √3)/2 = 0.634 and 2.366. Since ρ = r/(na), those sit at r = 1.90a and 7.10a — the two nodes of the 3s radial function, read off a quadratic.

ρ₀ = 2/(κa) = 2n gives κ = 1/(na), so E = −ħ²κ²/2μ = −ħ²/(2μa²n²), with a = 4πε₀ħ²/μe² = 0.0529 nm. With hydrogen's reduced mass that is −13.598 eV/n²; an infinitely heavy nucleus gives the Rydberg energy 13.606 eV, the value used in the examples here. Swap e² for Ze² and the same algebra returns a/Z and −13.606Z²/n² eV. In practice the useful rearrangement is ρ₀ = 2√(E_R/|E|) with E_R = 13.606 eV, which tests any trial energy for integer-ness in one line. But every step assumed E < 0. For E > 0, κ is imaginary, the runaway solution oscillates instead of growing, no truncation is needed, and every energy above the ionisation limit is allowed — which is why a discrete series has an edge with a continuum beyond it. The treatment is also non-relativistic, spinless, and point-nucleus, so fine structure, the Lamb shift and hyperfine splitting all sit below its resolution.