The Schrödinger equation for the Coulomb potential

For one electron bound to a fixed proton, −(ħ²/2m)∇²ψ − (ke²/r)ψ = Eψ with k = 1/(4πε₀). The potential depends on r = √(x² + y² + z²), which ties x, y and z together, so no product X(x)Y(y)Z(z) can work. Spherical coordinates untie them: ∇² = (1/r²)∂/∂r(r² ∂/∂r) + Λ/r², where Λ holds only θ and φ derivatives. Try ψ = R(r)Y(θ, φ) and multiply through by −2mr²/(ħ²RY). What remains is a function of r alone plus ΛY/Y, a function of the angles alone, summing to zero everywhere, so each must be a constant: ΛY = −λY, and λ passes into the radial equation. Notice where the Coulomb term went: only into the radial half. The angular problem is the same for every central potential. Dimensions fix the scales before anything is solved: ħ, m and ke² form exactly one length, a₀ = ħ²/(mke²) = 52.9 pm, and one energy, ke²/a₀ = 27.2 eV.

Split once more, Y = Θ(θ)Φ(φ). The φ part obeys d²Φ/dφ² = −m²Φ, so Φ = e^(imφ). The angles φ and φ + 2π name the same point, and ψ may take only one value there, so e^(2πim) = 1 and m is an integer. Try m = ½: then Φ(2π) = −Φ(0), a wavefunction with two values at one place. The θ part, written in x = cos θ, is the associated Legendre equation. Its power-series solutions diverge at x = 1 or x = −1, the two poles, unless the series terminates, and it terminates only when λ = l(l+1) with l a whole number and l ≥ |m|. So l = 2 allows m = −2, −1, 0, 1, 2: five states, 2l + 1 in general. Two things to notice. The Coulomb potential has not yet been used, so these integers belong to every central force. And m never appears in the radial equation, only λ does, so the energy cannot depend on m: with no field applied, nothing picks out the z axis.

Substitute u(r) = rR(r). Because (1/r²)d/dr(r² dR/dr) = (1/r)d²u/dr², the radial equation becomes −(ħ²/2m)u″ + U_eff(r)u = Eu, with U_eff = −ke²/r + l(l+1)ħ²/(2mr²): a particle on the half-line r > 0. The added term is the kinetic energy of angular motion, L²/(2mr²) with L² = l(l+1)ħ², the centrifugal term of classical orbit theory. Using ħ²/m = ke²a₀, U_eff = ke²[−1/r + l(l+1)a₀/(2r²)], and dU_eff/dr = 0 at r = l(l+1)a₀, where U_eff = −13.6 eV/[l(l+1)]. For l = 1 the floor is at 2a₀ = 106 pm and −6.80 eV; for l = 2 at 6a₀ = 318 pm and −2.27 eV. Raising l pushes the well outwards and makes it shallower. Only l = 0 has no wall: the Coulomb well deepens all the way in, and s states alone reach the nucleus. The substitution also tidies normalisation, since |ψ|² d³r = |u|² dr |Y|² dΩ, so ∫₀^∞ |u|² dr = 1.

Near r = 0 try a power law, u = rˢ. Then u″ and the centrifugal term both go as r^(s−2), while the Coulomb term goes only as r^(s−1) and the E term as rˢ, so as r → 0 the first two must balance on their own: u″ ≈ l(l+1)u/r², which needs s(s − 1) = l(l+1), so s = l + 1 or s = −l. The second is rejected: for l ≥ 1 it makes ∫|u|² dr diverge at the origin, and for l = 0 it gives R ∝ 1/r, whose Laplacian is a δ function that the 1/r Coulomb term cannot balance. So u(0) = 0 and u ∝ r^(l+1). Far out both potential terms fade and u″ ≈ κ²u with κ = √(−2mE)/ħ, giving e^(−κr) or e^(+κr); normalisation keeps only the first. At E = −13.6 eV, κ = 1/a₀, a decay length of 52.9 pm. Now count. A second-order equation has two independent solutions; the condition at r = 0 picks one and leaves only its scale free. Whether that solution also decays at infinity is not yours to choose, and for most E it does not. That mismatch, not a rule added by hand, is what makes the energy discrete.

Factor out both ends: u = r^(l+1) e^(−κr) v(r), with v = Σ cⱼ rʲ. Substituting gives cⱼ₊₁ = 2[κ(j + l + 1) − 1/a₀] cⱼ / [(j + 1)(j + 2l + 2)]. For large j the ratio tends to 2κ/j, the ratio in the series for e^(2κr), so a series that never ends makes u grow like e^(+κr) and fail at infinity. The only escape is a numerator that vanishes at some j = nᵣ, which needs κa₀(nᵣ + l + 1) = 1. Call n = nᵣ + l + 1. Then κ = 1/(na₀) and E = −ħ²κ²/(2m) = −ke²/(2a₀n²) = −13.6 eV/n², with v a polynomial of degree nᵣ and n ≥ l + 1. The ground state is nᵣ = 0, l = 0: Bohr's energy with zero angular momentum instead of Bohr's ħ. Correspondence holds at large n: from n = 100 to 101 the spacing is 2.68 × 10⁻⁵ eV, within 1.5% of h times the classical orbital frequency at that energy, 27.2 eV/n³ = 2.72 × 10⁻⁵ eV.

Four idealisations were built in, and each has a price. An infinitely heavy nucleus: in fact both particles orbit the centre of mass, and the exact fix is to replace m by the reduced mass μ = mM/(m + M). For hydrogen μ = 0.999456 mₑ, so E₁ = −13.598 eV rather than −13.606 eV, 7.4 meV shallower; deuterium's correction is half as large, which puts its Balmer-α line 0.18 nm short of hydrogen's 656 nm line, the shift by which deuterium was first identified. Non-relativistic motion: the electron moves at about αc = c/137, so relativistic corrections have relative size α² ≈ 5 × 10⁻⁵, of order 7 × 10⁻⁴ eV at n = 1, and for a hydrogen-like ion they grow as (Zα)². No spin: the equation cannot produce spin–orbit splitting or the factor of 2 in a shell's capacity; both arrive later in this section. A point nucleus: the proton's 0.84 fm radius is 1.6 × 10⁻⁵ a₀, and it lifts the 1s level by only about 5 × 10⁻⁹ eV. Each correction shifts or splits levels; none changes where n, l and m came from.