The radial equation and effective potential

After separation, the radial factor R(r) obeys −(ħ²/2μ)[R″ + (2/r)R′] + [V(r) + ħ²l(l+1)/2μr²]R = ER, and that middle term (2/r)R′ is what stops it being an ordinary one-dimensional problem. The fix is the identity (1/r²)(r²R′)′ = (1/r)(rR)″. Define u = rR, multiply through by r, and the first derivative vanishes: −(ħ²/2μ)u″ + V_eff u = Eu, with nothing left to distinguish it in form from a particle in a one-dimensional well. The same substitution settles the bookkeeping. Probability in three dimensions integrates over shells of volume 4πr² dr, so the norm of R is ∫|R|²r²dr; written in u this is ∫|u|²dr — the Jacobian has been absorbed, once, and must not be inserted again. Nothing here used the Coulomb form: any central potential — the 3D oscillator, a screened nucleus, the two nuclei of a diatomic molecule — reduces the same way, which is why this one page of algebra is reused for the rest of the course.

The centrifugal term is not modelling; it is the angular motion, already solved, sent to the radial equation as a bill. Kinetic energy in a central field splits into a radial piece plus L̂²/2μr², and acting on the factored state R(r)Yₗₘ the operator L̂² returns the exact number l(l+1)ħ². So an electron with angular momentum pays ħ²l(l+1)/2μr² of kinetic energy just to be at radius r, and the radial problem sees that cost exactly as a classical orbit sees L²/2μr² — the sole quantum edit is l(l+1)ħ² in place of L². Two things follow. Because 1/r² beats 1/r as r → 0, any l ≥ 1 walls the origin off behind a barrier that diverges faster than the Coulomb well dives. And the coefficient is l(l+1), not l²: even the smallest nonzero angular momentum costs 2ħ², which is why p states are pushed off the nucleus so decisively.

The half-line needs a condition at r = 0, and it is u(0) = 0 for every l. The quick reason: R = u/r must stay finite. The honest reason is sharper. Suppose u(0) = c ≠ 0 in an s state; then ψ ≈ cY₀₀/r near the origin, and ∇²(1/r) = −4πδ³(r), so the Schrödinger equation acquires a δ-function term that neither the Coulomb potential nor the energy can balance — the candidate fails exactly at the point being argued about. Near the origin the equation itself sorts the options: with u″ ≈ l(l+1)u/r² dominating, u behaves as r^(l+1) or r^(−l), and the second is discarded — for l ≥ 1 it is not normalisable, for l = 0 it is the δ-function impostor. So the physical solution rises off the origin as r^(l+1): each unit of l flattens the wavefunction near the nucleus by another power of r. And because u(0) = 0 is exactly the boundary condition of an infinite wall, 1D intuition must be borrowed in its half-line form: of the bound states of the full-line well −e²/4πε₀|x|, hydrogen's s spectrum keeps only the odd ones.

Before solving anything, nondimensionalise and stare. Measuring r in Bohr radii (ρ = r/a₀) and energy in units of 13.6 eV, hydrogen's effective potential becomes v(ρ) = −2/ρ + l(l+1)/ρ² — one curve per l, no other parameters. For l = 0 it is the bare Coulomb dive. For l ≥ 1 the barrier wins at small ρ, the well bottoms out at ρ = l(l+1), and the floor sits at −13.6/l(l+1) eV: −6.8 eV at 2a₀ for l = 1, −2.27 eV at 6a₀ for l = 2, −1.13 eV at 12a₀ for l = 3. That floor is already a theorem. A bound state must lie above the minimum of its well, so Eₙ = −13.6/n² eV with angular momentum l requires n² > l(l+1), i.e. n ≥ l + 1. The famous restriction l ≤ n − 1, which the series solution derives properly in the next topic, is visible here as bare energy bookkeeping: there is no 1p state because the p-well's floor at −6.8 eV sits above the −13.6 eV a 1-state would need.

Setting V_eff(r) = Eₙ gives the classical turning points — the radii where a classical electron of that energy and angular momentum would stall and swing back. In dimensionless form the condition is a quadratic, ρ² − 2n²ρ + n²l(l+1) = 0, so ρ± = n² ± n√(n² − l(l+1)). For the 2p state that is 4 ± 2√2: an allowed shell from 1.17a₀ to 6.83a₀ (0.062 nm to 0.361 nm), the quantum cousin of the perihelion and aphelion of a Kepler ellipse. The turning points organise the whole shape of u. Between them E > V_eff and u curves toward the axis, oscillating — its n − l − 1 radial nodes all live there. Outside them E < V_eff and u curves away, decaying as e^(−κr) at large r with κ = √(−2μE)/ħ: the orbital's soft exponential edge is one-dimensional tunnelling into the classically forbidden region, nothing more exotic.

Within the non-relativistic model — relativity and spin, which add the far larger fine structure, lie outside it — every term above is exact except one. The centrifugal barrier carries no approximation — l(l+1)ħ² is an eigenvalue, not an estimate. What is idealised is −e²/4πε₀r itself, which assumes a point proton. The real proton has a charge radius near 0.84 fm ≈ 1.6 × 10⁻⁵ a₀; inside it the potential flattens instead of diving. Which states notice follows from the small-r analysis: u ~ r^(l+1) means only l = 0 states put appreciable probability inside the nucleus, so finite nuclear size nudges s levels and barely touches the rest. The nudge is tiny in ordinary hydrogen but measurable — it separates isotopes, and in muonic hydrogen, where the roughly 200-times-smaller orbit magnifies the overlap, it is the effect behind the modern proton-radius measurements. Knowing which term to trust and which to correct is what it means to own the model.