Hydrogen wavefunctions and radial probability

The previous two topics separated the Coulomb problem and solved its angular part. The result is ψₙₗₘ = Rₙₗ(r)Yₗₘ(θ, φ) with n ≥ 1, 0 ≤ l ≤ n − 1 and −l ≤ m ≤ l, and each label leaves a readable fingerprint on Rₙₗ. Near the nucleus Rₙₗ ∝ rˡ, so only s states are non-zero at r = 0. Far out Rₙₗ ∝ e^(−κr) with κ = √(2mₑ|Eₙ|)/ħ, and Eₙ = −13.6 eV/n² makes κ = 1/(na₀): the decay length is na₀. In between, a polynomial of degree n − l − 1 supplies the oscillation. The first three are R₁₀ = 2a₀^(−3/2) e^(−r/a₀), R₂₀ = (2√2)⁻¹ a₀^(−3/2)(2 − r/a₀)e^(−r/2a₀) and R₂₁ = (2√6)⁻¹ a₀^(−3/2)(r/a₀)e^(−r/2a₀). The label m sits only in Yₗₘ, so nothing radial depends on it, and because ∫|Yₗₘ|² dΩ = 1 on its own, the radial function is normalised separately: ∫₀^∞ Rₙₗ² r² dr = 1.

|ψ|² is a probability per unit volume. For the ground state, ψ₁₀₀ = e^(−r/a₀)/√(πa₀³), so |ψ|² is largest at the nucleus, 1/(πa₀³) = 2.15 × 10³⁰ m⁻³, and e² = 7.4 times smaller at r = a₀. A small box at the nucleus is more likely to hold the electron than an equal box one Bohr radius away. The question “how far from the nucleus?” asks about something else: the whole thin shell from r to r + dr, of volume 4πr² dr. Integrating |ψ|² over that shell, or for any l integrating |Yₗₘ|² over the angles, gives P(r) dr = r²|Rₙₗ|² dr. The factor r² starts at zero and grows while |ψ|² decays, so P must rise from zero, peak where the two trade off, and fall. Its unit is m⁻¹: only P(r) dr, a pure number, is a probability, and a plot of |ψ|² labelled P(r) has lost a factor of 4πr².

For the ground state P(r) = (4r²/a₀³)e^(−2r/a₀). The most probable radius is where dP/dr = 0; differentiating ln P gives 2/r − 2/a₀ = 0, so r = a₀. The mean is a first moment, ⟨r⟩ = ∫₀^∞ rP dr = (4/a₀³)∫₀^∞ r³e^(−2r/a₀) dr, and the standard integral ∫₀^∞ rᵏe^(−βr) dr = k!/β^(k+1) turns it into (4/a₀³)(3!)(a₀/2)⁴ = 3a₀/2 = 79.4 pm. The mean lies beyond the mode because the distribution is lopsided: it rises as r² but decays over a long exponential tail. Probabilities are areas: ∫₀^a₀ P dr = 1 − 5e⁻² = 0.323, so the electron is found inside a₀ less than a third of the time. The second moment, ⟨r²⟩ = 3a₀², gives a spread Δr = √(3 − 2.25) a₀ = 0.87a₀, comparable with the radius itself. A ground-state atom has a size, not a boundary.

A node is a surface on which ψ = 0 and across which it changes sign. Radial nodes are spheres, the zeros of the polynomial in Rₙₗ at 0 < r < ∞, and there are n − l − 1 of them; the zero at r = 0 from the factor rˡ is not counted. The 3s function has two, at r = (9 ± √27)a₀/2 = 1.90a₀ and 7.10a₀; 3p has one, at 6a₀; 3d has none. Angular nodes come from Yₗₘ, l of them, as planes through the nucleus or cones about the z-axis. The 2p orbital with m = 0, ∝ cos θ, vanishes on the plane θ = 90°; the 3d orbital with m = 0, ∝ 3cos²θ − 1, vanishes on two cones at θ = 54.7° and 125.3°. The count l holds for the real orbitals chemists draw. In the complex e^(imφ) forms the |m| planes through the z-axis dissolve into a phase that winds around it, and |ψ|² keeps only the l − |m| cones. Every real orbital of level n has n − 1 nodal surfaces in total, and in hydrogen that total alone fixes the energy.

For any state ⟨r⟩ = (a₀/2)[3n² − l(l + 1)], so the size grows as n²: 1.5a₀ for 1s, 13.5a₀ for 3s, and 1.5 × 10⁴ a₀, about 0.8 μm, for n = 100 with l = 0. The states with the most angular momentum their level allows, l = n − 1, have a single radial lobe peaking at exactly n²a₀, the radius of Bohr's nth orbit, and a relative spread Δr/⟨r⟩ = 1/√(2n + 1) that falls from 0.58 at n = 1 to 0.22 at n = 10 and 0.07 at n = 100. That is the correspondence check: at large n the distribution closes onto a thin shell at the classical radius. At small n the classical picture fails outright. In the ground state the classical turning point, where −e²/(4πε₀r) equals E₁ = −e²/(8πε₀a₀), is r = 2a₀, yet the integral of P from 2a₀ to ∞ is 13e⁻⁴ = 0.238: almost a quarter of all position measurements land where no classical electron with that energy could be.

For level n, l takes n values and each has 2l + 1 orientations, so gₙ = Σ(2l + 1) = n² states share Eₙ; the spin found in the next topic doubles this to 2n², the shell capacities 2, 8, 18 used for many-electron atoms. The m-degeneracy is ordinary: any central potential is blind to orientation. The l-degeneracy is not. The radial equation carries the centrifugal term l(l + 1)ħ²/(2mₑr²), which differs between 2s and 2p, and in a generic central potential they would sit at different energies. In the Coulomb potential they do not, because of an extra conserved quantity, the Laplace–Runge–Lenz vector, the quantum form of the fact that a classical Kepler ellipse closes on itself and never precesses. Change the potential and the degeneracy goes: in sodium, whose valence electron sees a screened nucleus, 3s and 3p lie 2.10 eV apart, the energy of the 589 nm yellow line. Hydrogen's own n = 2 states differ by at most about 5 × 10⁻⁵ eV, through relativity, spin–orbit coupling and QED, all outside this model.