The central-field Hamiltonian and its commuting set

Two masses interacting through V(|r₁ − r₂|) have six coordinates. Change to R = (m₁r₁ + m₂r₂)/M and r = r₁ − r₂ with M = m₁ + m₂, and the kinetic term separates without approximation: H = P²/2M + p²/2μ + V(r), with 1/μ = 1/m₁ + 1/m₂. Nothing couples R to r, so Ψ = e^(iK⋅R)ψ(r) and the centre of mass is a free plane wave carrying no structure. Every energy level belongs to the relative operator p²/2μ + V(r), a genuine one-body problem in three dimensions. The price is that μ, not mₑ, sets the scale. For hydrogen mₚ/mₑ = 1836.153, so μ = 0.9994557 mₑ and every level is 0.0544% shallower than the infinite-nucleus value: −13.5983 eV rather than −13.6057 eV. For deuterium μ = 0.9997276 mₑ, and the ratio of the two reduced masses moves Balmer-α from 656.47 nm to 656.29 nm — the 0.18 nm gap that revealed deuterium in 1931. Positronium, with μ = mₑ/2, halves the entire spectrum.

Take L_z = xp_y − ypₓ and a potential depending only on r = |r|. Then [L_z, V] = x[p_y, V] − y[pₓ, V] = −iħ(x ∂V/∂y − y ∂V/∂x), and since ∂V/∂y = V′(r)⋅y/r and ∂V/∂x = V′(r)⋅x/r, the two terms are V′(r)xy/r and V′(r)yx/r, which cancel exactly. The kinetic term goes the same way: [L_z, pₓ] = iħp_y and [L_z, p_y] = −iħpₓ, so [L_z, p²] = iħ(p_ypₓ + pₓₚ_y) − iħ(pₓₚ_y + p_ypₓ) = 0. Nothing singled out z, so [H, Lᵢ] = 0 for all three components and therefore [H, L²] = 0 too. Two consequences follow. First, L is conserved — the quantum form of the classical central-force result. Second, because L_± = Lₓ ± iL_y commute with H, every state in a 2l + 1 multiplet shares one energy: the m-degeneracy is forced by rotational symmetry alone, in any central field. What is not forced is the equality of different l at the same n. That extra Coulomb degeneracy needs a separate symmetry, and it is the first thing screening destroys.

Commuting self-adjoint operators can be diagonalised together; that is the spectral theorem, and it is why ψ = R(r)Yₗ^m(θ, φ) works. The Hilbert space of one spinless particle factorises as L²(ℝ³) ≅ L²((0,∞), r²dr) ⊗ L²(S²), and L²(S²) breaks into rotation-irreducible blocks of dimension 2l + 1 spanned by the Yₗ^m. Because [H, L²] = [H, L_z] = 0, H is block-diagonal in that decomposition: it acts as one radial operator Hₗ on each block, identical for all 2l + 1 values of m. Writing a product function is therefore not an ansatz that happens to succeed; it is the shared eigenbasis expressed in the position representation. Two cautions. The set (H, L², L_z) is complete on the orbital factor — one state per (E, l, m) triple — but a general state is a superposition of products, not a product. And once the electron's spin factor C² is carried along, the space is L²(ℝ³) ⊗ C² and the Schrödinger H acts as H ⊗ I, which cannot fix mₛ at all; S_z must be appended by hand, doubling every hydrogen level to 2n² states.

Substituting u = rR turns ∫|R|²r²dr into ∫₀^∞|u|²dr and gives −(ħ²/2μ)u″ + V_eff u = Eu — a one-dimensional problem, but on 0 < r < ∞ rather than the whole line, and a half-line has an endpoint. Integrating by parts twice, ⟨u|Hₗ v⟩ − ⟨Hₗ u|v⟩ = −(ħ²/2μ)[u*v′ − u*′v]₀^∞; the piece at infinity dies for normalisable states, and the piece at r = 0 does not die by itself. Requiring it to vanish for every u and v in the domain makes Hₗ symmetric; taking that domain as large as the vanishing allows makes it self-adjoint, and u(0) = 0 is the condition that does it. For l ≥ 1 the centrifugal term ħ²l(l+1)/2μr² forces the behaviour anyway: the operator is in the limit-point case at the origin, essentially self-adjoint, with the Frobenius roots r^(l+1) and r^(−l) settling the matter. For l = 0 there is a genuine choice — a one-parameter family of self-adjoint extensions exists, and u(0) = 0 selects the Friedrichs member. Physically that is the demand that R = u/r carry no δ function, since ∇²(1/r) = −4πδ³(r) would hide a point source at the nucleus.

In one dimension p = −iħ d/dx is conjugate to x and a perfectly good observable; on the half-line the analogue fails. The symmetrised radial momentum pᵣ = ½(r̂⋅p + p⋅r̂) = −iħ(∂ᵣ + 1/r) obeys [r, pᵣ] = iħ, and in the u = rR picture it is just −iħ d/dr acting on functions with u(0) = 0, where the boundary term u*v at r = 0 vanishes — so it is symmetric. But symmetric is not self-adjoint. Solve pᵣ†χ = ±i(ħ/a)χ in that picture and χ ∝ e^(∓r/a); only e^(−r/a) is square-integrable on (0, ∞), so the deficiency indices are (1, 0). They are unequal, and von Neumann's theorem then says no self-adjoint extension exists — not several to choose between, none at all. So pᵣ has no real spectrum, no eigenbasis, no projection-valued measure and nothing to measure: a canonical commutator and a formal integration by parts do not make an observable, the domain does. What still works is pᵣ†pᵣ = −ħ²d²/dr² on u(0) = 0, positive and self-adjoint, so radial kinetic energy is fine even though radial momentum is not.

Every step above is an assumption you can name. The kinetic term is p²/2μ, so the treatment is non-relativistic; the leading correction is of relative order α² ≈ 5 × 10⁻⁵, the fine structure a good etalon resolves in Balmer-α. The nucleus is a point, which is why the proton's finite radius moves hydrogen levels by a few parts in 10¹⁰ but moves muonic hydrogen enormously, μ there being about 186 mₑ. Spin is imported rather than derived: it arrives as a tensor factor because the Schrödinger equation has no place for it. And the commuting set is only as stable as the Hamiltonian. Add the spin-orbit term ξ(r)L⋅S and L_z and S_z stop commuting with H, since L⋅S trades mₗ against mₛ; the surviving set is (H, L², S², J², J_z) and the good labels become n, l, j, mⱼ. Add a uniform magnetic field along z and only rotations about z survive, so L_z stays good while L² does not once the field is strong. Naming the commuting set is not decoration — it is the statement of which quantum numbers a given experiment is still entitled to use.