Quantum numbers and the spin tensor factor

A quantum number is never just a counter; it is the eigenvalue a state carries for one operator in a commuting set. For one electron in a central field that set is (H, L², L_z, S_z). The first two brackets vanish for any V(r): [H, L²] = [H, L_z] = 0. The third is free, because S acts on a different tensor slot from everything orbital, so S_z commutes with H, with L² and with L_z identically. Complete means the four eigenvalues leave no residual degeneracy: fix Eₙ, l(l + 1)ħ², mₗħ and mₛħ and you have named one ray in the Hilbert space rather than a subspace. Notice what is missing from that list. S² belongs to the set too, but for a single electron S² = (3/4)ħ²I on every state, so s = 1/2 labels the representation the electron carries, not the state it is in; it never varies and therefore distinguishes nothing. Only mₛ does any work.

n is the odd one out: it is not read off an independent operator but forced by normalisability. Put u = rR, scale ρ = 2r/na₀, and strip the asymptotics with u = ρ^(l+1)e^(−ρ/2)v(ρ). The ratio of successive coefficients in the power series for v is (k + l + 1 − n)/[(k + 1)(k + 2l + 2)], which for large k behaves as 1/k — the coefficients of e^ρ. If the series never stops, v grows like e^ρ, u like e^(+r/na₀), and the state leaves the Hilbert space. So the numerator must vanish at some integer k = nᵣ ≥ 0, giving n = nᵣ + l + 1. That one equation carries everything usually recited as separate rules: n starts at 1, l is at most n − 1, and the radial function of a given (n, l) has exactly nᵣ = n − l − 1 nodes. At n = 3 the subshells are 3s with two nodes, 3p with one and 3d with none — a nodeless radial function always sits at l = n − 1.

l and mₗ come from L²(S²) and know nothing about the Coulomb potential. L² and L_z are built from angles alone, so their eigenbasis — the spherical harmonics, with ⟨n̂|l m⟩ = Yₗ^m(n̂) — is shared by the screened atom, the three-dimensional oscillator and the nuclear shell model alike. The ladder operators fix the range: L₊ must annihilate the top state and L₋ the bottom, which forces mₗ to run in integer steps from −l to +l, a multiplet of 2l + 1 states. Parity is (−1)^l, and that is what later kills every dipole element with Δl = 0. The eigenvalues are also not a vector pointing somewhere: for l = 2, |L| = √6 ħ = 2.449ħ while the largest projection is only 2ħ, so cos θ = 2/√6 = 0.816 and θ = 35.3°. A non-zero angular momentum can never lie along its own quantisation axis, because Lₓ and L_y would then both be sharp and [Lₓ, L_y] = iħL_z forbids it.

The Schrödinger equation is one scalar partial differential equation; separating it in a central field yields two separation constants, l and mₗ, on top of the energy eigenvalue — three labels in all, with no fourth hiding in the boundary conditions. Spin enters by enlarging the space: replace L²(ℝ³) with L²(ℝ³) ⊗ C² and postulate S = (ħ/2)σ, with σ the Pauli matrices. The old Hamiltonian becomes H_orb ⊗ I₂. It commutes with all three components of S, its eigenvalues are exactly those of H_orb, and every eigenspace is doubled with no energy shift whatever — hydrogen's n = 3 level goes from 9 states to 18 and stays at −1.51 eV. Two honest footnotes. The evidence is experimental: a silver beam splits in two, sodium's D line is a doublet, the Zeeman pattern is anomalous. The derivation is relativistic: the Dirac equation forces four components and delivers gₛ = 2. And the axis is a choice — mₛ is the S_z eigenvalue, but n̂⋅S serves equally well, with |n̂,+⟩ = (cos(θ/2), e^(iφ)sin(θ/2)) in the z basis.

Fix n and sum the multiplets: 2l + 1 over l = 0 … n − 1 is 1 + 3 + 5 + … + (2n − 1), the sum of the first n odd numbers, which is n². Multiply — do not add — by the two states of the spin slot, and the level holds 2n². For n = 1, 2, 3, 4 that is 2, 8, 18, 32. The same number arrives from another direction in the next topic: rescaled on the negative-energy subspace, the Runge–Lenz vector closes so(4), and each level is one irreducible multiplet of dimension n². Read the count physically with care, because two other familiar sequences are not this one. The s, p, d and f blocks are 2, 6, 10 and 14 wide, since a block is one subshell's 2(2l + 1), not a whole shell's 2n². And the periodic table's rows run 2, 8, 8, 18, 18, 32, 32, because in a many-electron atom the l-degeneracy is already broken and 4s fills before 3d. So 2n² counts a hydrogenic level — it is the capacity of the K, L, M, N shells — and neither a block width nor a chemical period.

Add ξ(r)L⋅S and the list changes. L⋅S contains L₊S₋ + L₋S_+, which converts one unit of mₗ into one unit of mₛ, so [H, L_z] and [H, S_z] no longer vanish and mₗ and mₛ stop labelling eigenstates. What survives is J = L + S: the good set becomes (H, L², S², J², J_z) and the labels are n, l, j, mⱼ with j = l ± 1/2. Nothing is lost from the count, because 2(2l + 1) = (2l + 2) + 2l is exactly 2j + 1 summed over the two j values — six 2p states become four with j = 3/2 and two with j = 1/2. The scale is small: hydrogen's 2p interval is 10.97 GHz, or 4.54 × 10⁻⁵ eV, about 1.3 × 10⁻⁵ of |E₂| = 3.40 eV. Push a magnetic field past roughly 0.78 T, where μ_B B overtakes that interval, and the Paschen–Back regime hands mₗ and mₛ back. The label set is not a property of the atom; it is a property of the atom plus the terms you chose to keep.