Tensor products and adding angular momenta

An electron with orbital state |l mₗ⟩ and spin state |s mₛ⟩ lives in neither space but in their tensor product, spanned by the (2l+1)(2s+1) product kets |l mₗ⟩ ⊗ |s mₛ⟩. For l = 1 and s = ½ that is six states, labelled (mₗ, mₛ) = (1,↑), (1,↓), (0,↑), (0,↓), (−1,↑), (−1,↓). The total angular momentum on that space is J = L ⊗ 1 + 1 ⊗ S, written L + S once the tensor factors are understood. Two facts earn it the definition. Operators on different factors commute, [Lᵢ, Sⱼ] = 0, so in [Jᵢ, Jⱼ] the cross terms vanish and the two algebras simply add: [Jᵢ, Jⱼ] = iħ εᵢⱼₖ Jₖ. And J, not L, generates rotations of the whole electron, which is the symmetry a central Hamiltonian still has once spin–orbit coupling is switched on. Everything the ladder argument proved from the su(2) algebra therefore applies to J with no new work: its spectrum has to be ħ²j(j+1) and mⱼħ on towers of 2j+1 states.

Six states need four commuting labels either way, and there are two natural choices. The uncoupled set (L², S², L_z, S_z) labels a state by (l, s, mₗ, mₛ). The coupled set (L², S², J², J_z) labels it by (l, s, j, mⱼ). L² and S² sit in both, so l and s are good in either basis — the common worry that coupling destroys l is misplaced. What fails is L_z. Expand J² = L² + S² + 2L_zS_z + L₊S₋ + L₋S₊: those last two terms lower one factor while raising the other, connecting (mₗ, mₛ) to (mₗ ± 1, mₛ ∓ 1), and a short commutator gives [J², L_z] = ħ(L₋S₊ − L₊S₋) ≠ 0. The sum is untouched, which is why J_z = L_z + S_z commutes with J² and why mⱼ = mₗ + mₛ is the one magnetic label common to both bases. The rule for practice: in the coupled basis quote j and mⱼ and never mₗ or mₛ. How widely a product ket spreads over j is set by how many towers reach its mⱼ, so the two-term answer below is special to l ⊗ ½ and must not be generalised — couple 1 ⊗ 1 instead and the single ket |m₁ = 1⟩|m₂ = −1⟩ already carries all three of L = 0, 1 and 2, with probabilities 1/3, 1/2 and 1/6.

The construction is mechanical. The state of largest mⱼ is unique, because only one product ket has mₗ + mₛ = l + s, so it must be the top of the tallest tower: |3/2, 3/2⟩ = |mₗ = 1⟩|↑⟩ for a p electron. Apply J₋ = L₋ + S₋ to both sides and use ħ√(j(j+1) − m(m−1)) on each ladder. On the left that factor is ħ√(15/4 − 3/4) = √3 ħ. On the right, L₋|1⟩ = √2 ħ|0⟩ and S₋|↑⟩ = ħ|↓⟩. Dividing out gives |3/2, 1/2⟩ = √(2/3)|0,↑⟩ + √(1/3)|1,↓⟩ — a superposition whose two Clebsch–Gordan coefficients are unequal because a ladder, not a symmetry, fixed them. Two product kets reach mⱼ = 1/2, so the coupled basis needs a second state there, and it is the normalised orthogonal combination |1/2, 1/2⟩ = −√(1/3)|0,↑⟩ + √(2/3)|1,↓⟩, the top of the shorter tower. Lower that in turn, and repeat at each mⱼ until the states are used up.

Past two small momenta, stop doing it by hand. In the uncoupled basis of dimension (2j₁+1)(2j₂+1), build J₁z and J₂z as diagonal matrices of m₁ħ and m₂ħ and the ladder operators from ⟨m±1|J_±|m⟩ = ħ√(j(j+1) − m(m±1)), lift each to the product space with numpy.kron, and assemble J² = J₁² + J₂² + 2J₁z J₂z + J₁₊J₂₋ + J₁₋J₂₊. Then numpy.linalg.eigh returns eigenvalues ħ²j(j+1) with the right multiplicities — for l = 1 ⊗ s = ½, four values of 3.75ħ² and two of 0.75ħ² — and the eigenvector matrix is the Clebsch–Gordan transformation. One caution that costs people an afternoon: eigh fixes neither the sign of a column nor the mixing inside a degenerate eigenvalue, so the raw output is not the tabulated table. Because J² is block diagonal in mⱼ, diagonalise one mⱼ block at a time — here every block is 1×1 or 2×2 — and then only the phase convention is left to impose by hand.

The payoff of the coupled basis is the identity L⋅S = (J² − L² − S²)/2, whose eigenvalue is (ħ²/2)[j(j+1) − l(l+1) − s(s+1)]. For s = ½ it collapses to ⟨L⋅S⟩ = +ħ²l/2 when j = l + ½ and −ħ²(l+1)/2 when j = l − ½, so the two eigenvalues differ by ħ²(l + ½) — and the two energy levels by ζₙₗ(l + ½), once a radial constant is supplied to convert them. The upper level, though it holds more states, moves the smaller distance. The asymmetry is not an accident: weight each shift by its degeneracy and they cancel exactly, (2l+2)(l/2) + (2l)(−(l+1)/2) = 0, so the centre of gravity of the multiplet stays where the unsplit level was. Use that as an arithmetic check on every fine-structure calculation. Two spin-halves behave the same way: ⟨S₁⋅S₂⟩ = +ħ²/4 for the triplet and −3ħ²/4 for the singlet, a gap of ħ², which is the skeleton of both the exchange splitting in helium and hydrogen's 21 cm hyperfine line.

Everything above is a unitary change of basis and costs nothing. It also predicts nothing about size. The spin–orbit term is H = (ζₙₗ/ħ²) L⋅S with ζₙₗ = (ħ²/2mₑ²c²)⟨(1/r) dV/dr⟩, a radial expectation value you have to compute or measure. Its value separates cases that look identical on paper: hydrogen's 2p doublet splits by 4.5 × 10⁻⁵ eV, about 11 GHz, while sodium's 3p doublet — the D lines at 589.0 and 589.6 nm — splits by 2.13 meV, or 17.2 cm⁻¹, some fifty times wider, because ζ climbs steeply with the charge the electron sees. The scale also decides whether the coupled basis is the right one at all. Switch on a magnetic field and two terms compete; the coupled basis wins while ζₙₗ ≫ μ_B B, and the uncoupled basis returns in the Paschen–Back limit. With ζ₃p = 1.42 meV sodium needs roughly 25 T to cross over, but hydrogen's 2p, with ζ ≈ 3.0 × 10⁻⁵ eV, crosses near 0.5 T — a field you can reach on a bench.