Conserved operators in a central field: J and parity, not L

SI units throughout, with the metric diag(1, −1, −1, −1) wherever γ⁰ is read as a four-vector component. Take H = cα⋅p + βmc² + V(r), with V(r) an electrostatic potential energy such as −Zαħc/r multiplying the 4 × 4 identity. Two facts decide everything. V commutes with L, because [Lᵢ, V] = −iħ(r × ∇V)ᵢ and ∇V points along r. And V commutes with Σ, α and β, because it is the identity in spinor space. So V commutes with J = L + (ħ/2)Σ, and Unit 4's free results carry over untouched: [H, L] = −iħc α × p, [H, Σ] = 2ic α × p, and the combination with weight ħ/2 vanishes. J², J_z and H share eigenstates, so j and mⱼ are good labels, derived. Dimensions: ħc α × p is J s × m s⁻¹ × kg m s⁻¹ = J × J s, as [H, L] must be. Limits: V → 0 returns the free particle, and a field that is not central, such as a uniform E along z, keeps only [H, J_z] = 0.

Square the leak. Summing [H, Lᵢ]Lᵢ + Lᵢ[H, Lᵢ] over i gives [H, L²] = −iħc α⋅(p × L − L × p), and p × L = −L × p + 2iħp turns it into 2ħ²c α⋅p + 2iħc α⋅(L × p), derived. It is not zero. Take ψ = (φ(r)ξ, 0), an s wave in the upper slot with spinor ξ: L²ψ = 0, but Hψ has lower component cσ⋅p φξ, which is odd under inversion with j still ½, so it has l = 1 and [H, L²]ψ = −2ħ²(0, cσ⋅p φξ). Now S⋅L, with S = (ħ/2)Σ. Using [H, Σ] = 2ic α × p and the matrix identity Σ⋅(α × p) = 2iα⋅p, [H, S⋅L] = −ħ²c α⋅p − iħc α⋅(L × p), exactly −½[H, L²]. The check: J² = L² + 2S⋅L + (3/4)ħ² then commutes with H, as the first section demands. The culprit is the odd block cσ⋅p: σ⋅p is a scalar under rotations, so it keeps j, and a pseudoscalar under inversion, so it moves l by one.

Spatial inversion P₀, ψ(r) → ψ(−r), sends p to −p and leaves the matrices alone, so P₀HP₀⁻¹ = −cα⋅p + βmc² + V(r): the kinetic term flips sign. Unit 5's Dirac parity repairs it with γ⁰ = β = diag(1, 1, −1, −1). Because βαβ = −α, the operator P = γ⁰P₀ gives PHP⁻¹ = cα⋅p + βmc² + V(r) = H whenever V(−r) = V(r), which a central potential always satisfies, derived. P² = β² = 1, so its eigenvalues are η = ±1. P also commutes with J: P₀ commutes with L = r × p because both factors flip, and β commutes with Σ = diag(σ, σ). So H, J², J_z and P can be diagonalised together, and they fix the whole angular dependence; what remains is two radial functions, one per component. A Lorentz-scalar potential U(r), entering as β(mc² + U(r)), also commutes with J and with P, so the same labels would survive; it is named here and not treated.

Write ψ = (φ, χ). Each two-component piece with sharp j and mⱼ is built from Quantum Mechanics II's coupled states |l, ½, j, mⱼ⟩, and only l = j − ½ and l = j + ½ can make j; their parities (−1)^l are opposite. Pψ = ηψ reads φ(−r) = ηφ(r) and −χ(−r) = ηχ(r). So φ carries a single l with (−1)^l = η, and χ a single l′ with (−1)^l′ = −η. Then l′ ≠ l, hence l′ = 2j − l, derived. The spectroscopic letter names the upper l. For j = ½, S₁/₂ has η = +1, l = 0, l′ = 1, and P₁/₂ has η = −1, l = 1, l′ = 0. For j = 3/2, P₃/₂ has l = 1, l′ = 2, η = −1, and 3D₃/₂ has l = 2, l′ = 1, η = +1. Count check: each j offers two parities, exactly the two orbital choices l = j ± ½ of the Pauli atom, so n = 2 still holds 2 + 2 + 4 = 8 = 2n² states. Trading l for η loses no state and gains none.

If the lower component carries a fraction w of the norm, a measurement of L² returns l(l + 1)ħ² with probability 1 − w and l′(l′ + 1)ħ² with probability w. The two values differ by (2j + 1)ħ², so ΔL² = (2j + 1)ħ²√(w(1 − w)), derived. Symmetry fixes which two values; dynamics fixes w. To order (v/c)², Unit 6 gives w ≈ ⟨T⟩/(2mc²), so hydrogen 1s, with ⟨T⟩ = 13.6 eV, has w = 1.33 × 10⁻⁵, ⟨L²⟩ = 2.7 × 10⁻⁵ ħ² and ΔL² = 7.3 × 10⁻³ ħ²: the spread grows as √w, not w. For hydrogen-like uranium the exact 1s weight w = 0.129, quoted here and derived in topic 7.08, gives ⟨L²⟩ = 0.26ħ² and ΔL² = 0.67ħ². Limit: w → 0 makes L² sharp, which is how Quantum Mechanics II could label states by l. Unit 6's Foldy-Wouthuysen Hamiltonian, built from powers of p², functions of r and S⋅L, commutes with L² to order (v/c)²: the l it conserves belongs to the transformed two-component state, not to the Dirac spinor.

Good labels give selection rules. An electric-dipole operator is odd under P, so η must change, and j can change by at most one. Lyman-α therefore links 1S₁/₂ (η = +1) to 2P₁/₂ and 2P₃/₂ (η = −1), and is a doublet split by the measured 10.97 GHz between the two j values, 0.0054 Å at 121.6 nm. 2S₁/₂ cannot reach 1S₁/₂ with one electric-dipole photon, since both have η = +1, so it decays mainly by emitting two photons and lives about 0.12 s against 1.6 ns for 2P, both quoted. Symmetry stops short of energies: 2S₁/₂ and 2P₁/₂ share j and differ in η, so nothing here forces or forbids their degeneracy. Topic 7.06 shows the Dirac-Coulomb energies depend on n and j alone, so the two coincide to all orders in Zα, and the measured 1057.8 MHz Lamb shift between them is what the Dirac atom omits. Lamb and Retherford measured it with microwaves that drove 2S₁/₂ atoms to 2P₃/₂, a 9.91 GHz interval in zero field that they tuned with a magnetic field, a transition allowed precisely because η flips.