Spin from the equation: J = L + (ħ/2)Σ is conserved

SI units throughout, with the metric diag(1, −1, −1, −1) wherever a four-vector appears. Take H = cα⋅p + βmc² and Lᵢ = εᵢⱼₖxⱼpₖ. The mass term commutes with L, because β acts on the four components and L acts on position. In the kinetic term only [pₗ, xⱼ] = −iħδₗⱼ survives, so [H, Lᵢ] = −iħc εᵢⱼₖαⱼpₖ, which is [H, L] = −iħc α × p, derived. Dimensions: ħc α × p is J s × m s⁻¹ × kg m s⁻¹ = J × J s, as [H, L] must be. The Heisenberg equation turns it into dL/dt = cα × p, and since dr/dt = (i/ħ)[H, r] = cα this is just velocity × momentum. No torque acts; the velocity operator is simply not parallel to p, for each component of cα has eigenvalues ±c. Contrast the Schrödinger particle, whose velocity p/m is parallel to p, and the Klein-Gordon one, which meets p only through p²: both conserve L. The leak belongs to the first-order equation.

Dirac's requirements fixed only αᵢαⱼ + αⱼαᵢ = 2δᵢⱼ, αᵢβ + βαᵢ = 0 and β² = 1. From these alone, (αₓα_y)² = −αₓαₓα_yα_y = −1, so Σ_z = −iαₓα_y squares to 1. It is Hermitian, since (−iαₓα_y)† = iα_yαₓ = −iαₓα_y, and traceless, since Tr(αₓα_y) = Tr(α_yαₓ) = −Tr(αₓα_y). Cycling gives Σₓ = −iα_yα_z and Σ_y = −iα_zαₓ, compactly Σₖ = −(i/2)εₖᵢⱼαᵢαⱼ, and moving α_z through two factors gives [Σₓ, Σ_y] = 2iΣ_z. In the Dirac representation αₓα_y = diag(iσ_z, iσ_z), so Σ = diag(σ, σ), the Pauli matrices in both diagonal blocks; in the Weyl representation, where α = diag(−σ, σ), the same product gives the same Σ. Nothing here used relativity or a representation: any set of matrices obeying Dirac's algebra carries this Σ with it, derived.

The commutators of Σ_z follow from the algebra too. β and α_z each anticommute with both factors of αₓα_y, so they commute with Σ_z. αₓ commutes with its own factor and anticommutes with α_y, which gives αₓΣ_z − Σ_zαₓ = −2iα_y, and likewise α_yΣ_z − Σ_zα_y = 2iαₓ. Hence [H, Σ_z] = 2ic(αₓp_y − α_ypₓ) = 2ic(α × p)_z, and cycling gives [H, Σ] = 2ic α × p, derived. It is the operator of the L leak with the opposite sign, so [H, L + λΣ] = ic(2λ − ħ) α × p, which vanishes for every momentum only at λ = ħ/2: J = L + (ħ/2)Σ commutes with H. Checks: (ħ/2)Σ carries J s, like r × p; at p = 0 both leaks vanish; for a plane wave along z, (α × p)_z = 0, so L_z and Σ_z each commute with H there. In general the one spin component that survives alone is Σ⋅p, since (α × p)⋅p = 0, and topic 4.08 builds its label on it. That J generates rotations of ψ is assumed here and proved in Unit 5.

Set S = (ħ/2)Σ. The algebra of Σ gives [Sᵢ, Sⱼ] = iħεᵢⱼₖSₖ and S² = (ħ²/4)(Σₓ² + Σ_y² + Σ_z²) = (3/4)ħ² = s(s + 1)ħ² with s = 1/2. Distinct Σₖ anticommute, since ΣₓΣ_y = αₓα_y = −Σ_yΣₓ; hence for any unit vector n̂, (n̂⋅Σ)² = 1 with zero trace, so the spin along every axis is ±ħ/2 = ±5.27 × 10⁻³⁵ J s, each value twice among the four components. In Quantum Mechanics I spin was attached by hand: a two-component factor multiplied the Schrödinger wave function, S and L were separately conserved for a free particle, and J = L + S added two conserved things. Here neither part is conserved and the coefficient ħ/2 is forced. Relativity is not the source, for the Klein-Gordon equation is relativistic and spinless. Lévy-Leblond's Galilean first-order equation of 1967, named and quoted here, needs σ matrices too and even gives g = 2. The source is the first-order square root of p², which needs anticommuting matrices, as (σ⋅p)² = p² shows. Measured: Stern and Gerlach's silver beam splits in two, and the sodium D lines at 589.0 and 589.6 nm form a doublet: both count two spin states, as s = 1/2 requires.

Non-conservation does not make a free electron's spin precess as it flies. Take the 4.06 spinors u = N(ξ, cσ⋅p ξ/(E + mc²)) and a second spinor u′ at the same momentum with upper part ξ′. The cross terms give u′†αᵢu = N²c ξ′†(σᵢσ⋅p + σ⋅p σᵢ)ξ/(E + mc²) = 2N²cpᵢ ξ′†ξ/(E + mc²), while u′†u = 2N²E ξ′†ξ/(E + mc²). So at one momentum and positive energy, α acts as pc/E, the group velocity over c, and ⟨α × p⟩ = 0 in any packet built only from positive-energy spinors: ⟨L⟩ and ⟨Σ⟩ are each constant, derived. The operator statement [H, L] ≠ 0 lives in the matrix elements that join +E to −E; those oscillate at 2E/ħ, never slower than 2mc²/ħ = 1.55 × 10²¹ rad s⁻¹, the zitterbewegung that Unit 8 interprets. At positive energy alone the leak shows as unsharpness: a state of sharp E and sharp J_z generally has neither S_z nor L_z sharp.

Superpose u(p, ↑) over a cone of half-angle θ about z, with equal weight at every azimuth φ and one energy E. The lower components are κσ⋅p̂↑ = κ(cos θ ↑ + sin θ e^(iφ) ↓), with κ = pc/(E + mc²) and κ² = (E − mc²)/(E + mc²). Summed over φ, the e^(iφ)↓ piece carries mₗ = 1 with mₛ = −1/2 and everything else mₗ = 0 with mₛ = +1/2, so J_z = +ħ/2 and E are both sharp. The weights, normalised by 1 + κ², give ⟨S_z⟩ = (ħ/2)(1 + κ² cos 2θ)/(1 + κ²) = (ħ/2)[1 − (1 − mc²/E)sin²θ], derived, and ⟨L_z⟩ = ħ/2 − ⟨S_z⟩. At E = 2mc², where v = 0.866c, and θ = 30°, that is 0.4375ħ of spin and 0.0625ħ of orbit. The fraction is dimensionless, as it must be. Limits: E → mc² returns all of ħ/2 to spin, the Pauli result; θ → 0 does too, since p then lies along z; E ≫ mc² leaves (ħ/2)cos²θ, the massless form in which spin points only along ±p̂.