The Gordon decomposition: a g = 2 moment in the free current

Units throughout are SI with c kept, the metric is diag(1, −1, −1, −1), x^μ = (ct, x), p^μ = (E/c, p) and p̸ = γ^μ p_μ. Unit 4 gave the conserved current J^μ = qcψ̄γ^μ ψ = (cρ, J), with ρ = qψ†ψ and J = qcψ†αψ. Look at J. Each αᵢ has eigenvalues ±1 and does not commute with the Hamiltonian, so cαₓ can only ever return ±2.998 × 10⁸ m s⁻¹, even for an electron drifting at 10⁶ m s⁻¹; only its expectation value in a positive-energy state comes out as pc²/E. Nor does anything in qcψ†αψ say that the electron spins. The Klein-Gordon current of Unit 3, (iqħ/2m)[φ* ∂^μ φ − (∂^μ φ*)φ], speaks the language of motion: derivatives, which are momenta, divided by m. Gordon's identity rewrites Dirac's current in that language, and whatever is left over is the part of the current that the spin carries.

Take ū(p′)γ^μ u(p) with both spinors free and of positive energy. In momentum space the Dirac equation is p̸u(p) = mc u(p); its Dirac adjoint is ū(p′)p̸′ = mc ū(p′), because γ⁰γ^μ†γ⁰ = γ^μ. So γ^μ u = γ^μ p̸u/mc and ū′γ^μ = ū′p̸′γ^μ/mc, and averaging the two writes the matrix element as ū(p′)[γ^μ p̸ + p̸′γ^μ]u(p)/(2mc). Now split each product of two gammas into its symmetric and antisymmetric halves, γ^μ γ^ν = g^μν − iσ^μν. That gives γ^μ p̸ = p^μ − iσ^μν p_ν and p̸′γ^μ = p′^μ + iσ^μν p′_ν, the sign turning because σ^νμ = −σ^μν. Their sum is the identity, with Δ = p′ − p. Nothing was dropped and no speed assumed. Two checks: every term carries kg m s⁻¹, and at Δ = 0, with ūu = 2mc, it gives ū(p)γ^μ u(p) = 2p^μ; the time part 2E/c is the normalisation u†u itself.

Superpose positive-energy plane waves, or use the free Dirac equation directly, and momenta become derivatives: J^μ = (iqħ/2m)[ψ̄ ∂^μ ψ − (∂^μ ψ̄)ψ] + ∂_ν M^μν with M^μν = (qħ/2m)ψ̄σ^μν ψ. The first term is the Klein-Gordon current with ψ̄ in place of φ*; at low speed it is q times the Schrödinger current. The second is the divergence of an antisymmetric tensor, so ∂_μ ∂_ν M^μν = 0: it is conserved on its own, and being a divergence it integrates to no net charge. Its components are familiar from electromagnetism. Since σ^ij = ε^ijk Σ^k we may write M^ij = ε^ijk M^k and Mⁱ⁰ = cPⁱ; then ρₛₚᵢₙ = −∇⋅P and Jₛₚᵢₙ = ∇ × M + ∂P/∂t. These are the bound charge and bound current of a medium with magnetisation M = (qħ/2m)ψ̄Σψ and polarisation P. A free Dirac electron's current is that of a tiny magnetised, polarised body moving with it.

Write ψ = (φ, χ) in the Dirac representation. For a slow positive-energy state χ ≈ (σ⋅p/2mc)φ, smaller than φ by about v/2c. Since ψ̄Σψ = φ†σφ − χ†σχ, the magnetisation is M = (qħ/2m)φ†σφ up to corrections of relative order v²/c², while P, built from φ†σχ, is smaller than M/c by a factor of order v/c, like the electric dipole of a moving magnet. The moment of any localised current is μ = ½∫r × J d³x, and for J = ∇ × M an integration by parts turns this into ∫M d³x. So μ = (qħ/2m)⟨σ⟩ = (q/m)⟨S⟩, with S = ħσ/2. The convection term, the Schrödinger current, gives the orbital moment (q/2m)⟨L⟩. Spin carries twice the moment per unit angular momentum: g = 2, derived from the free current. For the electron the size is eħ/2m = μ_B = 9.274 × 10⁻²⁴ J T⁻¹, pointing against the spin because q = −e.

The 1/(2mc) in front of the spin term invites the reading that it is a small relativistic piece. Compare sizes instead. For a packet of size a and density n, the spin current is about (ħ/2m)n/a per unit charge and the convection current n v, so they are equal when v ≈ ħ/(2ma). For a = 1.0 nm that is 5.8 × 10⁴ m s⁻¹, or 1.9 × 10⁻⁴ c, a speed set by the packet's own momentum spread ħ/a and nothing relativistic. What is small is the spin term's share of the charge density, −∇⋅P, which is suppressed by v/c and again by ħ/(mca), the ratio of the reduced Compton wavelength 3.86 × 10⁻¹³ m to a. Momentum space says the same thing: the spin piece of the time component is of order β² times the convection piece, while in the spatial components the spin piece compares with the convection piece as |Δ| does with |p + p′|, with no power of β. At high speed even the time components become comparable, and the two pieces partly cancel.

Switch on a weak static field. To first order the energy is ⟨−qcα⋅A⟩ = −∫J⋅A d³x, with J the free current. With A = ½B × r the convection term gives −(q/2m)⟨L⟩⋅B; the spin term gives −∫(∇ × M)⋅A d³x = −∫M⋅B d³x, the surface term vanishing for a localised packet. That is the dipole energy −μ⋅B with g = 2, read off the free current with no reduction of the equation; Unit 6 reaches the same moment by reducing the Dirac equation in a field. Now the data, all quoted. The electron's gₑ = 2.002 319 304 36 lies 0.116% above 2, a radiative correction whose leading term α/2π = 1.161 × 10⁻³ belongs to quantum electrodynamics. The proton's g = 5.586, and the neutron's moment of −1.913 μ_N with no charge at all, show that g = 2 belongs to a point particle with minimal coupling, not to spin ½ as such: a Pauli term ψ̄σ^μν ψ F_μν is a Lorentz scalar by this unit's bilinear table, so covariance allows it, and it would shift g.