Dirac's requirements and the algebra of α and β

This lesson uses SI units with ħ and c written out, and the metric signature (+, −, −, −) with x^μ = (ct, x). Unit 3's Klein-Gordon equation gives every plane wave the right E² = p²c² + m²c⁴, but its conserved density ρ = (iħ/2mc²)(ψ*∂ψ/∂t − ψ∂ψ*/∂t) is (E/mc²)|ψ|² on a plane wave, negative whenever E is negative (quoted from Unit 3). The cause is the second time derivative: ψ and ∂ψ/∂t are independent initial data, and the density has to involve both. That sinks the equation only as a one-particle wave equation: as a field theory, with ρ read as a charge density, the Klein-Gordon equation describes spin-0 particles such as the charged pions correctly, quoted here. Dirac wanted a one-particle equation for the electron, and kept the Schrödinger form iħ ∂ψ/∂t = Hψ, first order, so that ψ alone fixes the future and a density with no time derivative in it becomes possible. Relativity sets ct beside x, so one ∂/∂t calls for one ∇: H must be linear in p = −iħ∇. With coefficients independent of x and t, which is what a free particle in uniform space-time allows, the most general choice is H = cα⋅p + βmc². Dimensions check at once: cp and mc² are both energies, so α and β are dimensionless. For an electron with p = 1.00 MeV/c the two terms are 1.00 MeV and 0.511 MeV, comparable, so neither can be dropped.

The first demand is the one every quantum Hamiltonian meets. From iħ ∂ψ/∂t = Hψ and its adjoint, d⟨ψ|ψ⟩/dt = (i/ħ)⟨ψ|(H† − H)|ψ⟩, which vanishes for every state only if H† = H; the same condition makes the energies real. Take the adjoint term by term. Each pᵢ = −iħ∂ᵢ is Hermitian, and constant matrices commute with derivatives, so (cαᵢpᵢ)† = c pᵢαᵢ† = cαᵢ†pᵢ. Then H† − H = c Σᵢ(αᵢ† − αᵢ)pᵢ + (β† − β)mc². On a plane wave this must vanish for every momentum: p = 0 kills the β bracket, and p along each axis in turn kills each α bracket. So αᵢ† = αᵢ and β† = β, derived. Two facts come free. Each coefficient has real eigenvalues and an orthonormal set of eigenvectors. And if one set αᵢ, β works, so does UαᵢU†, UβU† for any unitary U, because a change of basis among the components preserves both Hermiticity and every product: the demands fix an algebra, not one particular set of matrices.

The second demand is that every component of ψ obey the Klein-Gordon equation, so each free wave carries E² = p²c² + m²c⁴. H does not depend on t, so applying iħ ∂/∂t twice gives −ħ² ∂²ψ/∂t² = H²ψ, and the demand becomes H² = (p²c² + m²c⁴)1. Expand H² keeping every product in order: c²Σᵢⱼ αᵢαⱼpᵢpⱼ + mc³Σᵢ(αᵢβ + βαᵢ)pᵢ + β²m²c⁴. Since p₁p₂ = p₂p₁, the coefficient of p₁p₂ is α₁α₂ + α₂α₁, not 2α₁α₂. Now match powers of p, which is arbitrary: p₁² gives α₁² = 1, p₁p₂ gives α₁α₂ + α₂α₁ = 0, the linear term gives α₁β + βα₁ = 0, and the constant gives β² = 1. Over i = 1, 2, 3 that is ten conditions, all derived. The linear term is the one to watch. For an electron with cp = mc² = 0.511 MeV along the 3-axis the target is E² = 0.261 + 0.261 = 0.522 MeV², so E = ±0.723 MeV, √2 times the rest energy. Were β equal to α₃, the cross term would add 2 × 0.511 × 0.511 = 0.522 MeV², doubling E² and giving ±1.022 MeV = ±(cp + mc²), an energy linear in p.

For two ordinary numbers, real or complex, ab + ba = 2ab, which is zero only if a = 0 or b = 0, and α₁² = β² = 1 forbids both. So the coefficients are N × N matrices and ψ is a column (ψ₁, …, ψ_N) of functions: the Dirac equation is N coupled first-order equations. Because H² is the identity times p²c² + m²c⁴, each component separately obeys the Klein-Gordon equation, while H itself ties the components together. Nothing about spin was assumed; the components came from asking a first-order H to square correctly. The massless limit is a check. At m = 0, β drops out and only three anticommuting matrices are needed, which the Pauli matrices supply: σᵢσⱼ = δᵢⱼ + iεᵢⱼₖσₖ gives σᵢσⱼ + σⱼσᵢ = 2δᵢⱼ, so H = cσ⋅p, with two components, squares to p²c². That is Weyl's equation for a massless spin-half particle, quoted here. It is the mass, needing a fourth matrix that anticommutes with all three, that forces more components; the next topic proves N ≥ 4.

Some results need no representation. If a Hermitian matrix squares to 1 and Mv = λv, then v = M²v = λ²v, so λ = ±1: every αᵢ and β has eigenvalues +1 and −1 only. Nor can β be +1 or −1 throughout, for then αᵢβ + βαᵢ = ±2αᵢ, which is not zero. So β has both eigenvalues, and at rest, where H = βmc², the free particle has levels at +mc² and −mc², 2mc² = 1.022 MeV apart for an electron. Compare with measurement: the positive level is the rest energy 0.511 MeV, the energy of each photon when an electron and a positron annihilate at rest, the 511 keV line every PET scanner detects. The negative level cannot be argued away, since the anticommutator that fixed the dispersion is what forces it; Unit 8 gives it a meaning. The same argument gives α₃ the eigenvalues ±1, so cα₃ has eigenvalues ±c, a puzzle that returns when cα turns out to be the velocity operator. And for p ≪ mc the positive branch, mc²√(1 + p²/m²c²) ≈ mc² + p²/2m, hands back Schrödinger's kinetic energy.

Keep the ledger straight. Derived here: αᵢ and β are Hermitian, square to 1 and anticommute in pairs; they cannot be numbers; their eigenvalues are ±1, with β taking both. Imposed as requirements: the first-order form, linearity in p, Hermiticity and H² = p²c² + m²c⁴. Quoted: the relativistic E² = p²c² + m²c⁴ and Unit 3's indefinite density. Assumed, not yet proved: that the equation keeps its form under Lorentz transformations. Being first order in both ∂/∂t and ∇ is necessary for that but not sufficient, because the components of ψ must also mix among themselves under a boost, and Unit 5 builds the transformation that does it. Not settled by the algebra: the matrices themselves, fixed only up to a unitary change of basis, and their size, which topic 4.2 shows is at least 4 × 4. Still to come from these four matrices: a conserved density ψ†ψ that is never negative (topic 4.4), the spin of the particle (topic 4.7) and, once a field is added by minimal coupling (topic 4.9), g = 2 in Unit 6.