Why four components: the Dirac representation constructed

The previous topic turned Dirac's requirements into algebra: αₓ, α_y, α_z and β are Hermitian, each squares to the N × N identity, and any two anticommute. Nothing yet fixes N. Take αᵢ and multiply αᵢβ = −βαᵢ on the left by β; since β² = 1, βαᵢβ = −αᵢ. Now take the trace. A trace is unchanged when the last factor of a product is moved to the front, so Tr(βαᵢβ) = Tr(β²αᵢ) = Tr αᵢ, while βαᵢβ = −αᵢ says the same trace is −Tr αᵢ. A number equal to its own negative is zero: Tr αᵢ = 0, derived. Running the same steps with any αᵢ as the partner gives αᵢβαᵢ = −β and Tr β = 0. Only two facts were used, that the partner squares to 1 and anticommutes, and every one of the four matrices has three such partners. The trace of a finite matrix is the whole engine; it is the same argument that shows [x, p] = iħ has no finite matrices, since the left side would be traceless and the right side would have trace iħN.

A Hermitian matrix is diagonalised by a unitary and has real eigenvalues, and M² = 1 makes each one obey λ² = 1, so λ = +1 or −1. With n₊ of the first and n₋ of the second, the basis-independent trace is Tr M = n₊ − n₋, which the previous step set to zero: n₊ = n₋ and N = 2n₊, derived. One component is out, since a traceless 1 × 1 matrix is the number 0, and so is three, whose trace can only be ±1 or ±3. The same count applies to the Hamiltonian. Work in SI units, H = cα⋅p + βmc², keeping the metric diag(1, −1, −1, −1) for the four-vectors that arrive next topic; both terms carry joules, cp as (m s⁻¹)(kg m s⁻¹) and mc² as kg m² s⁻², because α and β are pure numbers. H is a sum of traceless matrices, so Tr H = 0, and the Klein-Gordon requirement gave H² = (p²c² + m²c⁴)1. So H/Eₚ, with Eₚ = √(p²c² + m²c⁴), is Hermitian, squares to 1 and is traceless: half the energies at every momentum are +Eₚ and half are −Eₚ, in any representation. No choice of matrices removes the negative branch; Unit 8 interprets it.

Evenness still allows N = 2, so test it. The identity and the three Pauli matrices span every 2 × 2 Hermitian matrix with real coefficients, a₀1 + a⋅σ, and tracelessness removes a₀. Quantum Mechanics I's identity σᵢσⱼ = δᵢⱼ1 + iεᵢⱼₖσₖ gives (a⋅σ)(b⋅σ) = (a⋅b)1 + i(a × b)⋅σ, and the cross-product term flips sign when the order is reversed, so the anticommutator is 2(a⋅b)1. Dirac's conditions become geometry: squaring to 1 means a unit vector, anticommuting means perpendicular vectors. Three mutually perpendicular unit vectors fit in three dimensions, which is why σₓ, σ_y and σ_z exist, but a fourth would have to be perpendicular to all three, hence zero, and the zero vector squares to 0, not 1. Two components hold at most three of Dirac's four matrices; with N even and N ≠ 2, N ≥ 4, derived. The limiting case shows what the fourth matrix pays for. At m = 0, β leaves H, three matrices suffice, the σᵢ serve, and H = ±cσ⋅p is a two-component equation, Weyl's. It is the mass term that costs two extra components.

Write 4 × 4 matrices in 2 × 2 blocks and try β = [[1, 0], [0, −1]], αᵢ = [[0, σᵢ], [σᵢ, 0]]. Each is Hermitian because 1 and σᵢ are. Block multiplication checks the algebra: αᵢαⱼ = [[σᵢσⱼ, 0], [0, σᵢσⱼ]], so αᵢαⱼ + αⱼαᵢ = 2δᵢⱼ times the 4 × 4 identity; αᵢβ = [[0, −σᵢ], [σᵢ, 0]] and βαᵢ = [[0, σᵢ], [−σᵢ, 0]] sum to zero; β² = 1 on sight, and all four traces vanish as they must. Why it works shows best if the block index gets Pauli matrices τ of its own: αᵢ = τₓ ⊗ σᵢ and β = τ_z ⊗ 1. The σ factor makes the α's anticommute among themselves, and τ_z, a Pauli matrix of the block index whose vector is perpendicular to that of τₓ, is the fourth direction the old space lacked. β is diagonal, so at rest H = βmc² = diag(mc², mc², −mc², −mc²). For the electron, with measured mₑc² = 0.51100 MeV, the upper pair sits at +0.511 MeV and the lower pair at −0.511 MeV, 1.022 MeV apart. That diagonal rest energy is why this is the representation for slow particles, and the one Unit 6 uses for the non-relativistic limit.

Nothing in the algebra prefers a diagonal β. If U†U = 1, then UαᵢU† and UβU† are still Hermitian, still square to 1 and still anticommute, since U†U = 1 lets products pass through. Take U = (1/√2)[[1, −1], [1, 1]] in blocks. Multiplying out, UβU† = [[0, 1], [1, 0]] and UαᵢU† = [[−σᵢ, 0], [0, σᵢ]], derived: the Weyl, or chiral, representation. Wavefunctions follow, ψ_W = Uψ_D, so Dirac halves φ and χ become ψ_L = (φ − χ)/√2 and ψ_R = (φ + χ)/√2, with ψ_W†ψ_W = ψ_D†ψ_D. Here H = [[−cσ⋅p, mc²], [mc², cσ⋅p]], and the mass is the only thing linking the halves. At m = 0 the equation falls apart into two Weyl equations, the two-component limit found above, which is why this representation suits fast or massless particles. The measured record fits that limit, quoted: in 1958 Goldhaber, Grodzins and Sunyar found the neutrino from electron capture in ¹⁵²ᵐEu to be left-handed, as one Weyl half alone describes, and tritium-decay data from KATRIN now bound the neutrino mass below 0.45 eV, so any mass linking its halves is more than a million times smaller than the electron's.

Could some other 4 × 4 set be genuinely different? Pauli's fundamental theorem, quoted here and not proved, says no: any two sets of 4 × 4 matrices obeying the algebra are related by M′ = SMS⁻¹ with one invertible S, unitary when both sets are Hermitian, and larger solutions are built from copies of the 4 × 4 one, so N is a multiple of four (also quoted). Measurement cannot tell the representations apart: a similarity transformation keeps every eigenvalue of H, and with ψ′ = Uψ every density ψ†ψ and expectation value ψ†Aψ is unchanged. Components are not physics. An electron at rest with spin up is (1, 0, 0, 0) in the Dirac representation and (1, 0, 1, 0)/√2 in the Weyl one, half its probability in each half. The freedom remains even with the α's held fixed: β = τ_y ⊗ 1 also squares to 1 and anticommutes with every τₓ ⊗ σᵢ, and the unitary e^(iπτₓ/4) ⊗ 1, a quarter turn of the block space about τₓ, carries τ_z ⊗ 1 into it while leaving each αᵢ alone. Pick the representation that shortens the calculation; nothing measured depends on the choice.