The Dirac field: Lagrangian, momentum and modes

SI units throughout, with x^μ = (ct, x), ∂_μ = ∂/∂x^μ and the metric (+, −, −, −). The density ℒ = ψ̄(iħcγ^μ∂_μ − mc²)ψ is c times topic 4.03's operator placed between ψ̄ and ψ. Since ψ̄γ⁰ = ψ† and γ⁰γᵏ = αᵏ, it is also ψ†(iħ∂ₜ − H_D)ψ, with H_D = cα⋅(−iħ∇) + βmc². Treat ψ and ψ̄ as independent, as z and z* may be for a complex variable. ℒ holds no derivative of ψ̄, so varying ψ̄ gives ∂ℒ/∂ψ̄ = 0, the Dirac equation; varying ψ gives iħ∂_μψ̄γ^μ + mcψ̄ = 0, topic 4.04's adjoint equation; both derived. ℒ is not real, but it differs from the real form (iħc/2)[ψ̄γ^μ∂_μψ − (∂_μψ̄)γ^μψ] − mc²ψ̄ψ by (iħc/2)∂_μ(ψ̄γ^μψ), a divergence that changes no field equation. Dimensions: ψ is in m^(−3/2), so ψ̄ψ is in m⁻³ and ℒ in J m⁻³, making the action ∫ℒ d³x dt a quantity in J s. The second form also shows ℒ = 0 on every solution.

The only time derivative in ℒ is iħψ†∂ₜψ, so π_ψ = ∂ℒ/∂(∂ₜψ) = iħψ†, in J s m^(−3/2), and ψ̄ has no momentum at all: derived. The momentum is the conjugate field, not a velocity, so ψ and iħψ† already form a canonical pair and the initial data are ψ alone. The oscillator has the same structure when written in a = √(mω/2ħ)(x + ip/mω): L = iħa*ȧ − ħωa*a equals pẋ − H up to a total time derivative, so iħa* is the momentum of a. The Legendre transform gives ℋ = π_ψ∂ₜψ − ℒ = ψ†H_Dψ, derived, and Hamilton's equation ∂ₜψ = δH/δπ_ψ returns iħ∂ₜψ = H_Dψ. Unlike topic 9.02's Klein–Gordon ℋ, this one holds no squared momentum, and H_D has eigenvalues of both signs, so nothing yet keeps H positive. The equal-time rule of topics 9.06 and 9.07, [ψⱼ(x), πₖ(y)] = iħδⱼₖ δ³(x − y) or its anticommutator, therefore reads [ψⱼ(x), ψₖ†(y)]∓ = δⱼₖ δ³(x − y): ħ cancels, and both sides are in m⁻³.

Topic 4.04's current j^μ = cψ̄γ^μψ, conserved and with the positive density ψ†ψ, is quoted; Noether's theorem from topic 9.01 now says where it comes from. ℒ is unchanged by ψ → e^(−iθ)ψ with ψ̄ → e^(iθ)ψ̄, and the Noether current (∂ℒ/∂(∂_μψ))δψ/θ = (iħcψ̄γ^μ)(−iψ) = ħcψ̄γ^μψ is ħ times j^μ. The charge is Q = q∫ψ†ψ d³x, with q = −e for the electron field, derived. Space translations give P = ∫ψ†(−iħ∇)ψ d³x in the same way, and time translation gives back H. Now compare integrands. ψ†ψ = |ψ₁|² + |ψ₂|² + |ψ₃|² + |ψ₄|² cannot be negative, so a field of numbers has a Q with the sign of q, whatever it does. ψ†H_Dψ has no sign: H_D has eigenvalues +Eₚ and −Eₚ, and a field built from negative-frequency solutions carries negative energy. The complex scalar of topic 9.03 is the mirror image, with an energy density that is a sum of squares and Unit 3's charge density, which takes either sign.

In a periodic box of volume V every solution is the mode sum on the formula card, with Eₚ = √(p²c² + m²c⁴) and topic 4.06's normalisation u†u = v†v = 2Eₚ/c, so the factor √(c/(2EₚV)) leaves b and d dimensionless. Nothing is assumed of b and d except that d† is the adjoint of d; for a classical field they are numbers and d† means d*. The v-term is a wave of momentum −p and energy −Eₚ, yet it carries the label p. Unit 8 fixed v(p, s) = iγ²u(p, s)*, and since iγ²v(p, s)* = u(p, s) in the Dirac representation, the charge conjugate iγ²ψ* is the same sum with b and d exchanged, label for label: (p, s) is the momentum and spin a d-quantum will carry if it proves to be a particle. Multiplying by u†(p, s) or v†(p, s) and integrating inverts the sum, b(p, s) = √(c/(2EₚV))∫e^{(i(Eₚ t − p⋅x)/ħ)}u†(p, s)ψ d³x and the same with v† and the opposite phase for d†(p, s), derived. b and d† are linear in ψ, so whatever rule is imposed on ψ and ψ† passes straight to them.

Put the sum into H = ∫ψ†H_Dψ d³x. H_D returns +Eₚ on each u-wave and −Eₚ on each v-wave, and ∫e^{(i(p′ − p)⋅x/ħ)} d³x = Vδₚₚ′ keeps only pairs of waves with the same spatial momentum. A u-wave of momentum p therefore meets the v-term labelled −p, whose wave also has momentum p, with a coefficient proportional to u†(p, s)v(−p, s′)e^{(2iEₚ t/ħ)} on b†(p, s)d†(−p, s′). That spinor product vanishes, derived in one line: u(p, s) and v(−p, s′) are eigenvectors of one Hermitian matrix, H_D(p) = cα⋅p + βmc², with eigenvalues +Eₚ and −Eₚ. Take pc = 0.75 mc² along z, so Eₚ = 1.25 mc². In units of √(mc), u(p, ↑) = (1.5, 0, 0.5, 0), v(p, ↓) = (−0.5, 0, −1.5, 0) and v(−p, ↓) = (0.5, 0, −1.5, 0). Then u†u = 2.5 mc = 2Eₚ/c and u†(p, ↑)v(−p, ↓) = 0.75 − 0.75 = 0, but u†(p, ↑)v(p, ↓) = −1.5 mc. Unit 4's ū(p, s)v(p, s′) = 0 is true and beside the point: H is built from ψ†, not ψ̄.

What survives is H = Σ Eₚ[b†b − d d†], P = Σ p[b†b − d d†] and Q = q Σ[b†b + d d†], summed over p and s and derived with no rule for the operators. The order d d† comes from ψ† standing left of ψ, the only place an operator order has entered. Each v-mode is a first-order oscillator of frequency −Eₚ/ħ, with coordinate d† and momentum iħd: the −Eₚ is no slip but Unit 4's negative-energy solution in Hamiltonian form. Read with numbers, H = Σ Eₚ(|b|² − |d|²) has no lower bound even at fixed charge, since a unit of |d|² in a mode of large p costs −Eₚ. Read with operators, ⟨d d†⟩ is the squared norm of d†|Ψ⟩, and the question is how large it may grow: topic 9.06 imposes commutators, topic 9.07 anticommutators. Two checks close here. For b-modes with p ≪ mc, Eₚ ≈ mc² + p²/2m, rest energy plus Schrödinger's kinetic energy per quantum. And one m serves both branches, so whatever the d-quanta are they have the electron's mass: |m(e⁺) − m(e⁻)|/m is measured below 8 × 10⁻⁹, quoted.