Dirac field with commutators: energy unbounded below

SI units with ħ and c kept, metric (+, −, −, −). Topic 9.05's expansion, quoted, is ψ(x, t) = Σₚ,ₛ √(c/2EₚV) [b(p, s)u(p, s)e^{(−i(Eₚt − p⋅x)/ħ)} + d†(p, s)v(p, s)e^{(i(Eₚt − p⋅x)/ħ)}] in a box of volume V. Let [b, b†] = A and [d, d†] = B for equal labels, every other pair commuting. The d-term reaches [ψ, ψ†] as [d†, d] = −B, so after p → −p in that term the equal-time bracket is Σₚ (c/2EₚV) e^{(ip⋅(x − y)/ħ)} [A Σₛ u u†(p) − B Σₛ v v†(−p)]. Unit 4's spin sums for u ū and v v̄, multiplied on the right by γ⁰, give Σₛ u u†(p) = (Eₚ + H_D(p))/c and Σₛ v v†(−p) = (Eₚ − H_D(p))/c, with H_D(p) = cα⋅p + βmc². Since δ_ab δ³(x − y) = (1/V)Σₚ e^{(ip⋅(x − y)/ħ)} δ_ab, every p needs (A − B)Eₚ + (A + B)H_D(p) = 2Eₚ I. H_D(p) has eigenvalues ±Eₚ and is no multiple of I, so A + B = 0 and A − B = 2: A = 1 and B = −1, derived.

The minus is an ordering sign. ψ multiplies v by d†, and ψ† multiplies v† by d, so [ψ, ψ†] meets the pair in the order [d†, d] = −[d, d†]; the canonical rule needs that term to enter with +1, so [d, d†] = −1. Topic 9.03's complex scalar had the same ordering, b† on its negative frequencies, and still gave [b, b†] = +1, because its rule pairs φ with π = (1/c²)∂φ†/∂t, and the time derivative gives the term carrying b the factor −iω where the term carrying a† gets +iω: a second minus sign, cancelling the first. The Dirac Lagrangian is first order in time, its momentum iħψ† carries no derivative, and nothing cancels. Nor can renaming help, since [d, d†] = −1 and [d†, d] = +1 are one statement. The rest frame shows the sign with the algebra stripped away: at p = 0, H_D = βmc², the u's fill the upper two components and the v's the lower two, and the lower block reads −B × 2mc = 2mc. Dimensions agree: √(c/2EₚV) u carries m^(−3/2), as ψ must, so both sides of the rule are in m⁻³.

Take b|0⟩ = d|0⟩ = 0, as the names suggest. Then ⟨0|d d†|0⟩ = ⟨0|(d†d − 1)|0⟩ = −1, and since d(d†)ⁿ = (d†)ⁿd − n(d†)ⁿ⁻¹, induction gives ⟨0|dⁿ(d†)ⁿ|0⟩ = (−1)ⁿ n!, derived. The energies are fine. With H = Σ Eₚ(b†b − d d†), [H, d†] = −Eₚ[d, d†]d† = +Eₚd†, so d† raises the energy by Eₚ, and [Q, d†] = −qd† gives each quantum the charge −q. Positive energy, the electron's mass, the opposite charge: every one-particle property of a positron except a positive norm, and that one is fatal. The state 0.6|0⟩ + 0.8d†|0⟩ has norm 0.36 − 0.64 = −0.28, and (|0⟩ + d†|0⟩)/√2 has norm zero without being the zero vector. Born-rule ratios built on such norms come out negative or above 1, so no measurement statistics can be read from them. An electron, b†|0⟩, still has norm +1: the failure lives entirely in the antiparticle sector, and the one-particle electron limit sees nothing wrong.

Positivity returns if the roles swap: let d†|0⟩ = 0. Then c = d† obeys [c, c†] = +1, an ordinary oscillator, and ⟨0|(d†)ⁿdⁿ|0⟩ = n!, all positive. Now H = Σ Eₚ(b†b − c†c) and Q = qΣ(b†b + c†c), normal ordered already, with no constants to drop. Each quantum d creates carries energy −Eₚ and charge q, the electron's own: Dirac's negative-energy electrons, now bosons. One mode holds any number n of them at energy −nEₚ, and Eₚ grows without limit with |p|, so H has no lowest eigenvalue, derived. Hole theory's cure is gone: without exclusion no state fills a mode, so Unit 8's sea cannot be built. Nothing of charge −q exists in this reading, against the measured positron, of charge +e and rest energy +0.511 MeV, whose annihilation at rest gives the observed pair of 511 keV photons. And [b, b†] = +1 in both readings, so b†(p, s)²|0⟩ ≠ 0: electrons would be bosons, a further failure that topic 9.07 removes.

The Heisenberg equation iħ∂ψₐ/∂t = [ψₐ, H] needs only [ψₐ(x), ψ†_b(y)X_b(y)] with X = H_Dψ. With commutators it is [ψₐ, ψ†_b]X_b + ψ†_b[ψₐ, X_b] = δ_ab δ³(x − y)X_b, because ψ commutes with ψ; with anticommutators the identity [A, BC] = {A, B}C − B{A, C} gives the same. Either way iħ∂ψ/∂t = −iħc α⋅∇ψ + βmc²ψ, derived. Mode by mode, [b, H] = Eₚb and [d, H] = −Eₚ[d, d d†] = +Eₚd, so b(t) = b e^{(−iEₚt/ħ)} and d†(t) = d† e^(iEₚt/ħ): the frequencies the expansion assumed. No reading enters, because the Heisenberg equation is an operator identity, blind to which state is called the vacuum. Extended to unequal times, the bracket [ψₐ(x), ψ†_b(y)] is the same c-number function that topic 9.07 shows vanishes at spacelike separation, quoted, so causality is intact as well. Neither the field equation nor microcausality can reject this rule; only the norm or the spectrum can.

It is tempting to picture reading 2's vacuum collapsing into d-quanta. In a free theory it cannot: H commutes with every b†b and d d†, so every Fock state is stationary and no occupation ever changes. What fails is existence, not stability. H has no lowest eigenvalue, so no state can serve as the vacuum, and no thermal state exists: one mode's partition sum Σₙ e^(nEₚ/kT) grows term by term by e^(Eₚ/kT) = e^{(1.98 × 10⁷)} at 300 K for p = 0, and diverges. Switch on any coupling to photons, as Unit 10 will, and conservation laws protect nothing: two d-quanta at rest, energy −2mc² and charge 2q, may become two at ±p plus two back-to-back photons of Eₚ − mc² each, allowed at every p, 1.10 MeV per photon at p = 3mc, so the energy released has no bound. Ordinary matter has a ground state and thermalises. For spin ½, commutators fail on positivity, the norm or the spectrum; topic 9.08 shows the scalar with anticommutators failing on causality instead.