Dirac field with anticommutators: positrons without a sea

Units are SI, with metric (+, −, −, −), x^μ = (ct, x), px = Eₚt − p⋅x for the four-vector product, and a periodic box of volume V so that momenta are discrete. Topic 9.05 fixed the dynamics and left the algebra open: π = iħψ† is the momentum conjugate to ψ, so the bracket of ψ with π must be iħδ_ab δ³(x − y). Take it to be an anticommutator, {A, B} = AB + BA, and divide out iħ: (ψₐ(t, x), ψ†_b(t, y)) = δ_ab δ³(x − y), with (ψₐ, ψ_b) = 0. This is an assumption, judged by what follows. To read it in modes, multiply the expansion by e^(ipx/ħ)u†(p, s) and integrate over the box. Only bₚₛ survives, because u†(p, s)u(p, s′) = (2Eₚ/c)δₛₛ′ and u†(p, s)v(−p, s′) = 0; v† isolates d†ₚₛ the same way. The rule then gives (bₚₛ, b†ₚ′ₛ′) = (dₚₛ, d†ₚ′ₛ′) = δₚₚ′δₛₛ′, every other pair anticommuting, derived. Dimensions agree: ψ†ψ is a number density, so ψ is in m^(−3/2), matching δ³ in m⁻³, and b and d are dimensionless.

The operator d† multiplies the negative-frequency wave v e^(ipx/ħ), and nothing in the field says whether d or d† should annihilate the vacuum. For a commutator the choice is costly: [d, d†] = −1 (topic 9.06), and swapping the roles of d and d† makes the norm positive only by sending the energy the wrong way. An anticommutator is symmetric, (d, d†) = (d†, d), so relabelling d̃ = d† gives (d̃, d̃†) = 1 again: for fermions the particle-hole swap is a canonical transformation, and for bosons it never is. Choose bₚₛ|0⟩ = dₚₛ|0⟩ = 0. Then ⟨0|d d†|0⟩ = ⟨0|(1 − d†d)|0⟩ = +1, and every product of creators acting on |0⟩ has norm 1, or 0 when it fills a mode twice. One mode pair (p, s) has exactly four states, none, an electron, a positron or both, so a box holding N momenta, each with two spins, has a Fock space of 4^(2N) dimensions whose norm is positive definite.

Topic 9.05 derived H = Σₚₛ Eₚ(b†b − d d†) before any algebra was chosen, the cross terms between b and d† having died by u†(p, s)v(−p, s′) = 0. Now reorder: d d† = 1 − d†d, so H = Σₚₛ Eₚ(b†b + d†d) − Σₚₛ Eₚ, derived. Because [d†d, d†] = d†(1 − d†d) = d† for fermions too, [H, d†ₚₛ] = +Eₚd†ₚₛ: a positron raises the energy by Eₚ ≥ mc², exactly as an electron does, and |0⟩ is the ground state. In topic 9.06, d d† = d†d − 1 left −Eₚd†d, and with it negative norms or no floor. Take one mode pair with p = 1.00 MeV/c, so Eₚ = 1.123 MeV. Before normal ordering its four states have energies Eₚ(n_b + n_d − 1): −1.123, 0, 0 and +1.123 MeV. After it they are 0, 1.123, 1.123 and 2.246 MeV, with the same spacings, because only a constant went. The momentum reorders the same way into P = Σₚₛ p(b†b + d†d), the constant Σ p vanishing by symmetry, so the positron carries +p although d† multiplies the wave of momentum −p.

The charge reorders too: Q = q Σₚₛ(b†b + d d†) = q Σₚₛ(b†b − d†d) + q Σₚₛ 1, derived, and since [Q, d†] = −q d†, each d-quantum carries −q = +e for the electron field, where q = −e. Now read the constants: −Σₚₛ Eₚ and q Σₚₛ 1 are the energy and charge of every negative-energy level filled once. Mode by mode, the unordered operators are hole theory's bookkeeping: the empty mode pair carries energy −Eₚ and charge q, one sea electron, and adding a positron brings both to zero, that electron removed. Normal ordering, which moves annihilators to the right with a minus sign for each fermion swap, :d d†: = −d†d, subtracts the sea. What Unit 8 lacked is now in hand: a vacuum that is one state of zero energy and charge, positrons as quanta in their own right, electrons and positrons treated alike, and no infinitely charged sea that must be declared unobservable. The measured check: b and d quanta share Eₚ, so their masses agree, and |m(e⁺) − m(e⁻)|/m is measured below 8 × 10⁻⁹; annihilation at rest is seen to yield two 511 keV photons.

Put both entries equal in (b†ₚₛ, b†ₚ′ₛ′) = 0 and (b†ₚₛ)² = 0: a second electron in a filled mode gives the zero vector, not a state, derived. The occupation n = b†b then obeys n² = b†(1 − b†b)b = n, so its only eigenvalues are 0 and 1. For two modes, moving ψ(x)ψ(y) through the creators gives ⟨0|ψ(x)ψ(y)b†₁b†₂|0⟩ = φ₂(x)φ₁(y) − φ₁(x)φ₂(y), a Slater determinant of the waves φₖ = √(c/2EₖV) uₖ e^{(−ipₖx/ħ)}. Quantum Mechanics II postulated that antisymmetry; here the anticommutator is the only assumption. Two limits check it: with one quantum, ⟨0|ψ(x)b†ₖ|0⟩ = φₖ(x), Unit 4's plane wave normalised to one electron in V, and for p ≪ mc the upper components of the b part become the non-relativistic electron field Unit 10 starts from. A thermal consequence follows at once. With K = H − μN and N = Σ(b†b − d†d), each mode is a two-state system, so ⟨b†b⟩ = 1/(eˣ + 1) with x = (Eₚ − μ)/k_B T, and ⟨d†d⟩ takes Eₚ + μ instead: Fermi-Dirac occupations capped at 1, plotted in the figure.

At unequal times, insert the expansion and Unit 4's spin sums Σₛ uū = p̸ + mc and Σₛ vv̄ = p̸ − mc: (ψ(x), ψ̄(y)) = Σₚ (c/2EₚV)[(p̸ + mc)e^(−ipz/ħ) + (p̸ − mc)e^(ipz/ħ)], with z = x − y. Since iħγ^μ∂_μ acting on e^(∓ipz/ħ) gives ±p̸, this is (iħγ^μ∂_μ + mc)[D(z) − D(−z)], with D the mode sum of topic 9.04's scalar amplitude up to a constant factor, derived. Topic 9.04 showed that D(z) = D(−z) for spacelike z in the continuum limit, so the bracket and its derivatives vanish there. At equal times the bracket returns γ⁰δ³(x − y), the canonical rule again. Without the d† term it would be (iħγ^μ∂_μ + mc)D(z) alone, which falls off over a reduced Compton wavelength, ħ/mc = 386 fm, but is not zero. The fields anticommute rather than commute at spacelike separation, which is harmless, because they change fermion number and are not observables. Observables are bilinears, and [WX, YZ] = W{X, Y}Z − WY{X, Z} + {W, Y}ZX − Y{W, Z}X writes their commutator through anticommutators, all zero there, so observables commute at spacelike separation, derived.