What hole theory cannot do: the case for quantised fields

Units are SI with c and ħ kept, the metric is diag(1, −1, −1, −1), and the electron's charge is −e. Quoted in one line from topics 8.03 to 8.07: filling every negative-energy state blocks the radiative fall, a hole is a positron of charge +e and mass m, and a pair costs 2mc² = 1.022 MeV. The price is a sea of real electrons, two per momentum state, of density Λ³/(3π²ħ³) below a momentum cutoff Λ. At Λ = mc one cubic micrometre of vacuum holds 5.87 × 10¹⁷ of them, a charge of −0.094 C whose Coulomb field at 1 m would be 8.4 × 10⁸ V m⁻¹. Hole theory declares that this charge makes no field, that the sea's energy makes no gravity, and that sea electrons exert no forces on one another: three subtractions made by hand, none produced by any rule inside the theory. Nor are the subtracted numbers properties of anything. Raise the cutoff to 10mc and the charge grows a thousandfold and the energy 6.0 × 10³-fold. A quantity that scales with an arbitrary cutoff as Λ³ is not being measured; it is being removed.

Unit 9 expands the Dirac field over Unit 4's spinors, ψ = Σ [b u e^{(−ip⋅x/ħ)} + d† v e^(ip⋅x/ħ)], with b destroying an electron and d a positron (stated; built in topic 9.05). Before any algebra is chosen its energy is H = Σ Eₚ (b†b − d d†), over momenta and both spins. With the anticommutator d d† + d†d = 1 of topic 9.07, H = Σ Eₚ (b†b + d†d) − Σ Eₚ. The constant −Σ Eₚ is hole theory's sea energy term for term, but it is now a number times the identity. It commutes with every operator, so it cancels from every energy difference and every transition amplitude: a pair at rest costs 2mc² = 1.022 MeV whatever the cutoff, the measured threshold. Discarding it, normal ordering, chooses an origin of energy and removes no electrons, because the vacuum, b|0⟩ = d|0⟩ = 0, contains none. One-mode check: H = E(b†b + d†d − 1) has levels −E, 0, 0 and +E, so a particle costs E and a pair 2E above the vacuum. The same algebra turns the sea's charge into the c-number −e Σ 1.

In hole theory an electron is an occupied positive-energy state and a positron a missing negative-energy one. Topic 8.05's ψ^c = iγ²ψ* maps every negative-energy solution of charge −e onto a positive-energy solution of charge +e, so the same algebra describes a sea of positrons whose holes are electrons. Both accounts predict the same physics, nothing in the theory selects one, and their unsubtracted vacuum charges are −eN and +eN. Heisenberg's 1934 remedy can be derived. Write Q = −e(Σ nₖ − 2cN), with nₖ = 0 or 1 over all 2N states, N in each band. Conjugation sends nₖ to 1 − nₖ′ for the conjugate state k′, and Q must become −Q; that forces c = ½, so Q = −e Σ(nₖ − ½): zero for the filled sea, −e for an electron, +e for a hole, at any cutoff. The symmetry it restores is measured: positron and electron g-factors differ by a fraction (−0.5 ± 2.1) × 10⁻¹² (quoted). Requirement: particles and antiparticles on equal terms, as b and d† enter ψ, with Q = −e Σ(b†b − d†d) after ordering (stated; topic 9.07).

Unit 3's Klein-Gordon particles also have negative energies with no floor (topic 3.09). A sea stops a fall only through exclusion, and the occupation factor shows what bosons do instead. Quantum Mechanics I's ladder algebra gives |⟨n + 1|a†|n⟩|² = n + 1, so the golden-rule rate into a mode already holding n bosons is (1 + n) times the rate into an empty one; for fermions c†c† = 0 makes it 1 − n, zero for a full state. Derived. Limit: n = 0 returns the bare rate, and 1 + n is the stimulated emission measured in every laser. A mode holding one negative-energy pion doubles the rate at which a second falls in, and each addition lowers the energy further: a boson sea is a drain, not a floor. Yet spin-0 particles have antiparticles: π⁺ and π⁻ share the mass 139.570 MeV, K⁰ and K̄⁰ have masses equal to 6 parts in 10¹⁹, and e⁺e⁻ collisions make π⁺π⁻ pairs above 279.14 MeV (all quoted). Requirement: energy bounded below without a sea. Quantised with commutators, the complex scalar field has H = Σ ħωₚ(a†a + b†b) plus a positive constant, with b† creating the antiparticle (stated; topic 9.03).

For spin 0 the equation is second order, so ψ and ∂ψ/∂z are continuous at a sharp step. With incident momentum p, momentum q beyond it and c²q² = (E − V₀)² − m²c⁴, matching gives R_B = ((p − q)/(p + q))² and T_B = 4pq/(p + q)², summing to 1 since the spin-0 flux is proportional to momentum. Derived. Above V₀ = E + mc² a packet leaves the step only if the group velocity c²q/(E − V₀) is positive, so q is negative: R_B = ((p + |q|)/(p − |q|))² exceeds 1, T_B is negative, and a pole sits at V₀ = 2E, where |q| = p. For E = 2mc² and V₀ = 5.5mc², R_B = 9.83. Beyond the step the density (E − V₀)|φ|²/mc² is negative: the transmitted wave is antiparticles leaving. Spin ½ under the same rule gives R_F = 0.159 (topic 8.02, quoted), never above 1, and for m = 0 zero at every step: the normal-incidence Klein tunnelling measured in graphene in 2009 (quoted). Checks: R_B is a ratio of momenta, and returns to 1 where the window opens and as V₀ → ∞. Quoted field reading (Hansen and Ravndal, 1981): R_B − 1 counts pairs the incident boson stimulates; exclusion denies a spin-½ step that stimulation.

Hole theory describes γ → e⁻e⁺ beside a nucleus with a one-particle equation by keeping the number of electrons, sea included, fixed and infinite: a pair is one sea electron lifted, an annihilation one dropped back. What is observed does change. Q = −e(N₋ − N₊) is conserved, and the electron's lifetime exceeds 6.6 × 10²⁸ years (quoted), but N₋ + N₊ changes by 2 with every pair, and the photon number, which never had a sea, changes with every emission. An amplitude between states of different particle number needs a state space containing them all. So the audit ends in five requirements. A Fock space, the sum of the N-particle spaces for N = 0, 1, 2, …, with operators that change N (topic 9.02). An empty vacuum whose infinite constants are ordering terms (9.02, 9.07). Particles and antiparticles on equal terms (9.03, 9.07). Energy bounded below without a sea: commutators leave the Dirac energy Σ Eₚ(b†b − d†d) unbounded below (9.06), anticommutators repair it (9.07). And a rule fixing which particles obey exclusion: the spin-statistics connection (9.09).