Holes as positrons: charge, energy, momentum and spin

Units are SI with c and ħ kept, the metric is diag(1, −1, −1, −1), and the electron's charge is −e, with e = 1.602 × 10⁻¹⁹ C. From topic 8.03, quoted in one line: the vacuum |sea⟩ has every negative-energy state filled and every positive-energy state empty. Its total charge and energy are infinite, so only differences from it can be observed, and the vacuum's values are defined to be zero. Treat the sea electrons as independent, so that charge, energy, momentum and J_z are sums of one-particle values oₖ over occupied states. Empty the state k and each sum loses exactly one term: ⟨O⟩ in |sea − k⟩ minus ⟨O⟩ in |sea⟩ is −oₖ. That line is the whole derivation. An emptied state of charge −e and energy −|E| leaves a hole of charge +e and energy +|E|, and since |E| ≥ mc², every hole costs at least 0.511 MeV: no state lies below the vacuum.

The rule gives the hole momentum Pₕ = −p. Velocity needs one more thought, because an absence travels with the packet it is missing. A negative-energy packet moves at the group velocity of its own branch, v = ∂(−√(p²c² + m²c⁴))/∂p = −pc²/|E|, against its own momentum. The hole therefore moves at −pc²/|E|, which is exactly Pₕ c²/Eₕ, the velocity of a positive-energy particle of momentum Pₕ and energy Eₕ. Its mass follows: Eₕ² − Pₕ²c² = |E|² − p²c² = m²c⁴. Take p = 0.75mc along +z. The emptied state has E = −1.25mc² and moves at 0.60c along −z, against p; the hole has Pₕ = 0.75mc along −z and Eₕ = 1.25mc² = 0.639 MeV, and moves at 0.60c along −z, along Pₕ. Limits: as p → 0 the hole sits at rest with energy mc², and for p ≪ mc, Eₕ ≈ mc² + Pₕ²/(2m), a free particle of mass m. All derived; measured positron and electron masses agree to better than 8 parts in 10⁹ (quoted).

Bookkeeping gave +e, and the equations of motion must agree. In fields ℰ and B each sea electron obeys dp/dt = −e(ℰ + v × B). Alone, a negative-energy electron is perverse: for p ≪ mc its velocity is −p/m, so dv/dt = +eℰ/m and it accelerates against the force on it, a negative inertial mass. The hole undoes both oddities. Its momentum Pₕ = −p changes as dPₕ/dt = +e(ℰ + v × B), where v is its own velocity Pₕ c²/Eₕ with Eₕ positive: a charge +e with positive inertia. Two minus signs, the negative energy and the removal, cancel. The current measured from the vacuum is −(−e)v = +ev, so in a magnetic field the hole's track curves the opposite way to an electron's at the same velocity. Derived; measured positron and electron charges cancel to 4 parts in 10⁸ (quoted). Anderson read his 1932 cloud-chamber track in exactly this way (quoted): the energy lost in a 6 mm lead plate, 63 MeV in and 23 MeV out, fixed the direction of travel, and the curvature then fixed the charge as positive.

Take p along z. Then Σ_z commutes with the free Hamiltonian (topic 4.08), the orbital L_z of a plane wave along z vanishes, and J_z = sħ with s = ±½ labels the state. The filled sea has J_z = 0, so removing sħ leaves −sħ: the hole has spin ½, in two states because either spin can be missing, with the projection reversed. Its momentum is reversed too, so the helicity 2S⋅P̂/ħ = 2(−S)⋅(−p̂)/ħ is unchanged: a hole left by an h = +1 state has h = +1. Unit 4's labels anticipated this. The plane wave v(p, s)e^{(i(Et − p⋅x)/ħ)}, with E = +√(p²c² + m²c⁴), carries momentum −p and energy −E, so its hole carries +p and +E, the positron's own values, which is why v is labelled by p. The spin label follows once a convention ties the two-spinor inside v to the hole's projection, the opposite of the state's own, and topic 8.05 makes that automatic with ψ^c = iγ²ψ*. Derived, apart from that convention, which is stated.

Lift a sea electron from (−|E₁|, p₁) into an empty state (E₂, p₂). Measured from the vacuum there are now an electron and a hole of energy |E₁| and momentum −p₁: charge zero, energy E₂ + |E₁| ≥ 2mc² = 1.022 MeV. In the frame where the pair's total momentum vanishes its energy is still at least 2mc², so (E₂ + |E₁|)² − |p₂ − p₁|²c² ≥ (2mc²)². A photon has (ħω)² − |ħk|²c² = 0 and no such frame, so one photon in free space cannot lift a sea electron: every state it could deliver it to lies on the cone E + |E₁| = c|p − p₁| from the emptied state, the lines of slope ±c in the figure, between the two branches, where there are no states. A body of mass M must take the momentum, and equating invariants gives ħωₜₕ = 2mc²(1 + m/M). For a ²⁰⁸Pb nucleus, Mc² = 193.7 GeV and the threshold is 1.022 MeV plus 2.7 eV; with a free electron as spectator it is 4mc² = 2.044 MeV. All derived.

Run the argument backwards. An electron dropping into a hole releases a free pair's energy and momentum, whose invariant mass is at least 2mc², so it cannot become one photon; from rest it most often gives two photons, back to back with mc² = 511 keV each, and a spin-triplet pair gives three (quoted). Measured, and quoted: PET scanners detect exactly that pair of 511 keV photons, and germanium spectra show the 2614.5 keV γ-ray of ²⁰⁸Tl with escape peaks at 2103.5 keV and 1592.5 keV, where a pair formed in the crystal and one or both annihilation photons left it. A bound electron can annihilate into one photon because its nucleus takes the recoil, which is observed but rare. The history, all quoted: Dirac (1930) read holes as protons; Oppenheimer (1930) replied that hydrogen's electron would then annihilate with its own proton within about 10⁻¹⁰ s; Weyl (1931) showed a hole has the electron's mass; Dirac (1931) proposed the anti-electron, and Anderson found it in 1932.