Dirac's sea and the radiative catastrophe

Units are SI, and where a four-vector appears the metric is diag(1, −1, −1, −1); α with an index or a dot is a Dirac matrix, α alone the fine-structure constant. Minimal coupling gives H = cα⋅(p − qA) + βmc² + qφ, so a radiation field in Coulomb gauge acts through H′ = −qcα⋅A. Now use βαᵢβ = −αᵢ. At rest H₀ = βmc², so β is the sign of the energy, and for two positive-energy rest spinors ⟨+|αᵢ|+⟩ = ⟨+|βαᵢβ|+⟩ = −⟨+|αᵢ|+⟩ = 0: every matrix element of the vertex from a resting positive-energy electron ends at negative energy. At momentum p, sandwich αₓ between the projectors Λ± = (1 ± H₀/E)/2. Topic 8.01's H₀αₓ + αₓH₀ = 2cpₓ gives Λ₊αₓΛ₊ = (cpₓ/E)Λ₊, and since αₓ² = 1, a fraction vₓ²/c² of the weight αₓ carries out stays at positive energy while 1 − vₓ²/c² lands in the sea. Schrödinger's vertex p⋅A/m has no route to negative energy; for a slow electron, Dirac's is almost all route.

Unit 7 puts the ground level at E₁ₛ = mc²√(1 − (Zα)²), and the negative-energy continuum runs from −mc² downwards without limit. A one-photon jump to a state of energy −|E′| releases ħω = E₁ₛ + |E′|: at least 1.0220 MeV less 13.6 eV for hydrogen, and still 0.890 MeV for Z = 92. A free electron cannot make the jump with one photon, because |p − p′| ≤ |p| + |p′| is less than (E + |E′|)/c = ħω/c, so the photon's momentum can never be matched; with two photons it can, so even a free electron would be unstable. In the atom the nucleus takes the excess, at least about mc, through its Coulomb field. Nothing stops the descent at the first landing: near the nucleus the electron at −|E′| can radiate again and land lower, and since |E′| has no maximum, neither does the energy one electron can give up. An atom in this theory has no ground state.

Count factors before doing any integral. The photon vertex costs e in amplitude, α in the rate. The momentum, at least mc, must come from one Coulomb exchange, Ṽ(Q) = −4πZαħ³c/Q², which at Q of order mc costs Zα in amplitude. The exchange happens where the electron is, which brings in the probability density at the nucleus, |ψ₁ₛ(0)|² = (Zαmc/ħ)³/π. Every other scale is mc, so Γ = C α(Zα)⁵ mc²/ħ, a scaling this lesson derives: a pure number times mc²/ħ, in s⁻¹, and vanishing as Z⁵ when Z → 0, as the free-electron argument above requires. The coefficient needs the full golden-rule sum, in which the Coulomb exchange may come before the photon or after it. Gauge invariance needs both orderings, and either alone gives a wrong answer. Summed numerically for this lesson, the two orderings reproduce the spectrum on the formula card, matching the quoted closed form f(w) to better than one part in 10⁴. Into the top band, |E′| from mc² to 2mc², the result is Γ = 0.192 α⁶mc²/ħ = 2.2 × 10⁷ s⁻¹ for hydrogen: a lifetime of 45 ns.

Set the numbers beside measurement. Hydrogen's 2p level decays in a measured 1.60 ns, as the golden rule's 1.596 ns predicts, so the top band alone would empty the ground state only 28 times more slowly than an excited level empties. Deeper bands make it worse: f(w) grows like w, so the band from mc² down to wmc² carries a rate that rises without limit, eventually as w²/π in units of α⁶mc²/ħ. The empty states down to −5.5mc² already drain the 1s level as fast as the 2p level decays, and the sum over the whole continuum diverges. Against that, hydrogen formed 13.8 billion years ago, 4.4 × 10¹⁷ s, is still most of the ordinary matter in the universe, and no 1 MeV glow rises from a flask of hydrogen gas. Even the top band alone misses by 25 orders of magnitude. Making the transitions slower would not rescue the theory; they have to be forbidden outright.

Dirac's 1930 proposal makes the vacuum the state in which every negative-energy solution, two spin states for each momentum, already holds one electron. The Pauli principle, used here from Quantum Mechanics II, then blocks every transition counted above, because every final state is full. The 1s level becomes the lowest level an extra electron can reach, and a free positive-energy electron is stable for the same reason. The sea is invisible to one-electron experiments because it is perfectly uniform. Its momenta pair p with −p, so it carries no momentum and, since a negative-energy state moves with velocity −c²p/|E|, no current; its spins pair, so it has no spin; and a uniform background pushes nothing in any particular direction. A semiconductor shows the same structure in miniature: a filled valence band carries no current, and only excitations across the gap are seen. Silicon's gap is 1.12 eV; Dirac's is 2mc² = 1.022 MeV.

Exclusion works only if the sea is really full, so count it. Each phase-space cell (2πħ)³ holds two electrons, so up to a momentum cutoff Λ the density is n = Λ³/(3π²ħ³): 5.9 × 10³⁵ m⁻³ for Λ = mc, a charge density of −9.4 × 10¹⁶ C m⁻³, and infinite as Λ → ∞. The energy density, −(1/π²ħ³)∫p²√(p²c² + m²c⁴) dp up to Λ, is −6.05 × 10²² J m⁻³ at Λ = mc and grows as −cΛ⁴/(4π²ħ³). Dirac removes both by definition: charge, energy and momentum are measured from the filled sea, Q_obs = Q − Q_vac, and Maxwell's equations are sourced by the difference. The one-electron equation now describes one electron moving above infinitely many, a many-body theory written as a one-body one. The sea stops being invisible once it is disturbed: its polarisation by a proton lowers hydrogen's 2s level by about 27 MHz of the measured 1058 MHz Lamb shift, quoted. Taking an electron out is topic 8.04; the full list of costs is topic 8.08.