The Klein paradox as pair production, and the critical field

SI units throughout; where a four-vector appears the metric is (+, −, −, −). In a region of constant potential energy V a free Dirac solution has E = V ± √(p²c² + m²c⁴), so the near side of a step offers electron states with E ≥ mc² and a sea with E ≤ −mc², while the far side, lifted by V₀, offers E ≥ V₀ + mc² and a sea with E ≤ V₀ − mc². Between mc² and V₀ − mc² the near side's positive continuum overlaps the far side's negative one: the Klein window, open only when V₀ exceeds 2mc² = 1.022 MeV. Topic 8.02's matching, quoted in one line, gives R = ((1 − r)/(1 + r))² and T = 4r/(1 + r)² with r = (q/p)(E + mc²)/(E − V₀ + mc²); for E = 2mc² and V₀ = 6mc², a positive transmitted momentum q gives r = −√5 = −2.24, R = 6.85 and T = −5.85. A Schrödinger step has no negative branch, hence no window. What is new here is not the algebra but the occupation: in Dirac's vacuum every far-side state in the window is filled and every near-side one is empty.

With q positive, the far-side wave has group velocity dE/dq = −qc²/|E − V₀|, pointing at the step: a sea electron arriving from the right. In the one-particle theory nothing could be arriving; in hole theory the far sea is full and its electrons strike the step all the time. In the window they find empty states beyond it and pass through, each leaving to the left as an ordinary electron and leaving a vacancy that topic 8.04 identifies as a positron, charge +e, rest plus kinetic energy V₀ − E, carried right by the same field that drives the electron left. The flux identity R + T = 1, charge conservation for the Dirac current, reads R − 1 = −T: the excess electrons on the incident side match the positrons beyond the step pair for pair, since an electron current running left is a positron current running right. Each pair costs E + (−E) = 0. And the sea electrons come whether or not anything is incident, so the step emits pairs from the vacuum: a rate the one-particle solution, which fixes only ratios of fluxes, cannot give.

To get a rate, count occupied states instead of solving for one wavefunction. In one dimension a filled band delivers 1/(2πħ) particles per second per unit energy per spin to the step, so sea electrons arriving from the right, each passing with the group-velocity transmission T_g = 4|r|/(1 + |r|)², between 0 and 1, give (1/2πħ)∫T_g dE pairs per second, the integral running over the window. The count needs every sea state occupied at once, a statement about infinitely many particles that no single Dirac wavefunction can make; Unit 9's quantised field reproduces it, quoted. Its size: mc²/(2πħ) = 1.24 × 10²⁰ s⁻¹, so a sharp step with a window a few mc² wide would release of order 10²⁰ pairs a second. An incident electron does not drive the process. By exclusion it can only occupy, and so block, the outgoing state a pair electron would have used. Klein's 1929 result was never really a paradox of reflection: it was the first sign that a one-particle wave equation was describing a many-particle vacuum, the case topic 8.08 makes in full.

No electrode is sharper than an atom, so ask what a real slope must do. Spread V₀ over a width w and the field is E = V₀/(ew); the window still needs V₀ ≥ 2mc², but the field now sets how far apart the two continua lie at fixed energy. The natural scale is the field that raises an electron's potential energy by mc² across its reduced Compton wavelength, ħ/mc = 0.386 pm: eE_c(ħ/mc) = mc², so E_c = m²c³/(eħ) = 1.32 × 10¹⁸ V m⁻¹. Units: kg² m³ s⁻³ over C J s is N C⁻¹, a field. Limits: as ħ → 0 it diverges, so a classical field never makes pairs, and it grows as m², so the muon's is 4.3 × 10⁴ times larger, 5.7 × 10²² V m⁻¹. Against it, hydrogen's field at the Bohr radius, 5.14 × 10¹¹ V m⁻¹, is α³E_c; the strongest static fields, at field-emission tips, are about 10¹⁰ V m⁻¹; and the most intense laser focus measured, about 10²³ W cm⁻², peaks near 9 × 10¹⁴ V m⁻¹, some 1500 times short.

Let a uniform field E raise the electron's potential energy as V(x) = eEx. At energy ε the squared Dirac equation fixes the local momentum by (ε − V)² = p²c² + m²c⁴, spin entering only at the next order in ħ. Where |ε − V| is below mc², p is imaginary: a forbidden stretch d = 2mc²/(eE) = 2(E_c/E)ħ/mc separates the near side's positive continuum from the far side's sea. Quantum Mechanics II's WKB exponent gives P ≈ exp(−(2/ħ)∫|p|dx); with u = ε − eEx the integral is (1/ceE)∫√(m²c⁴ − u²)du from −mc² to mc², a half-disc of area πm²c⁴/2, so P ≈ exp(−πm²c³/(eħE)) = exp(−πE_c/E), derived. Sauter's 1931 exact solution, quoted, agrees. Checks: the exponent is dimensionless, equals (π/2)d/(ħ/mc), and gives e^(−π) = 0.043 at E = E_c. A transverse momentum p_⊥ adds p_⊥²c² to m²c⁴; summing over p_⊥ and spin with the quantised field, Schwinger in 1951 obtained a rate Γ ≈ (eE)² exp(−πE_c/E)/(4π³ħ²c) pairs per unit volume per unit time, quoted, whose exponent is this one.

Coulomb's law divided by E_c gives E(r)/E_c = Zα(ħ/mcr)², derived, so a nucleus exceeds E_c inside r = √(Zα)ħ/mc: 33 fm for hydrogen, 299 fm for lead, and about 1800E_c at lead's surface. Yet a lone nucleus emits no positrons, and energy says why. A static field can only release a pair that costs nothing; the cheapest puts the electron in the lowest empty level, E(1s), and the positron at rest far away, at mc², so emission needs E(1s) + mc² ≤ 0, derived. The strong-field region is smaller than ħ/mc, and confining an electron inside it costs more kinetic energy than the potential returns: Unit 7's point-nucleus level, mc²√(1 − Z²α²), stays positive, and for lead the cheapest pair costs 1.80mc² = 920 keV. Topic 7.09's finite-nucleus 1s, quoted, reaches −mc² near Z = 172; beyond it an empty 1s would fill from the sea and release a positron, the Klein window in Coulomb form. Uranium-uranium collisions, a combined Z = 184 for about 10⁻²¹ s, gave no confirmed signal, quoted.