The Klein step: transmission through a step higher than 2mc²

Work in SI units, with the metric (+, −, −, −) wherever a four-vector appears. The step is V(z) = 0 for z < 0 and V₀ for z > 0, where V = −eΦ is the electron's potential energy, so V₀ > 0 means the electrostatic potential falls by V₀/e. In each region H = cα_z p_z + βmc² + V is Unit 4's free Hamiltonian plus a constant, and at normal incidence Σ_z commutes with α_z and β, so an electron with spin up along z stays spin up: no spin-flip amplitudes. The lower block row of Hψ = Eψ, (E − V + mc²)χ = cσ_z p_z φ, gives each plane wave at once: ψ ∝ e^{(ip_z z/ħ)}(1, 0, κ, 0)ᵀ with κ = cp_z/(E − V + mc²). The incident wave has κ₁ = cp/(E + mc²), the reflected wave −κ₁, and the transmitted wave κ₂ = cq/(E − V₀ + mc²), with c²q² = (E − V₀)² − m²c⁴. Nothing is yet said about the sign of q; that choice is where the paradox will sit.

Integrate Hψ = Eψ from z = −δ to +δ. The only derivative sits in −iħcα_z ∂ψ/∂z, so −iħcα_z[ψ(δ) − ψ(−δ)] = ∫(E − V − βmc²)ψ dz, which vanishes as δ → 0 for a finite step. Since α_z² = 1, ψ itself is continuous, and nothing is demanded of ∂ψ/∂z. With amplitudes a, b, d for the incident, reflected and transmitted waves, the upper and lower components give a + b = d and κ₁(a − b) = κ₂d. Write r = κ₂/κ₁: then d/a = 2/(1 + r) and b/a = (1 − r)/(1 + r). Unit 4's current, quoted, gives j_z = cψ†α_zψ = 2cκ|amplitude|² for each real-κ wave, so R = ((1 − r)/(1 + r))² and T = 4r/(1 + r)², with R + T = 1 identically. Schrödinger's second matching rule is hidden in the lower component: when every energy is small beside 2mc², κ ≈ p/2mc, the lower component is −iħ/2mc times ∂/∂z of the upper one, so its continuity becomes continuity of ∂ψ/∂z, and r becomes q/p.

Everything about the far side lives in c²q² = (E − V₀)² − m²c⁴. Take an electron with 1.00 MeV of kinetic energy, E = 1.511 MeV. For V₀ < E − mc² = 1.00 MeV, q is real and E − V₀ > mc²: ordinary transmission, 0 < T ≤ 1. For 1.00 MeV < V₀ < E + mc² = 2.022 MeV, |E − V₀| < mc², so q and κ₂ are imaginary, the transmitted current vanishes and R = 1; the wave decays over ħc/√(m²c⁴ − (E − V₀)²), shortest at V₀ = E, where it is ħ/mc = 386 fm. Above 2.022 MeV, q is real again. Draw the spectrum to see why: on the left the continua are E ≥ mc² and E ≤ −mc²; on the right they are lifted to E ≥ V₀ + mc² and E ≤ V₀ − mc². Once V₀ − mc² passes E, the electron's energy lies inside the far side's negative-energy continuum. Since E ≥ mc², that needs V₀ > 2mc² = 1.022 MeV at the very least, the measured rest energy of an electron–positron pair. No Schrödinger step has a third regime: its kinetic energy has no lower branch to overlap.

In the Klein regime q = +|q| and q = −|q| both solve the equations, and the scattering problem must keep the one that leaves the step. The phase of e^{(i(qz − Et)/ħ)} moves at E/q, so q > 0 looks outgoing. But the far-side branch is E = V₀ − √(c²q² + m²c⁴), whose slope v_g = ∂E/∂q = c²q/(E − V₀) has the opposite sign to q. Stationary phase decides: a packet built from these waves peaks where z = v_g t, so a transmitted packet present on the right after the collision needs v_g > 0, hence q < 0. Then κ₂ = cq/(E − V₀ + mc²) is a negative over a negative, r > 0 and 0 < T < 1: the transparency of a high step that Klein reported in 1929. The current agrees: for any of these plane waves j_z/ρ = c²q/(E − V₀) = v_g. Choose q > 0 instead and r < 0, so R > 1 and T < 0: the transmitted current flows toward the step, a wave arriving from z = +∞ that the problem never contained. R + T = 1 in both readings. Within one-particle theory the two differ only in the boundary condition; topic 8.06 gives the R > 1 reading a second life as pair creation.

With q chosen by group velocity, write u = V₀ − E, which exceeds mc² in the Klein regime. The spinor ratio becomes r² = [(u + mc²)(E + mc²)] / [(u − mc²)(E − mc²)], a product of two factors above 1, so r > 1 throughout and r falls steadily as V₀ rises: r → ∞ at the threshold V₀ = E + mc², where T = 0, and r → √((E + mc²)/(E − mc²)) as V₀ → ∞. Because 4r/(1 + r)² decreases with r for r > 1, T climbs monotonically from 0 to T∞ = 2cp/(E + cp). For E = 2mc², cp = √3 mc² and T∞ = 0.928; at V₀ = 4mc², u = 2mc², r = 3 exactly and T = 0.750. Even a 10 eV electron, with cp = 3.20 keV, has T∞ = 0.0124: a sharp step 1 GeV high still passes 1.2% of its flux. This is the opposite of tunnelling, where T falls exponentially as the barrier rises; here the far side is not forbidden at all, and a taller step only moves the electron deeper into a continuum it may enter.

Dimensions: r is a ratio of energies times a ratio of energies, and cψ†α_zψ is m s⁻¹ times m⁻³, a flux. Non-relativistic limit: with K and V₀ small beside mc², r → √((K − V₀)/K) = q/p and T → 4pq/(p + q)², Schrödinger's step, while the third regime retreats beyond V₀ = K + 2mc², out of the non-relativistic model's reach. Massless limit: with m = 0, r = 1 in both open regimes, the window closes and T = 1 for every V₀ ≠ E at normal incidence. Graphene's carriers obey a two-dimensional massless Dirac equation with v_F ≈ 1.0 × 10⁶ m s⁻¹ in place of c, and Klein tunnelling through graphene p–n junctions was observed in 2009 (quoted). The sharp step is the weak point. To rise 2mc² within ħ/mc = 386 fm takes a field of 2.6 × 10¹⁸ V m⁻¹, and Sauter showed in 1931 that a step spread over many ħ/mc transmits exponentially less (quoted; topic 8.06 takes it up). And the transmitted wave lies in the negative-energy continuum, its lower components larger than its upper ones: what such an electron is, is the question for Dirac's sea.