Plane-wave spinors u(p, s) and v(p, s) at any momentum

Units are SI with c and ħ kept; x^μ = (ct, x), the metric is diag(1, −1, −1, −1), and p_μx^μ = Et − p⋅x. Put ψ = u e^{(−ip_μxμ/ħ)} into iħ∂ψ/∂t = (−iħc α⋅∇ + βmc²)ψ: the derivatives return E and p, leaving Eu = (cα⋅p + βmc²)u. In the Dirac representation, with u = (u_A ; u_B), this is (E − mc²)u_A = cσ⋅p u_B and (E + mc²)u_B = cσ⋅p u_A. Multiply the first by E + mc² and use the second: (E² − m²c⁴)u_A = c²(σ⋅p)²u_A = c²p²u_A, so E = ±√(p²c² + m²c⁴), the Klein-Gordon dispersion recovered. For positive E solve the second block, never the first: E + mc² is at least 2mc², while E − mc² vanishes at rest, where the first block reads 0 = 0. The upper pair is then free, any two-spinor χₛ, and the lower pair is fixed by it. Two independent χₛ give two spin states at every momentum: derived, not assumed.

Because (σ⋅p)² = p², σ⋅p/p is unitary, so |u_B|/|u_A| = cp/(E + mc²) for either spin. With γ = E/mc² and β = v/c it is √((γ − 1)/(γ + 1)) = γβ/(γ + 1) = β/2 + β³/8 + …: 0.0501 at 0.1c. At 0.6c it is exactly 1/3, and for E ≫ mc² it tends to 1 − mc²/E, so an electron of 45.6 GeV, half the measured Z mass, has pairs equal to 1.1 parts in 10⁵. In rapidity η = artanh β it is tanh(η/2), a half angle Unit 5 traces to spinor boosts. The lower pair's share of the density is (E − mc²)/2E, near v²/4c² when slow. At v = αc, the speed scale of hydrogen, the share is α²/4 = 1.33 × 10⁻⁵. The measured n = 2 fine-structure splitting of hydrogen, 10.969 GHz or 4.54 × 10⁻⁵ eV, is 1.33 × 10⁻⁵ of the 3.40 eV binding energy: relativistic corrections are as small as the lower pair says. That fixes a scale, not the splitting, which Units 6 and 7 compute.

For the negative branch write ψ = v e^{(+ip_μxμ/ħ)} with E positive. Now iħ∂/∂t returns −E and −iħ∇ returns −p, so the equation is −Ev = (−cα⋅p + βmc²)v, with blocks (E + mc²)v_A = cσ⋅p v_B and (E − mc²)v_B = cσ⋅p v_A. The factor that never vanishes now multiplies the upper pair, so the lower pair is free: v_B = ηₛ and v_A = cσ⋅p ηₛ/(E + mc²). Nothing else changes: v is u with the roles of the pairs exchanged, and at p = 0 it reduces to √(2mc)(0 ; ηₛ), the negative-energy rest spinors of the last lesson. The labels need care. The wave carries momentum −p and energy −E, so v(p, s) is an eigenvector of H(−p), not of H(p). With the usual choice ηₛ = −iσ_y χₛ*, which sends (1, 0) to (0, 1), v(0, ↑) has Σ_z = −1. Writing p and s on v records the momentum and spin a positron will have, a convention justified only in Unit 8.

Any multiple of a solution is a solution, so normalisation is a decision. With u = N(χₛ ; cσ⋅p χₛ/(E + mc²)), u†u = N²[1 + c²p²/(E + mc²)²] = 2EN²/(E + mc²), since c²p² = (E − mc²)(E + mc²). Choosing u†u = 2E/c gives N = √((E + mc²)/c), and then ū u = |u_A|² − |u_B|² = 2mc at every p; with c = 1 the same choice reads u†u = 2E and ū u = 2m. Why E: a box holding one particle, seen moving, is contracted by γ = E/mc², so its density is γ times larger. The current agrees: c u†αu = 2pc, so flux over density is pc²/E, the particle's velocity, and (c u†u, c u†αu) = 2c(E/c, p) is a multiple of the four-momentum, covariance being assumed until Unit 5. Other choices rescale u by a constant k and every bilinear by k²: ū u = 1 takes k² = 1/(2mc) and has no m → 0 limit; u†u = 1 takes k² = c/2E, and Unit 5 will show that this u is no longer the boosted rest spinor. The factor returns wherever states are counted: one particle in a volume V is u e^{(−ip_μxμ/ħ)}/√(2EV/c).

Inside the u pair, u†(p, s)u(p, s′) = N²χₛ†[1 + c²(σ⋅p)²/(E + mc²)²]χₛ′, and (σ⋅p)² = p² turns the bracket into a number, so orthonormal χ↑, χ↓ give (2E/c)δₛₛ′; the same lines give v†v = (2E/c)δₛₛ′, ū u = 2mc δₛₛ′ and v̄ v = −2mc δₛₛ′. Across the pairs, ū(p, s)v(p, s′) = [N²c/(E + mc²)][χₛ†σ⋅p ηₛ′ − (σ⋅p χₛ)†ηₛ′], and σ⋅p is Hermitian, so the bracket vanishes for every s, s′, p and every choice of η: derived. The unbarred product adds the two terms instead: u†(p, s)v(p, s′) = 2χₛ†(σ⋅p)ηₛ′, which is 0.90 mc for spin up at cp = (0.45, 0, 0.60) mc². The orthogonality the spectral theorem guarantees is between eigenvectors of one Hermitian H(p), and those are u(p, s) and v(−p, s′), whose † product is zero. So u(p, ↑), u(p, ↓), v(−p, ↑) and v(−p, ↓) form an orthogonal basis of the four-component space, and the two u's alone span only half of it.

Dimensions: u†u = 2E/c is a momentum, so each entry of u carries √(kg m s⁻¹), and cp/(E + mc²) is a pure number, as the ratio of two parts of one column must be. With the measured electron mass, mₑc² = 0.51099895 MeV, the unit √(mc) used in the figure is 1.653 × 10⁻¹¹ √(kg m s⁻¹). The rest limit: p → 0 returns √(2mc)(χₛ ; 0) and √(2mc)(0 ; ηₛ), the four rest spinors of the last lesson, with u†u = 2mc. The slow limit: u_A → √(2mc)χₛ, a Pauli spinor whose density divided by 2mc is the one Quantum Mechanics II normalised to 1, and u_B → (σ⋅p/2mc)u_A, the small component Unit 6 eliminates to find g = 2. The massless limit: |u_A| = |u_B| and ū u = 0, while u†u = 2E/c stays finite, so the 2E/c convention has a massless limit where ū u = 1 has none; it is the limit Unit 5 reaches through the massless Weyl equations.