Large and small components: eliminating χ to order v/c

Units are SI throughout, and where a four-vector appears the metric is (+, −, −, −). Quote Unit 4's minimally coupled Hamiltonian, H = cα⋅π + βmc² + V, with π = p − qA and V = qΦ. In the Dirac representation β = diag(1, −1) and α carries σ in its off-diagonal blocks, so βmc² and V are even, block-diagonal, while cα⋅π is odd, the only term linking the upper pair of components to the lower. A positive-energy electron's spinor turns at mc²/ħ = 7.76 × 10²⁰ s⁻¹, a rate with no atomic physics in it, so write ψ = e^{(−imc²t/ħ)}(φ, χ). The block rows become iħ∂φ/∂t = cσ⋅πχ + Vφ and iħ∂χ/∂t = cσ⋅πφ + (V − 2mc²)χ. The phase cancelled mc² in the upper row and doubled it in the lower: χ now sits 2mc² = 1.022 MeV uphill of φ. That gap is measured, not assumed. It is the threshold for creating an electron-positron pair, and a pair annihilating at rest emits two photons of 511.0 keV each.

Collect the χ terms: (2mc² + iħ∂/∂t − V)χ = cσ⋅πφ, still exact. Now count powers of v/c. Acting on a state of energy E, iħ∂/∂t returns E′ = E − mc², and in an atom E′ and V are both of the size of the kinetic energy, ½mv²: tens of eV in hydrogen against 1.022 MeV. The right side is of order c × mv × φ, and the left is 2mc²χ up to a relative (v/c)². Keeping the leading balance gives χ ≈ (σ⋅π/2mc)φ, derived, correct through order v/c with errors of order (v/c)³. Three checks. Dimensions: σ is a pure number and π/mc a ratio of momenta, so χ shares φ's unit, m^(−3/2). Rest limit: a free electron at rest has a uniform φ, on which p gives zero, so χ = 0, Unit 4's rest spinor with its lower pair empty. Gauge: under A → A + ∇Λ the whole spinor gains e^(iqΛ/ħ), and π acting on e^(iqΛ/ħ)φ returns e^(iqΛ/ħ)πφ, so χ picks up exactly φ's phase, as the lower half of one spinor must.

With A = 0, σ⋅p is Hermitian and (σ⋅p)² = p², so the weight takes one line: ⟨χ|χ⟩ = ⟨φ|p²|φ⟩/(2mc)² = ⟨T⟩/2mc², derived. In amplitude that is p/2mc, half of v/c at this order. Hydrogen's 1s state makes it hold point by point. With φ = R(r)ξ, R ∝ e^(−r/a₀) and ξ a constant spinor, σ⋅pφ = −iħR′(σ⋅r̂)ξ = i(ħ/a₀)R(σ⋅r̂)ξ, and since ħ/(mca₀) = α, χ = i(α/2)(σ⋅r̂)φ. Because (σ⋅r̂)² = 1, |χ|/|φ| = α/2 = 3.65 × 10⁻³ at every radius, and χ holds α²/4 = 1.33 × 10⁻⁵ of φ's probability. Measurement agrees: by the virial theorem ⟨T⟩ is the measured ionisation energy, 13.6 eV, and 13.6 eV ÷ 1.022 MeV = 1.33 × 10⁻⁵. Notice what σ⋅r̂ does. It is odd under r → −r and carries one unit of orbital angular momentum, so χ is a p-wave while φ is an s-wave, both with j = ½: the two halves of a Dirac spinor carry opposite orbital parity, as Unit 4's parity operator βP requires.

Put χ into the upper row: iħ∂φ/∂t = cσ⋅π(σ⋅π/2mc)φ + Vφ = [(σ⋅π)(σ⋅π)/2m + V]φ, derived, a closed equation for two components. With A = 0 it is Schrödinger's equation for each spin component, the one-particle non-relativistic limit with nothing left over, as an order-v/c result should be. Follow where the kinetic energy came from. The coupling cσ⋅π is of order c × mv, χ is of order v/2c, and their product is of order ½mv², the size of V. The current says the same. Because α is odd, j = cψ†αψ = c(φ†σχ + χ†σφ) contains no φ†φ term; with χ inserted and A = 0 it becomes (ħ/2mi)(φ†∇φ − (∇φ†)φ) + (ħ/2m)∇×(φ†σφ), Schrödinger's current plus a divergence-free spin term. Every bit of it arrives through χ, which in hydrogen holds 1.3 × 10⁻⁵ of the norm. Expanding (σ⋅π)² when A ≠ 0, where the components of π fail to commute, is the next topic's work.

The leading result replaced 2mc² + E′ − V by 2mc², dropping a fraction (E′ − V)/2mc² whose average is ⟨T⟩/2mc², of order (v/c)²/4. Two exact solutions show its size, and that its sign is not what a glance at the bracket suggests. For a free electron both rows are algebraic: eliminating χ exactly gives E′(E′ + 2mc²) = c²p², which is E² = p²c² + m²c⁴, so elimination itself loses nothing, and |χ|/|φ| = cp/(E + mc²) lies below p/2mc. Unit 7's exact 1s state, quoted, has |χ|/|φ| = Zα/(1 + γ) with γ = √(1 − Z²α²), above Zα/2. Near the nucleus the true large component is steeper than Schrödinger's, but that extra gradient and the −V in the bracket offset exactly, leaving a bracket of 2mc² + E′, below 2mc² because a bound E′ is negative. Both obey one rule, exact = leading ÷ (1 + E′/2mc²), with E′ = +T when free and minus the binding energy when bound. So the (v/c)² terms must be computed, as topics 6.05 and 6.06 do. Their size is measured: hydrogen's 2P₃/₂ to 2P₁/₂ interval, 10.97 GHz, is 1.33 × 10⁻⁵ of n = 2's 822 THz binding, α²/4 again.

The bracket stops being 2mc² in four places. Heavy ions: for uranium Zα = 0.671, and Zα/2 = 0.336 falls 13% short of the quoted exact 0.386, so no order-v/c truncation is quantitative there. Close to a point nucleus: −V exceeds 2mc² inside r = Zαħ/2mc = Z × 1.41 fm. For hydrogen that is 1.41 fm, comparable to the proton's measured 0.84 fm charge radius, and the 1s electron spends 2.5 × 10⁻¹⁴ of its probability there: harmless at this order, although topic 6.08's Darwin term is a contact interaction at the nucleus, inside this very region. Fast fields: a field oscillating near 2mc²/ħ = 1.55 × 10²¹ s⁻¹ makes iħ∂/∂t as large as the gap, photons of 1.022 MeV can make pairs, and a one-particle equation is no longer honest. Negative energy: with E ≈ −mc² it is the upper row whose bracket is near −2mc², so there φ is the small component and the elimination runs in reverse.