The non-relativistic limit: Schrödinger plus −p⁴/(8m³c²)

Units are SI and the metric is diag(1, −1, −1, −1) with x^μ = (ct, x), so p^μ = iħ∂^μ sends E to iħ∂/∂t and p to −iħ∇, and p_μp^μ = m²c² becomes (1/c²)∂²φ/∂t² − ∇²φ + (mc/ħ)²φ = 0. Letting c → ∞ in that form is tempting and wrong: the first term seems to vanish, the last blows up, and what is left, −∇²φ + (mc/ħ)²φ = 0, has no time evolution at all. The flaw is that every solution oscillates at least as fast as mc²/ħ, so (1/c²)∂²φ/∂t² is of order (mc/ħ)²φ: it cancels the mass term rather than vanishing. Remove that oscillation first. With φ = e^{(−imc²t/ħ)}ψ, ∂²φ/∂t² = e^{(−imc²t/ħ)}[∂²ψ/∂t² − (2imc²/ħ)∂ψ/∂t − (m²c⁴/ħ²)ψ]. Divided by c², the last piece cancels (mc/ħ)²ψ exactly; multiplying what remains by −ħ²/2m gives iħ∂ψ/∂t = −(ħ²/2m)∇²ψ + (ħ²/2mc²)∂²ψ/∂t², derived with nothing dropped. Dimensions check: (ħ²/mc²)∂²/∂t² is (J s)² J⁻¹ s⁻² = J, an energy like the other two terms.

To leading order ψ obeys Schrödinger's equation, iħ∂ψ/∂t ≈ (p²/2m)ψ, so differentiating once more, ∂²ψ/∂t² ≈ −(p²/2m)²ψ/ħ². Inserted in the extra term, (ħ²/2mc²) × (−p⁴/4m²ħ²)ψ = −p⁴ψ/(8m³c²), and to order 1/c², iħ∂ψ/∂t = [p²/2m − p⁴/(8m³c²)]ψ. Quantum Mechanics II reached the same operator by Taylor-expanding √(p²c² + m²c⁴) − mc² and added it to the Hamiltonian by hand; here the wave equation produces it, so it is derived. Plane waves make the step exact. With ψ = e^{(i(p⋅x − Tt)/ħ)} the free equation becomes T = p²/2m − T²/(2mc²), a quadratic whose roots T = −mc² ± √(p²c² + m²c⁴) are the two energy branches with the rest energy removed. Iterating it from T₀ = p²/2m, T₁ = p²/2m − T₀²/(2mc²) carries the p⁴ term, and T₂ = p²/2m − T₁²/(2mc²) gets +p⁶/(16m⁵c⁴) right as well: each pass fixes one more order, and the iteration converges to the + root whenever T is below mc².

The exact free kinetic energy is mc²(√(1 + x) − 1) with x = p²/m²c², and its series x/2 − x²/8 + x³/16 − 5x⁴/128 + … has a branch point at x = −1, so it converges only for p below mc, that is T below (√2 − 1)mc² = 0.414 mc²: 57.8 MeV for a pion, 212 keV for an electron. Beyond it the terms eventually grow without limit: just past p = mc the first few orders still help, but by p = 1.6 mc, the figure's right edge, each added order makes the error larger. Schrödinger plus −p⁴/(8m³c²) also fails outright: its kinetic energy peaks at mc²/2 at p = √2 mc and then falls, so it cannot describe any particle with T above mc²/2. A measured check: the K conversion electrons of ¹³⁷Cs, T = 624.2 keV = 1.22 mc², sit at Bρ = 3.381 × 10⁻³ T m in magnetic spectrometers, pc = 1013.7 keV, exactly √(T² + 2Tmc²). Schrödinger predicts 798.7 keV, 21% low, and the corrected Hamiltonian has no solution at all. The free dispersion relation is spin-blind, since each component of a free Dirac spinor obeys the Klein-Gordon equation, so electrons test it fairly.

The extra term is small only when ψ varies slowly. For a plane wave its ratio to iħ∂ψ/∂t is exactly −T/(2mc²), in size 5.9% for a pion with pc = 70 MeV and about 1.3 × 10⁻⁵ for the electron in hydrogen's ground state, taking its mean kinetic energy of 13.6 eV. The negative-energy root breaks the condition outright. There T = −mc² − √(p²c² + m²c⁴), so ψ oscillates at 2mc²/ħ or faster and the ratio −T/(2mc²) is at least 1: the dropped term is as large as the kept ones. Dropping it also lowers the order in time. Klein-Gordon needs φ and ∂φ/∂t at t = 0, Schrödinger only ψ, and the reduction amounts to fixing ∂φ/∂t ≈ −(imc²/ħ)φ initially, which selects the positive-energy branch. Topic 3.02's density, quoted, confirms the reading: for a free stationary state with E = mc² + ε it is (1 + ε/mc²)|ψ|², positive and tending to Schrödinger's |ψ|² as ε/mc² → 0. The negative-energy part of a state has no Schrödinger limit, however small its momenta.

Let V(x) be a static potential energy, such as qΦ, entering as iħ∂/∂t → iħ∂/∂t − V, as topic 3.04 will justify. A stationary state φ = e^(−iEt/ħ)u obeys (E − V)²u = (p²c² + m²c⁴)u. Put E = mc² + ε, expand (mc² + ε − V)², cancel m²c⁴ and divide by 2mc²: εu = (p²/2m + V)u − (ε − V)²u/(2mc²), still exact. Let H₀ = p²/2m + V have eigenstate ψ₀ and energy ε₀, chosen, if the level is degenerate, to diagonalise (ε₀ − V)², as the nlm states do for a central V. The inner product with ψ₀, using H₀'s hermiticity, gives (ε − ε₀)⟨ψ₀|u⟩ = −⟨ψ₀|(ε − V)²|u⟩/(2mc²), so to first order ε⁽¹⁾ = −⟨ψ₀|(ε₀ − V)²|ψ₀⟩/(2mc²), and the normalisation of u never enters. Now H₀ψ₀ = ε₀ψ₀ says (ε₀ − V)ψ₀ = (p²/2m)ψ₀. Both ε₀ − V and p² are hermitian, so ⟨ψ₀|(ε₀ − V)²|ψ₀⟩ = ‖(ε₀ − V)ψ₀‖² = ‖(p²/2m)ψ₀‖² = ⟨p⁴⟩/4m², and ε⁽¹⁾ = −⟨p⁴⟩/(8m³c²) exactly, derived. Quantum Mechanics II used this identity as a trick for computing ⟨p⁴⟩; here the equation hands over (ε₀ − V)² itself. Two limits check it: with V = 0 it returns the free one-pass value −(p²/2m)²/(2mc²), and adding a constant to V shifts ε₀ by the same amount and leaves the correction unchanged, as it must.

For a hydrogen-like ψ₀, the quoted ⟨V⟩ = 2ε₀ and ⟨1/r²⟩ = 1/(n³(l + ½)a²), with a = ħ/(mcZα), turn the shift into −(mc²(Zα)⁴/2n⁴)(n/(l + ½) − 3/4), which depends on l. Being a squared norm, ‖(ε₀ − V)ψ₀‖², it contains no δ-function, and nothing else arrives at order 1/c². Dirac's 1/c² Hamiltonian, quoted in Quantum Mechanics II, adds a Darwin contact term, +mc²(Zα)⁴/2n³ for s states, and spin-orbit coupling. A one-component field has no σ, so every operator its expansion can build is made of p and x, and L⋅S cannot appear at any order. Hydrogen rejects the pattern: the Klein-Gordon shifts put 2s 29.2 GHz below 2p and leave 2p unsplit, while the measured 2p₃/₂–2p₁/₂ interval is 10.97 GHz and 2s₁/₂ lies only 1.06 GHz above 2p₁/₂. The Coulomb singularity does not spoil the expansion: |V| exceeds mc² only inside Zαħ/mc, 2.82 fm for hydrogen, where the 1s electron spends 2.0 × 10⁻¹³ of its probability. For spinless pions the structure is right, as topic 3.07 shows.