Klein-Gordon minimal coupling and gauge invariance

Units are SI, the metric is diag(1, −1, −1, −1), x^μ = (ct, r), A^μ = (Φ/c, A) and p^μ = iħ∂^μ (topic 3.01). Classically a charge has two momenta: the canonical p = γmv + qA, conjugate to r and quantised as −iħ∇, and the kinetic P = p − qA = γmv. The mass shell constrains the kinetic pair, (E − qΦ)²/c² − P² = m²c², quoted from relativistic Lagrangian mechanics. So the field enters by p^μ → p^μ − qA^μ (equivalently ∂_μ → D_μ = ∂_μ + (iq/ħ)A_μ), and (p_μ − qA_μ)(p^μ − qA^μ)φ = m²c²φ becomes, derived, (iħ∂/∂t − V)²φ = [c²(−iħ∇ − qA)² + m²c⁴]φ with V = qΦ. Checks: qA has units C × T m = kg m s⁻¹, a momentum, and V is in joules; with A^μ = 0 it is topic 3.01's free equation. V enters squared: for constant potentials (E − V)² = c²P² + m²c⁴, so V lifts both energy branches together, which is how a potential can push a positive-energy wave onto the lower one.

Change the potentials by any real Λ(r, t): A → A + ∇Λ and Φ → Φ − ∂Λ/∂t, which is A^μ → A^μ − ∂^μ Λ and leaves the electric and magnetic fields alone. Let φ → e^(iqΛ/ħ)φ. Acting on the phase, iħ∂^μ produces −q ∂^μ Λ, which cancels the +q ∂^μ Λ coming from the −qA^μ term, so D_μ φ → e^(iqΛ/ħ)D_μ φ and the equation, ρ and j are unchanged. Read backwards, demanding a free local phase forces ∂_μ to become D_μ, although it also permits extra gauge-invariant terms that minimal coupling leaves out. A constant Φ₀ is pure gauge: Λ = Φ₀t removes it and lowers E by qΦ₀, so the zero of energy is a choice and only differences of V between regions are physical. Around a closed loop the phases add to q∮A⋅dl/ħ = qΦ_B/ħ, which no Λ removes: superconducting rings trap flux in steps of h/2e = 2.07 × 10⁻¹⁵ Wb, measured in 1961, and electron holography confirmed the Aharonov–Bohm phase in 1986.

Replace ∂ by D in topic 3.02's current: j^μ = (iħ/2m)[φ*D^μ φ − φ(D^μ φ)*]. It is conserved because D_μ D^μ in the equation cancels between φ and φ* just as ∂_μ ∂^μ did, and gauge invariant because it is built from φ* times D φ. Its parts are ρ = (iħ/2mc²)(φ*∂φ/∂t − φ∂φ*/∂t) − (V/mc²)|φ|² and j = (1/m)Re[φ*(−iħ∇ − qA)φ]. For a stationary state e^(−iEt/ħ)u(r), derived, ρ = [(E − V)/mc²]|u|²: the energy that counts is the gauge-invariant E − V. Checks: the factor is dimensionless; at V = 0 it is topic 3.02's E/mc²; in the non-relativistic limit it tends to 1. For a plane wave in constant V and A, j = (P/m)|φ|², so j/ρ = c²P/(E − V), the group velocity. For a pion, atomic and nuclear fields never threaten the sign: the Coulomb potential at the centre of a lead nucleus, about 25 MeV for a π⁺, leaves a pion with 30 MeV of kinetic energy at ρ = (169.6 − 24.9)/139.6 = 1.04|φ|².

On a plateau the shell (E − V)² = c²P² + m²c⁴ sorts every wave into three regions, each derived from it. For V ≤ E − mc² it oscillates, with positive density, moving along P. For E − mc² ≤ V ≤ E + mc² it decays with ħκc = √(m²c⁴ − (E − V)²), and its density changes sign at V = E. For V ≥ E + mc² it oscillates again, with negative density and a velocity opposite to P. Complex conjugation explains the third region: φ* solves the equation with q → −q and frequency −E/ħ, so a region-three wave of charge q is a wave of charge −q with E − V = V − E ≥ mc², for which the plateau is a well. Reading that as an antiparticle waits for Unit 9, and the reflection and transmission at such a step for Unit 8. A wave arriving in region one reaches region three only by tunnelling across the gap, 2mc² of potential wide, so V must rise by 2mc² within about ħ/mc: 279 MeV over 1.41 fm for a pion, a field near 2 × 10²³ V m⁻¹, against 2.3 × 10²¹ V m⁻¹ at the surface of a lead nucleus.

Factor out the rest-energy oscillation, φ = e^{(−imc²t/ħ)}ψ. Then (iħ∂/∂t − V)φ = e^{(−imc²t/ħ)}(mc² + W)ψ with W = iħ∂/∂t − V, and squaring, the m²c⁴ cancels: 2mc²Wψ + W²ψ = c²(p − qA)²ψ. Where W, the local kinetic energy, is small against 2mc², drop W² to find, derived, iħ∂ψ/∂t = [(p − qA)²/2m + V]ψ: Quantum Mechanics II's Hamiltonian, with the same gauge rule, since Λ never touches the factor e^{(−imc²t/ħ)}. The conditions are |E − mc² − V| ≪ mc² and |p − qA| ≪ mc wherever ψ is appreciable, with fields that vary slowly over times of order ħ/mc². Kept to the next order, W² returns topic 3.03's correction with p replaced by p − qA. The size of what is dropped: in hydrogen's ground state, kinetic energy over 2mc² is 13.6 eV/1.022 MeV = 1.3 × 10⁻⁵, which is α²/4.

In a uniform B take A = ½B × r, for which ∇⋅A = 0, and expand: (p − qA)²/2m = p²/2m − (q/2m)B⋅L + (q²/8m)(B × r)². The linear term is an orbital moment μ = (q/2m)L with g = 1; the quadratic one raises the energy, which is diamagnetism. That is all, derived: a spinless particle in an s state has no linear Zeeman shift. Singlet atomic lines test exactly this, because S = 0 silences spin: the cadmium 643.8 nm line splits into Zeeman's normal triplet, its components μ_B B/h apart, 14.0 GHz at 1 T, as measured. An unpaired electron does not fit. Unit 6 reduces the Dirac equation and finds (σ⋅(p − qA))²/2m = (p − qA)²/2m − (qħ/2m)σ⋅B, an extra −(q/m)S⋅B with g = 2, quoted here; the measured g = 2.00231930436 differs from 2 only by a radiative correction beyond this course. The Klein-Gordon result is the baseline that makes g = 2 a prediction rather than an input.