The Pauli equation and g = 2 from minimal coupling

SI units throughout; where a four-vector appears the metric is (+, −, −, −). Topic 6.01 ended with iħ∂φ/∂t = [(σ⋅π)(σ⋅π)/2m + V]φ, π = p − qA, and the product must now be expanded. The textbook identity was proved for vectors of numbers, so rederive it assuming only that σ commutes with a and b, true because σ acts on spin and a, b on position. Then (σ⋅a)(σ⋅b) = σᵢσⱼ aᵢ bⱼ, and σᵢσⱼ = δᵢⱼ + iεᵢⱼₖσₖ, quoted from the Pauli algebra, gives aᵢ bᵢ + iσₖ εᵢⱼₖ aᵢ bⱼ = a⋅b + iσ⋅(a × b), derived, with (a × b)ₖ = εᵢⱼₖ aᵢ bⱼ keeping a to the left of b. No factor was moved past another, so the identity holds for operators. What changes is a × a: written out, (a × a)ₖ = εᵢⱼₖ aᵢ aⱼ = ½εᵢⱼₖ[aᵢ, aⱼ], zero for numbers and for p alone, but not in general. Quantum Mechanics I already met one such vector, L × L = iħL.

From [xᵢ, pⱼ] = iħδᵢⱼ, [pᵢ, f(r)] = −iħ∂ᵢ f, so [πᵢ, πⱼ] = −q[pᵢ, Aⱼ] − q[Aᵢ, pⱼ] = iħq(∂ᵢAⱼ − ∂ⱼAᵢ) = iħq εᵢⱼₖ Bₖ, derived; hence π × π = iħqB. Dimensions: ħqB is J s × C × T, and a tesla is kg s⁻¹ C⁻¹, so the product is kg² m² s⁻², a momentum squared, as a commutator of two momenta must be. Gauge: A → A + ∇Λ adds ∂ᵢ∂ⱼΛ − ∂ⱼ∂ᵢΛ = 0, so only B survives, as Unit 4's gauge covariance demands of any energy. Limits: nothing assumed B uniform or static, so the result holds pointwise in any field, and it vanishes as ħ → 0, where p and A commute. Scale: in 1.00 T, ħeB = (ħ/ℓ_B)² with ℓ_B = √(ħ/eB) = 25.7 nm, so for an electron confined to about 26 nm, whose momenta are near 4.1 × 10⁻²⁷ kg m s⁻¹, the two transverse components already fail to commute at order one.

Put a = b = π: (σ⋅π)² = π² + iσ⋅(iħqB) = π² − qħσ⋅B, derived. The upper components therefore obey the Pauli equation, iħ∂φ/∂t = [π²/2m − (qħ/2m)σ⋅B + V]φ. With S = ħσ/2 the new term is −(q/m)S⋅B = −μ⋅B for a moment μ = (q/m)S, twice the q/2m that relates an orbital moment to its angular momentum: g = 2, exactly, because the coefficient came from an identity, not a fit. For an electron q = −e, so μ = −(e/m)S, antiparallel to the spin, and its component along any axis is ±μ_B, with μ_B = eħ/2m = 9.274 × 10⁻²⁴ J T⁻¹. It is the moment topic 5.07 found as a curl in the free current, now appearing as an energy. Two features follow from the algebra. The term carries no c, so it is as large as the orbital Zeeman term, not a relativistic correction to it. And (σ⋅π)² is the square of a Hermitian operator, so π²/2m − (qħ/2m)σ⋅B is never negative: the spin energy can at most cancel the transverse kinetic energy, as it does in the lowest Landau level, the degeneracy topic 6.03 makes exact.

In a uniform field take A = ½B × r. Its divergence vanishes, so A⋅p = p⋅A, and A⋅p = ½(B × r)⋅p = ½B⋅(r × p) = ½B⋅L. Expanding π² = p² − 2qA⋅p + q²A² and adding the spin term gives H = p²/2m + V − (q/2m)(L + 2S)⋅B + (q²/8m)|B × r|², derived. The orbital moment, g_L = 1, comes from the cross term of π², the classical ratio of a circulating charge; the spin moment, gₛ = 2, from the commutator; the last term is the diamagnetic energy, second order in B and 2 × 10⁻⁶ of μ_B B for hydrogen's 1s in 1 T. Because L + 2S is not parallel to J = L + S, the moment operator is not a multiple of J, and projecting it onto each spin-orbit-coupled level gives a g_J that differs from level to level: Quantum Mechanics II's projection gives g_J = 2/3 for ²P₁/₂, 4/3 for ²P₃/₂ and 2 for ²S₁/₂, which split sodium's D1 line into four Zeeman components and D2 into six, the anomalous pattern observed. With gₛ = 1 every g_J would be 1, and each line would show only the normal triplet.

With B = 0, (σ⋅p)² = p², so a non-relativistic theory may write its kinetic energy either way, and Quantum Mechanics II borrowed (σ⋅π)²/2m because that form reproduced the observed moment. Minimal coupling alone cannot choose: π²/2m gives no spin coupling, (σ⋅π)²/2m gives g = 2, and adding −(g/2)(q/m)S⋅B by hand gives any g at all, with the same field-free limit every time. The Dirac reduction removes the choice, because its kinetic operator arrives as a product of the two off-diagonal blocks, cσ⋅π acting twice across the 2mc² gap, so it is (σ⋅π)(σ⋅π), never π². Is relativity the reason? No. Lévy-Leblond's 1967 Galilean wave equation, quoted, is first order in ∂/∂t and ∇ with two pairs of components, (σ⋅π)φ + 2mχ = 0 and (iħ∂/∂t − V)φ + (σ⋅π)χ = 0; eliminating χ gives exactly the Pauli equation, g = 2 included, with no c anywhere. Nor does covariance force the 2: topic 6.04 adds a Pauli term that respects every symmetry and changes g.

Count powers of c. Both π²/2m and qħB/2m carry c⁰, so the spin term belongs to the leading order with the kinetic energy, and everything topic 6.01 dropped is smaller by (v/c)². Some of those dropped terms do contain σ⋅B: in a uniform field the next term, −(σ⋅π)⁴/8m³c², contains +(qħ/2m)(π²/2m²c²)σ⋅B, which shrinks a moving electron's spin energy by the factor 1 − v²/2c², the same 1/γ that slows its cyclotron motion. The ratio of the two, which is what g measures, stays 2, and topic 6.03's exact Landau levels confirm it at every order. Binding is different, because in a 1s state ⟨v²⟩/c² = (Zα)². Breit's exact Dirac result for 1s, quoted, is g = (2/3)[1 + 2√(1 − (Zα)²)] ≈ 2 − (2/3)(Zα)², which lowers hydrogen's g by 3.55 × 10⁻⁵, and the measured 1s value, 2.002 284, sits that far below the free electron's 2.002 319. That free value, quoted, is what lies outside the equation altogether: an excess of 2.32 × 10⁻³ from the quantised field, α/π at leading order, which topic 6.04 quotes and Unit 13 begins to derive.