Dirac Landau levels: g = 2 as an exact degeneracy

Work in SI units, with the metric (+, −, −, −) wherever four-vectors appear. Take Unit 4's Hamiltonian H = cα⋅π + βmc² with π = p − qA, a uniform B = Bẑ and no scalar potential. Squaring gives H² = c²(α⋅π)² + m²c⁴ + mc³(αᵢβ + βαᵢ)πᵢ, and the last term vanishes because β anticommutes with every αᵢ. In the Dirac representation αᵢαⱼ = diag(σᵢσⱼ, σᵢσⱼ), so H² = m²c⁴ + c²(σ⋅π)² on each 2 × 2 block. Elimination says the same thing: the lower row gives χ = cσ⋅πφ/(E + mc²) with no approximation, because with V = 0 the denominator is a number rather than an operator, and the upper row then reads (E² − m²c⁴)φ = c²(σ⋅π)²φ. Topic 6.01 had to expand E − V + mc² in v/c; here nothing is expanded, which is why this problem, unlike hydrogen, is solved exactly in a few lines. Each eigenvalue K² of (σ⋅π)² gives E = ±√(m²c⁴ + c²K²), the 2mc² gap intact.

Topic 6.02's identity gives (σ⋅π)² = π² + iσ⋅(π × π), and π × π = iħqB because [πₓ, π_y] = iħqB_z. So (σ⋅π)² = πₓ² + π_y² + p_z² − qħBσ_z. The transverse part is the Landau operator that Quantum Mechanics II and Unit 3 diagonalised, with eigenvalues (2n_L + 1)ħ|q|B for n_L = 0, 1, 2, …, quoted here. σ_z commutes with it and p_z with everything, so the eigenvalues of (σ⋅π)² are p_z² + ħ|q|B(2n_L + 1 + s), where s = −σ_z q/|q| is +1 when the spin moment points against B. Writing n = n_L + (1 + s)/2 gives E² = m²c⁴ + p_z²c² + 2nħ|q|Bc², derived exactly. Dimensions: with T = kg s⁻¹ C⁻¹, ħ|q|B carries (J s)(C)(kg s⁻¹ C⁻¹) = kg² m² s⁻², a momentum squared, and times c² it is J², as E² must be. Unit 3's spinless particle has E² = m²c⁴ + p_z²c² + (2n_L + 1)ħ|q|Bc², quoted: spin moves each odd integer to the even integer just above or just below it.

Read the ladder level by level. n = 0 needs n_L = 0 and s = −1, a single spin state, with E = √(m²c⁴ + p_z²c²): free motion along B and nothing else. At p_z = 0 it sits at mc² exactly, whatever B is, because the spin energy −ħω_c/2 cancels the zero-point cyclotron energy +ħω_c/2 to all orders, not just the first. Unit 3's spinless ground state, quoted, sits instead at mc²√(1 + B/B_c), already 212 keV above mc² at B = B_c. Each level n ≥ 1 holds two states, orbit n with s = −1 and orbit n − 1 with s = +1, and each carries Unit 3's Landau degeneracy |q|B/h per unit area, 2.42 × 10¹⁴ m⁻² at 1 T, so each level above the lowest holds twice as many states as the lowest. At p_z = 0 the label s is exact for the whole four-component state: βΣ_z, which is σ_z on the upper components and −σ_z on the lower, commutes with H, because it commutes with β and with αₓ and α_y. At p_z ≠ 0 it fails to commute with α_z, so s then labels the upper components only, and the energies are unchanged.

Give the spin a general g and work non-relativistically: E − mc² = (n_L + ½)ħω_c + s(g/4)ħω_c + p_z²/2m, with ω_c = |q|B/m. Orbit n − 1 with s = +1 and orbit n with s = −1 then differ by (g/2 − 1)ħω_c, which vanishes only at g = 2: one spin flip must cost exactly one cyclotron quantum. Topic 6.02's Pauli equation showed that much. The Dirac solution adds that the coincidence survives every relativistic correction, because spin and orbit sit together inside (σ⋅π)², so the energy depends on n_L and s only through n_L + (1 + s)/2, at every order in B/B_c and every p_z. A spin flip at fixed orbit and a step up in orbit at fixed spin both cost E(n + 1) − E(n), even at large n where that has fallen to ħω_c/γ. The classical counterpart, quoted, is the Bargmann-Michel-Telegdi result that for motion across a pure magnetic field the spin turns relative to the momentum at a|q|B/m, whatever γ is. The measured electron sits close: its spin resonance at 1 T is 28.0250 GHz against a cyclotron frequency of 27.9925 GHz, equal to within 0.116%.

At p_z = 0, E(n) = mc²√(1 + 2nB/B_c), with B_c = m²c²/(|q|ħ) = 4.41 × 10⁹ T for the electron, since ħω_c/mc² = B/B_c. Then E(n) − mc² = nħω_c − (nħω_c)²/(2mc²) + …, so level n falls below nħω_c by the fraction nB/(2B_c), and the line from n to n + 1 sits below ħω_c by (n + ½)(ħω_c)²/mc². In a 5.00 T laboratory field ħω_c = 0.579 meV and the fraction for n = 1 is 5.7 × 10⁻¹⁰, tiny, but above the precision of Penning-trap g − 2 measurements, which must carry it. At B = B_c the n = 1 level sits at (√3 − 1)mc² = 374 keV, not 511 keV, and at 10¹¹ T, a field inferred for magnetars, it is 2.97 MeV where nħω_c says 11.6 MeV: the spacing grows as √B. Through all of this the pairs stay whole, because relativity acts on n. The limits close the check. B → 0 leaves a free Dirac particle along z; B ≪ B_c returns topic 6.02's (n_L + ½ + s/2)ħω_c + p_z²/2m; and m → 0 in two dimensions, with c replaced by v_F ≈ 1.0 × 10⁶ m s⁻¹ and spin by sublattice pseudospin, gives graphene's ±v_F√(2nħeB), whose unpaired n = 0 level produces the measured half-integer Hall plateaus σ_xy = ±4(N + ½)e²/h, quoted.

Topic 6.04 introduces Pauli's term, which covariance and gauge invariance allow but minimal coupling leaves out. Quoted here, in a pure magnetic field it adds −μ′βΣ⋅B to H, where μ′ = aqħ/2m is the anomalous part of the moment. At p_z = 0, βΣ_z commutes with the rest of H and takes the value −s q/|q| on each state, so the term adds exactly s aħω_c/2: E(n, s) = mc²√(1 + 2nB/B_c) + s aħω_c/2, derived from the quoted term. Each pair opens by aħω_c, the n = 0 level drops to mc² − aħω_c/2, and any transition that keeps s is not moved. The pair splitting in frequency is νₐ = aν_c: Schwinger's a = α/2π = 0.0011614, quoted, would give 32.51 MHz per tesla, and the measured a = 0.00115965218 gives 32.46 MHz per tesla. A Penning trap measures exactly this, the splitting inside a pair against the spacing between pairs: a = νₐ/ν_c, with the relativistic shift corrected in ν_c and absent from νₐ, since both members of a pair share n. That is the sense in which g − 2 is measured as the breaking of a degeneracy the Dirac equation makes exact.