
Advanced Quantum Mechanics course sequence · Interactive lesson
Anomalous moments: Pauli's term and g = 2(1 + a)
Minimal coupling fixed g at exactly 2, and the electron misses it in the third decimal place. Here we write down Pauli's term, the further coupling relativity and gauge invariance allow, follow it into every spin-dependent term of the reduction, and see why the trap that measures the miss is among the most precise instruments in physics.
UAQM-06.04 · Lesson 4 of 9 in this unit. Position is not a completion score.Build the model
Covariance allows any g. Minimal coupling chooses 2, and the electron misses by 0.116%.
The key idea
An anomalous magnetic moment is the part of a spin-½ particle's moment beyond the g = 2 that minimal coupling gives; writing g = 2(1 + a), it enters the Dirac equation through Pauli's term, proportional to σ(μν)F(μν), with coefficient μ′ = aqħ/2m.
Example. The electron's moment is 1.00115965 μB where minimal coupling gives exactly μB, so a = 0.00116 and g = 2.00232; the proton's is 2.793 μN against the 1 μN of a minimally coupled Dirac proton, so a = 1.793 and g = 5.586.
The complete physical model and assumptions
Unit 4 coupled the Dirac equation to the field in the simplest way, by replacing ∂μ with Dμ, and topics 6.02 and 6.03 showed that this choice alone makes g = 2. It is a choice. Gauge invariance asks only that the potential enter through Dμ or through F(μν), and covariance asks only for a Lorentz scalar, so ψ̄σ(μν)ψF(μν) is allowed too, with a coefficient μ′ that has the dimensions of a magnetic moment and that neither symmetry fixes. That is Pauli's term. In Hamiltonian form it is −μ′β(Σ⋅B − iα⋅E/c): an even half that adds μ′ to the moment, so g = 2(1 + a) with μ′ = aqħ/2m, and an odd half that meets cα⋅π in the reduction and multiplies the spin-orbit and Darwin coefficients by 1 + 2a. The same a opens topic 6.03's degenerate pairs by aħωc, which lets a Penning trap read a as a ratio of two frequencies from which B cancels, and it parts hydrogen's 2S(1/2) and 2P(1/2) by about 68 MHz. The term describes; it does not predict. Here a is an input, from Schwinger's α/2π or from measurement, and its origin is the radiative correction Unit 13 begins. A constant μ′ also treats the particle as a point: the proton's a = 1.79 and the neutron's entire moment signal internal structure, which Pauli's term can carry at low momentum transfer and no further.
1Two couplings survive gauge invariance and covariance
Units are SI with xμ = (ct, x) and the metric (+, −, −, −), as in Unit 5. Gauge invariance lets the potential enter only through Dμ = ∂μ + (iq/ħ)Aμ or through F(μν) = ∂μAν − ∂νAμ, and covariance asks for a Lorentz scalar. Among Unit 5's bilinears, terms linear in the field with at most one derivative of A leave two candidates that conserve parity: ψ̄γμDμψ, which is minimal coupling, and the tensor bilinear contracted with the field, ψ̄σ(μν)ψF(μν). The Dirac-Pauli equation, (iħγμDμ − mc − (μ′/2c)σ(μν)F(μν))ψ = 0, keeps both. Its coefficient is new: μ′/2c must carry momentum per tesla, so μ′ is a magnetic moment in J T⁻¹, and neither symmetry ties it to q or m. A third invariant, ψ̄iσ(μν)γ⁵ψF(μν), gives an electric dipole moment; it is odd under parity and time reversal and is named only. So covariance and gauge invariance allow any g; Dirac's g = 2 is the content of coupling through Dμ alone.
2Pauli's term in Hamiltonian form: an even and an odd half
Multiply the equation by cγ⁰. With γ⁰ = β and γi = βαᵢ, σ(0i) = iγ⁰γi = iαᵢ and σ(ij) = εᵢⱼₖΣₖ, while F(0i) = Eᵢ/c and F(ij) = −εᵢⱼₖBₖ. So σ(μν)F(μν) = 2σ(0i)F(0i) + σ(ij)F(ij) = −2(Σ⋅B − iα⋅E/c), and H = cα⋅π + βmc² + qφ − μ′β(Σ⋅B − iα⋅E/c), derived. The new piece is Hermitian, since (iβαᵢ)† = −iαᵢβ = iβαᵢ, and both halves are energies: μ′B is, and so is μ′E/c, because V m⁻¹ divided by m s⁻¹ is a tesla. Sort it into even and odd parts, as topic 6.06 will sort everything. −μ′βΣ⋅B commutes with β, so it is even and acts on the large component directly. iμ′βα⋅E/c anticommutes with β, so it is odd and links large to small components, as cα⋅π does. The odd half is what a magnetic moment becomes when it moves through an electric field, and covariance forces it in: a boost turns part of E into B, so a term in B alone would not take the same form in every frame.
3The even half: g = 2(1 + a)
On the upper block β = 1, so the even half adds −μ′σ⋅B to the large component's Hamiltonian. Topic 6.02's (σ⋅π)²/2m supplied −(qħ/2m)σ⋅B, so the spin energy is −(qħ/2m + μ′)σ⋅B. Writing μ′ = a(qħ/2m) makes it −(1 + a)(qħ/2m)σ⋅B, a moment (g/2)(q/m)S with g = 2(1 + a), derived; only the sum is measurable, so g − 2 measures μ′. Limits: a = 0 returns Dirac, and q = 0 removes the Dirac moment and leaves μ′ alone, the neutron's case. In a uniform B at pz = 0, βΣz commutes with the whole Hamiltonian, so the even half shifts each of topic 6.03's exact levels by exactly ±aħωc/2, with ωc = |q|B/m and the sign set by the level's eigenvalue of βΣz, at every field strength: the result that topic quoted. Measured, quoted: a = 0.00115965218, so g = 2.00231930436. Schwinger's a = α/2π = 0.0011614, quoted, is 0.15% high, and the next order, −0.328(α/π)² = −1.77 × 10⁻⁶, also quoted, leaves a gap of only 1.5 × 10⁻⁸.
4The odd half: spin-orbit and Darwin grow by 1 + 2a
Now O = cα⋅π + iμ′βα⋅E/c, whose upper-right block is cσ⋅π + (iμ′/c)σ⋅E and lower-left block its adjoint. Every reduction, topic 6.01's elimination or 6.06's transformation, starts from βO²/2mc², whose upper block is (cσ⋅π + (iμ′/c)σ⋅E)(cσ⋅π − (iμ′/c)σ⋅E)/2mc². Its part linear in μ′ is (iμ′/2mc²)[(σ⋅E)(σ⋅π) − (σ⋅π)(σ⋅E)]. The Pauli identity, kept in order, turns the bracket into E⋅π − π⋅E + iσ⋅(E×π − π×E), and E⋅π − π⋅E = iħ∇⋅E. So the anomalous terms are −(aqħ/4m²c²)[σ⋅(E×π − π×E) + ħ∇⋅E], derived: 2a times the spin-orbit and Darwin terms topic 6.06 derives, so the totals carry 1 + 2a. Dimensions: aqħ²∇⋅E/m²c² is C J² s² V m⁻² ÷ kg² m² s⁻² = J. Nothing else enters at this order: the double commutator gives a-dependent terms only at relative order (v/c)², and renormalising the large component gives none linear in a. For spin-orbit, 1 + 2a = g − 1: g from the moment in the rest-frame field, −1 from Thomas precession, which is kinematics.
5Splitting topic 6.03's pairs: how a Penning trap reads a
Minimal coupling made orbit n with spin moment along B degenerate with orbit n − 1 with spin moment against B. The even half opens every such pair by aħωc and leaves spin-keeping transitions alone, so the trap offers two frequencies: νc = |q|B/2πm between pairs, and the anomaly frequency νₐ = aνc within one. Equivalently νₛ = (g/2)νc flips the spin and νₐ = νₛ − νc. Their ratio a = νₐ/νc contains no B, so the electron is its own magnetometer, and measuring the difference directly buys a factor 1/a ≈ 862: a to one part in 10⁶ needs νₐ and νc to about 10⁻⁶ each, but νₛ/νc to 1.2 × 10⁻⁹. The relativistic bending of the ladder, about a part in 10⁹ of νc, cancels from νₐ, as topic 6.03 showed. The trap's electrostatic quadrupole, absent from this model, adds axial and magnetron motions and shifts the observed cyclotron frequency; the Brown-Gabrielse invariance theorem that undoes this, and the magnetic-bottle detection of single quantum jumps, are quoted, as is the 2023 result a = 0.00115965218059 ± 1.3 × 10⁻¹³.
6Hydrogen's n = 2: the anomaly's share of the Lamb shift
Quantum Mechanics II's expectation values turn the 2a terms into energies. For s states spin-orbit vanishes and Dirac's Darwin energy α⁴mc²/2n³ becomes, times 2a, aα⁴mc²/n³: 2S(1/2) rises by aα⁴mc²/8. For p states the Darwin term vanishes with ψ(0), and the spin-orbit energy (α⁴mc²/4n³)[j(j + 1) − l(l + 1) − ¾]/[l(l + ½)(l + 1)] is −α⁴mc²/48 for 2P(1/2), so the anomaly lowers it by aα⁴mc²/24. The two levels Unit 7's Dirac energies keep degenerate therefore part by aα⁴mc²/6, derived. With α⁴mc²/h = 350.38 GHz and the measured a, that is 50.8 MHz up plus 16.9 MHz down, 67.7 MHz in all, or 67.8 MHz with Schwinger's a, when it reads α⁵mc²/12π. Against the measured Lamb shift of 1057.8 MHz, quoted, it is 6.4%, with the right sign: 2S(1/2) on top. The dominant part is the self-energy Unit 13 computes, about 1040 MHz in Bethe's estimate; vacuum polarisation, about −27 MHz, and the covariant remainder are quoted there. Recoil, neglected here, lowers the 67.7 MHz by about 0.15%: the S share scales as (μ/m)³ and the P share as (μ/m)².
Key equations
Read the units, assumptions and limits before substituting.
Pauli's term in the Dirac equation
(iħγμDμ − mc − (μ′/2c)σ(μν)F(μν))ψ = 0
A Lorentz scalar built from F alone, so covariance and gauge invariance allow it with any μ′. Minimal coupling is the choice μ′ = 0, not a consequence of either.
μ′ in J T⁻¹; σ(μν) = (i/2)[γμ, γν]; F(μν) = ∂μAν − ∂νAμ in T; metric (+, −, −, −).Pauli's term in Hamiltonian form (derived)
H = cα⋅π + βmc² + qφ − μ′β(Σ⋅B − iα⋅E/c)
−μ′βΣ⋅B commutes with β and acts on the large component; iμ′βα⋅E/c anticommutes with β, so it couples large to small components, as cα⋅π does.
π = p − qA in kg m s⁻¹, E in V m⁻¹, B in T, φ in V. From σ(0i) = iαᵢ and σ(ij) = εᵢⱼₖΣₖ.The moment: g = 2(1 + a) (derived)
−(qħ/2m)σ⋅B − μ′σ⋅B = −(g/2)(qħ/2m)σ⋅B, with g = 2(1 + a)
The first term is topic 6.02's, from (σ⋅π)²/2m; the second is the upper block of the even half. Only their sum is measured, so g − 2 is a direct measure of μ′.
a = 2mμ′/(qħ), dimensionless. Electron, measured and quoted: a = 0.00115965218, g = 2.00231930436.Anomalous spin-orbit and Darwin terms (derived)
ΔHₐ = −(aqħ/4m²c²)[σ⋅(E×π − π×E) + ħ∇⋅E]
From the cross term of cα⋅π with the odd half in O²/2mc². Totals carry 1 + 2a; for spin-orbit that is g − 1, since the Thomas −1 is kinematic and does not scale with g.
E in V m⁻¹, π in kg m s⁻¹, ΔHₐ in J. Exactly 2a times Dirac's own two terms, which topic 6.06 derives.What a Penning trap measures
a = νₐ/νc, with νₐ = νₛ − νc
B cancels from the ratio, and measuring the difference νₐ directly gains a factor 1/a ≈ 862 in precision over measuring νₛ and νc separately.
νc = |q|B/2πm, 27.992 GHz T⁻¹ for the electron; νₛ = (g/2)νc, the spin-flip frequency; all in Hz.Anomalous splitting of 2S(1/2) from 2P(1/2)
ΔE(2S(1/2)) − ΔE(2P(1/2)) = aα⁴mc²/8 + aα⁴mc²/24 = aα⁴mc²/6
The Darwin term raises 2S and spin-orbit lowers 2P(1/2): about 6.4% of the measured 1057.8 MHz Lamb shift, and with the right sign.
α⁴mc²/h = 350.38 GHz. With the measured a: 50.8 MHz up plus 16.9 MHz down, 67.7 MHz; recoil neglected.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Set a to 0 and every bar vanishes: minimal coupling leaves S(1/2) and P(1/2) degenerate. At the measured 0.00116 with n = 2 they part by 67.7 MHz, 6.4% of the measured 1057.8 MHz Lamb shift. Raise n and all three shifts fall as 1/n³, keeping the ratio 6 : −2 : 1.
Live results
- S(1/2) SHIFT
- 50.8 MHz
- P(1/2) SHIFT
- -16.9 MHz
- S(1/2) − P(1/2) SPLITTING
- 67.7 MHz
- g = 2(1 + a)
- 2.0023
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Anomaly a = (g − 2)/2: 0.00116; Principal quantum number n: 2
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The tempting reading is that g = 2 is a prediction of relativity: Dirac made the equation covariant and out came g = 2. Covariance by itself says nothing about g. Pauli's term ψ̄σ(μν)ψF(μν) is exactly as covariant and as gauge invariant as minimal coupling, and adding it produces any g at all. What fixed 2 was the decision to let the field in only through Dμ, the simplest coupling the symmetries permit, the same decision topic 5.07's Gordon decomposition exposed as a g = 2 moment in the free current. The non-relativistic Hamiltonian (σ⋅π)²/2m, minimally coupled, gives g = 2 as well, so relativity is neither necessary nor sufficient. The electron's 0.116% miss is therefore not relativity failing but the radiative correction minimal coupling leaves out, and the proton's g = 5.59 shows how far the simplest coupling misses for a particle with structure.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
An electron sits in a Penning trap's uniform 4.80 T field; ignore the trap's electric field and the relativistic shifts. Using the measured a = 0.00115965218, find νc, the anomaly frequency νₐ and the spin-flip frequency νₛ, the energy that splits each of topic 6.03's pairs, and how precisely νₛ/νc would have to be known to give a to one part in 10⁶.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Cyclotron frequency: νc = eB/2πm = 27.99249 GHz T⁻¹ × 4.80 T = 134.364 GHz.
- The Pauli term splits each pair by aħωc, so νₐ = aνc = 0.00115965218 × 134.364 GHz = 155.815 MHz, and the spin flips at νₛ = νc + νₐ = (g/2)νc = 134.520 GHz.
- Pair splitting: hνₐ = 4.1357 × 10⁻¹⁵ eV s × 1.55815 × 10⁸ s⁻¹ = 6.44 × 10⁻⁷ eV, or 0.644 μeV, set against a cyclotron quantum hνc = 0.556 meV.
- From νₛ and νc separately, a = νₛ/νc − 1, so an error δ in the ratio is an error δ in a itself. One part in 10⁶ of a is 1.16 × 10⁻⁹, the precision the ratio would need. From a = νₐ/νc, νₐ and νc each need only about 10⁻⁶: a gain of 1/a = 862.
- Checks: a → 0 gives νₛ = νc, topic 6.03's degeneracy. Schwinger's a = α/2π, quoted, would put νₐ at 156.052 MHz, 236 kHz higher, a difference such a trap resolves at once.
Answer. νc = 134.364 GHz, νₐ = 155.815 MHz and νₛ = 134.520 GHz; each pair splits by 0.644 μeV; νₛ/νc would need 1.16 × 10⁻⁹, against about 10⁻⁶ for νₐ and νc measured directly.
MediumExample 2
Hydrogen's 2P(3/2)–2P(1/2) interval is 10969.0 MHz, from the measured Lamb shift, 1057.8 MHz, and the measured 2S(1/2)–2P(3/2) interval, 9911.2 MHz, both quoted. Show that at order α⁴ the interval is pure spin-orbit, compute it with g = 2 and with the measured a = 0.00115965218, include the proton's recoil through μ/m = 0.999456, and say what the comparison tests.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Both levels have n = 2 and l = 1, so the kinetic correction is common, and the Darwin term vanishes for both because ψ(0) = 0. Only spin-orbit differs: (α⁴mc²/4n³)[j(j + 1) − l(l + 1) − ¾]/[l(l + ½)(l + 1)] gives +α⁴mc²/96 for j = 3/2 and −α⁴mc²/48 for j = 1/2, an interval of α⁴mc²/32.
- With α⁴mc²/h = 350377 MHz, the g = 2 interval is 10949.3 MHz, made of 3649.8 MHz of spin-orbit raising 2P(3/2) and 7299.5 MHz lowering 2P(1/2).
- The Pauli term multiplies spin-orbit by 1 + 2a = g − 1: 2P(3/2) rises by 2a × 3649.8 MHz = 8.46 MHz and 2P(1/2) falls by 2a × 7299.5 MHz = 16.93 MHz, widening the interval by 25.40 MHz.
- Recoil: the Dirac interval scales with the reduced mass, quoted, 10949.3 MHz × 0.999456 = 10943.3 MHz. The anomalous part gains (μ/m)³ from ⟨1/r³⟩, but the moving proton adds its own magnetic field to the motional field the moment sees, multiplying the coefficient by m/μ, so it scales as (μ/m)²: 25.40 MHz × 0.998912 = 25.37 MHz. Total: 10968.7 MHz.
- Measured: 10969.0 MHz. With g − 1 = 1 the prediction falls 25.7 MHz short; with g − 1 = 1 + 2a, 0.3 MHz short, and the order-α⁶ part of Unit 7's exact Dirac energies, +0.38 MHz, quoted, brings it to 10969.1 MHz, within the 0.1 MHz to which the interval is quoted here. Unit 13's self-energy shifts both P levels alike at this order, so the interval tests the spin-orbit coefficient g − 1 directly.
Answer. 2P(3/2) rises 8.46 MHz and 2P(1/2) falls 16.93 MHz; the interval becomes 10968.7 MHz against 10969.0 MHz measured, where g − 1 = 1 would give 10943.3 MHz.
HardExample 3
A neutron has no charge, yet its measured moment is −1.913 μN, quoted. Treat it as a Dirac particle carrying only Pauli's term, reduce to order (v/c)², find the spin-independent contact interaction it has with an electron, and turn it into a neutron-electron scattering length to compare with the measured values, which lie between −1.3 × 10⁻³ fm and −1.6 × 10⁻³ fm.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- With q = 0, π = p and minimal coupling gives no moment at all, so the whole moment is anomalous: μ′ = −1.913 × 5.051 × 10⁻²⁷ J T⁻¹ = −9.66 × 10⁻²⁷ J T⁻¹, and H = cα⋅p + βmnc² − μ′β(Σ⋅B − iα⋅E/c).
- The even half gives −μ′σ⋅B. The cross term of cα⋅p with the odd half in O²/2mnc² gives −(μ′/2mnc²)[σ⋅(E×p − p×E) + ħ∇⋅E], the electron's anomalous terms with aqħ/2m replaced by μ′, while Dirac's own spin-orbit and Darwin terms carry q and vanish. The spin-independent part, U = −(μ′ħ/2mnc²)∇⋅E, is Foldy's term.
- An electron at the origin has ∇⋅E = −eδ³(r)/ε₀, so U = (μ′ħe/2mnc²ε₀)δ³(r): a contact interaction, attractive because μ′ is negative, with volume integral −3.82 × 10⁵ eV fm³.
- Matching U to Fermi's pseudopotential (2πħ²/mₙ)bδ³(r), as the Born approximation does, gives bF = (mₙ/2πħ²)∫U d³r = μ′e/(4πε₀ħc²). With μ′ = κₙₑħ/2mₚ and κₙ = −1.913, bF = κₙαħc/(2mpc²) = −1.913 × 1.440 MeV fm ÷ 1876.5 MeV = −1.47 × 10⁻³ fm. The neutron mass cancels, and αħc/mpc² is a length.
- Checks: κₙ → 0 removes U, and for a charged particle U is the 2a part of the Darwin term. The measured values, which disagree among themselves, lie close to bF, so the Foldy term is nearly all of the neutron-electron interaction; the remainder is the neutron's charge distribution, quoted. In lead, 82 electrons add −0.120 fm to a nuclear scattering length near 9.4 fm, a 1.3% effect.
Answer. U = (μ′ħe/2mnc²ε₀)δ³(r), attractive; bF = κₙαħc/(2mpc²) = −1.47 × 10⁻³ fm, inside the measured range of −1.3 × 10⁻³ fm to −1.6 × 10⁻³ fm.
Quick check
Select an answer and explain it before checking.
Review & continue
An anomalous magnetic moment is the part of a spin-½ particle's moment beyond the g = 2 that minimal coupling gives; writing g = 2(1 + a), it enters the Dirac equation through Pauli's term, proportional to σ(μν)F(μν), with coefficient μ′ = aqħ/2m.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.