University Quantum Field Theory I · The One-Loop Vertex: the Anomalous Magnetic Moment and Infrared Divergences · 14.3
F₂(0) = α/2π: Schwinger's anomalous moment derived
This is the loop that made QED's radiative corrections believable: one integral with no counterterm to hide behind, and a number, 0.0011614, that a 1948 experiment could already test. Pull the magnetic coefficient out of the reduced vertex, show that neither regulator touches it, and integrate it for every spacelike momentum transfer.
Build the model
Connect the measurement to the mechanism.
Dirac's equation gives g = 2 exactly, and topic 14.01 showed that the vertex fixes g = 2(F₁(0) + F₂(0)), so any departure from 2 must come from the coefficient of iσ(μν)qν/2m in the loop-corrected vertex. That coefficient is special. The Pauli operator ψ̄σ(μν)ψF(μν) has mass dimension 5, QED's counterterms δ₁, δ₂, δ₃ and δₘ contain nothing like it, and so the one-loop diagram must deliver F₂ finite, or QED would not be renormalisable.
It does. Pulling out the factor qν leaves a numerator with no loop momentum, an integrand falling as ℓ⁻⁶, and a factor z(1 − z) that vanishes in the soft-photon corner where F₁ collects its infrared logarithm. So d = 4 and μγ = 0 may be set before integrating, the Feynman-parameter simplex collapses to one parameter, F₂(q²) = (α/2π)∫₀¹ dξ m²/[m² − ξ(1 − ξ)q²], and at q² = 0 the integrand is 1.
Schwinger's a = α/2π = 0.0011614 follows with no trace of the lepton mass, which is why electron, muon and tau share it at this order. It stops being the whole story in three ways. It is the first term of a series in α/π, and the measured aₑ sits 0.15% below it; loops containing other particles push the muon's 0.39% above; and at Q ≫ m the moment fades as (α/π)(m²/Q²)ln(Q²/m²), so it belongs to the lepton seen from a Compton wavelength away, not closer.
- Simple definition
- Schwinger's anomalous moment is the one-loop value of the vertex's magnetic form factor at zero momentum transfer, F₂(0) = α/2π, which raises a charged lepton's g from Dirac's 2 to 2 + α/π.
- Example
- With α = 1/137.036, F₂(0) = 0.0011614 and g = 2.0023228, so the electron's moment is 1.0011614 μB; the measured 1.0011597 μB lies 0.15% lower in a, the size of the quoted two-loop term.
The coefficient of iσ(μν)qν/2m in topic 14.02's reduced vertex, integrated over ℓ in d = 4 with μγ = 0, both safe for this term alone.
x, y weight the electron propagators, z the photon; m in MeV, q² = (p′ − p)² in MeV²; F₂ is dimensionless.
Times the diagram's 2ie² and the 2 from combining three denominators it gives e²/(8π²Δ) = α/(2πΔ): the α/2π in front of every F₂ formula.
Δ = m²(1 − z)² − xyq² + zμγ² in MeV²; the result carries MeV⁻², and no pole in ε appears.
All the q² dependence in one line. At q² = 0 the integrand is 1 for every ξ, and the lepton mass has cancelled.
x = (1 − z)ξ, y = (1 − z)(1 − ξ), Jacobian 1 − z; the z integral is ∫₀¹ 2z dz = 1. Real for q² below 4m².
F₂ at any scattering kinematics. At Q = 2m it gives 0.6232 of α/2π, which neither limiting form reproduces.
Q² = −q² in MeV², β dimensionless. Small Q: F₂ ≈ (α/2π)(1 − Q²/6m²), a series that converges only for Q² below 4m².
σ(μν) flips chirality, so F₂ needs a mass and vanishes as m → 0; the logarithm comes from the tails near each end of the ξ range, where the integrand falls as m²/(ξQ²) between ξ ≈ m²/Q² and ξ of order 1.
From β → 1 + 2m²/Q². It is 1.5% above the exact form at Q² = 100m² and 0.02% above at Q² = 10⁴m².
The same for e, μ and τ at one loop. The measured aₑ = 0.00115965 is 0.15% lower: the two-loop term of topic 14.04.
α = 1/137.036; F₁(0) = 1, the charge condition topic 14.05 checks; a and g are dimensionless.
Read the magnetic coefficient off the reduced vertex
Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout. Topic 14.02 shifted the loop momentum to ℓ = k + yq − zp and reduced the numerator; quote its result. The γμ term carries −½ℓ² plus finite pieces, the qμ term is odd under x ↔ y and integrates to zero, and the term mz(z − 1)(p + p′)μ, rewritten with the Gordon identity of UAQM-05.07, (p + p′)μ = 2mγμ − iσ(μν)qν between on-shell spinors, leaves (iσ(μν)qν/2m) × 2m²z(1 − z). That coefficient contains no ℓ, so the only momentum integral it meets is ∫d⁴ℓ/(2π)⁴ (ℓ² − Δ)⁻³ = −i/(32π²Δ). Collect the factors: the diagram's 2ie², the 2 from combining three denominators and −i/(32π²Δ) give 4e²/(32π²Δ) = e²/(8π²Δ), which with e² = 4πα is α/(2πΔ). So F₂(q²) = (α/2π)∫ 2m²z(1 − z)/Δ over the simplex, with Δ = m²(1 − z)² − xyq² + zμγ². Dimensions: m²/Δ is a pure number, as F₂ must be, because iσ(μν)qν/2m already has the dimensions of γμ.
Why the ultraviolet cannot reach it
Count powers. The vertex has superficial degree of divergence 0, and the whole divergence sits in the ℓ² term: symmetric integration turns ℓαℓβ into gαβℓ²/d, which can only multiply γμ, and ∫ddℓ ℓ²/(ℓ² − Δ)³ carries the Γ(2 − d/2) pole that topic 14.05 cancels against δ₂. The magnetic term has paid one power of momentum to the explicit qν, so its integrand falls as ℓ⁻⁶, and ∫d⁴ℓ ℓ⁻⁶ converges like ∫dℓ/ℓ³. It had to. The operator ψ̄σ(μν)ψF(μν) has mass dimension 3/2 + 3/2 + 2 = 5, its coefficient dimension −1, and QED's δ₁, δ₂, δ₃ and δₘ contain no such term, so a divergent F₂ would have had nowhere to go. Because nothing diverges, the O(ε) changes the d-dimensional Dirac algebra makes to 2m²z(1 − z) multiply a finite integral and vanish as ε → 0, so d = 4 is safe. Dimensional regularisation, Pauli–Villars and a hard cutoff all give the same F₂.
Why the soft photon cannot reach it either
Set q² = 0 and keep the photon mass: Δ = m²(1 − z)² + zμγ². The danger is the corner z → 1, where the photon takes all the Feynman weight, both electron lines sit on shell, and Δ falls to μγ². Integrating y from 0 to 1 − z supplies a factor 1 − z, so with λ = μγ/m the magnetic integrand is 2z(1 − z)²/[(1 − z)² + zλ²], which tends to 2z and stays bounded as λ → 0. The γμ coefficient has no such protection: its 1 − 4z + z² equals −2 at z = 1, so ∫₀¹ dz (1 − z)(1 − 4z + z²)/[(1 − z)² + zλ²] grows like −2 ln(1/λ), the infrared logarithm F₁ carries into topics 14.05 and 14.06. The magnetic term's extra factor 1 − z is the whole difference. In numbers, λ = 10⁻³ lowers F₂(0) by only 0.31%, a shift that vanishes like πλ, while the γμ integral has reached −11.3 and is still growing. F₂ has no entry in the regulator ledger, so μγ = 0 is safe.
Collapse the simplex to one parameter
With d = 4 and μγ = 0, Δ = (1 − z)²m² − xyq². Trade x and y for ξ: x = (1 − z)ξ and y = (1 − z)(1 − ξ) satisfy x + y = 1 − z automatically, and the measure becomes dz dξ (1 − z). Then xy = (1 − z)²ξ(1 − ξ), so Δ = (1 − z)²[m² − ξ(1 − ξ)q²]. The integrand 2m²z(1 − z)/Δ times the Jacobian 1 − z is 2z × m²/[m² − ξ(1 − ξ)q²]: the z dependence has separated, and ∫₀¹ 2z dz = 1. So F₂(q²) = (α/2π)∫₀¹ dξ m²/[m² − ξ(1 − ξ)q²]. For spacelike q² = −Q² the ξ integral is elementary: with β = √(1 + 4m²/Q²), F₂ = (α/2π)(2m²/(βQ²)) ln[(β + 1)/(β − 1)]. Since ∫₀¹ ξ(1 − ξ) dξ = 1/6, small Q gives F₂ ≈ (α/2π)(1 − Q²/6m²). For timelike q² the denominator stays positive only while q² is below 4m²; above the pair threshold q² = 4m², F₂ acquires an imaginary part, which this lesson leaves alone.
The moment fades at short distance
For Q² ≫ m², β → 1 + 2m²/Q² and the closed form becomes F₂ → (α/π)(m²/Q²) ln(Q²/m²). The ξ-integrand shows where that comes from: m²/[m² + ξ(1 − ξ)Q²] stays near 1 only within ξ ≲ m²/Q² of either end and beyond that falls as m²/(ξQ²), mirrored about ξ = ½, so each half of the range contributes ∫ m²dξ/(ξQ²) from ξ ≈ m²/Q² to ξ of order 1, which is (m²/Q²) ln(Q²/m²); the region ξQ² ≲ m² itself adds only about m²/Q², with no logarithm. The power m²/Q² passes the massless-limit check: σ(μν) joins left- to right-handed spinors, so the helicity flip needs a mass, and at fixed Q the magnetic amplitude, of order F₂Q/m, vanishes like (m/Q) ln(Q²/m²) as m → 0. In numbers, F₂ has halved at Q² = 6.94m², which for the electron is Q = 1.35 MeV, a distance ħc/Q = 147 fm against its Compton wavelength ħ/mₑc = 386 fm. At Q = 100 MeV the electron's F₂ is 6.4 × 10⁻⁷, 0.055% of α/2π, while the muon, with m = 105.7 MeV, keeps 87%. The anomalous moment is a property of the lepton seen from about a Compton wavelength away.
Schwinger's number, and what it is the first term of
At q² = 0 the integrand is 1 for every ξ, so F₂(0) = α/2π = 0.0011614 with α = 1/137.036, and with F₁(0) = 1, g = 2(F₁(0) + F₂(0)) = 2.0023228. This is Schwinger's 1948 result, quoted in UAQM-06.04 and now derived. Restoring units, the electron's moment is (1 + a)eħ/2mₑ = 1.0011614 μB, with μB = 5.788 × 10⁻⁵ eV T⁻¹; Kusch and Foley's atomic-beam measurement that year gave 1.00119 ± 0.00005 μB, within one standard deviation. The mass dropped out, so the muon and tau get the same a at this order. Today's measurements, quoted here, show what one loop leaves out: aₑ = 0.00115965218 lies 0.151% below α/2π, the size of the quoted two-loop term −0.328(α/π)², and aμ = 0.00116592 lies 0.388% above, mainly through electron loops in the photon line of the muon's vertex. Both comparisons belong to topic 14.04; the point here is that the first term already fixes a to 0.4%.
Change one variable at a time
Make the relationship visible.
Start with the electron at Q = 2 MeV and raise Q: the integrand hollows out until only the ends of the ξ range survive, and the marker slides down the e curve. Then switch to the muon and tau at the same Q: same starting height, but each keeps α/2π until Q passes its own mass.
LEPTON MASS m0.511 MeV
Q / m3.905
F₂(q²) / F₂(0)0.3321
F₂(q²)0.3857 × 10⁻³
Live interpretationLEPTON MASS m: 0.511 MeV. Q / m: 3.905. F₂(q²) / F₂(0): 0.3321. F₂(q²): 0.3857 × 10⁻³
Catch the common trap
Explain before calculating.
An electron and a muon each scatter at one loop with the same spacelike momentum transfer, Q² = −q² = 100 MeV², so Q = 10 MeV. Using F₂(q²) = (α/2π)∫₀¹ dξ m²/[m² + ξ(1 − ξ)Q²], what are their magnetic form factors?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA storage ring holds muons in a uniform 1.450 T magnetic field, where the spin turns relative to the momentum at ωₐ = a eB/mμ (quoted from the Thomas–BMT equation for a purely magnetic field). Predict νₐ = ωₐ/2π with Schwinger's a, and say how the prediction for an electron in the same field differs.
- At one loop a = F₂(0) = α/2π = 0.0011614 for the muon exactly as for the electron: the mass cancelled from the ξ-integrand at q² = 0, so no new loop is needed.
- The mass enters only through the cyclotron factor: eB/mμ = (1.6022 × 10⁻¹⁹ C)(1.450 T)/(1.8835 × 10⁻²⁸ kg) = 1.2334 × 10⁹ s⁻¹, so eB/(2πmμ) = 196.30 MHz. Units: C T kg⁻¹ = s⁻¹, and a is a pure number.
- νₐ = a × 196.30 MHz = 0.0011614 × 196.30 MHz = 227.98 kHz, or 228.0 kHz to the precision of a.
- With the measured aμ = 0.00116592, quoted, the same field gives 228.87 kHz: one loop falls 0.89 kHz, or 0.39%, short, a gap topic 14.04 traces to loops containing other particles.
- For an electron a is the same, but mₑ is 206.77 times smaller: νₐ = 0.0011614 × 1.450 T × 27.992 GHz T⁻¹ = 47.14 MHz.
Answerνₐ = 228.0 kHz for the muon with a = α/2π (228.9 kHz with the measured aμ). An electron in the same field has the same a but νₐ = 47.14 MHz, 206.8 times higher, because only the cyclotron frequency carries the mass.
MediumAn electron scatters with spacelike momentum transfer Q = 1.022 MeV, so that Q² = 4m². Evaluate F₂(q²) exactly from the closed form, then test the small-Q series and the large-Q formula against it. Which, if either, could have been used?
- Q²/m² = 4, so β = √(1 + 4m²/Q²) = √2 = 1.41421, (β + 1)/(β − 1) = 2.41421/0.41421 = 5.8284, and its logarithm is 1.76275.
- Prefactor 2m²/(βQ²) = 2/(4 × 1.41421) = 0.35355, so F₂/F₂(0) = 0.35355 × 1.76275 = 0.62323 and F₂ = 0.62323 × 0.0011614 = 7.238 × 10⁻⁴. A direct numerical ξ integral of 1/[1 + 4ξ(1 − ξ)] gives the same 0.62323.
- Small-Q series: 1 − Q²/6m² = 1 − 4/6 = 0.333, 47% low. The series in q² converges only within 4m² of the origin, the distance to the pair threshold, and Q² = 4m² sits on that boundary.
- Large-Q form: (2m²/Q²) ln(Q²/m²) = 0.5 × ln 4 = 0.693, 11% high; its corrections fall only as m²/Q², which is 0.25 here.
- Dimensions: only the ratio Q²/m² appears, so F₂ is dimensionless as it must be. Physically Q = 2m probes ħc/Q = 197.3 MeV fm/1.022 MeV = 193 fm, inside the electron's 386 fm Compton wavelength, where the moment is already partly resolved.
AnswerF₂(−4m²) = 0.6232 α/2π = 7.24 × 10⁻⁴. Neither limit works here: the small-Q series gives 0.333 (47% low) and the large-Q form 0.693 (11% high), so at Q ≈ 2m only the closed form will do.
HardKeep a photon mass μγ = λm in Δ and set q² = 0. Show that F₂(0) still tends to α/2π as λ → 0, find the leading correction, and evaluate it for λ = 0.010. Then integrate the γμ coefficient's term (1 − z)(1 − 4z + z²)m²/Δ with the same λ, and say which of the two form factors must keep the photon mass until something cancels it.
- At q² = 0, Δ = m²[(1 − z)² + zλ²]. After the y integration supplies 1 − z, F₂(0) = (α/2π)∫₀¹ dz 2z(1 − z)²/[(1 − z)² + zλ²]. With t = 1 − z the integrand is 2(1 − t)t²/[t² + (1 − t)λ²].
- Write t²/[t² + (1 − t)λ²] = 1 − (1 − t)λ²/[t² + (1 − t)λ²]. The 1 gives ∫₀¹ 2(1 − t) dt = 1, Schwinger's value. The remainder, 2λ²∫₀¹ (1 − t)² dt/[t² + (1 − t)λ²], is dominated by t ≲ λ, where it is close to 2λ²∫₀^∞ dt/(t² + λ²) = πλ.
- So F₂(0) ≈ (α/2π)(1 − πλ): finite, continuous at λ = 0, and corrected linearly in μγ, not logarithmically. For λ = 0.010 the estimate gives 0.9686; a numerical integral gives 0.9703, the difference being of order λ² ln(1/λ).
- The γμ term: with t = 1 − z, 1 − 4z + z² = −2 + 2t + t², so ∫₀¹ t(−2 + 2t + t²) dt/[t² + (1 − t)λ²] ≈ −2∫₀¹ t dt/(t² + λ²) + ∫₀¹ (2 + t) dt = −2 ln(1/λ) + 2.5.
- For λ = 0.010 that is −2(4.605) + 2.5 = −6.71, against −6.76 numerically, and it diverges as λ → 0. The logarithm sits in the γμ coefficient, so it is F₁'s: F₁ must keep μγ until δ₂ removes it at q² = 0 (topic 14.05) and soft bremsstrahlung at finite q² (topic 14.08). F₂ never meets it and may set μγ = 0 at once.
AnswerF₂(0) = (α/2π)(1 − πλ + …) = 0.9703 α/2π = 1.127 × 10⁻³ at λ = 0.010, tending to α/2π linearly as μγ → 0. The γμ term gives −6.76 in units of α/2π and grows as −2 ln(1/λ): an infrared logarithm F₁ must carry and F₂ never meets.