University Quantum Field Theory I · The One-Loop Vertex: the Anomalous Magnetic Moment and Infrared Divergences · 14.4

g − 2 against experiment: aₑ and the series in α/π

Topic 14.03 gave you α/(2π). Here you hold it against the most precisely measured property of any elementary particle, see what each further loop and each heavier particle buys, and learn why the same comparison, run backwards, measures α to a tenth of a part per billion, and so needs an α from elsewhere before it tests anything.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The anomaly aₑ = (g − 2)/2 is F₂(0), a pure number fixed by the on-shell vertex, and QED predicts it as a power series in α/π ≈ 1/430.5. The coefficients come graph by graph: 1/2 from the single one-loop graph topic 14.03 evaluated, then −0.328 from 7 graphs, 1.181 from 72, −1.912 from 891 and about 6.7 from 12 672. None of these depends on a mass, so they are the same for every lepton.

Graphs containing a muon, a tau or hadrons add pieces suppressed by (mₑ/M)², and W and Z loops a smaller one still. A Penning trap measures aₑ as νₐ/ν_c to 1.3 × 10⁻¹³, and the loops close the gap step by step until, after five, the muon and hadron terms account for the last 4.5 × 10⁻¹². Because aₑ ≈ α/(2π), the comparison reads both ways: inverted, it gives α to 0.11 parts per billion, rivalling the best atom-recoil measurement; with an α measured independently, from the recoil of caesium or rubidium atoms, it tests QED at about 10⁻¹².

The model stops in three places. The hadronic part is taken from measured e⁺e⁻ cross-sections, not computed; the five-loop coefficient has two evaluations that disagree; and the two recoil values of α disagree with each other by 5.5 standard deviations, so the test is now limited by α, not by QED. Dyson argued that the series is only asymptotic, but its terms would stop shrinking only after something like 1/α orders, far beyond any calculation.

Simple definition: The electron's anomaly aₑ = (g − 2)/2 = F₂(0) is predicted as a series in α/π with computed coefficients, plus small terms from loops of muons, tau leptons and hadrons and from weak bosons, and is measured in a Penning trap as the ratio νₐ/ν_c of two frequencies.
Example: Schwinger's α/(2π) = 0.001 161 410 overshoots the measured 0.001 159 652 by 1.76 × 10⁻⁶; adding −0.328(α/π)² = −1.772 × 10⁻⁶ leaves 1.5 × 10⁻⁸, and three more loops with the heavy-particle terms leave, with rubidium's α, 3.4 × 10⁻¹³.

The prediction, term by termaₑ = A₁(α/π) + aₑ(μ, τ loops) + aₑ(hadrons) + aₑ(weak)

A graph with no other particle can depend on no mass ratio, so A₁ is the same for every lepton; each other term knows a heavier particle and is small for the electron.

All dimensionless. A₁ comes from electron and photon lines alone; with α⁻¹ = 137.035 999 166, α/π = 0.002 322 819.

The n-loop term and its sizeCₙ(α/π)ⁿ = 1.161 × 10⁻³, −1.772 × 10⁻⁶, 1.480 × 10⁻⁸, −5.567 × 10⁻¹¹, 4.56 × 10⁻¹³ for n = 1 to 5

Computed. Each order is about 430 times smaller than the last, times a coefficient of order one; the fifth is still 3.5 times the experimental uncertainty.

C₁ = 1/2, derived in topic 14.03; C₂ to C₅ quoted, from 7, 72, 891 and 12 672 graphs; C₅ = 6.737 is used here.

The two-loop coefficient (quoted)C₂ = 197/144 + π²/12 − (π²/2) ln 2 + (3/4)ζ(3) = −0.328 478 966

Quoted, and checked here by adding the four numbers. It takes α/(2π)'s 0.15 per cent overshoot down to 1.3 parts in 10⁵.

ζ(3) = 1.202 057; Petermann and Sommerfield, 1957, correcting the value Karplus and Kroll published in 1950.

A heavy lepton loop decouplesaₑ(M loop) = (1/45)(mₑ/M)²(α/π)² + O((mₑ/M)⁴ ln(M/mₑ))

Derived at leading order by the dispersive route below: heavier means smaller, as decoupling requires. Measured R(s) in the same integral gives about 1.85 × 10⁻¹² from hadrons; higher-order hadronic graphs, which lower it, and light-by-light scattering bring their total to 1.693 × 10⁻¹², quoted.

M the loop lepton's mass, mₑ = 0.511 MeV. Muon: 2.804 × 10⁻¹²; tau: 0.0099 × 10⁻¹²; trap uncertainty 0.13 × 10⁻¹².

Reading α from aₑδα/α = (d ln aₑ/d ln α)⁻¹ δaₑ/aₑ, with d ln aₑ/d ln α = 0.9985

Computed. Because aₑ ≈ α/(2π), a measurement of aₑ is a measurement of α at almost the same relative precision.

δaₑ = 1.3 × 10⁻¹³ on aₑ = 1.160 × 10⁻³ is 0.11 ppb, so α⁻¹ = 137.035 999 166 ± 15 × 10⁻⁹.

The test needs an independent αΔaₑ = aₑ(measured) − aₑ(theoryα_X) ≈ aₑ (α_X⁻¹ − α⁻¹(aₑ))/α⁻¹

Computed. This, not the agreement after inverting, is the test of QED: 3.9 and 2.1 standard deviations, with the recoil α's uncertainty comparable to the trap's.

α_X from atom recoil. Caesium: −1.02(26) × 10⁻¹²; rubidium: +0.34(16) × 10⁻¹², the trap's 0.13 × 10⁻¹² included.

One number, measured to 0.11 parts per billion

Natural units, ħ = c = 1, metric (+, −, −, −), as throughout; the anomaly is a pure number, so nothing needs restoring. Topic 14.01 showed g = 2(F₁(0) + F₂(0)) and topic 14.05 confirms F₁(0) = 1, so aₑ = (g − 2)/2 = F₂(0): a property of the on-shell vertex at q² = 0, with no scheme or scale in it. UAQM-06.04's Penning trap reads it as a = νₐ/ν_c, anomaly over cyclotron frequency, so the magnetic field cancels; with ν_c near 150 GHz, νₐ is near 174 MHz, and the trap and cavity shifts are quoted corrections. The 2023 measurement of Fan, Myers, Sukra and Gabrielse, quoted, is aₑ = 0.001 159 652 180 59(13): an absolute uncertainty of 1.3 × 10⁻¹³, 0.11 parts per billion of aₑ. Schwinger's α/(2π) = 0.001 161 410, derived in topic 14.03, overshoots it by 1.758 × 10⁻⁶: 0.15 per cent, but 13.5 million standard deviations. At α → 0 every term vanishes and Dirac's g = 2 returns, the tree-level limit.

Loop by loop: the series in α/π

A graph built only from electron lines and photons gives a pure number times (α/π)ⁿ, with α/π = 0.002 322 819, about 1/430.5. C₁ = 1/2 is topic 14.03's. The rest are quoted: C₂ = −0.328 478 966 from 7 two-loop graphs, C₃ = 1.181 241 457 from 72 (Laporta and Remiddi, 1996, in closed form), C₄ = −1.912 245 765 from 891 (Laporta, 2017, to 1100 digits) and C₅ = 6.737 from 12 672, numerically. Subtract the partial sums from the measured aₑ, computed with α⁻¹ = 137.035 999 166: −1.758 × 10⁻⁶ after one loop, +1.475 × 10⁻⁸ after two (1.3 parts in 10⁵), −5.07 × 10⁻¹¹ after three, +4.92 × 10⁻¹² after four and +4.47 × 10⁻¹² after five. Each of the next three loops cancels most of the gap the previous one left. After the fourth, what remains is no longer the next electron loop: the fifth changes it by 0.46 × 10⁻¹², 3.5 standard deviations, and leaves 34. A sixth cannot help, since (α/π)⁶ = 1.6 × 10⁻¹⁶. The last gap needs particles other than the electron.

Heavier particles: decoupling, and hadrons from data

The first graph that knows another mass puts a muon loop in the photon line of topic 14.03's vertex. Unit 13's dispersion relation writes that loop as a spread of massive photons: 1/q² gains ∫ds (α/3π)(R(s)/s)/(q² − s), with R = (1 + 2M²/s)√(1 − 4M²/s) for a lepton pair of mass M. Topic 14.03's integral with a photon of mass √s ≫ mₑ gives (α/π)(mₑ²/s)/3, so aₑ gains (α/π)²(mₑ²/9)∫ds R(s)/s². Setting s = 4M²/v turns the integral into (1/4M²)∫₀¹(1 + v/2)√(1 − v)dv = 1/(5M²), so aₑ(M) = (1/45)(mₑ/M)²(α/π)², derived at leading order. The muon gives 2.804 × 10⁻¹² and the tau lepton 0.0099 × 10⁻¹²: heavier, smaller, as decoupling demands. Hadrons enter the same integral with R(s) measured in e⁺e⁻ → hadrons, not computed, giving about 1.85 × 10⁻¹²; higher-order hadronic graphs, which lower it, and light-by-light scattering bring their total to 1.693(12) × 10⁻¹², quoted. W and Z loops add 0.031 × 10⁻¹², quoted, a quarter of the experimental uncertainty.

Running the series backwards: aₑ measures α

Since aₑ ≈ α/(2π), the trap pins α almost as tightly as it pins aₑ: d ln aₑ/d ln α = 1 + (C₂/C₁)(α/π) + … = 0.9985, computed, so δα/α = 1.0015 δaₑ/aₑ. Solving aₑ(theory; α) = aₑ(measured) with every term gives α⁻¹ = 137.035 999 166, computed, and the trap's 1.3 × 10⁻¹³ becomes ±15 × 10⁻⁹: the published 137.035 999 166(15), 0.11 parts per billion. The theory budget is smaller. The hadrons' ±0.012 × 10⁻¹² is ±1.4 × 10⁻⁹; the five-loop coefficient has a second evaluation, 5.891, that disagrees with 6.737, and using it lowers α⁻¹ by 6.8 × 10⁻⁹. Truncating shows what each piece is worth: inverting with Schwinger's term alone gives α⁻¹ = 137.244, with two loops 137.034 25, with three 137.036 005, with four 137.035 998 58. The corrections shrink by factors of about 120 and 270, and the last, 5.8 × 10⁻⁷, is nine-tenths muon, tau and hadron loops.

The test: an α that never saw g − 2

Atom interferometry measures h/m_X for a caesium or rubidium atom; with the Rydberg constant and the mass ratio m_X/mₑ, α² = (2R∞/c)(m_X/mₑ)(h/m_X), quoted, makes no use of the g − 2 series. In SI both sides are dimensionless: R∞/c in s m⁻² times h/m_X in m² s⁻¹. Berkeley's caesium value (2018) is α⁻¹ = 137.035 999 046(27) and Paris's rubidium value (2020) 137.035 999 206(11), quoted. Put each into the full series and subtract, computed: aₑ(measured) − aₑ(theory) = −1.02 × 10⁻¹² with caesium, 3.9 combined standard deviations, and +0.34 × 10⁻¹² with rubidium, 2.1. The two atomic values differ from each other by 160 × 10⁻⁹ in α⁻¹, 5.5 standard deviations. So QED is tested at about 10⁻¹², near a part per billion of aₑ, and the open question is α, not the graphs or the trap. The muon shares A₁, but heavy-particle terms grow as the lepton mass squared, (m_μ/mₑ)² = 4.3 × 10⁴, so its hadronic and weak terms are far larger and decide its g − 2: named only.

Change one variable at a time

Make the relationship visible.

Interactive model

Five-loop coefficient C₅6.74

Hadronic term ÷ 1.693 × 10⁻¹²1.00

Trap uncertainty on aₑ0.13 × 10⁻¹²

Set the hadronic term to 0 and the zero crossing slides 200 × 10⁻⁹ to the left, past caesium's line: leave hadrons out and aₑ points to a different α. Then move C₅ between 5.89 and 6.74, its two evaluations, and the crossing shifts by 7 × 10⁻⁹, about half the trap's ±15 × 10⁻⁹.

α⁻¹ − 137.035 999 FROM aₑ166.3 × 10⁻⁹

± FROM THE TRAP15.4 × 10⁻⁹

CAESIUM RESIDUAL-3.9 σ

RUBIDIUM RESIDUAL2.1 σ

Live interpretationα⁻¹ − 137.035 999 FROM aₑ: 166.3 × 10⁻⁹. ± FROM THE TRAP: 15.4 × 10⁻⁹. CAESIUM RESIDUAL: −3.9 σ. RUBIDIUM RESIDUAL: 2.1 σ

Catch the common trap

Explain before calculating.

Misconception check

Inserting α⁻¹ = 137.035 999 166 makes the full series agree with the measured aₑ to 0.002 × 10⁻¹², and it is tempting to call that a test of QED at 10⁻¹³ or better. It is not a test at all. That α was obtained by inverting this series against this measurement, so agreement is built in: leave out the five-loop term or the hadrons and re-inverting simply moves α⁻¹ by 54 or 200 × 10⁻⁹ and restores perfect agreement. The trap's 1.3 × 10⁻¹³ measures how well aₑ fixes α, not how well QED is checked. A test needs an α obtained without g − 2, and with the caesium and rubidium recoil values the residuals are −1.02 and +0.34 × 10⁻¹², 3.9 and 2.1 standard deviations: QED is confirmed at about 10⁻¹², and what limits that test now is the disagreement between the recoil experiments.

With α⁻¹ = 137.035 999 166, the series, with five electron loops plus the muon and tau loops and the hadron and weak terms, reproduces the measured aₑ to 0.002 × 10⁻¹², while the trap's uncertainty is 0.13 × 10⁻¹². What does this agreement show about QED?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyThe measured aₑ carries an absolute uncertainty of 1.3 × 10⁻¹³. With α/π = 0.002 322 819, find (α/π)ⁿ for n = 4, 5 and 6, decide how many loops a prediction must include, and find how large the six-loop coefficient would have to be before it mattered.

Powers: (α/π)⁴ = 2.911 × 10⁻¹¹, (α/π)⁵ = 6.762 × 10⁻¹⁴ and (α/π)⁶ = 1.571 × 10⁻¹⁶. All are pure numbers, like aₑ itself.
Four loops: with C₄ = −1.912 the term is 5.57 × 10⁻¹¹, about 430 times the uncertainty. It cannot be left out.
Five loops: the bare power is 0.52 of the uncertainty, which looks negligible, but C₅ ≈ 6.7 makes the term 4.5 × 10⁻¹³, 3.5 times the uncertainty. The coefficient, not the power, decides.
Six loops: 1.3 × 10⁻¹³ ÷ 1.571 × 10⁻¹⁶ = 830, so the term reaches one standard deviation only if |C₆| is near 800; the five known coefficients lie between 0.3 and 7 in size.

AnswerFive loops are needed and four are not enough; a six-loop term would matter only if its coefficient were near 800, over a hundred times the largest known.

MediumTake α⁻¹ = 137.035 999 166, aₑ = 0.001 159 652 180 59(13) and C₁ to C₅ = 0.5, −0.328 478 965 58, 1.181 241 457, −1.912 245 765 and 6.737. Form the partial sums through five loops, then add the muon and tau loops (2.7476 × 10⁻¹²), the hadrons (1.6927 × 10⁻¹²) and the weak term (0.0305 × 10⁻¹²). Give each residual, aₑ minus the sum, in standard deviations, and check the muon loop against (1/45)(mₑ/m_μ)²(α/π)².

α/π = 1/(π × 137.035 999 166) = 0.002 322 819 464. The five terms are 0.001 161 409 732, −1.772 305 × 10⁻⁶, 1.480 420 × 10⁻⁸, −5.566 80 × 10⁻¹¹ and 4.556 × 10⁻¹³.
Residuals after one and two loops: −1.757 6 × 10⁻⁶, or −1.35 × 10⁷ standard deviations, and +1.475 × 10⁻⁸, or +1.13 × 10⁵.
After three and four: −5.074 × 10⁻¹¹ (−390) and +4.924 × 10⁻¹² (+37.9). After five: +4.469 × 10⁻¹² (+34.4); the fifth loop moved it by only 3.5.
Muon check: (mₑ/m_μ)² = 1/206.768² = 2.339 × 10⁻⁵, so (1/45)(2.339 × 10⁻⁵)(α/π)² = 5.198 × 10⁻⁷ × 5.395 × 10⁻⁶ = 2.804 × 10⁻¹². The tau adds 0.010 × 10⁻¹² and higher orders −0.066 × 10⁻¹², giving 2.748 × 10⁻¹², the quoted 2.7476 × 10⁻¹² to three decimals.
Adding the muon and tau loops leaves +1.721 × 10⁻¹² (+13.2); the hadrons leave +0.028 × 10⁻¹² (+0.22); the weak term leaves −0.002 × 10⁻¹² (−0.02).
The final near-zero is no test, because this α was itself obtained from this aₑ. What the ladder does show is which terms the trap's precision demands.

AnswerResiduals −1.76 × 10⁻⁶, +1.48 × 10⁻⁸, −5.07 × 10⁻¹¹, +4.92 × 10⁻¹² and +4.47 × 10⁻¹², then +1.72, +0.03 and −0.002 × 10⁻¹²: the hadrons are needed at 13 standard deviations, the weak term not yet.

HardInvert the full series for α. From the trial α⁻¹ = 137.036, evaluate the theory's aₑ, correct α⁻¹ using the slope d aₑ/d(α⁻¹) from the first two terms, and propagate the trap's 1.3 × 10⁻¹³, the hadrons' ±0.012 × 10⁻¹² and the alternative five-loop value C₅ = 5.891. Then test the result against caesium, α⁻¹ = 137.035 999 046(27), and rubidium, 137.035 999 206(11).

At α⁻¹ = 137.036 the full series, with C₅ = 6.737 and 4.471 × 10⁻¹² of heavy-particle terms, gives aₑ = 0.001 159 652 173 545, so measured minus theory is +7.045 × 10⁻¹².
Slope: d aₑ/dα = (1/π)(C₁ + 2C₂ α/π) = (1/π)(0.5 − 0.001 526) = 0.158 669, and d(α⁻¹) = −dα/α², so d aₑ/d(α⁻¹) = −α² × 0.158 669 = −5.325 14 × 10⁻⁵ × 0.158 669 = −8.449 × 10⁻⁶.
Newton step: δ(α⁻¹) = 7.045 × 10⁻¹² ÷ (−8.449 × 10⁻⁶) = −8.338 × 10⁻⁷, so α⁻¹ = 137.035 999 166. The series is so nearly linear here that a second step changes nothing at this precision.
Budget: 1.3 × 10⁻¹³ ÷ 8.449 × 10⁻⁶ = 15.4 × 10⁻⁹ from the trap and 0.012 × 10⁻¹² ÷ 8.449 × 10⁻⁶ = 1.4 × 10⁻⁹ from the hadrons, 15.5 × 10⁻⁹ in quadrature. C₅ = 5.891 removes 0.846(α/π)⁵ = 5.72 × 10⁻¹⁴ and lowers α⁻¹ by 6.8 × 10⁻⁹.
Caesium's α⁻¹ is 120 × 10⁻⁹ lower, so the theory's aₑ rises by 120 × 10⁻⁹ × 8.449 × 10⁻⁶ = 1.01 × 10⁻¹²; the exact residual is −1.016 × 10⁻¹². Its uncertainty is √(0.228² + 0.13²) × 10⁻¹² = 0.26 × 10⁻¹², so −3.9 standard deviations.
Rubidium's is 40 × 10⁻⁹ higher: residual +0.336 × 10⁻¹², uncertainty √(0.093² + 0.13²) × 10⁻¹² = 0.16 × 10⁻¹², so +2.1 standard deviations. The two recoil values differ by 160 × 10⁻⁹ against √(27² + 11²) × 10⁻⁹ = 29 × 10⁻⁹: 5.5 standard deviations.

Answerα⁻¹ = 137.035 999 166(15), with a possible −7 × 10⁻⁹ from the five-loop disagreement; aₑ then sits 3.9 standard deviations from caesium's α and 2.1 from rubidium's, which disagree with each other by 5.5.