University Quantum Field Theory I · Vacuum Polarisation, Charge Renormalisation and the Running of alpha · 13.5
Running α: the effective charge at large momentum transfer
Every collider number, from a Bhabha luminosity count to the Z line shape, needs α at the momentum its photon carries, not at the q² = 0 where Thomson scattering fixed it. Here the bubble chain becomes that coupling, its logarithm is derived, and it is run to the Z: leptons by calculation, hadrons by data.
Build the model
Connect the measurement to the mechanism.
Topic 13.03 fixed α = 1/137.036 at q² = 0 and left a finite Π̂₂(q²); now ask what coupling a photon of virtuality q² transmits. Every one-photon exchange carries the chain of bubble insertions, Unit 12's geometric series sums it to e²/(q²[1 − Π̂₂(q²)]), and so αeff(q²) = α/[1 − Π̂₂(q²)] is the coupling at that momentum transfer: α itself at q² = 0, and the same for every probe, because the Ward identity keeps the vertex out of the charge. For a scattering photon, q² = −Q², Π̂₂ is real and positive, so the vacuum of virtual pairs screens, and a probe resolving distances inside 1/m, 386 fm for the electron, sees more charge.
For Q ≫ m the Feynman-parameter integral collapses to αeff = α/[1 − (α/3π) ln(Q²/(A m²))] with A = e⁵⁄³, derived, and because the chain is geometric it carries the leading logarithm at every order, as topic 6.09's running λ did. Each lepton adds its own logarithm once Q passes about 2.3 times its mass and almost nothing below, so at Q = 91.19 GeV the electron alone takes 1/α to 134.65 and the three leptons to 132.73, computed. There the calculation stops: a quark loop needs a quark mass that confinement does not supply, so the hadronic share, Δαhad ≈ 0.0276, comes from topic 13.04's dispersion integral over measured R(s) and brings 1/α to about 128.95, quoted.
Beyond that, the chain misses two-loop insertions, about 8 × 10⁻⁵ in Δα.
- Simple definition
- The effective charge αeff(q²) = α/[1 − Π̂₂(q²)] is the coupling a photon of virtuality q² transmits once every chain of vacuum-polarisation bubbles on it is summed: exactly 1/137.036 at q² = 0, and growing logarithmically with large spacelike momentum transfer.
- Example
- At spacelike Q = 91.19 GeV the electron loop alone gives Π̂₂ = (α/3π)[ln(Q²/mₑ²) − 5/3] = (7.743 × 10⁻⁴)(22.52) = 0.01743, so 1/αeff = 137.036 × (1 − 0.01743) = 134.65: that photon couples 1.8% more strongly than Coulomb's law.
Derived: the geometric series of 1PI insertions turns e²/q² into e²/(q²[1 − Π̂₂]), so every one-photon amplitude is its tree form with α replaced by αeff(q²).
α = 1/137.036 and Π̂₂ are pure numbers; q² in GeV², with q² = −Q² for a scattering photon; Π̂₂(0) = 0, so αeff(0) = α
Topic 13.03's Π̂₂ at q² = −Q², derived. Its sign is screening: αeff exceeds α at every spacelike Q and rises with it.
Q² = −q² in GeV²; m the loop lepton's mass in GeV; the integrand is positive at every x
Derived from ∫x(1 − x)dx = 1/6 and ∫x(1 − x) ln[x(1 − x)]dx = −5/18. The logarithm crosses zero at Q/m = e⁵⁄⁶ = 2.30, the lepton's effective onset.
Q and m in GeV; α/3π = 7.743 × 10⁻⁴; the dropped terms add 6(m/Q)² to the bracket, a power with no logarithm
Derived. A loop too heavy to be excited moves the charge only by a power of Q/m: it neither screens nor antiscreens, which is why heavy leptons leave atomic physics alone.
a pure number, from ln(1 + u) ≈ u and ∫x²(1 − x)²dx = 1/30; a τ loop at Q = 1.00 GeV gives 4.9 × 10⁻⁵ (exact 4.7 × 10⁻⁵)
Computed. 1/αeff is linear in ln Q, falling 2/(3π) = 0.212 per e-fold for each lepton, so A and the masses cancel from any difference between two large Q.
sum over e, μ, τ with Q ≫ each mass; at Q = 91.19 GeV the terms are 2.389, 1.258 and 0.659, so 1/αeff = 132.73
Gives 1/α(MZ²) ≈ 128.95, quoted. Hadrons supply nearly half the running, and no quark loop with a physical quark mass reproduces their share.
Δαₗₑₚ = 0.0314 at one loop, computed; Δαhad ≈ 0.0276 from the dispersion integral over measured R(s), quoted
One coupling for each momentum transfer
Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout. Take the photon exchanged in e⁻μ⁻ → e⁻μ⁻ and dress it with every chain of one-particle-irreducible bubbles. Unit 12's resummation, quoted, sums the chain to −igμν/(q²[1 − Π̂₂(q²)]) once the photon counterterm has made Π̂₂ finite, and the qμ qν terms drop against the conserved currents at either end. The amplitude is therefore the tree amplitude with e² replaced by e²/[1 − Π̂₂(q²)], which defines αeff(q²) = α/[1 − Π̂₂(q²)], derived. Two facts make it a coupling rather than a correction. At q² = 0 it is α = 1/137.036, because topic 13.03 set Π̂₂(0) = 0 so that Thomson scattering defines the charge. And it belongs to the photon, not to the process: the Ward identity Z₁ = Z₂, quoted, keeps vertex and leg corrections from changing the charge, so every charged particle exchanging this photon sees the same αeff, while vertex and box graphs remain process-specific corrections on top.
The sign of Π̂₂ is screening
Set q² = −Q² for a scattering photon. Topic 13.03's Π̂₂ = −(2α/π)∫x(1 − x) ln[m²/(m² − x(1 − x)q²)] becomes (2α/π)∫₀¹dx x(1 − x) ln[1 + x(1 − x)Q²/m²], derived, whose integrand is positive at every x. So Π̂₂ is real, positive and rising with Q², and αeff(−Q²) exceeds α at every spacelike momentum. The picture is a dielectric. Virtual pairs spread over about one Compton wavelength, 1/mₑ = 386 fm for the electron, around any charge; Coulomb's law and Thomson scattering see the charge from outside that cloud, and the screened value they see is what 1/137.036 names. A photon with Q = 1.00 GeV resolves 0.197 fm, deep inside the cloud, and sees less of the screening: the electron loop alone gives αeff = α/(1 − 0.0104), about 1% more charge. No sign was chosen here. The α/3π multiplying the logarithm is the one topic 13.04 tied to a positive pair-creation rate, so screening is forced by unitarity.
Large Q: a logarithm and the constant A = e⁵⁄³
For Q² ≫ m² split the logarithm: ln[1 + x(1 − x)Q²/m²] = ln[x(1 − x)Q²/m²] + ln[1 + m²/(x(1 − x)Q²)]. Away from the ends the second piece is about m²/(x(1 − x)Q²), so against the weight x(1 − x) it adds about m²/Q² at every x, and the end slivers of width m²/Q² add less still: dropping it costs 6(m/Q)² in the bracket below, a power with no logarithm. The first piece splits into ln(Q²/m²)∫x(1 − x)dx = (1/6) ln(Q²/m²) and ∫x(1 − x) ln[x(1 − x)]dx, which the symmetry x ↔ 1 − x makes twice ∫x(1 − x) ln x dx = 2(−1/4 + 1/9) = −5/18. Times 2α/π: Π̂₂ = (α/3π)[ln(Q²/m²) − 5/3] = (α/3π) ln(Q²/(A m²)), A = e⁵⁄³ = 5.29, derived. Checks: Q²/(A m²) is a pure number; at Q/m = 10 the bracket is 2.938 against the exact integral's 2.995, a gap of 0.057 close to 6(m/Q)² = 0.060, and at Q/m = 100 it is 7.5437 against 7.5443. Below Q/m = e⁵⁄⁶ = 2.30 the logarithm turns negative, while the exact Π̂₂ stays positive and falls to αQ²/(15πm²) for Q ≪ m. That limit is decoupling: a lepton contributes its logarithm only above about twice its mass, and nothing measurable well below it.
The geometric series sums the leading logarithms
Write L = ln(Q²/(A m²)). Expanding αeff = α/[1 − (α/3π)L] gives α[1 + αL/3π + (αL/3π)² + …]: the chain of n bubbles supplies αⁿ⁺¹Lⁿ, the largest logarithm at each order, and the resummed form keeps every one, which is topic 6.09's lesson for λ in a new coupling. It also makes 1/αeff = 1/α − L/(3π) exactly linear in ln Q, falling 2/(3π) = 0.212 per e-fold of Q, or 0.489 per decade, for each lepton. Between two scales far above m the constants cancel: 1/αeff(Q₁) − 1/αeff(Q₂) = (2/3π) ln(Q₂/Q₁), the scheme-free slope that topic 13.06 turns into a beta function. The resummation is not cosmetic. For the three leptons at the Z, αL/3π summed is 0.0314; the one-loop truncation α(1 + 0.0314) gives 1/α = 132.86, the geometric sum 132.73, a gap thirteen times the 0.01 to which 1/α(MZ²) is known. What the chain cannot hold is the two-loop self-energy, of order α²L, a next-to-leading logarithm; it adds about 8 × 10⁻⁵ to Δαₗₑₚ, quoted.
Three leptons at the Z
Run the leptons to Q = MZ = 91.19 GeV with mₑ = 0.511 MeV, mμ = 105.66 MeV and mτ = 1776.9 MeV, quoted. The electron term is (1/3π)[ln(Q²/mₑ²) − 5/3] = (24.184 − 1.667)/9.425 = 2.389; the muon's is (13.521 − 1.667)/9.425 = 1.258, and the tau's (7.876 − 1.667)/9.425 = 0.659, which the exact closed form raises by about 6(mτ/Q)²/(3π) = 2.4 × 10⁻⁴. So 1/αeff = 137.036 − 2.389 = 134.65 with the electron alone and 132.73 with all three, computed, a Δαₗₑₚ of 4.306/137.036 = 0.0314. The electron dominates only logarithmically: each term is a lever arm ln(Q/m), and heavier leptons have shorter arms. The same arithmetic shapes the figure: flat at 137.036 below about 1 MeV, then a slope of 0.212 per e-fold from the electron, doubled above a few hundred MeV and tripled above about 4 GeV, as each heavier lepton passes 2.3 times its mass. LEP's small-angle Bhabha data, at Q² of a few GeV², saw this growth directly, quoted.
What the hadrons add, and why only data can say
The value used at the Z is 1/α(MZ²) ≈ 128.95, quoted, so after the leptons the hadrons must still lower 1/α by 132.73 − 128.95 = 3.78, a Δαhad of 0.0276. A quark loop is the lepton graph with charge Qq e in three colours, lowering 1/α by Nc Qq²(1/3π) ln(Q²/(A mq²)) per flavour, but it needs a quark mass, and for u, d and s confinement supplies none: their spectral weight sits in the ρ, ω and φ and the two-pion continuum from 280 MeV, not at a free-quark threshold. Current masses of a few MeV overshoot the running by more than a unit of 1/α, constituent masses near 330 MeV fall short by 0.7, and the mass that fits, about 68 MeV, is a fit, not a prediction. So the hadronic share is taken from topic 13.04's once-subtracted dispersion relation over measured R(s): Δαhad ≈ 0.0276, quoted, with compilations differing by about 10⁻⁴, and that spread, not the leptons, limits 1/α(MZ²). Tables also list 1/α̂(MZ) ≈ 127.95 in the MS-bar scheme; the two differ by which constants and heavy loops each convention absorbs, topic 13.06's business, not by physics.
Change one variable at a time
Make the relationship visible.
Start at the Z: the leptons stop at 132.73 and the faint quark curve misses the square. Lower mq until it lands, near 68 MeV, a fit rather than a quark mass. Then drag Q to 1 GeV and compare the two lepton readouts: below the τ's mass its leading log has the wrong sign.
1/αeff, ELECTRON LOOP134.65
1/αeff, e + μ + τ132.73
LEADING LOG, e + μ + τ132.73
WITH FREE-QUARK MODEL129.58
Live interpretation1/αeff, ELECTRON LOOP: 134.65. 1/αeff, e + μ + τ: 132.73. LEADING LOG, e + μ + τ: 132.73. WITH FREE-QUARK MODEL: 129.58
Catch the common trap
Explain before calculating.
Counting only the three charged leptons and using the leading-log form, how does 1/αeff change when the spacelike momentum transfer rises from Q = 20.0 GeV to Q = 91.2 GeV?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyIn small-angle Bhabha scattering the exchanged photon carries Q² = 6.0 GeV². Using the electron loop alone in the leading-log form, find Π̂₂, 1/αeff and the percentage by which αeff exceeds α = 1/137.036, and compare the distance 1/Q with the electron's Compton wavelength 1/mₑ.
- Q = √6.0 GeV = 2.449 GeV and mₑ = 0.511 MeV, so Q²/mₑ² = 6.0 GeV²/(5.110 × 10⁻⁴ GeV)² = 2.30 × 10⁷ and ln(Q²/mₑ²) = 16.950. The argument is a ratio of squared masses, a pure number, as a logarithm demands.
- Subtract 5/3 to bring in A: ln(Q²/(A mₑ²)) = 16.950 − 1.667 = 15.283, so Π̂₂ = (α/3π)(15.283) = (7.743 × 10⁻⁴)(15.283) = 0.01183.
- 1/αeff = (1 − Π̂₂)/α = 137.036 × 0.98817 = 135.41, the same as 137.036 − 15.283/(3π) = 137.036 − 1.622. The ratio αeff/α = 1/(1 − 0.01183) = 1.0120, so the photon couples 1.20% more strongly than at q² = 0.
- Distances: 1/Q = ħc/Q = (0.197327 GeV fm)/(2.449 GeV) = 0.0806 fm, against 1/mₑ = 386 fm, a factor 4.8 × 10³ inside the polarisation cloud. Since Q/mₑ = 4.8 × 10³ is far above 2.3, the leading log is safe: the exact integral agrees to better than 10⁻⁷ in 1/αeff.
AnswerΠ̂₂ = 0.0118, 1/αeff = 135.41 and αeff exceeds α by 1.20%; 1/Q = 0.0806 fm, 4.8 × 10³ times smaller than 1/mₑ = 386 fm. The muon loop, at Q/mμ = 23, would lower 1/αeff by a further 0.49.
MediumCounting leptonic vacuum polarisation only, at what spacelike Q has 1/αeff fallen from 137.036 to 136.000? Solve with the electron's leading log alone, then test the muon there with the closed form of topics 13.02 and 13.04, (α/3π)[−5/3 + 4 m²/Q² + (1 − 2 m²/Q²)β ln((β + 1)/(β − 1))], β = √(1 + 4 m²/Q²), quoted, and correct the answer.
- Electron alone: 1.036 = (1/3π) ln(Q²/(A mₑ²)), so ln(Q²/(A mₑ²)) = 3π × 1.036 = 9.764 and Q = mₑ√A e(4.882) = 0.511 MeV × 2.301 × 131.9 = 155.1 MeV. Q/mₑ = 303, so the electron's logarithm is safe.
- The muon is not obviously negligible: Q/mμ = 155.1/105.66 = 1.468, inside neither limit. The leading log gives (1/3π)(2 ln 1.468 − 5/3) = −0.095, a rise in 1/α that a positive integrand forbids; the small-Q form Q²/(5mμ²)/(3π) gives 0.046.
- The closed form: β = √(1 + 4/1.468²) = √2.857 = 1.690, and the bracket is 0.355, so the muon lowers 1/αeff by 0.355/(3π) = 0.0376 at 155.1 MeV. The electron alone therefore overshoots the target.
- Solve electron plus muon (plus τ) = 1.036 numerically, by bisection in Q: Q = 134.8 MeV, where the electron supplies 1.0062, the muon at Q/mμ = 1.276 supplies 0.0297, and the τ at Q/mτ = 0.076 adds 1.2 × 10⁻⁴ through αQ²/(15πm²).
- The muon, still below its effective onset 2.3 mμ = 243 MeV, moves the answer down by 13%: thresholds are smooth, and the closed form is needed within a few times each mass. Hadronic loops, left out here, also begin to count in this range.
AnswerQ ≈ 135 MeV (134.8 MeV), against 155.1 MeV from the electron alone; the muon, at Q/mμ = 1.28, supplies 0.030 of the 1.036 drop.
HardTake 1/α(MZ²) = 128.95, quoted, at MZ = 91.19 GeV. Find what the hadrons must add after the three leptons, as a shift in 1/α and as Δαhad. Then model the hadrons as free quark loops with Nc = 3, mc = 1.5 GeV, mb = 4.7 GeV and one effective mass mq for u, d and s: find the mq that fits, in the leading-log form, and the 1/α(MZ²) that 5 MeV and 330 MeV would give.
- Leptons: 1/αeff(MZ²) = 132.730 from the three logarithms. The hadrons must lower it by 132.730 − 128.95 = 3.780, a Δαhad of 3.780/137.036 = 0.0276, matching the dispersion-integral value quoted from topic 13.04.
- Each quark flavour is a lepton loop times Nc Qq². Charm: (4/3)(1/3π)[ln(MZ²/mc²) − 5/3] = (4/3)(8.215 − 1.667)/9.425 = 0.926. Bottom: (1/3)(5.931 − 1.667)/9.425 = 0.151. Together they give 1.077.
- Light quarks: Nc ΣQq² = 3(4/9 + 1/9 + 1/9) = 2, so (2/3π)[ln(MZ²/mq²) − 5/3] must equal 3.780 − 1.077 = 2.703. Then ln(MZ²/mq²) = 2.703 × 3π/2 + 5/3 = 12.736 + 1.667 = 14.403, MZ/mq = e(7.202) = 1342, and mq = 91.19 GeV/1342 = 68 MeV.
- Current masses, mq = 5 MeV: ln(MZ²/mq²) = 19.622, the light term is (2/3π)(17.956) = 3.810, and 1/α(MZ²) = 132.730 − 1.077 − 3.810 = 127.84, too low by 1.1. Constituent masses, 330 MeV: ln = 11.243, the light term is (2/3π)(9.577) = 2.032, and 1/α(MZ²) = 129.62, too high by 0.67.
- Neither physical mass works, and the 68 MeV that does is a number fitted to the answer. The light-quark spectral weight lies in the ρ, ω, φ and the ππ continuum, so the hadronic share must come from measured R(s) through the dispersion relation; free quark loops serve only as a check well above a few GeV.
AnswerThe hadrons must lower 1/α by 3.78, Δαhad = 0.0276. The free-quark model fits only with mq ≈ 68 MeV for u, d and s; 5 MeV gives 127.84 and 330 MeV gives 129.62.