University Quantum Field Theory I · Vacuum Polarisation, Charge Renormalisation and the Running of alpha · 13.3
Charge renormalisation: α = 1/137.036 defined at q² = 0
Topic 13.02 left the photon's self-energy with a pole in ε. Here you decide what the charge of QED means by naming the measurement that fixes it, and watch the pole disappear into a bare charge nobody measures, leaving every exchanged photon with a finite correction you can compute at any momentum transfer.
Build the model
Connect the measurement to the mechanism.
The e₀ written in the QED Lagrangian is never measured. Experiments measure how strongly charges act on one another, and topic 12.09's resummed propagator says what that is: between conserved currents a photon carries e₀²/(q²[1 − Π(q²)]). As q² → 0, the regime of Coulomb's law between distant charges and of Thomson scattering, exact to all orders by topic 10.07's low-energy theorem, this becomes Z₃e₀²/q² with Z₃ = 1/(1 − Π(0)), and Z₁ = Z₂ (topic 12.07) keeps vertex and leg corrections out of it.
So the charge is defined by e² = Z₃e₀², and α = e²/4π = 1/137.036. At one loop Π₂(0) is negative and carries topic 13.02's pole, so Z₃ = 1 − (α/3π)[2/ε − γE + ln(4πμ²/m²)] lies below 1: the bare charge is larger than the measured one, screened by virtual pairs. Dividing through by 1 − Π(0) then writes every exchange as e²/(q²[1 − Π̂(q²)]), and at one loop Π̂₂ = Π₂(q²) − Π₂(0) is a convergent integral with no ε, μ or cutoff left in it: the divergence has been traded for one measured number.
The definition has limits. Π̂₂ is the first term of a series in α/π; above q² = 4m² it turns complex (topic 13.04); every charged species adds its own loop, and the hadronic one must be taken from data; and with a gauge-invariant cutoff the electron loop's inverse bare coupling, 1/α₀ = 1/α − (1/3π) ln(Λ²/m²), reaches zero at Λ ≈ 10²⁷⁷ GeV, where the bare charge diverges: the Landau pole of topic 13.06.
- Simple definition
- Charge renormalisation defines the physical charge by photon exchange at zero momentum transfer, e² = Z₃e₀² with Z₃ = 1/(1 − Π(0)), so α = e²/4π = 1/137.036 is the constant of long-range Coulomb forces and of Thomson scattering, not the coefficient in the Lagrangian.
- Example
- With a gauge-invariant cutoff at Λ = 1.00 TeV the electron loop gives Π₂(0) = −(α/3π) ln(Λ²/m²) = −0.0224, so Z₃ = 0.978 and 1/α₀ = 133.96 against the measured 137.036, while Π̂₂ at q² = −(51.1 MeV)² is 5.84 × 10⁻³ whatever Λ is.
Derived from the resummed propagator as q² → 0, with Z₁ = Z₂ (topic 12.07) keeping vertex and leg factors out. One Z₃ serves every charged species.
e and e₀ dimensionless in natural units; α = e²/4π = 1/137.036; Π(q²) dimensionless, the 1PI photon self-energy
Derived as 1 + Π₂(0) from topic 13.02's loop. Π₂(0) is negative with any regulator, so Z₃ lies below 1, as topic 4.03's 0 ≤ Z₃ ≤ 1 allows; Λ = 1 TeV gives 0.978.
d = 4 − ε; μ and m = 0.511 MeV in GeV; γE = 0.5772. A gauge-invariant cutoff Λ gives 1 − (α/3π) ln(Λ²/m²)
Derived. α₀ exceeds α for every Λ above m: vacuum polarisation screens. Λ = 1 TeV gives 1/α₀ = 133.96; it reaches zero only near 10²⁷⁷ GeV (topic 13.06).
α₀ = e₀²/4π, dimensionless; Λ the gauge-invariant cutoff in GeV; electron loop only
Derived by dividing numerator and denominator by 1 − Π(0). Since Π̂(0) = 0, the long-range exchange is exactly e²/q².
q² in GeV²; the factor between conserved currents, in GeV⁻²; exact, before any expansion in α
Derived from the condition of residue 1 at q² = 0. The renormalised one-loop Π = Π₂(q²) − δ₃ is then Π̂₂, reached without ever writing e₀.
δ₃ dimensionless, the coefficient in topic 12.02's vertex −i(q²gμν − qμqν)δ₃; fixed by Π(0) = 0
Derived. No ε, μ or Λ: every regulator gives this, up to q²/Λ² terms. At q² = −(51.1 MeV)² it is 5.84 × 10⁻³.
q² in GeV²; positive for spacelike (negative) q²; for |q²| ≪ m², Π̂₂ ≈ −αq²/(15πm²)
No experiment measures e₀
Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. The Lagrangian carries a coefficient e₀, but no apparatus couples to a Lagrangian: what is measured is how strongly charges act on one another, loops included. Two measurements sit at zero momentum transfer. Charges held a distance r apart exchange photons with |q| of order 1/r, and once r is far beyond the electron's reduced Compton wavelength, 1/m = 386 fm, the force is Coulomb's α/r². A photon of vanishing energy scattering off a free electron gives σT = 8πα²/(3m²) = (8π/3)(αħc/mc²)² = 0.6652 b, exact to all orders when e and m are the physical charge and mass: topic 10.07's low-energy theorem, quoted. Either fixes α, known today as 1/137.035999 to about 0.1 parts per billion from the electron's g − 2 and from atom-recoil experiments, quoted. So the question is not what e₀ is. It is how e₀, which carries the regulator, relates to the e those experiments return.
The residue of the dressed photon is the charge
Between conserved currents topic 12.09's resummed propagator acts as −igμν/(q²[1 − Π(q²)]), so one-photon exchange carries e₀²/(q²[1 − Π(q²)]) times the two currents. Vertex corrections and the electrons' wave-function factors could also change the strength, but with Z₁ = Z₂ (topic 12.07) they cancel at zero momentum transfer, where the on-shell vertex is exactly −ieγμ, quoted. As q² → 0 the exchange tends to e₀²/(q²[1 − Π(0)]) = Z₃e₀²/q², with Z₃ = 1/(1 − Π(0)) the residue topic 12.09 found. That is what the Coulomb and Thomson measurements return, so the physical charge is defined by e² = Z₃e₀², derived. Z₃ belongs to the photon, so every species' charge is rescaled by the same factor, and the proton and electron charges, equal and opposite to 1 part in 10²¹ (quoted), stay so after renormalisation. In four dimensions e₀, e and Π are pure numbers, so the definition is dimensionally consistent at every order.
At one loop Z₃ is below 1 and the bare charge is larger
To order α, Z₃ = 1/(1 − Π₂(0)) = 1 + Π₂(0) + O(α²). At q² = 0 topic 13.02's Δ is m², so Π₂(0) = −(α/3π)[2/ε − γE + ln(4πμ²/m²)] and Z₃ = 1 − (α/3π)[2/ε − γE + ln(4πμ²/m²)], derived. A Pauli-Villars regulator of mass Λ, four-dimensional and gauge invariant, replaces 2/ε − γE + ln 4πμ² by ln Λ² (topic 13.02, quoted), giving Z₃ = 1 − (α/3π) ln(Λ²/m²). Either way Π₂(0) is negative and Z₃ is below 1, as topic 4.03's 0 ≤ Z₃ ≤ 1 allows, so e₀² = e²/Z₃ exceeds e². At Λ = 1.00 TeV, ln(Λ²/m²) = 28.97 and Z₃ = 0.978, and since 1/α₀ = Z₃/α, 1/α₀ = 137.036 − 28.97/3π = 133.96, derived. The picture is screening: virtual pairs polarise the vacuum around a bare charge and partly cancel it as seen from far away. Nothing observes α₀. Raise Λ to 10 TeV and 1/α₀ becomes 133.47 while α stays 1/137.036, because α was the input and α₀ the output.
Away from q² = 0: divide by 1 − Π(0)
The definition pays off at every other q². Divide numerator and denominator of e₀²/(q²[1 − Π(q²)]) by 1 − Π(0): the numerator becomes Z₃e₀² = e² and the denominator 1 − Z₃[Π(q²) − Π(0)]. So every exchanged photon carries e²/(q²[1 − Π̂(q²)]) with Π̂ = Z₃[Π(q²) − Π(0)], exact and derived. To order α, Π̂₂(q²) = Π₂(q²) − Π₂(0) = −(2α/π)∫₀¹ dx x(1 − x) ln[m²/(m² − x(1 − x)q²)], derived. The pole, γE, ln 4π and ln μ² are gone: topic 13.02 showed they multiply the same ∫₀¹ x(1 − x) dx = 1/6 at every q², so subtracting the q² = 0 value removes them, and a Pauli-Villars Λ cancels the same way, up to terms of order q²/Λ². What survives is a convergent integral of m and q² alone, so the prediction is regulator-independent. It is 1.404 × 10⁻⁴ at q² = −m² and 3.327 × 10⁻³ at q² = −(10.0 MeV)², and positive for every spacelike q²: a probe at nonzero Q sees inside part of the screening cloud, so the exchange is stronger than e²/q².
The counterterm route reaches the same Π̂₂
Topic 12.02 wrote QED in renormalised fields, A₀ = √Z₃ A, so −¼Z₃F² splits into −¼F² and a counterterm vertex −i(q²gμν − qμqν)δ₃ with δ₃ = Z₃ − 1, quoted. Since e₀Z₂√Z₃ = Z₁e and Z₁ = Z₂, the combination e₀A₀ = eA that couples to the electron is unchanged: the measured charge already sits in the Feynman rules. The counterterm adds −δ₃ to the one-loop Π, so Π(q²) = Π₂(q²) − δ₃. The on-shell condition, residue 1 at q² = 0, is Π(0) = 0, so δ₃ = Π₂(0) = −(α/3π)[2/ε − γE + ln(4πμ²/m²)] and Π(q²) = Π₂(q²) − Π₂(0) = Π̂₂, both derived: the same function as before. The two routes are one calculation kept in two ledgers. Bare perturbation theory puts the divergence into the relation between e₀ and e, renormalised perturbation theory puts it into δ₃, and in both only α and Π̂₂ reach an observable. Changing μ shifts δ₃ and Π₂(q²) together by −(α/3π) ln(μ′²/μ²) and leaves Π̂₂ alone.
Checks, limits and where the definition stops
Dimensions: Π̂₂ is a pure number, its logarithm's argument a ratio of masses squared. Free-field limit: as α → 0, Z₃ → 1 and e₀ = e. For |q²| ≪ m² the logarithm expands to Π̂₂ ≈ −αq²/(15πm²), so Π̂ vanishes at q² = 0 and the force between distant charges is α/r², the correction carrying a factor e(−2mr), 10⁻¹¹⁹ at the Bohr radius, where 2mr = 274. The q²/m² term acts as a contact interaction that lowers hydrogen's 2s level by 27.1 MHz (UAQM-13.09, quoted; topic 13.08 derives it). A heavier lepton's Π̂₂ is smaller by (mₑ/m_ℓ)², 2.3 × 10⁻⁵ for the muon, though its share of Z₃ is comparable to the electron's: heavy loops decouple from everything but α itself. For −q² ≫ m², Π̂₂ → (α/3π)[ln(−q²/m²) − 5/3], quoted from topic 13.02 and growing without bound (topics 13.05 and 13.06), and above q² = 4m² the logarithm's argument turns negative and Π̂₂ gains an imaginary part (topic 13.04). One loop stops being enough at relative order α/π, and the hadronic loop must come from data.
Change one variable at a time
Make the relationship visible.
Drag the cutoff from 10² to 10⁸ GeV: the lower curve and the dashed line fall together, 3.57 × 10⁻³ per decade, and Z₃ with them, yet the gap between them, Π̂₂, never moves. Then slide Q below m and watch Π̂₂ vanish: that is why α fixed at q² = 0 is the charge of Coulomb's law.
Z₃ AT ONE LOOP0.9776
BARE 1/α₀133.96
REGULATED Π₂(−Q²)-16.59 × 10⁻³
FINITE Π̂₂(−Q²)5.841 × 10⁻³
Live interpretationZ₃ AT ONE LOOP: 0.9776. BARE 1/α₀: 133.96. REGULATED Π₂(−Q²): −16.59 × 10⁻³. FINITE Π̂₂(−Q²): 5.841 × 10⁻³
Catch the common trap
Explain before calculating.
The electron loop is regulated with a gauge-invariant cutoff Λ = 1.00 TeV, and α = 1/137.036 is fixed by Thomson scattering. The cutoff is then doubled and the charge renormalised again under the same condition. What changes at one loop?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyThomson scattering fixes α. Using α = 1/137.036, m = 0.511 MeV and ħc = 197.327 MeV fm, evaluate σT = 8πα²/(3m²) in barns. Then, with the electron loop regulated by a gauge-invariant cutoff Λ = 1.00 TeV, find Z₃ and the bare α₀. A student puts α₀ into the Thomson formula: by how much is the prediction off, and why is the formula with α the right one?
- Restore ħ and c: σT = (8π/3)rₑ² with rₑ = αħc/(mc²) = 197.327 MeV fm/(137.036 × 0.511 MeV) = 2.818 fm, so σT = 8.378 × 7.941 fm² = 66.52 fm² = 0.6652 b, since 1 b = 100 fm².
- Z₃ at one loop: ln(Λ²/m²) = 2 ln(1.00 × 10⁶ MeV/0.511 MeV) = 28.97 and α/3π = 7.743 × 10⁻⁴, so Π₂(0) = −0.02243 and Z₃ = 1 + Π₂(0) = 0.9776.
- Bare coupling: 1/α₀ = Z₃/α = 137.036 × 0.9776 = 133.96, so α₀ is larger than α by the factor 1/Z₃ = 1.0229.
- σT goes as α², so α₀ would predict (1.0229)² = 1.0464 times as much, 0.6961 b: 4.6 per cent too large, and a number that changes whenever Λ does.
- Why α: in Compton scattering both photons are external, and LSZ gives each a factor √Z₃, so the two vertices carry e₀²Z₃ = e². Topic 10.07's theorem then makes 8πα²/(3m²) exact at ω → 0. Checks: α²/m² is an area, as a cross-section must be, and as α → 0, Z₃ → 1 and the two predictions agree.
AnswerσT = 0.6652 b. At Λ = 1.00 TeV, Z₃ = 0.978 and 1/α₀ = 133.96; using α₀ would overstate σT by 4.6 per cent, at 0.696 b, and make it depend on the regulator.
MediumElectrons scatter from a heavy nucleus at Q = 3.00 MeV, so q² = −(3.00 MeV)². In dimensional regularisation with the renormalised Lagrangian, find δ₃ at μ = m and at μ = 1.00 GeV, show that the renormalised one-loop Π(q²) is the same for both, evaluate it and the fractional rise it gives the one-photon-exchange cross-section, and compare the small-q² and large-Q² approximations.
- Π(0) = 0 fixes δ₃ = Π₂(0) = −(α/3π)[2/ε − γE + ln 4π + ln(μ²/m²)]. With α/3π = 7.743 × 10⁻⁴ and −γE + ln 4π = 1.9538: at μ = m, δ₃ = −1.549 × 10⁻³/ε − 1.513 × 10⁻³. At μ = 1.00 GeV, ln(μ²/m²) = 15.16 adds −1.174 × 10⁻², so δ₃ = −1.549 × 10⁻³/ε − 1.325 × 10⁻².
- Π₂(q²) has the same pole and constants with Δ = m² − x(1 − x)q² in place of m². So Π(q²) = Π₂(q²) − δ₃ = −(2α/π)∫₀¹ dx x(1 − x) ln(m²/Δ): 1/ε, γE, ln 4π and ln μ² cancel at either μ, leaving Π̂₂ with no trace of the scale.
- With Q/m = 5.871, numerical integration, or topic 13.02's closed form at U = 4m²/Q² = 0.1161, gives Π̂₂ = 1.570 × 10⁻³.
- The cross-section goes as |e²/(q²[1 − Π̂₂])|², so it rises by the factor (1 − Π̂₂)⁻² = 1.00315, or 0.315 per cent. The vertex correction and soft-photon emission of Unit 14 enter at the same order and are not included here.
- Approximations: αQ²/(15πm²) gives 5.34 × 10⁻³, 3.4 times too large, because Q/m = 5.9 is not small; (α/3π)[ln(Q²/m²) − 5/3] gives 1.450 × 10⁻³, 8 per cent low. At Q of a few m only the full integral is reliable.
Answerδ₃ = −1.549 × 10⁻³/ε − 1.513 × 10⁻³ at μ = m and −1.549 × 10⁻³/ε − 1.325 × 10⁻² at μ = 1.00 GeV; in both, Π = Π̂₂ = 1.570 × 10⁻³, raising the cross-section by 0.315 per cent.
HardAdd the muon (105.66 MeV) and τ lepton (1776.86 MeV) loops to the electron's, all regulated by a gauge-invariant cutoff Λ = 1.00 TeV. Find each lepton's share of 1 − Z₃, then Z₃ and 1/α₀. Find each lepton's contribution to Π̂₂ at q² = −(10.0 MeV)². Explain why the loops that dominate Z₃ are not the ones that dominate Π̂₂, and what a low-energy measurement can therefore see.
- Each lepton carries charge −e and adds its own loop with the same coefficient, so 1 − Z₃ = (α/3π) Σ ln(Λ²/m_ℓ²). The logarithms are 28.97 for the electron, 18.31 for the muon and 12.67 for the τ, giving shares 0.02243, 0.01418 and 0.00981.
- Summing, 1 − Z₃ = 0.04642, so Z₃ = 0.9536, and 1/α₀ = Z₃/α = 137.036 − (28.97 + 18.31 + 12.67)/3π = 137.036 − 6.361 = 130.68. The two heavy leptons together screen more than the electron does.
- Electron at Q = 10.0 MeV: Q/m = 19.6, so the full integral is needed. It gives 3.327 × 10⁻³, within 0.4 per cent of the large-Q form (α/3π)[ln(Q²/m²) − 5/3] = 3.315 × 10⁻³; the small-q² form would give 5.93 × 10⁻², 18 times too large.
- Muon and τ: Q ≪ m_ℓ, so Π̂₂ ≈ αQ²/(15πm_ℓ²) = 1.39 × 10⁻⁶ and 4.90 × 10⁻⁹, matching the full integral to 0.1 per cent. The total, 3.328 × 10⁻³, is the electron's alone to 0.04 per cent.
- Why: Z₃ collects each loop's q²-independent part, a logarithm of Λ/m_ℓ of comparable size for every species, and it is absorbed into the measured α. Π̂₂ is what remains after the subtraction at q² = 0, and a loop with m_ℓ ≫ Q can only leave (Q/m_ℓ)² times a number: it decouples. A low-energy experiment sees the electron loop and is blind to heavier ones, except through α itself.
- Checks: every term is dimensionless; a lepton's Z₃ share vanishes as Λ → m_ℓ; and the muon-to-electron ratio of small-q² slopes is (mₑ/mμ)² = 2.34 × 10⁻⁵.
Answer1 − Z₃ = 0.0224 (e) + 0.0142 (μ) + 0.0098 (τ) = 0.0464, so Z₃ = 0.954 and 1/α₀ = 130.68; Π̂₂(−(10.0 MeV)²) = 3.33 × 10⁻³ (e), 1.39 × 10⁻⁶ (μ) and 4.90 × 10⁻⁹ (τ).