University Quantum Field Theory I · Vacuum Polarisation, Charge Renormalisation and the Running of alpha · 13.6

QED's one-loop beta function and its Landau pole

Topic 6.08 turned scale independence into a beta function for φ⁴. Here the same four lines run on QED, with one input: the photon's own counterterm. Out come e³/(12π²), a sign that a cross-section forces on you, and a Landau pole so far away that it says more about QED's limits than about nature.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

QED's charge runs for the reason φ⁴'s λ did: in d = 4 − ε the bare charge has mass dimension ε/2, so writing it as μ^(ε/2) times a dimensionless e brings in a scale μ that e₀ cannot know about, and the 1/ε pole of the renormalisation constant then fixes μ de/dμ. What is new is which constant. The bare charge collects the vertex Z₁, the electron's Z₂ and the photon's Z₃, and the Ward identity's Z₁ = Z₂ cancels the first two, leaving e₀ = μ^(ε/2)e/√Z₃. The running is the photon's business alone: gauge independent, the same for every charged particle, and fixed by one number, the pole −(α/3π)(2/ε) of topic 13.02's electron loop.

Four lines turn it into β(e) = e³/(12π²), or μ dα/dμ = 2α²/(3π) per charged lepton. The sign is not a convention. Z₃ below 1 is screening, and the same α/3 sets the pair-creation rate of topic 13.04, so a positive cross-section forces α to grow at short distances. Integrated, β returns topic 13.05's logarithm with the same slope, which is why the one-loop coefficient belongs to no scheme.

The model ends at its own prediction. From 1/137.036 the electron loop alone drives 1/α to zero 646 e-folds above mₑ, near 10²⁸⁶ eV, but one loop fails once α nears 1, and gravity and every other charged particle arrive long before. The pole says QED cannot be the whole story at arbitrarily short distances, not what happens there.

Simple definition: QED's beta function β(e) = μ de/dμ is the rate at which the renormalised charge must change with the scale μ so that the bare charge e₀ = μ^(ε/2)e/√Z₃ stays fixed; at one loop it is +e³/(12π²) for each charged lepton.
Example: At α = 1/137.036, e = 0.3028 and β = e³/(12π²) = 2.34 × 10⁻⁴. So 1/α falls by 2/(3π) = 0.212 per e-fold of μ, or 0.489 per factor of 10 in μ, while the electron is the only lepton lighter than μ.

Bare charge once Z₁ = Z₂e₀ = μ^(ε/2) e Z₃^(−1/2), Z₃ = 1 + δ₃

From e₀ = μ^(ε/2)eZ₁/(Z₂√Z₃) and topic 12.07's Z₁ = Z₂, quoted: the vertex and the electron legs cancel, so only the photon's factor can make the charge run.

e₀ has mass dimension ε/2 (GeV^(ε/2)); e and Z₃ are pure numbers; μ is an arbitrary scale in GeV; d = 4 − ε

Minimal-subtraction photon countertermδ₃ = −e²/(6π² ε)

Quoted from topics 13.02 and 13.03. It is negative, so Z₃ is below 1 and the measured charge e = √Z₃ e₀ is the screened one.

the pole of topic 13.02's Π₂, −(α/3π)(2/ε), with α = e²/4π; a pure number, −1.549 × 10⁻³/ε at α = 1/137.036

Reading β off the polee₀ = μ^(ε/2) e (1 + c e²/ε) ⟹ β(e) = −εe/2 + c e³

Derived from μ de₀/dμ = 0: the ε/2 carried by μ^(ε/2) meets the 1/ε of the pole and leaves a finite coefficient. The same move gave φ⁴ its 3λ²/(16π²).

c is a pure number; here Z₃^(−1/2) = 1 + e²/(12π²ε), so c = 1/(12π²); true to order e³

QED's one-loop beta functionβ(e) ≡ μ de/dμ = −εe/2 + e³/(12π²)

Derived. Positive, so e grows at short distances. The −εe/2 is the tree-level dimension of e₀ and vanishes at d = 4; as e → 0, β → 0 and nothing runs.

β is a pure number, like e; at d = 4 and α = 1/137.036, e = 0.3028 gives β = 2.34 × 10⁻⁴; one charged lepton

The running of α, fermion by fermionμ dα/dμ = (2α²/3π) Σ_f N_c Q_f², sum over fermions lighter than μ

Derived for the electron, 1.13 × 10⁻⁵ at α = 1/137.036. Each loop is the electron's with e → Q_f e, so it adds Q_f²; below its mass a loop decouples from the physical charge.

α = e²/4π; Q_f the charge in units of e; N_c = 1 for a lepton, 3 for a quark, whose low-μ share needs a dispersion integral

Landau pole of the electron loopμ_L = mₑ exp(3π/(2α)) ≈ 1.45 × 10²⁸⁶ eV

Computed: the scale at which 1/α(μ) = 1/α − (2/3π) ln(μ/mₑ) reaches zero. It lies 258 decades beyond the Planck energy, 1.22 × 10²⁸ eV, and far beyond one loop's reach.

mₑ = 0.511 MeV; α = 1/137.036 at μ = mₑ; 3π/(2α) = 645.8 e-folds, which is 280.45 decades

Only the photon's Z₃ is left in the bare charge

Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout, with d = 4 − ε. Two results are quoted in one line each. Topic 6.08: a coupling defined at a scale μ runs at β = μ dg/dμ, fixed by holding the bare coupling μ-independent. Topic 6.09: β = bg² integrates to 1/g(μ) = 1/g(μ₀) − b ln(μ/μ₀). In QED the bare charge sits in e₀ψ̄₀A̸₀ψ₀, with ψ₀ = √Z₂ ψ and A₀ = √Z₃ A, while the renormalised vertex is μ^(ε/2)eZ₁, so e₀ = μ^(ε/2)eZ₁/(Z₂√Z₃). Topic 12.07's Ward identity gives Z₁ = Z₂, quoted, and leaves e₀ = μ^(ε/2)e/√Z₃. That is the difference from φ⁴, whose δ_λ came from the vertex bubbles. QED's vertex correction is real, and Unit 14 computes it, but its pole cancels the electron's own Z₂ and never reaches the charge. The running is therefore the photon's business: gauge independent, because the one-loop Z₃ has no photon inside it, and the same for a muon or a proton as for the electron.

Four lines from δ₃ to e³/(12π²)

Topic 13.02's electron loop has the pole −(α/3π)(2/ε), and minimal subtraction sets δ₃ equal to it: δ₃ = −e²/(6π²ε), quoted. To this order Z₃^(−1/2) = 1 − δ₃/2 = 1 + e²/(12π²ε), so e₀ = μ^(ε/2)e[1 + e²/(12π²ε)]. Apply μ d/dμ and set the result to zero: (ε/2)e[1 + e²/(12π²ε)] + β[1 + e²/(4π²ε)] = 0. Try β = −εe/2 + be³. The order-e³ terms are e³/(24π²) from the first bracket, −e³/(8π²) from −εe/2 times e²/(4π²ε), and be³, so b = 1/(12π²) and, derived, β(e) = −εe/2 + e³/(12π²). At d = 4, with α = e²/4π, μ dα/dμ = (e/2π)β = 2α²/(3π), derived. Dimensions check: e is a pure number at d = 4, and so is μ de/dμ. At α = 1/137.036, e = 0.3028, β = 2.34 × 10⁻⁴ and μ dα/dμ = 1.13 × 10⁻⁵. As e → 0 the loop term vanishes and only the tree-level −εe/2 is left, the dimension of e₀ at work.

The sign is the sign of a cross-section

A positive β says the charge grows at short distances. Z₃ = 1 + δ₃ is below 1, so the measured e = √Z₃ e₀ is smaller than the bare charge: virtual pairs screen it, and a probe that comes closer sees more of it. The sign is no convention, because topic 13.04 fixes it. Above threshold, Im Π̂₂ tends to −α/3, with the sign set by the −i0, and by the optical theorem −Im Π̂₂ is proportional to the rate for a virtual photon to make a pair, a positive number. A once-subtracted dispersion relation, Π̂₂(−Q²) = −(Q²/π)∫ ds Im Π̂₂(s)/[s(s + Q²)], turns that rate into the real part at spacelike momentum, and the asymptotic −α/3 alone fixes its large-Q² growth, (α/3π) ln(Q²/m²) plus a constant, shown. So the slope that makes 1/α fall is the pair-creation rate read through a dispersion integral, and in QED every term in that sum is positive. A coupling that falls with the scale needs self-interacting gauge fields, the non-abelian theories of a later course, named.

Integrated, it is topic 13.05's logarithm

Integrate μ dα/dμ = 2α²/(3π) as topic 6.09 did: 1/ᾱ(μ) = 1/ᾱ(μ₀) − (2/3π) ln(μ/μ₀). For a start, the MS-bar remainder of Π₂(0) is (α/3π) ln(m²/μ²), which vanishes at μ = m, so ᾱ(m) is Thomson's α at one loop and 1/ᾱ(μ) = 137.036 − (2/3π) ln(μ/m). Topic 13.05's on-shell effective charge, which contains no μ, is 1/α_eff(Q) = 137.036 − (1/3π)[ln(Q²/m²) − 5/3] for Q ≫ m. Both fall by 2/(3π) per e-fold. They differ by the constant 5/(9π) = 0.177, which vanishes for μ = Q exp(−5/6), shown. The constant belongs to the scheme and the slope to the electron loop, as topic 6.08's α′ = α + cα² argument requires. The rule has a domain: for Q ≪ m, Π̂₂(−Q²) falls to αQ²/(15πm²) and the physical running stops, so in leading logarithms each lepton counts only above its mass. With all three, 1/α at 91.19 GeV is about 132.7 (topic 13.05) against the 128.95 extracted from data, quoted; hadrons supply the rest.

Where the Landau pole lies, and why it is not a prediction

Run the electron loop alone from 1/α = 137.036 at μ = mₑ. 1/α reaches zero when (2/3π) ln(μ_L/mₑ) = 1/α, so μ_L = mₑ exp(3π/(2α)), derived from the leading-log solution. The exponent is 645.8, or 280.45 decades, so μ_L = 1.45 × 10²⁸⁶ eV, computed, a length ħc/μ_L near 10⁻²⁹³ m. The Planck energy, 1.22 × 10²⁸ eV, quoted, is 258 decades lower. Like φ⁴'s λ, α grows with the scale, and the pole's place is exponentially sensitive to the slope: switching on the muon and tau at their masses brings it down to about 10¹⁰¹ eV (Example 3), and a factor 2 in β moves it by 140 decades. The formula also fails before its own pole. The two-loop term, β(e) = e³/(12π²) + e⁵/(64π⁴), quoted, is 3α/(4π) = 0.17% of the first at α = 1/137, but 24% at α = 1, which 1/α reaches only 3π/2 = 4.7 e-folds before the pole. The pole marks where perturbative QED stops being a closed theory; whether QED exists at all distances is the non-perturbative question of triviality, named.

Change one variable at a time

Make the relationship visible.

Interactive model

Inverse coupling 1/α at μ = m137

Unit-charge fermions N, each of mass m1

Probe scale log₁₀(μ/m)22 decades

Start at 1/α = 137 with N = 1: the pole sits 280 decades above mₑ, far right of the Planck line. Raise N and it moves left as 1/N, yet even N = 9 needs 1/α(m) at 98 or below to pull it left of the Planck line. Slide the probe to read 1/α at any scale.

SLOPE d(1/α)/d ln μ-0.212

POLE log₁₀(μ_L/m)280.4 decades

POLE log₁₀(μ_L/eV) IF m = mₑ286.1

1/α AT THE PROBE126.25

Live interpretationSLOPE d(1/α)/d ln μ: −0.212. POLE log₁₀(μ_L/m): 280.4 decades. POLE log₁₀(μ_L/eV) IF m = mₑ: 286.1. 1/α AT THE PROBE: 126.25

Catch the common trap

Explain before calculating.

Misconception check

Carrying φ⁴'s habit over, students take the charge's counterterm from the vertex. In φ⁴ that worked because the field renormalisation is 1 at one loop, so λ₀ = μ^ε(λ + δ_λ) involved the vertex alone. In QED it does not: the electron's Z₂ has a pole at one loop. Writing e₀ = μ^(ε/2)eZ₁ with Feynman gauge's δ₁ = −(α/4π)(2/ε) gives β = −e³/(8π²), a negative slope that would make QED asymptotically free, and in Landau gauge, where δ₁ has no pole at one loop, it gives zero. A gauge-dependent answer for a measurable rate is the sign of the error. The bare charge is e₀ = μ^(ε/2)eZ₁/(Z₂√Z₃); the Ward identity's Z₁ = Z₂ removes the vertex and the legs together, and the gauge-independent photon factor alone gives β = +e³/(12π²).

In d = 4 − ε, minimal subtraction gives δ₃ = −e²/(6π²ε), and Z₁ = Z₂. Keeping the bare charge independent of μ, with the electron loop alone and α = 1/137.036 at μ = mₑ, which statement is right?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyWith the electron loop alone, how far above μ = mₑ must the scale rise for 1/α to fall from 137.036 to 136.036? Give that scale in MeV and as a distance, check that the muon can still be left out, and give β(e) at the start.

At α = 1/137.036, e = √(4πα) = 0.30282 and β = e³/(12π²) = 2.345 × 10⁻⁴, a pure number like e. Equivalently μ dα/dμ = 2α²/(3π) = 1.130 × 10⁻⁵.
The leading-log solution says 1/α falls by 2/(3π) = 0.21221 per e-fold of μ, whatever its starting value. A drop of 1.000 needs ln(μ/mₑ) = 1/0.21221 = 4.7124, which is exactly 3π/2.
So μ = mₑ exp(4.7124) = 0.51100 MeV × 111.32 = 56.88 MeV. That is below m_μ = 105.66 MeV, so the electron is still the only charged lepton lighter than μ and the one-loop slope used is the right one.
As distances, ħc/mₑ = 197.327 MeV fm ÷ 0.51100 MeV = 386.2 fm and ħc/μ = 197.327 ÷ 56.88 = 3.469 fm. Closing in by a factor 111 raises α from 1/137.036 to 1/136.036, an increase of 0.735%.

Answerμ ≈ 56.9 MeV, or 3.47 fm, a factor exp(3π/2) = 111 above mₑ; β(e) = 2.34 × 10⁻⁴ at the start, and α has grown by 0.735%.

MediumAt momentum transfer Q = 90.0 MeV, with the electron loop alone, compare topic 13.05's on-shell effective charge with the MS-bar coupling obtained by integrating β up to μ = Q. Show that the two have the same slope, find the constant between them, and find the μ at which they coincide.

On shell, for Q ≫ m: 1/α_eff(Q) = 1/α − (1/3π)[ln(Q²/m²) − 5/3]. Here ln(Q²/m²) = 2 ln(90.0/0.51100) = 10.342, so 1/α_eff = 137.036 − 0.10610 × 8.6757 = 136.1155. The exact Feynman-parameter integral differs by 2 × 10⁻⁵, so the large-Q form is safe.
In MS-bar, ᾱ(mₑ) = α at one loop, because the finite part (α/3π) ln(m²/μ²) of Π₂(0) vanishes at μ = m. Integrating μ dα/dμ = 2α²/(3π) then gives 1/ᾱ(Q) = 137.036 − 0.21221 × ln(90.0/0.51100) = 137.036 − 0.21221 × 5.1712 = 135.9386.
Slope: both fall by 2/(3π) = 0.2122 per e-fold. Done exactly at 60.0 and 90.0 MeV, the on-shell 1/α_eff falls by 0.0860, and 0.21221 × ln 1.5 = 0.0860 is the MS-bar fall. The coefficient belongs to the electron loop, not to the scheme.
Constant: 136.1155 − 135.9386 = 0.177, which is 5/(9π) = 0.1768. It disappears if μ = Q exp(−5/6) = 90.0 MeV × 0.4346 = 39.1 MeV, where 1/ᾱ = 136.1155, the on-shell value.

Answer1/α_eff(90.0 MeV) = 136.1155 on shell and 1/ᾱ(90.0 MeV) = 135.9386 in MS-bar: the same slope, 0.212 per e-fold, and a scheme constant 5/(9π) = 0.177 that vanishes at μ = 39.1 MeV.

HardRun 1/α up from 137.036 at mₑ, switching the muon (105.66 MeV) and the τ (1776.86 MeV) on at their masses, and find the three-lepton Landau pole. Compare it with the electron-alone pole, 1.45 × 10²⁸⁶ eV, and with the Planck energy, 1.22 × 10²⁸ eV.

Each unit-charge lepton lighter than μ adds 2/(3π) = 0.21221 to the fall of 1/α per e-fold. From mₑ to m_μ only the electron runs: ln(m_μ/mₑ) = 5.3316, so 1/α(m_μ) = 137.036 − 1.1314 = 135.905.
From m_μ to m_τ two leptons run: ln(m_τ/m_μ) = 2.8224, a fall of 2 × 0.21221 × 2.8224 = 1.1979, so 1/α(m_τ) = 134.707.
Above m_τ all three run, so 1/α falls by 3 × 2/(3π) = 2/π = 0.63662 per e-fold and reaches zero after ln(μ_L/m_τ) = 134.707 × π/2 = 211.60 e-folds, which is 91.895 decades.
So μ_L = 1.7769 × 10⁹ eV × 10^91.895 = 1.4 × 10¹⁰¹ eV, computed. The slope only tripled, yet the pole fell by 185 decades, because ln(μ_L/m_τ) is 1/α divided by the slope: a third of the e-folds.
It still lies 73 decades above the Planck energy. The sensitivity is extreme: a 1% change in 1/α(m_τ) moves the pole by 0.92 decades. Quarks, whose low-energy share needs topic 13.04's dispersion integral, would lower it further.

Answer1/α(m_τ) = 134.71 and μ_L ≈ 1.4 × 10¹⁰¹ eV: 185 decades below the electron-alone pole and still 73 decades above the Planck energy.