University Quantum Field Theory I · Renormalising QED: the Electron Self-Energy and the Ward-Takahashi Identity · 12.1

Power counting in QED: D = 4 − B − (3/2)F

Before computing a single QED loop you can know which ones will diverge. Count powers of loop momentum, let the vertex fix the topology, and the answer depends only on the external legs: one short list, the same at every order. Build that list, then find the symmetry that strikes off or softens each entry.

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Build the model

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Power counting asks, before any integral is done, how a loop behaves when all its momenta grow together. In QED four facts settle it: each loop brings d⁴k, an internal electron falls as 1/k, an internal photon as 1/k², and the vertex −ieγ^μ carries no momentum while joining two fermion ends to one photon end. That vertex ties lines, vertices and loops so tightly that the order of perturbation theory cancels from the count, leaving D = 4 − B − (3/2)F for B external photons and F external fermions.

The cancellation is the statement that e is dimensionless, and it makes QED renormalisable, because the same short list is superficially divergent at every order: the electron self-energy, the photon self-energy, the vertex and the one-, three- and four-photon functions. Symmetry prunes it. Charge conjugation kills the odd photon functions, gauge invariance makes the four-photon function finite, and chiral symmetry and transversality soften the two self-energies to logarithms.

The count is only superficial: an upper bound on the overall divergence that misses factors a symmetry forces into the numerator, blind to divergent subgraphs, which Weinberg's theorem handles, and to the infrared, where the massless photon brings its own trouble. Give a coupling negative mass dimension, as a Pauli term or Fermi's contact interaction does, and D grows with order: the list never closes.

Simple definition: The superficial degree of divergence D of a QED amplitude is the power of the cutoff its loop integral grows as when every loop momentum is scaled up together; for B external photons and F external fermions it is D = 4 − B − (3/2)F at every order.
Example: The one-loop vertex has D = 4 × 1 − 2 − 2 × 1 = 0, as 4 − 1 − 3 = 0 says, so it grows only as ln Λ: raising a cutoff from 1 GeV to the Planck scale, 1.22 × 10¹⁹ GeV, takes ln(Λ²/m²), m the electron mass, from 15.2 to 103, where a linear divergence would grow 10¹⁹-fold.

Superficial degree from the integrandD = 4L − Pₑ − 2P_γ

The vertex −ieγ^μ carries no momentum, so this is the whole count: the power of Λ the integral grows as when every loop momentum scales up together. Derived.

L loops, Pₑ internal electron and P_γ internal photon lines, all pure numbers; d⁴k adds 4, 1/k̸ takes 1, 1/k² takes 2

Ends at the vertex2V = 2Pₑ + F and V = 2P_γ + B

Each vertex has two fermion ends and one photon end; internal lines use two ends, external lines one. These fix Pₑ and P_γ in terms of V.

V vertices, F external fermions, B external photons; with L = Pₑ + P_γ − V + 1 from momentum conservation (topic 3.05)

QED's superficial degreeD = 4 − B − (3/2)F

Also the mass dimension of the amputated 1PI function, since [A] = 1, [ψ] = 3/2 and [e] = 0: when Λ dominates, every power must come from the loop.

B, F pure numbers; V cancels, so D is the same at every order; derived

With couplings of other dimensionsD = 4 − B − (3/2)F − Σᵢ [gᵢ] Vᵢ

Each vertex of a negative-dimension coupling raises D, by one for a Pauli insertion, so with enough insertions every function diverges and the list never closes. Derived.

[gᵢ] the mass dimension of coupling i and Vᵢ its vertices; [e] = 0, a Pauli coupling eκ/4m has [g] = −1

The divergent part is a polynomial of degree DD(∂G/∂p) = D(G) − 1

Each derivative, by p or by a mass m, adds a power of 1/k to one propagator, so only Taylor terms up to order D in them can diverge. Derived at one loop.

G any amputated 1PI function and D(G) its degree, a pure number; p any external momentum, in GeV

Light-by-light scattering for ω ≪ mσ(γγ → γγ) = (973/(10125π)) α⁴ω⁶/m⁸

The ω⁶ is derived: one power of momentum per photon in M, from gauge invariance, squared and divided by s. The coefficient is quoted (Euler, Karplus and Neuman).

ω each photon's centre-of-mass energy and m the electron mass, in GeV; σ in GeV⁻², × 0.3894 mb GeV²

Count powers of loop momentum

Natural units, ħ = c = 1, and the metric (+, −, −, −) throughout. Scale every loop momentum of a QED diagram by a large factor. Each loop ∫d⁴k brings four powers; each internal electron, i(k̸ + m)/(k² − m²), falls as 1/k and takes one; each internal photon, −ig_μν/k² in Feynman gauge, takes two; the vertex −ieγ^μ takes none. The general-ξ part, (1 − ξ)k_μk_ν/k⁴, also falls as 1/k², so the count is gauge-independent. The integral then grows as Λ^D with D = 4L − Pₑ − 2P_γ, topic 6.01's count with the fermion propagator's slower 1/k fall-off built in. At one loop the electron self-energy, one electron and one photon line, has D = 4 − 1 − 2 = 1; vacuum polarisation, two electron lines, D = 2; the vertex, two electrons and a photon, D = 0, a logarithm; the e⁻e⁻ → e⁻e⁻ box, two of each, D = −2. Positive D means a power of the cutoff, zero a logarithm, negative a convergent overall integral.

The vertex fixes the topology and the order drops out

Every vertex has exactly two fermion ends and one photon end. An internal electron line uses two fermion ends and an external one uses one, so 2V = 2Pₑ + F; likewise V = 2P_γ + B. Momentum conservation leaves L = Pₑ + P_γ − V + 1 loops. Rewrite D = 4L − Pₑ − 2P_γ as 3Pₑ + 2P_γ − 4V + 4, put in Pₑ = V − F/2 and P_γ = (V − B)/2, and V cancels: D = 4 − B − (3/2)F, derived. The two-loop rainbow self-energy, with V = 4, Pₑ = 3 and P_γ = 2, gives 8 − 3 − 4 = 1, the one-loop value. The cancellation is dimensional analysis: with [A] = 1 and [ψ] = 3/2, the amputated 1PI function has mass dimension 4 − B − (3/2)F, and because e is dimensionless, α = e²/4π = 1/137.036, all of it must come from the loop once Λ dominates. In d dimensions [e] = (4 − d)/2, so D falls with order in three dimensions and rises in six.

A finite list, the same at every order

D ≥ 0 needs B + (3/2)F ≤ 4, and charge conservation makes F even, so the list can be written down once for all orders. With F = 2: the electron self-energy Σ(p), D = 1, and the vertex Γ^μ with D = 0; adding a second photon gives Compton scattering, D = −1. With F = 0: the one-, two-, three- and four-photon functions, D = 3, 2, 1 and 0. Anything with four fermion legs has D ≤ −2. The vacuum bubbles, B = F = 0 with D = 4, cancel from every amplitude in topic 2.06's Gell-Mann-Low ratio. So six functions, and only six, can need a counterterm. Weinberg's theorem (quoted in topic 6.01) adds that every other divergence sits inside a subgraph that is one of these six, so their counterterms, inserted order by order, cure it: stated here, not proved. A theory with a coupling of negative dimension has no such list: a Pauli term (eκ/4m)ψ̄σ^μν ψF_μν raises D by one per insertion.

Furry's theorem strikes out the odd photon functions

Charge conjugation is a symmetry of QED that leaves the vacuum alone and flips the photon field, A_μ → −A_μ (topic 11.04). A vacuum expectation of a time-ordered product of an odd number of photon fields is therefore equal to minus itself and vanishes to all orders: Furry's theorem, topic 11.07. At one loop the two orientations of the electron loop cancel, the trace changing by (−1)ⁿ for n vertices. The one-photon function, D = 3, is ⟨0|A_μ|0⟩, which Lorentz invariance also kills, since a constant four-vector would pick a direction. The three-photon function, D = 1, would need an A³ counterterm that is neither C-even nor gauge-invariant. Neither appears. Note what Furry does not do: for an even number of photons (−1)ⁿ = +1, the orientations add, and the two- and four-photon functions survive. The argument is a symmetry statement, so it holds with any C-invariant regulator.

Gauge invariance makes light-by-light finite

The four-photon function has D = 0, so each of the six one-loop diagrams, the orderings of four photons round an electron loop, diverges logarithmically, and by the polynomial rule the divergence is a constant tensor, c(g^μν g^ρσ + g^μρ g^νσ + g^μσ g^νρ) ln Λ. The Ward identity k₁_μ M^μνρσ = 0 must hold at every momentum, and the constant tensor contracted with k₁ is not zero, so c = 0: the six logarithms cancel in the sum, shown here in outline and seen in any gauge-invariant regulator. What survives is built from one field strength F_μν = ∂_μA_ν − ∂_νA_μ per photon, so M carries four powers of external momenta and has degree 0 − 4 = −4: finite, with no (A_μA^μ)² counterterm. At low energy dimensions then fix M ∼ α²ω⁴/m⁴, the Euler-Heisenberg interaction with coefficient 2α²/45m⁴ (quoted), and the cross-section rises as ω⁶, vanishing as ω → 0, where Maxwell's linear theory returns.

What remains, and what symmetry still does to it

Three functions remain, and each keeps a divergence. Expand Σ in p and m to order D = 1: a Λ term with no γ matrices, plus p̸ ln Λ and m ln Λ. At m = 0 QED is invariant under ψ → e^(iθγ⁵)ψ, which forbids ψ̄ψ, so the Λ term must carry a factor m and sinks to m ln Λ: topic 12.04 derives δm = (3α/4π)m ln(Λ²/m²), against the linearly growing electromagnetic mass of Advanced QM's Unit 13. Transversality, q_μ Π^μν = 0 (topic 12.09), forces Π^μν = (q² g^μν − q^μ q^ν)Π(q²), two powers of q out front, so Π is logarithmic and a Λ²g^μν photon mass is forbidden. The vertex stays logarithmic and multiplies γ^μ alone. Yukawa theory, gφψ̄ψ (topic 7.08), counts exactly alike with B external scalars, but its scalar coupling has no Furry theorem and no gauge invariance, so its φ³ and φ⁴ functions need couplings and counterterms of their own.

Change one variable at a time

Make the relationship visible.

Interactive model

External photons B0

External fermions F2

Loops L1

Start at the electron self-energy, B = 0 and F = 2, and drag L from 1 to 4: the loop bar climbs by 4 per loop, the two subtractions climb to match, and D = 1 never moves. Then set B = 4, F = 0: D = 0, and the open bar drops it to −4.

VERTICES V, ORDER e^V2

ELECTRON LINES Pₑ1

PHOTON LINES P_γ1

D = 4L − Pₑ − 2P_γ1

Live interpretationVERTICES V, ORDER e^V: 2. ELECTRON LINES Pₑ: 1. PHOTON LINES P_γ: 1. D = 4L − Pₑ − 2P_γ: 1

Catch the common trap

Explain before calculating.

Misconception check

“D = 1 for the electron self-energy means the mass shift grows linearly with the cutoff, and D = 0 for the four-photon function means QED needs an (A_μA^μ)² counterterm.” Both read the superficial degree as the actual one. D counts powers assuming the numerator is as large as dimensions allow; a symmetry can force a mass or an external momentum into it, and each such factor lowers the true degree by one. Chiral symmetry puts an m in front of Σ's unit-matrix part, so δm grows as m ln Λ; gauge invariance puts one momentum per photon into the four-photon amplitude, leaving degree −4. The error also runs the other way: D below zero does not make a diagram finite. A two-loop Compton diagram whose loop carries a self-energy insertion has D = −1 overall, yet diverges through that subgraph, and is cured by the counterterm for Σ, not by one of its own.

QED's one-loop four-photon function, light-by-light scattering through an electron loop, has superficial degree D = 4 − 4 − 0 = 0. Why does the renormalised QED Lagrangian need no (A_μA^μ)² counterterm?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA three-loop correction to the electron-photon vertex has V = 7 vertices. Find its internal electron and photon lines, confirm the loop count, and compute D from the integrand and from D = 4 − B − (3/2)F. Repeat for the one-loop light-by-light diagram, V = 4, and give each amplitude's order in e.

The vertex function has B = 1 and F = 2. Fermion ends: Pₑ = V − F/2 = 7 − 1 = 6. Photon ends: P_γ = (V − B)/2 = (7 − 1)/2 = 3.
Loops: L = Pₑ + P_γ − V + 1 = 6 + 3 − 7 + 1 = 3, as stated.
From the integrand, D = 4L − Pₑ − 2P_γ = 12 − 6 − 6 = 0; from the formula, 4 − 1 − 3 = 0. The one-loop vertex (V = 3, Pₑ = 2, P_γ = 1) also has D = 0: the order does not move it.
Light-by-light has B = 4 and F = 0: Pₑ = 4 − 0 = 4, P_γ = (4 − 4)/2 = 0, L = 4 + 0 − 4 + 1 = 1, and D = 4 − 4 − 0 = 0 both ways.
Orders: e⁷ for the three-loop vertex, a relative correction of order (α/π)³ = 1.25 × 10⁻⁸ to the tree vertex e; e⁴ for light-by-light. Neither power of e changes D, because e is dimensionless.

AnswerThree-loop vertex: Pₑ = 6, P_γ = 3, L = 3, D = 0. Light-by-light: Pₑ = 4, P_γ = 0, L = 1, D = 0. Both have D = 0 at every order, superficially logarithmic, though gauge invariance makes the light-by-light sum finite.

MediumUse D = 0 and gauge invariance to find how the low-energy cross-section for γγ → γγ depends on the centre-of-mass photon energy ω ≪ m. Then, with the quoted coefficient 973/(10125π), evaluate it for two head-on 2.0 eV laser photons and for two 50 keV photons, and compare each with the Thomson cross-section, 0.665 b.

D = 0 is the mass dimension of the amplitude, so M is dimensionless. Gauge invariance makes it quartic in field strengths: one power of each photon's momentum, so M ∝ e⁴ω⁴/m⁴, with m the only other scale below threshold.
For 2 → 2 in the centre-of-mass frame, σ = (1/2)∫|M|² dΩ/(64π²s), the 1/2 for identical photons, with s = 4ω², so σ ∝ α⁴ω⁸/(m⁸ω²) = α⁴ω⁶/m⁸, derived up to the number. It has dimension −2, a cross-section, and vanishes as ω → 0.
Restore units: 1/m² = 1/(5.110 × 10⁻⁴ GeV)² = 3.830 × 10⁶ GeV⁻² = 1.491 × 10⁶ mb = 1.491 × 10⁻²¹ cm². With α⁴ = 2.836 × 10⁻⁹ and 973/(10125π) = 0.03059, the prefactor is 1.294 × 10⁻³¹ cm².
At 2.0 eV, ω/m = 3.914 × 10⁻⁶ and (ω/m)⁶ = 3.59 × 10⁻³³, so σ = 4.6 × 10⁻⁶⁴ cm², 7.0 × 10⁻⁴⁰ of the Thomson cross-section.
At 50 keV, ω/m = 0.0978 and (ω/m)⁶ = 8.78 × 10⁻⁷, so σ = 1.1 × 10⁻³⁷ cm², 1.7 × 10⁻¹³ of Thomson, near the edge of ω ≪ m. The ratio of the two, 2.4 × 10²⁶, is (25 000)⁶: power counting, not the coefficient, sets the scale.

Answerσ ∝ α⁴ω⁶/m⁸; σ ≈ 4.6 × 10⁻⁶⁴ cm² at 2.0 eV and 1.1 × 10⁻³⁷ cm² at 50 keV, 7.0 × 10⁻⁴⁰ and 1.7 × 10⁻¹³ of the Thomson cross-section.

HardThe one-loop vertex Γ^μ has D = 0. Show that its divergence can multiply only γ^μ and hence that the magnetic form factor F₂, the coefficient of iσ^μν q_ν/2m, is finite. Then add a Pauli term (eκ/4m)ψ̄σ^μν ψF_μν: find its coupling's mass dimension and the new D. Finally compare Schwinger's quoted aₑ = α/2π with the measured 1.15965218 × 10⁻³.

Γ^μ has mass dimension 0 = D. By the polynomial rule only its zeroth Taylor term in external momenta and masses can diverge: a constant matrix with one index, γ^μ or γ^μ γ⁵, and parity, conserved by QED, leaves γ^μ. The divergence is cγ^μ ln Λ, the vertex counterterm of topic 12.02.
F₂'s structure carries one power of q, and each power of an external momentum or of m in the numerator replaces a power of loop momentum, lowering the degree to −1 or below. So F₂ comes from convergent integrals; the 1/m it carries is set by the finite loop. F₂ is finite at every order and no counterterm can shift it: QED predicts it.
The Pauli term: [ψ̄σ^μν ψF_μν] = 3/2 + 3/2 + 2 = 5, so [eκ/4m] = −1. Its vertex has the QED vertex's two fermion ends and one photon end plus one power of the photon's momentum, so D = 4L − Pₑ − 2P_γ + V_P, and the same topology gives D = 4 − B − (3/2)F + V_P.
One Pauli insertion in the vertex function gives D = 1, so terms linear in q, σ^μν q_ν among them, can diverge: a Pauli counterterm is needed and κ becomes an input. Two insertions make the four-fermion function D = −2 + 2 = 0, divergent too: the list no longer closes.
Numbers: α/2π = 1/(2π × 137.035999) = 1.16141 × 10⁻³, above the measured 1.15965 × 10⁻³ by 1.76 × 10⁻⁶, or 0.15%. The quoted two-loop term, −0.3285(α/π)² = −1.77 × 10⁻⁶, closes the gap to 1.5 × 10⁻⁸, the size of the three-loop term, 1.181(α/π)³ = 1.48 × 10⁻⁸, also quoted.

AnswerThe vertex divergence is cγ^μ ln Λ only, so F₂ is finite and predicted; a Pauli coupling has dimension −1 and gives D = 4 − B − (3/2)F + V_P. α/2π = 1.16141 × 10⁻³ sits 0.15% above the measured value, the size of the quoted α² term.