University Quantum Field Theory I · Renormalising QED: the Electron Self-Energy and the Ward-Takahashi Identity · 12.3

The electron self-energy Σ₂(p) at one loop

The electron self-energy is the first QED loop you carry all the way to a formula, and it is where dimensional regularisation stops being a rule for integrals and becomes a rule for the Dirac algebra too. Get the d-dimensional contractions and the μ^(ε/2) right here, and the vertex of Unit 14 runs on the same rails.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Σ₂(p) is the course's first divergent QED loop, and in it the Dirac structure and the divergence have to be regularised together. Continuing the loop momentum to d = 4 − ε continues every index it is contracted with, so the γ matrices keep (γ^μ, γ^ν) = 2g^μν but with g^μ_μ = d, and the four-dimensional contractions of Unit 7 acquire O(ε) terms; the coupling acquires mass dimension ε/2 and is written eμ^(ε/2). With those two rules the Feynman rules of Units 7 to 9 reduce the diagram, through one Feynman parameter and topic 6.03's master integral, to Σ₂(p) = (α/4π)∫₀¹dx [(4 − ε)m − (2 − ε)xp̸][2/ε − γ_E + ln 4π − ln(Δ/μ²)].

The pole multiplies 4m − p̸, and the O(ε) terms of the algebra, multiplied by that pole, leave a finite (α/4π)(p̸ − 2m) as large as any logarithm in the answer, which is why they cannot be thrown away. The result has clear edges. Off shell it depends on the gauge parameter ξ, so Σ₂ at a general momentum is bookkeeping, not an observable: only its value and slope at p̸ = m feed the mass and field renormalisation, and the slope stays finite only with the photon mass μ_γ in place.

Above p² = (m + μ_γ)² it turns complex, the electron able to shed a real photon. It is the order-α term of a series, and the γ⁵ that chiral theories need, which cannot keep all its four-dimensional properties in d dimensions, never enters it.

Simple definition: The one-loop electron self-energy Σ₂(p) is the one-particle-irreducible diagram in which the electron emits and reabsorbs one virtual photon: a matrix of the form A(p²) + B(p²)p̸ that, inside the propagator i/(p̸ − m₀ − Σ), moves its pole and its residue.
Example: At p² = 0, in Feynman gauge with μ_γ → 0, μ = m and the MS-bar poles removed, Σ₂ = (α/4π)(2m − ½p̸): a scalar part of 2 × 0.2967 = 0.593 keV for the electron and a p̸ coefficient of −2.90 × 10⁻⁴.

The coupling in d dimensionse₀ = eμ^(ε/2), since [e₀] = (4 − d)/2 = ε/2

Keeps α a pure number and Σ₂ a mass for every ε. Every logarithm then arrives as ln(Δ/μ²) with the same coefficient as 2/ε, because μ^ε(2/ε) = 2/ε + ln μ² + O(ε).

d = 4 − ε; [ψ] = (d − 1)/2 and [A] = (d − 2)/2 in mass units; e, α = e²/4π dimensionless; μ in MeV. Derived.

Three gammas in d dimensionsγ^μγ^νγ_μ = −(2 − ε)γ^ν

Gives the self-energy numerator γ^μ(k̸ + m)γ_μ = (4 − ε)m − (2 − ε)k̸. At ε = 0 it is Unit 7's −2γ^ν.

From (γ^μ, γ^ν) = 2g^μν with g^μ_μ = d, so γ^μγ_μ = d; gamma matrices dimensionless. Derived.

Four gammas in d dimensionsγ^μγ^νγ^ργ_μ = 4g^νρ − εγ^νγ^ρ

Tracing both sides gives 4d g^νρ, as cyclicity demands of the left side, and only with the −εγ^νγ^ρ term: a check every O(ε) term must pass.

ν, ρ free indices; g^νρ the metric (+, −, −, −) continued to d dimensions; tr 1 = 4 kept. Derived.

Five gammas in d dimensionsγ^μγ^νγ^ργ^σγ_μ = −2γ^σγ^ργ^ν + εγ^νγ^ργ^σ

Unit 14's vertex numerator γ^α ℓ̸ γ^μ ℓ̸ γ_α needs it; with ℓ^νℓ^σ → g^νσℓ²/d its ε term reaches the coefficient of the pole.

Order reversed in the first term, kept in the O(ε) term; all indices run over d values. Derived.

Σ₂ in Feynman gauge, dimensionally regularisedΣ₂(p) = (α/4π) ∫₀¹ dx [(4 − ε)m − (2 − ε)xp̸][2/ε − γ_E + ln 4π − ln(Δ/μ²)]

Has the dimension of a mass, vanishes as α → 0, and is real while p² stays below (m + μ_γ)², where Δ is positive for every x.

Δ = (1 − x)m² + xμ_γ² − x(1 − x)p² in MeV²; m, μ, μ_γ, p̸ and Σ₂ in MeV; x weights the photon line; γ_E = 0.5772 is Euler's constant. Derived.

What the O(ε) terms leave(α/4π) ∫₀¹ dx (−εm + εxp̸)(2/ε) = (α/4π)(p̸ − 2m)

Finite, and the size of a logarithm: at p̸ = m it is −αm/4π = −0.297 keV for the electron. Do the algebra in four dimensions and it silently disappears.

The rest of the pole term is (α/4π)(2/ε)(4m − p̸); both in MeV, exact as ε → 0. Derived.

The diagram, and why the Dirac algebra follows the loop

Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. Unit 7's propagator i(k̸ + m)/(k² − m² + i0), topic 8.08's Feynman-gauge photon −ig_μν/(q² − μ_γ² + i0) and Unit 9's vertex −ieγ^μ give −iΣ₂(p) = (−ie)²∫d^dk/(2π)^d γ^μ i(k̸ + m)γ_μ (−i)/[(k² − m² + i0)((p − k)² − μ_γ² + i0)], and (−ie)² × i × (−i) = −e². Topic 12.01 gave this function superficial degree 1, so it needs a regulator, and dimensional regularisation continues the integral to d = 4 − ε. The numerator cannot stay behind. The photon's g_μν and the loop momentum live in the same d-dimensional space, and symmetric integration, ℓ^μℓ^ν → g^μνℓ²/d, is consistent only if g^μ_μ = d. So keep (γ^μ, γ^ν) = 2g^μν and tr 1 = 4, but let every index run over d values. At this order the bare mass in the propagator may be written m, since m − m₀ is itself of order α.

Contractions in d dimensions, derived

Everything follows from the anticommutator. First, γ^μγ_μ = ½g_μν(γ^μ, γ^ν) = g^μ_μ = d. Then γ^μγ^νγ_μ = (2g^μν − γ^νγ^μ)γ_μ = 2γ^ν − dγ^ν = −(2 − ε)γ^ν. Move γ_μ one slot further each time: γ^μγ^νγ^ργ_μ = 2γ^ργ^ν + (d − 2)γ^νγ^ρ, and with 2γ^ργ^ν = 4g^νρ − 2γ^νγ^ρ this is 4g^νρ − εγ^νγ^ρ. Once more: γ^μγ^νγ^ργ^σγ_μ = 2γ^σγ^νγ^ρ − (4g^νρ − εγ^νγ^ρ)γ^σ = −2γ^σγ^ργ^ν + εγ^νγ^ργ^σ. At ε = 0 these are Unit 7's identities. Each is a polynomial in d, so it can be tested at any integer d, and the trace of the γ^μγ^νγ^ργ_μ identity, 4d g^νρ on both sides, works only because the O(ε) term is there. The self-energy uses the simplest case: γ^μ(k̸ + m)γ_μ = (4 − ε)m − (2 − ε)k̸. One warning for later units, stated rather than proved: a γ⁵ cannot both anticommute with every γ^μ and keep a non-zero trace with four of them once d differs from 4. Σ₂ contains no γ⁵, so nothing here depends on how that is resolved.

The coupling carries μ^(ε/2)

The action is dimensionless, so in d dimensions the Lagrangian density has mass dimension d. The kinetic terms ψ̄i∂̸ψ and −¼F² then fix [ψ] = (d − 1)/2 and [A] = (d − 2)/2, and the vertex e₀ψ̄γ^μψA_μ leaves [e₀] = d − (d − 1) − (d − 2)/2 = ε/2. Write e₀ = eμ^(ε/2), with μ an arbitrary mass, so that e = 0.30282 and α = e²/4π = 1/137.036 stay pure numbers. Two things follow. The loop integral alone has dimension d + 1 − 4 = 1 − ε; the factor μ^ε restores 1, so Σ₂ is a mass for every ε, as a mass shift must be. And μ never appears by itself: the master integral delivers Γ(ε/2)(4πμ²/Δ)^(ε/2) = 2/ε − γ_E + ln 4π − ln(Δ/μ²) + O(ε), quoted from topic 6.04, so the coefficient of ln μ² is always the coefficient of 2/ε. Doubling μ therefore shifts the part of Σ₂ along m by (α/4π) × 4 ln 4 = 5.545(α/4π) and the part along p̸ by −1.386(α/4π), in the ratio of the pole coefficients 4m and −p̸.

One Feynman parameter, one shift, one master integral

Topic 6.02's 1/(AB) = ∫₀¹dx [xA + (1 − x)B]⁻², with x on the photon line, turns the denominators into x[(p − k)² − μ_γ²] + (1 − x)(k² − m²) = (k − xp)² − Δ, where Δ = (1 − x)m² + xμ_γ² − x(1 − x)p². Shift to ℓ = k − xp. The piece −(2 − ε)ℓ̸ is odd in ℓ and integrates to zero, leaving the numerator (4 − ε)m − (2 − ε)xp̸. Topic 6.03's Minkowski integral ∫d^dℓ/(2π)^d (ℓ² − Δ)⁻² = iΓ(2 − d/2)Δ^{(d/2 − 2)}/(4π)^(d/2), quoted, gives −iΣ₂ = −ie²μ^ε∫dx [numerator]Γ(ε/2)Δ^(−ε/2)/(4π)^{(2 − ε/2)}. With e² = 4πα and topic 6.04's expansion, Σ₂(p) = (α/4π)∫₀¹dx [(4 − ε)m − (2 − ε)xp̸][2/ε − γ_E + ln 4π − ln(Δ/μ²)], derived. Three checks. Every term has the dimension of a mass. As α → 0 the free propagator returns. And at m = 0 the scalar part vanishes identically, so a massless electron gains no mass at this order. Δ equals m² at x = 0 and μ_γ² at x = 1 and stays positive between while p² is below (m + μ_γ)²; beyond that ln Δ has a branch cut and Σ₂ turns complex.

What the O(ε) terms leave behind

The expansion is where care is needed. Multiply out: (4 − ε)m × 2/ε = 8m/ε − 2m, and (2 − ε)xp̸ × 2/ε = 4xp̸/ε − 2xp̸, with ∫₀¹x dx = ½. So Σ₂ = (α/4π)(2/ε − γ_E + ln 4π)(4m − p̸) + (α/4π)(p̸ − 2m) − (α/4π)∫₀¹dx (4m − 2xp̸)ln(Δ/μ²) + O(ε), derived. The middle term is the O(ε) remainder. It does not shrink as ε → 0, because the −ε that produced it was multiplied by 2/ε. At p̸ = m it is −αm/4π; with α/4π = 5.807 × 10⁻⁴ that is −0.297 keV for the electron, whose mass is known to 3 parts in 10¹⁰. Write the finite part as (α/4π)(a m + b p̸) after removing the MS-bar pole: the remainder is −2 in a and +1 in b at every momentum. Setting d = 4 in the Dirac algebra while integrating in d dimensions is dimensional reduction, a legitimate scheme at this order if every diagram and counterterm uses it, but its finite parts differ from these by exactly that remainder.

Off shell it depends on ξ; on shell it is a value and a slope

Feynman gauge was a choice. Unit 8's general propagator adds i(1 − ξ)q_μq_ν/q⁴, which changes Σ₂ off shell; its pole part becomes (α/4π)(2/ε)[3m − ξ(p̸ − m)], stated here, and expanding the extra term at large q reproduces it in a few lines. Writing q̸ = (p̸ − m) − (k̸ − m) and using (k̸ − m)(k̸ + m) = k² − m² shows why every ξ-dependent piece carries a factor p̸ − m: at p̸ = m the value of Σ₂ is gauge independent while its slope is not, and in Landau gauge, ξ = 0, the p̸ pole is absent altogether. That is why topics 12.04 and 12.05 use only Σ₂ and dΣ₂/dp̸ at p̸ = m. The photon mass guards the second. On shell Δ = (1 − x)²m² + xμ_γ², so near x = 1 only μ_γ² keeps it from zero. The value survives μ_γ → 0, because ∫₀¹ln[(1 − x)²]dx = −2 converges; the slope, which brings down x(1 − x)/Δ, diverges logarithmically without it. Off-shell Σ₂ is never compared with a measurement: being gauge dependent, it is not one.

Change one variable at a time

Make the relationship visible.

Interactive model

O(ε) Dirac terms: 1 kept, 0 dropped1

Scale μ in units of m1.00

Marker at p²/m²0.5

Set the O(ε) switch to 0: a rises by 2 and b falls by 1 at every p², the remainder four-dimensional algebra throws away. Double μ: a gains 8 ln 2 = 5.55 and b loses 2 ln 2 = 1.39, the ratio −4 of the pole coefficients 4m and −p̸. Near p² = m², b steepens: topic 12.05 needs μ_γ for that slope.

SCALAR PART a3.227 × αm/4π

VECTOR PART b-0.921 × α/4π

SLOPE db/d(p²/m²)-1.090 × α/4π

Live interpretationSCALAR PART a: 3.227 × αm/4π. VECTOR PART b: −0.921 × α/4π. SLOPE db/d(p²/m²): −1.090 × α/4π

Catch the common trap

Explain before calculating.

Misconception check

The tempting shortcut is to treat the O(ε) terms of the Dirac algebra as negligible, do the contractions in four dimensions and continue only the loop integral. They are negligible only until they meet the pole: −ε × 2/ε = −2, not zero. In Σ₂ that product leaves (α/4π)(p̸ − 2m), which at p̸ = m is −0.297 keV for the electron, as large as any logarithm in the result and about two million times the precision to which mₑ is known. Four-dimensional algebra is not an arithmetic slip but a different scheme, dimensional reduction, whose finite parts differ by exactly that remainder. The real error is mixing schemes: an MS-bar mass or counterterm taken from one and a loop computed in the other give a pole mass wrong by α/4π = 5.8 × 10⁻⁴ of itself. On-shell observables agree between schemes only when every diagram and every counterterm follows the same rules.

Regulate the infrared with a photon mass μ_γ = 0.050 MeV and take m = 0.511 MeV. In Feynman gauge, above which value of √p² does Σ₂(p) first acquire an imaginary part?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyThe rule γ^μγ^νγ_μ = (2 − d)γ^ν is claimed for any d. Derive it from the anticommutator, test it with explicit 2 × 2 matrices in d = 2 and d = 3, and write the self-energy numerator γ^μ(k̸ + m)γ_μ in each case and at d = 4 − ε.

γ^μγ_μ = ½g_μν(γ^μ, γ^ν) = g_μνg^μν = d. Then γ^μγ^νγ_μ = (2g^μν − γ^νγ^μ)γ_μ = 2γ^ν − γ^ν(γ^μγ_μ) = (2 − d)γ^ν. Nothing assumed d = 4, so the rule is a polynomial in d.
d = 2: take γ⁰ = σ₁ and γ¹ = iσ₂, so (γ⁰)² = 1, (γ¹)² = −1 and they anticommute. Lowering the index flips the sign of γ¹, so γ^μγ_μ = (γ⁰)² − (γ¹)² = 2 = d, and γ^μγ⁰γ_μ = γ⁰γ⁰γ⁰ − γ¹γ⁰γ¹ = γ⁰ + γ⁰(γ¹)² = 0 = (2 − 2)γ⁰.
d = 3: take γ⁰ = σ₃, γ¹ = iσ₁ and γ² = iσ₂. Each γⁱγ⁰γⁱ = −γ⁰(γⁱ)² = γ⁰, so γ^μγ⁰γ_μ = γ⁰ − γ⁰ − γ⁰ = −γ⁰ = (2 − 3)γ⁰, and γ^μγ_μ = 1 + 1 + 1 = 3. Multiplying the matrices out confirms both, and the same holds for γ¹ and γ².
Numerator: γ^μ(k̸ + m)γ_μ = (2 − d)k̸ + dm. At d = 2 it is 2m, with no k̸ at all; at d = 3 it is 3m − k̸; at d = 4 it is Unit 7's 4m − 2k̸; at d = 4 − ε it is (4 − ε)m − (2 − ε)k̸.
Integer dimensions fix the polynomial, so its continuation to d = 4 − ε is unambiguous: the −ε pieces are as exact as the 4 and the 2, and must be carried until they meet the 2/ε pole.

Answerγ^μγ^νγ_μ = (2 − d)γ^ν, confirmed at d = 2 (zero) and d = 3 (−γ^ν). The numerator is 2m at d = 2, 3m − k̸ at d = 3 and (4 − ε)m − (2 − ε)k̸ at d = 4 − ε, each with the dimension of a mass.

MediumEvaluate Σ₂ at the spacelike point p² = −m² in Feynman gauge with μ_γ → 0. Remove the MS-bar pole (α/4π)(2/ε − γ_E + ln 4π)(4m − p̸), set μ = m, and write what is left as (α/4π)(a m + b p̸). Then find a and b again with the Dirac algebra done in four dimensions, and with μ = 2m.

With μ_γ = 0 and p² = −m², Δ = (1 − x)m² + x(1 − x)m² = (1 − x²)m². It is positive on 0 ≤ x ≤ 1 and vanishes only at x = 1, where ln(1 − x²) is integrable, so this value needs no photon mass.
With the O(ε) terms kept, a = −2 − 4∫₀¹ln(1 − x²)dx. Split ln(1 − x²) = ln(1 − x) + ln(1 + x); the integrals are −1 and 2 ln 2 − 1, so a = −2 − 4(2 ln 2 − 2) = 6 − 8 ln 2 = 0.4548.
b = 1 + 2∫₀¹x ln(1 − x²)dx. With u = x², ∫₀¹x ln(1 − x²)dx = ½∫₀¹ln(1 − u)du = −½, so b = 1 − 1 = 0.
So Σ₂ minus its pole is (α/4π)(6 − 8 ln 2)m = 5.807 × 10⁻⁴ × 0.4548 m = 2.641 × 10⁻⁴ m, or 0.1350 keV for the electron, with no p̸ term at this momentum.
Four-dimensional algebra loses the −2 in a and the +1 in b: a = 8 − 8 ln 2 = 2.455 and b = −1, a scalar part 5.4 times larger and a p̸ term that was absent. Raising μ to 2m adds 4 ln 4 = 5.545 to a and −ln 4 = −1.386 to b: a = 6.000 and b = −1.386.

Answera = 6 − 8 ln 2 = 0.455 and b = 0, so Σ₂ minus its pole is 2.64 × 10⁻⁴ m, 0.135 keV for the electron. Four-dimensional algebra gives a = 2.455, b = −1; μ = 2m gives a = 6.000, b = −1.386.

HardA virtual electron with p² = 4m² runs inside a hard process. In Feynman gauge with μ_γ → 0, find the imaginary part of Σ₂ as a combination of m and p̸, and its value on a spinor with p̸u = 2mu. Check it against the cutting rule for e* → eγ with the −g_μν polarisation sum, and explain its sign.

With μ_γ = 0, Δ = (1 − x)(m² − xp²) = (1 − x)(1 − 4x)m², negative for x between ¼ and 1. There ln(Δ − i0) = ln|Δ| − iπ, the −i0 inherited from the propagators' +i0, so the bracket gains +iπ and Im Σ₂ = (α/4π)π∫(4m − 2xp̸)dx from x = ¼ to 1. The O(ε) pieces are real and carry no iπ.
∫(4m − 2xp̸)dx from ¼ to 1 = 3m − (15/16)p̸, so Im Σ₂ = (α/4)[3m − (15/16)p̸] = 5.473 × 10⁻³ m − 1.710 × 10⁻³ p̸. On the spinor p̸ → 2m, giving Im Σ₂ = (α/4)(3 − 15/8)m = (9/32)αm = 2.052 × 10⁻³ m.
Cutting rule: −2 Im Σ₂ = Γ_eff = (1/2W)∫dΦ₂ Σ|M|² with W = 2m and ūu = 2W. Two-body phase space with a massless photon is ∫dΦ₂ = (1 − m²/W²)/(8π) = 3/(32π), and the −g_μν sum, over the outgoing electron's spins, gives Σ|M|² = −e²ū(p)γ^μ(k̸ + m)γ_μu(p) = e²(4p⋅k − 8mW).
In p's rest frame p⋅k = (W² + m²)/2 = 2.5m², so Σ|M|² = e²(10 − 16)m² = −6e²m². Then Γ_eff = (1/4m)(3/32π)(−6e²m²) = −(9/64)(e²/π)m = −(9/16)αm, and −Γ_eff/2 = (9/32)αm, matching step 2.
The rate is negative because off shell the current J^μ = ū(k)γ^μu(p) is not conserved: with q = p − k, q_μJ^μ = (W − m)ū(k)u(p), so the timelike, negative-norm photon inside −g_μν does not cancel. At W = m the window from x = m²/W² to 1 closes, J is conserved and Im Σ₂ → 0: the absorptive part is as gauge dependent as Σ₂ itself.

AnswerIm Σ₂ = (α/4)[3m − (15/16)p̸]; on the spinor it is (9/32)αm = 2.05 × 10⁻³ m, 1.05 keV for the electron, equal to −Γ_eff/2 with Γ_eff = −(9/16)αm from the cut: negative, a Feynman-gauge artefact of an off-shell line.