University Quantum Field Theory I · Vacuum Polarisation, Charge Renormalisation and the Running of alpha · 13.2

The electron loop: Π₂(q²) in dimensional regularisation

This is the loop the rest of the unit lives on. Take the electron loop in the photon propagator from its Dirac trace to a formula you can evaluate, in d dimensions, so that gauge invariance survives the calculation instead of being rescued after it, and learn to read which part of the answer is regulator and which is physics.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The first quantum correction to a photon's propagation is a closed electron loop: the photon turns into a virtual pair, which annihilates back into a photon. Topic 7.04's rule makes the loop −1 times a Dirac trace, and power counting (topic 12.01) says the integral diverges quadratically, which is why topic 13.01's sharp cutoff left a photon mass behind. In d = 4 − ε dimensions the same integral is finite and short: one Feynman parameter, the shift l = k + xq, symmetric integration with l^μl^ν → g^μν l²/d, and topic 6.03's master integrals.

The term a cutoff turned into Λ², (2/d − 1) times the l² integral, equals in d dimensions exactly −Δ times the integral without l², because Γ(1 − d/2) = Γ(2 − d/2)/(1 − d/2). Added to the rest it leaves 2x(1 − x)(q²g^μν − q^μq^ν) at every value of the Feynman parameter, the transverse form topic 12.09's Ward identity demands, now visible term by term. Expanding near four dimensions gives Π₂(q²) = −(2α/π)∫₀¹ dx x(1 − x)[2/ε − γ_E + ln 4π − ln(Δ/μ²)]: a single pole, standing where a gauge-invariant cutoff puts ln Λ², and a finite logarithm carrying all the dependence on q² and m.

The result is not yet a prediction. The pole has to be absorbed into the charge (topic 13.03), the logarithm turns complex above q² = 4m² (topic 13.04), and two-loop diagrams, with a photon inside the loop, correct it at relative order α/π.

Simple definition: Π₂(q²) is the scalar function that multiplies (q²g^μν − q^μq^ν) in the one-loop electron-loop correction to the photon propagator; computed in d = 4 − ε dimensions it is finite except for a single pole in ε.
Example: For the electron loop at spacelike q² = −(1.00 GeV)² with μ = 1.00 GeV, the pole part is −(2α/3π)/ε = −1.55 × 10⁻³/ε and the MS-bar finite part is −5α/9π = −1.29 × 10⁻³; the electron mass enters only at order m²/Q² ≈ 3 × 10⁻⁷.

The electron loop, with its signiΠ₂^μν(q) = −(−ie)² μ^ε ∫ d^d k/(2π)^d Tr[γ^μ S(k) γ^ν S(k + q)]

Topic 7.04's −1 and trace, quoted. No symmetry factor and a single orientation: the external photons pin the two vertices.

S(k) = i(k̸ + m)/(k² − m² + i0); e = 0.3028, dimensionless; μ the scale in GeV; Π₂^μν in GeV²; d = 4 − ε

The trace, valid in d dimensionsTr[γ^μ(k̸ + m)γ^ν(k̸ + q̸ + m)] = 4[k^μ(k + q)^ν + k^ν(k + q)^μ − g^μν(k⋅(k + q) − m²)]

Quoted from Advanced QM. It uses only anticommutation and cyclicity, never g^μ_μ = 4, so continuing to d dimensions adds no order-ε term to it.

Tr 1 = 4 kept at every d; k and q four-momenta in GeV, m the electron mass in GeV

The l² term joins the rest(2/d − 1) ∫ d^d l/(2π)^d l²/(l² − Δ)² = −Δ ∫ d^d l/(2π)^d 1/(l² − Δ)²

Derived from Γ(1 − d/2) = Γ(2 − d/2)/(1 − d/2). It turns the g^μν coefficient into −Δ + m² + x(1 − x)q² = 2x(1 − x)q², transverse at each x.

l = k + xq in GeV; Δ = m² − x(1 − x)q² in GeV²; x the Feynman parameter, 0 to 1; exact for every d

Transverse at every dΠ₂(q²) = −(8e²μ^ε/(4π)^(d/2)) ∫₀¹ dx x(1 − x) Γ(2 − d/2) Δ^{(d/2 − 2)}

Derived, before any expansion in ε: the whole loop is one scalar function times the structure that topic 12.09's Ward identity requires.

defined by iΠ₂^μν = (q²g^μν − q^μq^ν) iΠ₂(q²); Π₂ is dimensionless, μ^ε cancelling the GeV^(−ε) of Δ^(−ε/2)

Π₂ near four dimensionsΠ₂(q²) = −(2α/π) ∫₀¹ dx x(1 − x)[2/ε − γ_E + ln 4π − ln(Δ/μ²)]

Derived. One pole, the same for every q² and m, which one counterterm absorbs; all dependence on q² and m sits in the finite logarithm.

α = e²/4π = 1/137.036; γ_E = 0.5772; ∫₀¹ x(1 − x) dx = 1/6, so the pole is −(α/3π)(2/ε)

High-energy limit, −q² ≫ m²Π₂(q²) → −(α/3π)[2/ε − γ_E + ln 4π − ln(−q²/μ²) + 5/3]

Derived. The mass drops out, leaving a massless fermion's loop; its logarithm of −q² is what topic 13.05 turns into the running of α.

q² spacelike, in GeV²; corrections of order m²/q²; the 5/3 comes from ∫₀¹ x(1 − x) ln[x(1 − x)] dx = −5/18

One diagram, a minus sign and a trace

Natural units ħ = c = 1, the metric (+, −, −, −) and d = 4 − ε throughout. At order e² the photon's two-point function has one loop: a photon of momentum q makes an electron-positron pair that recombines. The vertices are −ieμ^(ε/2)γ^μ, the μ^(ε/2) keeping e dimensionless off four dimensions (topic 12.03), the propagators S(k) = i(k̸ + m)/(k² − m² + i0), and topic 7.04 supplies −1 and a trace for the closed line. There is no symmetry factor: the two external photons pin the two vertices, and a loop with two vertices has only one orientation. A tadpole, a loop with one photon attached, vanishes by Furry's theorem (topic 11.07). The trace needs only (γ^μ, γ^ν) = 2g^μν and cyclicity, so with Tr 1 = 4 it is Advanced QM's four-dimensional result, quoted. Since (−ie)² i² = +e², the loop's −1 leaves iΠ₂^μν = −4e²μ^ε ∫ d^d k/(2π)^d [k^μ(k + q)^ν + k^ν(k + q)^μ − g^μν(k⋅(k + q) − m²)]/[(k² − m²)((k + q)² − m²)], derived.

Feynman parameter, shift, and a numerator not yet transverse

Topic 6.02's 1/(AB) = ∫₀¹ dx [xA + (1 − x)B]⁻², quoted, with A = (k + q)² − m² and B = k² − m², turns the denominator into [(k + xq)² − Δ]², Δ = m² − x(1 − x)q². Shift to l = k + xq, which dimensional regularisation permits (topic 6.03). Then k = l − xq and k + q = l + (1 − x)q; terms odd in l integrate to zero, k^μ(k + q)^ν + k^ν(k + q)^μ becomes 2l^μl^ν − 2x(1 − x)q^μq^ν, and k⋅(k + q) − m² becomes l² − x(1 − x)q² − m². Symmetric integration, l^μl^ν → g^μν l²/d, leaves the numerator (2/d − 1)g^μν l² − 2x(1 − x)q^μq^ν + g^μν[m² + x(1 − x)q²], derived. Look at it before integrating. The q^μq^ν coefficient is −2x(1 − x), so transversality needs the g^μν coefficient to be +2x(1 − x)q², and it is not: it contains l², which is not q². Transversality is not a property of the integrand. It has to come out of the integral, which is exactly where topic 13.01's cutoff spoiled it.

The l² term becomes −Δ, and the loop is transverse

After the Wick rotation, topic 6.03's master integrals, quoted, give ∫ d^d l/(2π)^d (l² − Δ)⁻² = iΓ(2 − d/2)Δ^{(d/2 − 2)}/(4π)^(d/2) and ∫ d^d l/(2π)^d l²(l² − Δ)⁻² = −i(d/2)Γ(1 − d/2)Δ^{(d/2 − 1)}/(4π)^(d/2). The second carries the quadratic divergence as a pole of Γ(1 − d/2) at d = 2. Multiply it by (2/d − 1): since (2/d − 1)(d/2) = 1 − d/2 and (1 − d/2)Γ(1 − d/2) = Γ(2 − d/2), the l² term is exactly −Δ times the first integral, at every d. The g^μν coefficient becomes −Δ + m² + x(1 − x)q² = 2x(1 − x)q², and the numerator is 2x(1 − x)(q²g^μν − q^μq^ν) at each x separately, derived. At x = 0.3 and q² = −2m², for instance, Δ = 1.42m² and −1.42m² + m² − 0.42m² = −0.84m² = 2(0.21)(−2m²). Contracting with g_μν instead, using topic 12.03's γ^μγ_μ = d and γ^μγ^νγ_μ = −(2 − ε)γ^ν, quoted, gives (d − 1)q²Π₂ as it must. Keep the 1/d as 1/d until this cancellation is done: ε times a pole is finite.

Near four dimensions: one pole and a logarithm

Collecting, Π₂(q²) = −(8e²μ^ε/(4π)^(d/2)) ∫₀¹ dx x(1 − x)Γ(2 − d/2)Δ^{(d/2 − 2)}, exact in d. Put d = 4 − ε: Γ(ε/2) = 2/ε − γ_E + O(ε), and (4πμ²/Δ)^(ε/2) = 1 + (ε/2) ln(4πμ²/Δ) + O(ε²), whose order-ε term meets the pole and survives. With 8e²/16π² = 2α/π, Π₂(q²) = −(2α/π)∫₀¹ dx x(1 − x)[2/ε − γ_E + ln 4π − ln(Δ/μ²)], derived. Since ∫₀¹ x(1 − x) dx = 1/6, the pole is −(α/3π)(2/ε) = −1.549 × 10⁻³/ε, the same for every q² and every mass, so one counterterm, −i(q²g^μν − q^μq^ν)δ₃, absorbs it (topic 13.03). Pauli-Villars regularisation, which adds loops of heavy regulator fermions with coefficients chosen so that every divergence cancels, is four-dimensional and gauge invariant, and gives the same bracket with 2/ε − γ_E + ln 4π + ln μ² replaced by ln M², M the regulator mass, quoted: the pole is the image of ln Λ², and a sharp cutoff's Λ² has no counterpart in either. The scale μ enters only beside the pole, so ∂Π₂/∂ ln μ² = −α/3π at every q². A trace normalisation other than Tr 1 = 4 would shift Π₂ by a q²-independent constant only.

Checks: dimensions, small q² and large q²

Π₂ is a pure number: e is dimensionless at d = 4 and μ^ε cancels the dimension of Δ^(−ε/2). For |q²| ≪ m², expanding ln Δ with ∫₀¹ x²(1 − x)² dx = 1/30 gives Π₂(q²) = Π₂(0) − αq²/(15πm²) + O(q⁴/m⁴), derived: a heavy loop decouples into a constant, which the charge absorbs, and a q²/m² term that is a contact interaction in position space. In hydrogen that term lowers the 2s level by 27.1 MHz, part of the measured 1057.8 MHz Lamb shift, quoted from UAQM-13.09 and derived in topic 13.08. For −q² ≫ m², Δ → x(1 − x)(−q²) and ∫₀¹ x(1 − x) ln[x(1 − x)] dx = −5/18 give Π₂ → −(α/3π)[2/ε − γ_E + ln 4π − ln(−q²/μ²) + 5/3], derived: the mass drops out. At Q = 91.19 GeV the electron loop's Π₂(−Q²) − Π₂(0) = (α/3π)[ln(Q²/m²) − 5/3] = 0.0174, computed, the shift topic 13.05 turns into the running of α; the measured 1/α near 129 at that scale also needs the muon, the tau and hadrons, quoted. Between the limits, near Q ≈ 2m, neither form holds and the integral must be done.

Change one variable at a time

Make the relationship visible.

Interactive model

Momentum transfer log₁₀(Q/m)1.0

Scale log₁₀(μ/m)0.0

Drag log₁₀(μ/m) from 0 to 2: the curve and both dashed lines drop together by (α/3π) ln 10⁴ = 7.1 × 10⁻³, while the rise above q² = 0 does not move. Then sweep Q through 2m and watch the curve leave the flat line for the sloped one.

MS-BAR Π₂ AT Q2.319 × 10⁻³

MASSLESS FORM2.275 × 10⁻³

RISE ABOVE q² = 02.319 × 10⁻³

Q FOR AN ELECTRON LOOP5.110 MeV

Live interpretationMS-BAR Π₂ AT Q: 2.319 × 10⁻³. MASSLESS FORM: 2.275 × 10⁻³. RISE ABOVE q² = 0: 2.319 × 10⁻³. Q FOR AN ELECTRON LOOP: 5.110 MeV

Catch the common trap

Explain before calculating.

Misconception check

It is tempting to read the calculation as dimensional regularisation throwing away topic 13.01's Λ² by fiat and keeping a convenient logarithm. Nothing is thrown away. The l² integral's quadratic divergence is still there, as the pole of Γ(1 − d/2) at d = 2; at d = 4 the residue of that integral is proportional to Δ, and that piece is exactly what joins the m² and q² terms to make the loop transverse. The Λ² was never a property of QED: it came from a regulator that is not invariant under the shift l = k + xq. Pauli-Villars, a four-dimensional regulator that respects gauge invariance, finds no Λ² either, and its ln M² carries the same coefficient, −α/3π, as dimensional regularisation's 2/ε. The pole stands for the logarithm; it is not where the power went into hiding.

A student evaluating the electron loop replaces l^μl^ν by g^μν l²/4 rather than g^μν l²/d, then uses the d-dimensional master integrals and lets ε → 0, doing everything else correctly. What does the error leave in Π₂^μν?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAt q² = 0, write the pole part and the MS-bar finite part of Π₂(0) for an electron loop (m = 0.511 MeV) and for a muon loop (m = 105.66 MeV), both at μ = 1.000 GeV. Show that the difference between the two loops is finite and does not depend on μ.

At q² = 0, Δ = m² for every x, so the x-integral is ∫₀¹ x(1 − x) dx = 1/6 and Π₂(0) = −(α/3π)[2/ε − γ_E + ln 4π − ln(m²/μ²)].
α/3π = 7.743 × 10⁻⁴. The pole part is −(2α/3π)/ε = −1.549 × 10⁻³/ε for both loops: it does not know which mass is running round.
MS-bar keeps −(α/3π) ln(μ²/m²). Electron: ln(μ²/m²) = 2 ln(1000/0.511) = 15.158, giving −1.174 × 10⁻². Muon: 2 ln(1000/105.66) = 4.495, giving −3.480 × 10⁻³.
Difference, electron minus muon: −(α/3π) ln(m_μ²/mₑ²) = −7.743 × 10⁻⁴ × 10.663 = −8.256 × 10⁻³. The ln μ² cancelled, and so did the pole.
Checks: each number is dimensionless, as Π₂ must be, and each finite part would vanish at μ = m, where its logarithm does.

AnswerPole −1.549 × 10⁻³/ε for both; MS-bar parts −1.174 × 10⁻² (electron) and −3.48 × 10⁻³ (muon); their difference, −8.26 × 10⁻³, is finite and μ-independent.

MediumEvaluate the MS-bar finite part of the electron loop's Π₂ at the spacelike point q² = −4m² (Q = 2m = 1.022 MeV) with μ = m, using the closed form 6∫₀¹ x(1 − x) ln(1 + ax(1 − x)) dx = −5/3 + 4/a + (1 − 2/a) v ln((v + 1)/(v − 1)), v = √(1 + 4/a), quoted and checkable numerically. Compare with the small-q² and massless forms.

With μ = m, the MS-bar part is −(2α/π)∫₀¹ x(1 − x)[−ln(Δ/m²)] dx = (α/3π) × 6∫₀¹ x(1 − x) ln(1 + ax(1 − x)) dx, with a = Q²/m² = 4.
v = √2 = 1.41421 and ln((v + 1)/(v − 1)) = ln 5.8284 = 1.76275, so the bracket is −1.66667 + 1 + 0.5 × 1.41421 × 1.76275 = −0.66667 + 1.24645 = 0.57978.
Π₂ finite part = 7.7427 × 10⁻⁴ × 0.57978 = 4.489 × 10⁻⁴. A Simpson's-rule check of the integral with 20 000 steps gives the same 0.57978.
Small-q² form: the bracket becomes a/5 = 0.800, giving 6.19 × 10⁻⁴, 38% too high, because q² = −4m² is not small against m².
Massless form: ln a − 5/3 = 1.3863 − 1.6667 = −0.2804, giving −2.17 × 10⁻⁴, the wrong sign, because m² is not small against Q².

AnswerΠ₂ finite part = 4.49 × 10⁻⁴ at q² = −4m², μ = m. The small-q² form gives 6.19 × 10⁻⁴ and the massless form −2.17 × 10⁻⁴: near Q = 2m neither limit holds.

HardContract the electron loop with g_μν. Using topic 12.03's γ^μγ_μ = d and γ^μγ^νγ_μ = −(2 − ε)γ^ν, find the contracted numerator, integrate it, and show that g_μνΠ₂^μν = (d − 1)q²Π₂. A student then extracts Π₂ by dividing by 3q². By how much is the result off, and does the error reach the Medium example's 4.49 × 10⁻⁴, or Π₂(q²) − Π₂(0)?

Traces of an odd number of γ's vanish, so Tr[γ^μ(k̸ + m)γ_μ(k̸ + q̸ + m)] = −(2 − ε) × 4k⋅(k + q) + 4dm² = 4[(2 − d)k⋅(k + q) + dm²], since −(2 − ε) = 2 − d.
After the shift, k⋅(k + q) → l² − x(1 − x)q², and (2 − d)(d/2)Γ(1 − d/2) = dΓ(2 − d/2) makes the l² term −dΔ times the scalar integral.
The bracket is −dΔ − (2 − d)x(1 − x)q² + dm² = 2(d − 1)x(1 − x)q², so g_μνΠ₂^μν = (d − 1)q²Π₂, which is g_μν(q²g^μν − q^μq^ν) = (d − 1)q² times Π₂, as transversality requires.
Dividing by 3q² gives (d − 1)Π₂/3 = (1 − ε/3)Π₂. The pole −(2α/3π)/ε times −ε/3 leaves +2α/9π = 5.162 × 10⁻⁴, the same at every q².
So the MS-bar value at Q = 2m, μ = m becomes 4.489 × 10⁻⁴ + 5.162 × 10⁻⁴ = 9.651 × 10⁻⁴, more than twice the true one, while Π₂(q²) − Π₂(0) loses the constant and is unchanged.
Checks: 2α/9π is dimensionless, and g_μνΠ₂^μν vanishes at q² = 0, as it must if the photon stays massless.

Answerg_μνΠ₂^μν = (d − 1)q²Π₂. Dividing by 3q² adds 2α/9π = 5.16 × 10⁻⁴ at every q², more than doubling the MS-bar value at Q = 2m to 9.65 × 10⁻⁴ but leaving Π₂(q²) − Π₂(0) untouched.