University Quantum Field Theory I · One Loop in φ⁴: Dimensional Regularisation, Counterterms and the Running Coupling · 6.5
The one-loop tadpole: a mass shift and δZ = 0
This is the friendliest loop in the course: one vertex, one propagator, no external momentum and no Feynman parameter. Use it to see what a regulator keeps and what it throws away, why the mass moves while the field's normalisation stays put, and why a light scalar ought to surprise you.
Build the model
Connect the measurement to the mechanism.
At order λ the only one-particle-irreducible two-point diagram of φ⁴ theory closes two legs of a single vertex on each other. The external momentum passes straight through that vertex, so the loop is the same number at every p: Σ = (λ/2)⟨0|φ²|0⟩, the quartic interaction with one pair of fields replaced by their vacuum fluctuations, which is a mass term. A constant self-energy turns i/(p² − m₀² − Σ) into a free propagator with its pole moved to m² = m₀² + Σ and its residue untouched, so Z = 1 and δZ = 0 at one loop.
The size of the shift depends on how the quadratically divergent integral is regulated. A Euclidean cutoff gives λΛ²/(32π²) less a logarithm, a shift set by the highest scale the theory admits; dimensional regularisation keeps only the logarithm, as a pole in ε proportional to m², and nothing at all when m = 0; Pauli-Villars keeps a square of its regulator mass with a coefficient of its own. All three agree on the coefficient of the logarithm, and every difference among them is a constant that m₀² absorbs, which is why the measured mass is all that survives and why topic 6.06's counterterm can cancel the tadpole exactly.
The picture ends at two loops, where the sunset carries p through its loops and Z first moves from 1, and it leaves open why a scalar is light when nothing shields its mass from the scales above it.
- Simple definition
- In φ⁴ theory the one-loop self-energy is the tadpole Σ = (λ/2)⟨0|φ²|0⟩, a regulator-dependent constant that moves the propagator's pole to m² = m₀² + Σ and leaves its residue Z = 1.
- Example
- With λ = 1, m = 1 GeV and a Euclidean cutoff Λ = 10 GeV, Σ = (1/(32π²))[100 − ln 101] GeV² = 0.302 GeV², so a particle measured at 1 GeV needs m₀² = 0.698 GeV², while the residue of its pole stays exactly 1.
From −iΣ = (−iλ/2)∫ d⁴l/(2π)⁴ i/(l² − m² + iε) after l⁰ = i lE⁰. No external momentum appears, so Σ is one number for every p, with a positive Euclidean integrand.
λ dimensionless; lE the Euclidean loop momentum and m the mass, in GeV; Σ in GeV²; ½ the symmetry factor; derived
A constant insertion resums to a free propagator with a moved pole, so at order λ the whole effect of the loop is a change of mass.
m the measured mass and m₀ the Lagrangian mass, in GeV; the zero of p² − m₀² − Σ(p²), topic 3.08; derived
Z is set by the slope of the inverse propagator, not by where it crosses zero. The tadpole moves the zero and keeps the slope, as topic 4.05 anticipated.
Z dimensionless, the weight of the one-particle pole (topic 4.03); δZ ≡ Z − 1; holds at order λ; derived
Λ ≪ m gives λΛ⁴/(64π²m²) → 0; Λ ≫ m leaves λΛ²/(32π²) less a logarithm, so m₀² must cancel a quantity set by the highest scale kept.
Λ the cutoff in GeV; ΣΛ in GeV²; derived from (λ/(16π²)) ∫₀Λ l³ dl/(l² + m²)
Proportional to m², so zero for m = 0. The logarithm survives as the 2/ε pole; the quadratic divergence has moved to the pole of Γ(1 − d/2) at d = 2.
ε dimensionless; μ in GeV, from λ → λμε; γE = 0.5772; derived from Γ(−1 + ε/2) = −(2/ε + 1 − γE) + O(ε)
Lorentz invariant and finite, yet still growing as M². The square's coefficient is 1 for the cutoff and 2 ln 2 here; the −m² ln(scale²/m²) is common to every regulator.
M in GeV; regulator weights −2 + m²/M² and 1 − m²/M² cancel both Λ² and ln Λ²; derived
A loop the external momentum never enters
Work in natural units, ħ = c = 1, with the metric (+, −, −, −). In (λ/4!)φ⁴ the only order-λ diagram with two external legs joins both to one vertex and closes the other two fields on each other. Twelve of the 4! = 24 ways of attaching the legs give this picture, so the symmetry factor is 1/2, and in topic 3.08's convention −iΣ(p²) = (−iλ/2) ∫ d⁴l/(2π)⁴ i/(l² − m² + iε). Momentum conservation at the single vertex sends p straight through: l circulates on its own and p appears nowhere in the integrand. Doing the l⁰ integral around the pole at l⁰ = ωₗ − iε leaves ∫ d³l/((2π)³ 2ωₗ) = ⟨0|φ²(x)|0⟩, so Σ = (λ/2)⟨0|φ²|0⟩, derived. Read it as the quartic term with one pair of fields replaced by their vacuum average, (λ/4!) × 6⟨φ²⟩φ² = ½(λ⟨φ²⟩/2)φ²: a mass term. The particle moves through the zero-point fluctuations of its own field and is heavier for it, by the same amount at every momentum. Dimensions agree: [λ] = 0 and [φ] = 1, so Σ is a mass squared.
The pole moves, the residue stays
Topic 3.08 summed the 1PI insertions into i/(p² − m₀² − Σ(p²) + iε). With Σ a constant this is exactly a free propagator, i/(p² − m² + iε) with m² = m₀² + Σ: the pole moves and nothing else happens, so at order λ every effect of the loop is a change of mass. Topic 4.03's residue, 1/Z = 1 − dΣ/dp² at p² = m², is then exactly 1, because a constant has no slope. The field needs no rescaling: Z = 1, and when topic 6.06 writes Z = 1 + δZ, δZ = 0 at one loop, derived, as topic 4.05 anticipated in putting √Z = 1 on every leg. Picture the inverse propagator p² − m₀² − Σ against p²: a straight line of slope 1 that the tadpole lowers by Σ. That moves its zero, the mass, and leaves its slope, 1/Z. Only a self-energy that tilts the line can change Z, and in φ⁴ none does before two loops. The free-field limit λ → 0 returns m = m₀ and Z = 1 together, as it must.
Wick rotation, then a cutoff at Λ
Write l⁰ = i lE⁰. The Feynman poles sit in the second and fourth quadrants of the l⁰ plane, so the contour swings anticlockwise onto the imaginary axis without crossing either (topic 6.02). Then d⁴l = i d⁴lE and l² = −lE², and Σ = (λ/2) ∫ d⁴lE/(2π)⁴ 1/(lE² + m²): a positive integrand and an integral of mass dimension 4 − 2 = 2, quadratically divergent, as topic 6.01's D = 4 − E says for E = 2. Cut it off at |lE| = Λ. The unit 3-sphere has area 2π², so d⁴lE/(2π)⁴ = l³ dl/(8π²), and with u = l², ∫₀Λ l³ dl/(l² + m²) = ½[Λ² − m² ln(1 + Λ²/m²)]. Hence ΣΛ = (λ/(32π²))[Λ² − m² ln(1 + Λ²/m²)], derived. For Λ ≪ m it falls as λΛ⁴/(64π²m²), since no modes survive; at Λ/m = 1 the bracket is 0.307 m², and at Λ/m = 10 it is 95.4 m², of which the logarithm is only 4.6 m². For m = 0 it is λΛ²/(32π²), not zero: with a cutoff, a massless scalar does not stay massless.
The same loop in d = 4 − ε dimensions
Topic 6.03's master integral with n = 1 gives ∫ dd lE/(2π)d (lE² + m²)⁻¹ = Γ(1 − d/2)(m²)(d/2 − 1)/(4π)(d/2). Put d = 4 − ε and write the coupling as λμε so it stays dimensionless. The gamma function now sits by its pole at −1: Γ(−1 + x) = Γ(x)/(x − 1) = −(1/x − γE)(1 + x) + O(x) = −(1/x + 1 − γE) + O(x), with x = ε/2. Expand the rest, (4π)(−d/2) = [1 + (ε/2) ln 4π]/(16π²) and με (m²)(1 − ε/2) = m²[1 − (ε/2) ln(m²/μ²)], keep each product of 2/ε with an O(ε) term, and Σ = −(λm²/(32π²))[2/ε + 1 − γE + ln 4π − ln(m²/μ²)], derived. It is proportional to m² and vanishes at m = 0, where the integral is scaleless. No Λ² survives: the quadratic divergence is the pole of Γ(1 − d/2) at d = 2, which the limit d → 4 never meets. What survives is the logarithm with the cutoff's coefficient, −λm²/(32π²) per unit of ln Λ² there and per unit of ln μ² here.
What each regulator keeps
A hard Euclidean cutoff is O(4) symmetric after the rotation, but shifting the loop momentum of a worse-than-logarithmic integral moves a finite surface term, so answers depend on how momenta are routed, and in QED it would give the photon a mass (topic 13.01, quoted). Pauli-Villars subtracts the same loop with heavy masses and keeps Lorentz and shift invariance, and in QED gauge invariance. One regulator only turns Λ² into M² ln Λ², so the tadpole needs two, at M and √2 M, with weights −2 + m²/M² and 1 − m²/M² chosen so that the weights, and the weights times the masses squared, each add to zero; the regulator at M enters with the wrong sign, as no physical particle's propagator does. The result, (λ/(32π²))[2 ln 2 (M² − m²) − m² ln(M²/m²)], derived, is finite and still grows as M². Dimensional regularisation keeps Lorentz, shift and gauge invariance and drops every power divergence, at the price of the scale μ and of care with γ⁵, an intrinsically four-dimensional matrix, named. The square's coefficient, 1 or 2 ln 2, is regulator-made; the logarithm is common to all three.
Fine-tuning now, field renormalisation at two loops
Read the cutoff as the scale where φ⁴ stops being the right theory, and m₀² = m² − λΛ²/(32π²) + … must cancel a huge number to leave a small one. For a scalar of 125 GeV, the mass of the one elementary scalar yet observed, with λ = 1 and Λ at the Planck mass, MPl c² = √(ħc⁵/G) = 1.22 × 10¹⁹ GeV, the shift is 4.7 × 10³⁵ GeV² against m² = 1.56 × 10⁴ GeV²: m₀² must match it to one part in 3 × 10³¹. This is named the fine-tuning, or hierarchy, problem, and is not solved here. A fermion is spared by chiral symmetry, its self-energy growing only as m ln Λ (Unit 12, quoted); nothing shields a scalar. The field's normalisation waits one order. At λ² the sunset, two vertices joined by three propagators with symmetry factor 1/6, routes p through its loops, so dΣ/dp² ≠ 0 and Z first differs from 1, giving the field an anomalous dimension γ = λ²/(12(16π²)²) ≈ 3.3 × 10⁻⁶ λ², quoted. The other two-loop graphs, tadpoles hung on tadpoles, still ignore p.
Change one variable at a time
Make the relationship visible.
Drag s and both curves climb as s²; the solid one crosses Σ = m² near s ≈ 18/√λ, where the bare mass squared turns negative. Change M₂²/M₁² and only the dashed curve steepens: the square's coefficient is regulator-made. The MS-bar value in the bottom row never reaches 0.05 in size.
CUTOFF Σ/m²0.302
PAULI-VILLARS Σ/m²0.420
DIM REG, MS-BAR, μ = Λ: Σ/m²-0.0177
BARE m₀²/m², CUTOFF0.698
Live interpretationCUTOFF Σ/m²: 0.302. PAULI-VILLARS Σ/m²: 0.420. DIM REG, MS-BAR, μ = Λ: Σ/m²: −0.0177. BARE m₀²/m², CUTOFF: 0.698
Catch the common trap
Explain before calculating.
In φ⁴ theory with λ = 0.60 and a Euclidean cutoff Λ/m = 20, the order-λ self-energy is the tadpole, Σ = 0.749 m², with m the measured mass. Taking the full propagator as i/(p² − m₀² − Σ(p²) + iε), what does the tadpole do?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA φ⁴ theory with λ = 0.40 is regulated by a Euclidean cutoff Λ = 30 GeV, and its particle is measured at m = 2.0 GeV. Find the one-loop tadpole, the Lagrangian mass m₀ that yields the measured mass, and Z. What is the shortest distance the regulated theory still describes?
- At order λ the self-energy is the tadpole, ΣΛ = (λ/(32π²))[Λ² − m² ln(1 + Λ²/m²)]. Using the measured m inside the loop rather than m₀ changes Σ only at order λ².
- Λ²/m² = 900/4.0 = 225 and ln 226 = 5.4205, so the bracket is 900 − 4.0 × 5.4205 = 878.32 GeV². With λ/(32π²) = 0.40/315.83 = 1.2665 × 10⁻³, Σ = 1.112 GeV².
- The pole sits at m² = m₀² + Σ, so m₀² = 4.000 − 1.112 = 2.888 GeV² and m₀ = 1.699 GeV: the Lagrangian mass is 15% below the measured one.
- Σ contains no p², so dΣ/dp² = 0 and Z = 1 exactly: δZ = 0. Checks: Σ is in GeV², as a mass squared must be, and Λ → 0 gives Σ → 0 and m₀ → m.
- Restoring ħc, the cutoff discards modes shorter than ħc/Λ = 197.327 MeV fm / (3.0 × 10⁴ MeV) = 6.58 × 10⁻³ fm.
AnswerΣ = 1.11 GeV², m₀ = 1.70 GeV (m₀² = 2.89 GeV²) and Z = 1; the regulated theory says nothing below about 6.6 × 10⁻³ fm.
MediumDo the l⁰ integral of the tadpole exactly and cut off the spatial loop momentum at |l| = K instead. Derive ΣK, evaluate it for λ = 1.0, m = 1.0 GeV and K = 10 GeV, and compare it with the Euclidean-cutoff value 0.302 GeV² at Λ = 10 GeV. Which part of the result does not depend on the regulator?
- Closing the l⁰ contour around l⁰ = ωₗ − iε gives 1/(2ωₗ), so ΣK = (λ/2) ∫ d³l/((2π)³ 2ωₗ) over |l| ≤ K, which is (λ/(8π²)) ∫₀K l² dl/√(l² + m²).
- The primitive is ½[l√(l² + m²) − m² ln((l + √(l² + m²))/m)], so ΣK = (λ/(16π²))[K√(K² + m²) − m² ln((K + √(K² + m²))/m)], derived.
- Numbers: K√(K² + m²) = 10 × √101 = 100.499 GeV², and (K + √(K² + m²))/m = 20.050, whose logarithm is 2.9982; the bracket is 97.501 GeV². With λ/(16π²) = 6.3326 × 10⁻³, ΣK = 0.617 GeV², 2.04 times the Euclidean 0.302 GeV².
- For K ≫ m, ΣK → (λ/(32π²))[2K² + m² − m² ln(4K²/m²)], while ΣΛ → (λ/(32π²))[Λ² − m² ln(Λ²/m²)]. The coefficient of the square is 2 against 1: an artefact of the regulator's shape.
- The logarithm carries −λm²/(32π²) per unit of ln(scale²) in both, the coefficient dim reg puts on ln μ². That is the universal piece, the one the 2/ε pole records.
AnswerΣK = 0.617 GeV², about twice ΣΛ = 0.302 GeV². The coefficient of the square depends on the regulator; the −(λm²/(32π²)) ln(scale²) term does not.
HardA light scalar φ, measured at m = 125 GeV, couples to a heavy scalar χ of mass M = 10 TeV through −(κ/4)φ²χ² with κ = 0.10. In dimensional regularisation, find χ's one-loop contribution to φ's self-energy and its MS-bar remainder at μ = m, compare it with m², and say what doubling M does. Does the vanishing of dim reg's massless tadpole protect m?
- Read the rule off the Lagrangian (topic 3.07): the 2! × 2! ways of attaching φφχχ cancel the 1/4, so the vertex is −iκ. Closing the two χ lines on each other gives symmetry factor 1/2, so the loop is φ⁴'s tadpole with λ → κ and m → M.
- Hence Σχ = −(κM²/(32π²))[2/ε + 1 − γE + ln 4π − ln(M²/μ²)]. MS-bar removes 2/ε − γE + ln 4π and leaves Σχ = (κM²/(32π²))[ln(M²/μ²) − 1].
- Numbers: κM²/(32π²) = 0.10 × 1.0 × 10⁸ GeV²/315.83 = 3.166 × 10⁴ GeV², and ln(M²/m²) = 2 ln 80 = 8.764, so Σχ = 3.166 × 10⁴ × 7.764 = 2.46 × 10⁵ GeV².
- Against m² = 1.5625 × 10⁴ GeV² that is 15.7 times larger. Neglecting φ's own small tadpole, the MS-bar mass parameter must be 1.56 × 10⁴ − 2.46 × 10⁵ = −2.30 × 10⁵ GeV² to leave the measured 125 GeV: a cancellation to one part in 16.
- Doubling M to 20 TeV gives 1.16 × 10⁶ GeV², 4.7 times more: M² times a slowly growing logarithm. Checks: GeV² throughout, and Σχ → 0 as M → 0, where the χ loop becomes scaleless.
AnswerΣχ ≈ 2.5 × 10⁵ GeV² in MS-bar at μ = 125 GeV, 16 times m², and almost five times more if M doubles: dim reg's zero massless tadpole gives no protection once a heavier scale is present.