University Quantum Field Theory I · One Loop in φ⁴: Dimensional Regularisation, Counterterms and the Running Coupling · 6.2
The φ⁴ bubble: Feynman parameters and the Wick rotation
Every one-loop amplitude in this course reaches a number by three moves: merge the propagators with Feynman parameters, shift the loop momentum to complete the square, and turn the energy contour into Euclidean space. Learn them on φ⁴'s bubble, where nothing else intrudes, and the QED loops of Units 12 to 14 become bookkeeping.
Build the model
Connect the measurement to the mechanism.
At order λ² the φφ → φφ amplitude acquires a loop. Momentum conservation at two vertices leaves one four-momentum k unfixed, and the amplitude integrates two Feynman propagators over it. As written that integral cannot be judged: its poles sit on the real k⁰ axis, nudged by iε, and k² has no sign, so a large k need not mean a large denominator. Two exact moves fix this.
A Feynman parameter x merges the two propagators into one squared denominator, and the shift l = k + xp completes the square, leaving (l² − Δ + iε)⁻² with every external momentum and the mass packed into Δ = m² − x(1 − x)p². The Wick rotation then turns the l⁰ contour a quarter-turn counterclockwise, through the two quadrants that hold no pole, and returns i times a Euclidean integral of (lE² + Δ)⁻², a positive function of a positive lE². Both moves are derived here. The result, Mₛ = (λ²/2)∫₀¹dx ∫d⁴lE/(2π)⁴ (lE² + Δ)⁻², is real below threshold and grows as ln Λ under a cutoff: the logarithm power counting predicted.
The rewriting stops short in three places. It organises the divergence without removing it, so a regulator is still needed. The shift is safe only because the integral is no worse than logarithmic. And once p² exceeds 4m², Δ is negative over a window of x; there the rotated result is reached by continuing in p², and the imaginary part that unitarity demands appears.
- Simple definition
- Feynman parameters and the Wick rotation are exact rewritings that turn a one-loop Minkowski integral over k into i times a Euclidean integral of (lE² + Δ)⁻², averaged over a parameter x, with all external momenta inside Δ = m² − x(1 − x)p².
- Example
- In forward φφ scattering the t-channel bubble has p² = t = 0, so Δ = m² at every x. With a cutoff at Λ/m = 100, 16π² times the Euclidean integral is ln(10001) − 10000/10001 = 8.21, and for λ = 0.50 that gives Mₜ = 6.50 × 10⁻³ beside the tree value −0.50.
Derived from the order-λ² Dyson term with topic 3.02's counting. The four factors of i multiply to +1, leaving iMₛ = (λ²/2) ∫ d⁴k/(2π)⁴ (AB)⁻¹: λ² times one Minkowski integral.
λ dimensionless; k, p four-momenta in GeV; p = p₁ + p₂, so p² = s (t or u in the other channels); ½ from swapping internal lines
Derived: the antiderivative −1/[(A − B)(xA + (1 − x)B)] changes by (1/B − 1/A)/(A − B) = 1/(AB). Two separate poles become one squared denominator.
A, B: propagator denominators in GeV²; x dimensionless. Holds if xA + (1 − x)B never vanishes, which +iε on both guarantees
Derived from Schwinger's 1/A = ∫₀^∞ dα e(−αA): insert 1 = ∫dρ δ(ρ − Σαᵢ), put αᵢ = ρxᵢ, and ∫₀^∞ dρ ρⁿ⁻¹ e(−ρX) = (n − 1)!/Xⁿ.
xᵢ dimensionless with Σxᵢ = 1; each Aᵢ in GeV²; n = 2 gives the card above, n = 3 is topic 14.02's QED vertex
Derived. The cross term 2xk⋅p goes into l, so the integrand depends on l² alone; d⁴k = d⁴l, a safe shift because the integral is no worse than logarithmic.
l in GeV, Δ in GeV². Positive p²: Δ least at x = ½, m² − p²/4. Spacelike p²: Δ ≥ m². Δ goes negative only if p² exceeds 4m²
Derived for positive Δ: the counterclockwise quarter-turn sweeps quadrants 1 and 3, which are empty. Where Δ is negative the result is reached by continuing p² → p² + iε.
l⁰ = i lE⁰, so l² = −lE² and d⁴l = i d⁴lE; poles at l⁰ = ±(Eₗ − iε′), Eₗ = √(|l|² + Δ), in quadrants 2 and 4
Derived. Every e-fold of lE adds 1/(8π²): the logarithm that D = 0 predicted. The ln Λ² carries no x and no p², so one constant can absorb it.
I = ∫ d⁴lE/(2π)⁴ (lE² + Δ)⁻² over |lE| ≤ Λ, dimensionless; Λ, √Δ in GeV. For Λ² ≫ Δ, 16π²I → ln(Λ²/Δ) − 1
One loop momentum left free
Take L = ½(∂φ)² − ½m²φ² − (λ/4!)φ⁴, with ħ = c = 1 and the metric (+, −, −, −) throughout. At order λ² the connected, amputated φφ → φφ function has three diagrams: p₁ and p₂ enter one vertex and p₃, p₄ leave the other (the s channel), or the pairs are p₁, p₃ (t) and p₁, p₄ (u). Loops on external legs are not among them; LSZ (topic 4.05) removes those. In the s channel, conservation at both vertices fixes the total internal momentum at p = p₁ + p₂ and leaves one loop momentum k free. Swapping the two internal lines returns the same diagram, so topic 3.02's counting gives ½: 576 contractions, each weighted by (1/2!)(1/4!)² = 1/1152. The vertices and propagators bring (−iλ)²i² = λ², so iMₛ = (λ²/2)∫d⁴k/(2π)⁴ (AB)⁻¹ with A = (k + p)² − m² + iε and B = k² − m² + iε. Lorentz invariance makes it a function of p² alone. At s = 10m² and 90° in the centre-of-mass frame, t = u = −3m². The integrand falls as k⁻⁴ against d⁴k: D = 0, as topic 6.01 counted.
Feynman's identity, derived twice
Two propagators put their poles in different places. Feynman's identity trades them for one squared denominator and an integral over a parameter. The function −1/[(A − B)(xA + (1 − x)B)] has derivative [xA + (1 − x)B]⁻², and from x = 0 to 1 it changes by (1/B − 1/A)/(A − B) = 1/(AB). Test it with A = 3, B = 1: ∫₀¹ (1 + 2x)⁻² dx = ½ − 1/6 = 1/3. The identity fails only if the segment from B to A in the complex plane passes through zero, and with +iε on both every point of it has a positive imaginary part. For n denominators start from Schwinger's 1/Aᵢ = ∫₀^∞ dαᵢ e(−αᵢAᵢ), or −i∫₀^∞ dαᵢ e(iαᵢAᵢ) for Minkowski denominators, convergent for the same reason. Insert 1 = ∫₀^∞ dρ δ(ρ − Σαᵢ) and put αᵢ = ρxᵢ: the Jacobian ρⁿ and δ(ρ − ρΣxᵢ) = δ(1 − Σxᵢ)/ρ leave ∫₀^∞ dρ ρⁿ⁻¹ e(−ρΣxᵢAᵢ) = (n − 1)!/(ΣxᵢAᵢ)ⁿ, the third card, derived. Topic 14.02 uses n = 3.
Complete the square, then shift
Feed A and B into the identity: x[(k + p)² − m²] + (1 − x)(k² − m²) = k² + 2xk⋅p + xp² − m². The cross term disappears into l = k + xp, leaving l² − Δ + iε with Δ = m² − x(1 − x)p²; the iε rides along because x + (1 − x) = 1. The external momenta and the mass now sit in one number per x: iMₛ = (λ²/2)∫₀¹dx ∫d⁴l/(2π)⁴ (l² − Δ + iε)⁻². For positive p², x(1 − x) peaks at ¼, so Δ is least at x = ½, where it is m² − p²/4. At s = 10m² that is −1.5m², and Δ is negative for 0.113 ≤ x ≤ 0.887, where x(1 − x) exceeds m²/s. For spacelike p² the sign flips and Δ ≥ m²: at t = −3m² the t-channel Δ runs from m² at the ends to 1.75m² at x = ½. The shift has unit Jacobian, but moving the variable of a divergent integral is safe only when the divergence is at most logarithmic, as here. A linearly divergent integral changes by a finite surface term when shifted; topic 6.03's dimensional regularisation makes every shift legitimate.
The Wick rotation: a quarter-turn past no pole
At fixed l the l⁰ integrand ((l⁰)² − Eₗ² + iε)⁻², with Eₗ = √(|l|² + Δ), has double poles at l⁰ = ±(Eₗ − iε′): the positive one just below the real axis, the negative one just above it, where topic 1.06 put DF's poles to encode time ordering. They occupy quadrants 4 and 2. The integrand falls as |l⁰|⁻⁴, so arcs at infinity give nothing, and turning the contour counterclockwise from the real axis onto the imaginary axis sweeps only quadrants 1 and 3, which are empty: the two integrals are equal. Now put l⁰ = i lE⁰. Then dl⁰ = i dlE⁰ and l² = −lE², so (l² − Δ)² = (lE² + Δ)² and the fifth card follows, derived: Mₛ = (λ²/2)∫₀¹dx ∫d⁴lE/(2π)⁴ (lE² + Δ)⁻², real and positive while Δ is positive. Check it at fixed l by residues: closing below around Eₗ − iε′ gives ∫dl⁰/2π = i/(4Eₗ³), and the Euclidean side gives (i/2π)(π/2Eₗ³), the same number.
A Euclidean logarithm, as power counting promised
The rotated integrand depends only on lE² = (lE⁰)² + |l|², which is positive, so four-dimensional spherical coordinates apply: the unit 3-sphere has area 2π², and d⁴lE = 2π²lE³ dlE. With u = lE² the integral up to a cutoff Λ is elementary, and the sixth card follows, derived: I = (1/8π²)∫₀Λ lE³(lE² + Δ)⁻² dlE = [ln(1 + Λ²/Δ) − Λ²/(Λ² + Δ)]/(16π²). Its large-Λ form [ln(Λ²/Δ) − 1]/(16π²) is good to 0.02 in the bracket already at Λ = 10√Δ, where the two brackets are 3.625 and 3.605. Far out the integrand lE³/(lE² + Δ)² falls as 1/lE, so each e-fold of lE adds the same 1/(8π²): the logarithm D = 0 predicted. Dimensions check: d⁴lE brings mass⁴ and (lE² + Δ)⁻² mass⁻⁴, so I and M are pure numbers, as a 2 → 2 amplitude in four dimensions must be. The ln Λ² carries neither x nor p², so over three channels the divergence is the constant (3λ²/32π²)ln Λ², while ∂/∂p² adds a power of (lE² + Δ)⁻¹ and converges.
Above threshold: continue in p², then check
For s above 4m² the s-channel Δ is negative for x₋ ≤ x ≤ x₊, the roots of x(1 − x) = m²/s, and at small |l| the poles ±√(|l|² + Δ − iε) swing round to hug the imaginary axis: still in quadrants 2 and 4, but pinching the rotated contour as ε → 0. So evaluate below threshold, where the result is analytic in p², and continue p² → p² + iε. Since x(1 − x) ≥ 0 this sends Δ → Δ − iε, and inside the window −ln(Δ − iε) = −ln|Δ| + iπ. With x₊ − x₋ = √(1 − 4m²/s) this gives Im Mₛ = (λ²/32π)√(1 − 4m²/s), derived, which equals 2Ecm pcm σₜₒₜ for the quoted tree cross-section λ²/(32πs): topic 5.07's optical theorem holds. More checks: λ → 0 leaves the tree value −λ; at p² = −Q² with m → 0, ∫₀¹ ln[x(1 − x)]dx = −2 gives the bubble (λ²/32π²)[ln(Λ²/Q²) + 1], finite as m vanishes; and over three channels dM/d ln Λ = 3λ²/16π², the β coefficient quoted from topic 6.08. No φ⁴ coupling has been measured, but the same moves give Unit 14's a = α/2π = 0.001161, quoted, against the measured 0.0011597.
Change one variable at a time
Make the relationship visible.
Push p² past 4m², put x at 0.5 and drag |l| to zero: Δ turns negative and the poles swing up against the imaginary axis, yet never leave quadrants 2 and 4. That is the case reached by continuation; the bar's width is 32π Im Mₛ/λ².
Δ/m² AT THIS x0.580
RIGHT POLE, Re l⁰/m0.936
RIGHT POLE, Im l⁰/m-0.214
WINDOW x₊ − x₋ = 32π Im Mₛ/λ²0.000
Live interpretationΔ/m² AT THIS x: 0.580. RIGHT POLE, Re l⁰/m: 0.936. RIGHT POLE, Im l⁰/m: −0.214. WINDOW x₊ − x₋ = 32π Im Mₛ/λ²: 0.000
Catch the common trap
Explain before calculating.
The s-channel bubble at s = 6m² has Δ(x) = m² − x(1 − x)s, which is negative over part of 0 ≤ x ≤ 1. What is Im Mₛ at one loop?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyTwo scalars of mass m = 0.40 GeV in φ⁴ theory collide at √s = 1.20 GeV and scatter through 60° in the centre-of-mass frame. Find t and u, then the least value of Δ(x) = m² − x(1 − x)p² over 0 ≤ x ≤ 1 for each of the s-, t- and u-channel bubbles, and say which of them can be Wick-rotated directly at every x.
- s = (1.20 GeV)² = 1.44 GeV², and each particle has pcm² = s/4 − m² = 0.36 − 0.16 = 0.20 GeV².
- t = −2pcm²(1 − cos 60°) = −0.20 GeV² and u = −2pcm²(1 + cos 60°) = −0.60 GeV². Check: s + t + u = 0.64 GeV² = 4m², as it must be for four equal masses.
- For positive p², x(1 − x) peaks at ¼ at x = ½, so the s channel's least Δ is m² − s/4 = 0.16 − 0.36 = −0.20 GeV². It is negative where x(1 − x) exceeds m²/s = 1/9, between the roots x = (1 ± √(1 − 4/9))/2 = 0.127 and 0.873.
- For spacelike p², Δ = m² + x(1 − x)|p²| is never below m²: its least value is 0.16 GeV², at x = 0 and 1, rising at x = ½ to 0.21 GeV² in the t channel and 0.31 GeV² in the u channel.
- So the t- and u-channel bubbles rotate directly at every x, while the s-channel bubble needs the continuation in p² for 0.127 ≤ x ≤ 0.873.
Answert = −0.20 GeV², u = −0.60 GeV². Least Δ: −0.20 GeV² in the s channel (negative for 0.127 ≤ x ≤ 0.873) and 0.16 GeV² = m² in the t and u channels, so only the s channel needs the continuation.
MediumThe ψ self-energy loop of topic 2.01's scalar Yukawa model carries two different masses. Combine 1/([(k + p)² − m₁² + iε][k² − m₂² + iε]) with one Feynman parameter, find Δ(x), and show that Δ can vanish inside 0 ≤ x ≤ 1 only once p² reaches (m₁ + m₂)². For m₁ = 0.20 GeV, m₂ = 0.50 GeV and p² = 0.64 GeV², find where Δ is negative and the width of that window.
- Put x on the m₁ line: x[(k + p)² − m₁²] + (1 − x)(k² − m₂²) = k² + 2xk⋅p + xp² − xm₁² − (1 − x)m₂². With l = k + xp this is l² − Δ, Δ = xm₁² + (1 − x)m₂² − x(1 − x)p². At x = 0 it is m₂², at x = 1 it is m₁², and m₁ = m₂ = m returns m² − x(1 − x)p².
- As a quadratic, Δ = p²x² − (p² + m₂² − m₁²)x + m₂². For p² ≤ 0 no term of xm₁² + (1 − x)m₂² − x(1 − x)p² is negative, so Δ stays positive on [0, 1]. For positive p², real roots need (p² + m₂² − m₁²)² ≥ 4p²m₂², which for m₂ ≥ m₁, as here, means |√p² − m₂| ≥ m₁. The branch √p² ≤ m₂ − m₁ puts the parabola's vertex beyond x = 1, where Δ(1) = m₁² is positive, so inside [0, 1] Δ first vanishes at p² = (m₁ + m₂)² = 0.49 GeV², the two-particle threshold.
- At p² = 0.64 GeV², Δ = 0.64x² − 0.85x + 0.25 in GeV². Its roots are x = (0.85 ± √(0.7225 − 0.64))/1.28 = (0.85 ± 0.2872)/1.28, so Δ is negative for 0.440 ≤ x ≤ 0.888, with its least value −0.032 GeV² at x = 0.664.
- The width is x₊ − x₋ = √[(p² − m₁² − m₂²)² − 4m₁²m₂²]/p² = 0.2872/0.64 = 0.449. That is 2pcm/√s for the pair at √s = 0.80 GeV, where pcm = 0.180 GeV: the window measures two-body phase space, and for m₁ = m₂ it becomes √(1 − 4m²/s).
AnswerΔ = xm₁² + (1 − x)m₂² − x(1 − x)p², which can vanish in [0, 1] only for p² ≥ (m₁ + m₂)² = 0.49 GeV². At 0.64 GeV² it is negative for 0.440 ≤ x ≤ 0.888, a window of width 0.449 = 2pcm/√s.
HardCheck the Wick rotation without rotating. For ∫ d⁴l/(2π)⁴ (l² − Δ + iε)⁻², do the l⁰ integral by residues at fixed l, then cut the spatial momentum off at |l| = Λ. Compare with the four-dimensional Euclidean cutoff for Λ/m = 100 at Δ = m² and at Δ = 4m², and say what the two regulators agree on.
- At fixed l, with Eₗ = √(|l|² + Δ), the integrand (l⁰ − Eₗ + iε′)⁻²(l⁰ + Eₗ − iε′)⁻² has a double pole at Eₗ − iε′, below the axis. Close the contour below, clockwise: the residue there is d/dl⁰ (l⁰ + Eₗ)⁻² = −1/(4Eₗ³), so ∫dl⁰/2π = (−2πi)(−1/(4Eₗ³))/2π = i/(4Eₗ³). The rotated form gives (i/2π)(π/2Eₗ³), the same, so the rotation's factor +i is confirmed.
- With d³l/(2π)³ = |l|²d|l|/(2π²), the whole integral is (i/8π²)∫₀Λ |l|²(|l|² + Δ)(−3/2) d|l| = (i/8π²)[arsinh(Λ/√Δ) − Λ/√(Λ² + Δ)]. For large Λ, arsinh(Λ/√Δ) → ln(2Λ/√Δ), giving (i/16π²)[ln(Λ²/Δ) + 2 ln 2 − 2].
- The Euclidean cutoff gave (i/16π²)[ln(Λ²/Δ) − 1]. The ln Λ² and the −ln Δ agree; only the constant differs, by 2 ln 2 − 1 = 0.386.
- Numbers for Λ/m = 100, quoting 16π² times the integral divided by i: at Δ = m² the spatial cutoff gives 2[arsinh 100 − 100/√10001] = 8.597 and the Euclidean one 8.211; at Δ = 4m² they give 7.211 and 6.825. Each pair differs by 0.386.
- Between the two values of Δ, both regulators give 1.386, the ln 4 that −ln Δ predicts. Whatever depends on Δ only through differences, as the bubble's momentum dependence does, is blind to the choice of cutoff; the constant is what topic 6.06's counterterm absorbs, quoted here.
- The spatial cutoff is not Lorentz invariant, and the agreement is special to a logarithmic divergence. The four-dimensional Euclidean cutoff keeps the invariance, and topic 6.03's dimensional regularisation keeps it without a cutoff at all.
AnswerResidues give i/(4Eₗ³) for the l⁰ integral, as the rotation does. 16π² times the integral over i is 8.597 against 8.211 at Δ = m² and 7.211 against 6.825 at Δ = 4m²: the regulators differ by a constant 0.386 and agree on the Δ-dependence, 1.386 = ln 4.