University Quantum Field Theory I · One Loop in φ⁴: Dimensional Regularisation, Counterterms and the Running Coupling · 6.7
A finite prediction: the one-loop φφ → φφ amplitude
This is the first loop amplitude in the course you could hand to an experimenter. Measure the scattering amplitude once, at threshold, and the one-loop φφ → φφ amplitude at every other energy and angle follows from m alone, with no ε, no μ and no cutoff anywhere in the answer.
Build the model
Connect the measurement to the mechanism.
Each one-loop bubble in φ⁴ diverges, but only by a constant. Power counting gives the four-point function D = 0, so the divergent part of a bubble cannot depend on the external momenta: it is the same 2/ε − γE + ln 4π in every channel in dimensional regularisation, the same ln Λ² with a cutoff. A momentum-independent infinity is exactly what one counterterm vertex, −iδλ, can absorb.
Fixing δλ so that M = −λ at the threshold point s = 4m², t = u = 0 subtracts each channel at its own reference momentum, and what remains is a difference of two logarithms of the same kind, −(λ²/(32π²)) ∫dx ln[Δ(p²)/Δ(p₀²)] per channel. Nothing in it remembers how the integral was regulated: ε, μ, γE and ln 4π have cancelled, and a cutoff or a Pauli-Villars mass gives the same function up to terms that fall as 1/Λ² and cancel among the channels. The bare coupling λ₀ = λ + δλ does depend on the regulator, grows without limit as the regulator is removed, and is never measured.
The result is a relation between measurements: the threshold amplitude fixes λ, and the amplitude at any other s and angle is predicted, imaginary part and all. It stops being trustworthy where its own logarithms grow, once (3λ/(32π²)) ln(s/m²) is no longer small; there the running coupling of topics 6.08 and 6.09 takes over.
- Simple definition
- The renormalised one-loop φφ → φφ amplitude is −λ plus the s-, t- and u-channel bubbles, each subtracted at its own threshold momentum by the counterterm δλ, leaving a finite function of s, t and u in which no regulator appears.
- Example
- For λ = 1 at √s = 6m and θ = 90°, the bubbles lift −Re M from 1 to 1.0183 and add Im M = 0.0094; for m = 0.80 GeV the differential cross-section rises from 26.8 to 27.7 nb sr⁻¹.
No ε, μ or cutoff survives. Given m and the threshold value M = −λ, the amplitude at every energy and angle is fixed.
Σ over p² = s, t, u with p₀² = 4m², 0, 0; M and λ dimensionless in four dimensions, m and √s in GeV; −i0 is Feynman's prescription
The +2 is the s channel's threshold subtraction. It makes the bracket vanish at s = 4m², t = u = 0, though the integrand there has an integrable log singularity at x = ½.
F(p²) = ∫₀¹ dx ln[1 − x(1 − x)p²/m² − i0]; F(0) = 0, and F(4m²) = ∫₀¹ ln(1 − 2x)² dx = −2
Pole, scheme constants and μ all sit in δλ, so in the bare λ₀ = λ + δλ, which is never measured and grows without limit as the regulator is removed.
d = 4 − ε; μ is the mass scale in λμε; γE = 0.5772. A Euclidean cutoff Λ gives 3 ln(Λ²/m²) − 1 in the bracket instead
It equals sβσₜᵣₑₑ with σₜᵣₑₑ = λ²/(32πs), the optical theorem for identical scalars quoted from Unit 5: a check the counterterm can neither spoil nor fake.
β is each particle's speed in the centre-of-mass frame; only the s channel is above threshold, and the real δλ cannot touch it
Each channel differs by a term falling as 1/Λ², and the three cancel: the regulator first shows itself at order (s² + t² + u² − 16m⁴)/Λ⁴.
Λ is a Euclidean cutoff on the shifted loop momentum, in GeV; Σ(p² − p₀²) = s + t + u − 4m², which is zero on shell
Each logarithm grows as 2 ln √s at fixed θ, so −Re M rises by 3λ²/(16π²) per unit of ln √s: the coefficient topic 6.08 obtains as β(λ).
s, −t, −u ≫ m², with t = −(s − 4m²)(1 − cos θ)/2 and u = −(s − 4m²)(1 + cos θ)/2
One constant cancels a momentum-independent infinity
Work in natural units, ħ = c = 1, with the metric (+, −, −, −). Topic 6.04 left each bubble adding (λ²/(32π²)) ∫₀¹ dx [2/ε − γE + ln 4π − ln(Δ/μ²)] to M, with Δ = m² − x(1 − x)p² − i0 and p² = s, t or u; that result is quoted here. Its divergent and scheme-dependent part, 2/ε − γE + ln 4π + ln μ², is one number, the same in all three channels and at every momentum. It has to be: the four-point function has D = 0 (topic 6.01), so its divergence carries no power of the external momenta. The counterterm vertex −iδλ is also a momentum-independent constant, and topic 6.06's condition, M = −λ at s = 4m², t = u = 0, makes δλ equal to the three bubbles evaluated there. Subtract, and each channel becomes the same bracket at p² minus the same bracket at p₀². The 2/ε, γE, ln 4π and μ cancel identically, before ε → 0 is taken, and −(λ²/(32π²)) ∫dx ln[Δ(p²)/Δ(p₀²)] per channel is all that is left. That subtraction is the derivation.
Doing the subtractions: F(4m²) = −2 and F(0) = 0
Write F(p²) = ∫₀¹ dx ln[1 − x(1 − x)p²/m² − i0], so each channel contributes −(λ²/(32π²))[F(p²) − F(p₀²)]. The t and u channels are subtracted at p₀² = 0, where F vanishes. The s channel is subtracted at threshold, where Δ₀ = m²(1 − 2x)² touches zero at x = ½: the logarithm is singular there but integrable, and ∫₀¹ ln(1 − 2x)² dx = −2. Hence M = −λ − (λ²/(32π²))[F(s) + F(t) + F(u) + 2]. The x integral is elementary. For spacelike p² = −Q², F = 2√(1 + 4m²/Q²) arsinh(Q/2m) − 2, real and positive. Above threshold, Re F(s) = 2β arcosh(√s/2m) − 2 and Im F(s) = −πβ, with β = √(1 − 4m²/s); the imaginary part comes from the strip of x on which Δ is negative, where the −i0 chooses the branch. At √s = 6m and 90°, −t = −u = 16m², F(t) = F(u) = 1.2281 and Re F(s) = 1.3239, so the bracket is 5.780.
Change the regulator and only δλ notices
Redo the calculation with a sharp Euclidean cutoff Λ on the shifted loop momentum. Each bubble's bracket becomes ln(1 + Λ²/Δ) − Λ²/(Λ² + Δ) = ln(Λ²/Δ) − 1 + 2Δ/Λ² + O(Λ⁻⁴), and the threshold condition gives δλ = (λ²/(32π²))[3 ln(Λ²/m²) − 1]. The 6/ε has become 3 ln Λ²; for λ = 1 the bare coupling is 1.13 at Λ/m = 10³ and 1.74 at Λ/m = 10¹⁷. In M the logarithms of Λ cancel exactly as the poles did, and the remainder differs from the dimensional result by 2∫(Δ − Δ₀)dx/Λ² = −(p² − p₀²)/(3Λ²) per channel, because ∫₀¹ x(1 − x) dx = 1/6. Pauli-Villars, each i/(k² − m²) replaced by i/(k² − m²) − i/(k² − MPV²), leaves −(5/6)(p² − p₀²)/MPV² instead, by the same expansion. Either remainder is linear in p², so summed over channels it is proportional to s + t + u − 4m², which vanishes on shell. The regulator first appears at order 1/Λ⁴ and disappears as it is removed.
Checks: dimensions, threshold and the optical theorem
In four dimensions a 2 → 2 amplitude is dimensionless, and so are λ and every logarithm's argument, a ratio of squared masses; there is nowhere for Λ or μ to hide. Setting λ² → 0 returns the tree amplitude −λ. At threshold the result is −λ at every angle, because t = u = 0 there. Just above it, with β the speed of each particle in the centre-of-mass frame, Re F(s) + 2 ≈ 2β² and F(t) ≈ −t/(6m²), so the real correction grows as β² while Im M = λ²β/(32π) grows as β: in the non-relativistic limit the loop is mostly the imaginary, unitarity part. That imaginary part survives renormalisation untouched, because δλ is real, and it equals sβσₜᵣₑₑ with σₜᵣₑₑ = λ²/(32πs), the optical theorem for identical scalars. No elementary scalar's φφ → φφ amplitude has been measured, so the measured test of this logic is the Lamb shift, taken up below.
What it predicts, and where it stops
Once m and λ are measured nothing is free, and λ is itself a measurement, the threshold amplitude. Everything else is prediction. For λ = 1 at √s = 6m and 90°, −Re M = 1.0183 and Im M = 0.0094, so dσ/dΩ = |M|²/(64π²s) is 3.7% above tree level, 27.7 against 26.8 nb sr⁻¹ for m = 0.80 GeV (1 GeV⁻² = 0.3894 mb). At high energy the bracket tends to ln(s/m²) + ln(−t/m²) + ln(−u/m²) − 4, so at fixed angle −Re M climbs by 3λ²/(16π²) = 0.0190λ² per unit of ln √s, the rate topic 6.08 identifies as β(λ). The massless limit is singular: as m → 0 the reference point s = 4m², t = u = 0 slides into the infrared, which is why topic 6.08 subtracts at a scale μ instead. And the loop is small only while (3λ/(32π²)) ln(s/m²) is; for λ = 1 at 90° it would equal the tree term near √s/m ≈ 2 × 10²³, but the logarithms need resumming long before, the job of topic 6.09.
Bethe's subtraction, re-read as a mass counterterm
Bethe's 1947 estimate (UAQM-13.04, quoted) found that a free electron's non-relativistic self-energy grows linearly with the photon cutoff K, read it as electromagnetic mass, and subtracted it so that a free electron's energy is p²/2m with the measured m. That is an on-shell mass counterterm: δm fixed by the pole condition, as topic 6.06 fixes δm in φ⁴. The bound-state remainder diverged only as ln K, and with K = mc² it gave about 1040 MHz against the measured 1057.8 MHz. What the programme adds is a counterterm for every superficially divergent function, here δm, δZ and δλ, each fixed by a measurement or a normalisation, and an answer from which the regulator has gone, as in the amplitude above. Bethe had one counterterm and kept his cutoff: a non-relativistic theory misdescribes photons above mc², so ln K could not cancel. Unit 15 finishes the job in relativistic QED, where the same steps remove ln K.
Change one variable at a time
Make the relationship visible.
Slide √s: both curves leave zero at threshold, where M = −λ defines λ, and Im M saturates while −Re M keeps climbing. Swing θ to 0° and the solid curve drops as the t channel switches off; 180° matches 0°, as identical particles must. Double λ and both percentages double.
s CHANNEL Re F(s) + 25.956
t CHANNEL F(t)3.352
u CHANNEL F(u)3.352
Im M = λ²β/(32π)0.0396
Live interpretations CHANNEL Re F(s) + 2: 5.956. t CHANNEL F(t): 3.352. u CHANNEL F(u): 3.352. Im M = λ²β/(32π): 0.0396
Catch the common trap
Explain before calculating.
You renormalise the one-loop φφ → φφ amplitude twice, once in dimensional regularisation and once with a Euclidean cutoff Λ/m = 10⁴, imposing M = −λ at s = 4m², t = u = 0 both times. What differs between the two calculations?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyTake λ = 0.80 and scatter just above threshold, at √s = 2.10m and θ = 90°. Using the near-threshold forms Re F(s) + 2 ≈ 2β² and F(p²) ≈ −p²/(6m²) for the t and u channels, find the one-loop shift of −Re M and the value of Im M, and say which dominates.
- Kinematics: |p|² = s/4 − m² = 0.1025m², so β² = |p|²/(s/4) = 0.1025/1.1025 = 0.0930 and β = 0.305. At 90°, −t = −u = 2|p|²(1 − cos θ) = 0.205m².
- s channel: Re F(s) + 2 ≈ 2β² = 0.1859. The t and u channels: F ≈ 0.205/6 = 0.0342 each. The bracket is 0.1859 + 0.0683 = 0.2543; the exact closed forms give 0.2590.
- Real shift: (λ²/(32π²)) × 0.2543 = 0.64 × 0.2543/315.8 = 5.15 × 10⁻⁴, so −Re M = 0.8005.
- Imaginary part: Im M = λ²β/(32π) = 0.64 × 0.305/100.5 = 1.94 × 10⁻³, about 3.8 times the real shift.
- Limiting case: as β → 0 the real shift falls as β² and Im M only as β, so close enough to threshold the imaginary part always wins, and both vanish at β = 0, where M = −λ by construction.
Answer−Re M ≈ 0.8005, a one-loop shift of 5.2 × 10⁻⁴, and Im M ≈ 1.9 × 10⁻³: near threshold the imaginary part, linear in β, dominates the real correction, quadratic in β.
MediumA scalar of mass m = 0.80 GeV has λ = 1.0, measured at threshold. Predict M at √s = 6m and θ = 90° from the closed forms of F, then the differential cross-section dσ/dΩ = |M|²/(64π²s) at tree level and at one loop, in nb sr⁻¹.
- Kinematics: s = 36m², and at 90° t = u = −(s − 4m²)/2 = −16m².
- s channel: β = √(1 − 4/36) = 0.9428 and arcosh 3 = 1.7627, so Re F(s) = 2(0.9428)(1.7627) − 2 = 1.3239 and Im F(s) = −πβ = −2.962.
- t and u channels, Q² = 16m²: F = 2√(1 + 4/16) arsinh 2 − 2 = 2(1.1180)(1.4436) − 2 = 1.2281 each.
- Bracket: 1.3239 + 2(1.2281) + 2 = 5.780, so −Re M = 1 + 5.780/315.83 = 1.0183, and Im M = λ²β/(32π) = 0.9428/100.53 = 0.00938.
- Cross-section: s = 36 × 0.64 = 23.04 GeV². Tree: 1/(64π² × 23.04) = 6.871 × 10⁻⁵ GeV⁻² sr⁻¹, which is 26.76 nb sr⁻¹ with 1 GeV⁻² = 0.3894 mb. To order λ³, |M|² = λ²(1 + 2 × 0.0183), so one loop gives 27.74 nb sr⁻¹.
AnswerM = −1.0183 + 0.0094i, and dσ/dΩ rises from 26.8 nb sr⁻¹ at tree level to 27.7 nb sr⁻¹ at one loop, a 3.7% increase predicted from m and λ alone.
HardRegulate the same bubbles with a sharp Euclidean cutoff Λ on the shifted loop momentum. Find δλ for large Λ, and the bare coupling λ₀ = λ + δλ for λ = 1 at Λ/m = 10³ and 10¹⁷. Then, for the t channel at −t = 16m² with Λ/m = 20, predict how far the renormalised cutoff bracket sits from the dimensional value −1.2281, and show why that difference drops out of M.
- Each bubble adds λ²/2 times the integral of (lE² + Δ)⁻² over |lE| ≤ Λ, measure d⁴lE/(2π)⁴, to M. That integral is (1/(16π²))[ln(1 + Λ²/Δ) − Λ²/(Λ² + Δ)], so M gains (λ²/(32π²)) times a bracket that for Λ² ≫ |Δ| is ln(Λ²/Δ) − 1 + 2Δ/Λ² + O(Λ⁻⁴).
- δλ is (λ²/(32π²)) times the three brackets at threshold: Σ∫dx [ln(Λ²/Δ₀) − 1] = 3 ln(Λ²/m²) − 3 − F(4m²) − 2F(0) = 3 ln(Λ²/m²) − 1, so δλ = (λ²/(32π²))[3 ln(Λ²/m²) − 1].
- For λ = 1: at Λ/m = 10³ the bracket is 6 ln 10³ − 1 = 40.45, δλ = 40.45/315.83 = 0.128 and λ₀ = 1.128; at Λ/m = 10¹⁷ it is 233.86, δλ = 0.740 and λ₀ = 1.740. The bare coupling keeps growing while M stays put.
- Renormalised, each channel's bracket is −∫ln(Δ/Δ₀)dx + 2∫(Δ − Δ₀)dx/Λ². With Δ − Δ₀ = −x(1 − x)(p² − p₀²) and ∫₀¹ x(1 − x) dx = 1/6, the cutoff adds −(p² − p₀²)/(3Λ²).
- t channel: p² − p₀² = −16m², so the shift is +16/(3 × 400) = +0.01333 and the bracket should be −1.22807 + 0.01333 = −1.21474. Numerical integration of the exact cutoff bracket gives −1.21486; the 1/Λ⁴ term, −(3/2)∫(Δ² − Δ₀²)dx/Λ⁴ = −0.00013, accounts for the rest.
- Summed over channels the 1/Λ² pieces give −[(s − 4m²) + t + u]/(3Λ²), and s + t + u = 4m² on shell, so they cancel: the cutoff amplitude agrees with the dimensional one to order 1/Λ⁴.
Answerδλ = (λ²/(32π²))[3 ln(Λ²/m²) − 1], so λ₀ = 1.128 at Λ/m = 10³ and 1.740 at 10¹⁷. The t-channel bracket shifts by +0.0133, to −1.2147 (exact −1.2149), and the 1/Λ² shifts cancel among s, t and u on shell.