University Quantum Field Theory I · One Loop in φ⁴: Dimensional Regularisation, Counterterms and the Running Coupling · 6.4
Poles in ε, the scale μ and the bubble evaluated
This is where dimensional regularisation stops being a formal trick and hands back numbers. Expanding about d = 4 splits every one-loop bracket into a pole all schemes remove, constants they may keep, and a logarithm that carries the physics; the φ⁴ bubble then comes out in closed form, imaginary part included.
Build the model
Connect the measurement to the mechanism.
Topic 6.03 left the logarithmic master integral as Γ(2 − d/2) Δ(d/2 − 2)/(4π)(d/2), finite below four dimensions and singular at d = 4. Setting d = 4 − ε and expanding each factor to first order separates it into a pole, constants and a logarithm, (1/16π²)[2/ε − γE + ln 4π − ln Δ], and the constants exist only because an O(ε) correction multiplied by 2/ε survives as a finite number. A logarithm of Δ alone, a quantity in GeV², is not yet physics.
The repair is already in the Lagrangian: in d dimensions λ has mass dimension ε, so it must be written λμε with λ dimensionless, and the spare με a loop carries over the tree turns ln Δ into ln(Δ/μ²). Minimal subtraction removes the pole; MS-bar removes −γE + ln 4π with it, which is the same as rescaling μ by 2.66. Carried through the s-channel bubble, the bracket becomes a function of p² that is real below threshold and acquires, from the Feynman i0 alone, an imaginary part πβ above p² = 4m², λ²β/(32π) in M, that no scheme and no μ can touch: the one unitarity demanded from topic 5.08's tree cross-section.
The expansion has two limits. Truncated at O(ε⁰) it is safe at one loop, but at two loops an O(ε) remainder meets a second 1/ε and must be kept, which is quoted here; and it does not remove the pole, which stays in the amplitude until the counterterms of topics 6.06 to 6.08 absorb it into λ.
- Simple definition
- Near four dimensions a dimensionally regulated one-loop integral becomes a Laurent series in ε = 4 − d: a pole 2/ε, the constants −γE + ln 4π, and a logarithm ln(Δ/μ²) whose scale μ enters because λ has mass dimension ε.
- Example
- At ε = 0.2, so d = 3.8, the exact Γ(ε/2)(4π)(ε/2) = Γ(0.1)(4π)0.1 = 12.254 sits 0.300 above the truncation 2/ε − γE + ln 4π = 11.954; at ε = 0.1 the gap is 0.143. It is the O(ε) remainder, and it vanishes at d = 4.
Derived from Γ(z) = Γ(1 + z)/z and Γ(1 + z) = 1 − γE z + O(z²) with z = ε/2. The residue is 2, not 1, because the argument is ε/2.
ε = 4 − d, dimensionless; γE = −Γ′(1) = 0.5772, Euler's constant, quoted from analysis
Derived. A one-loop term carries one more με than the tree, and that spare factor, against 2/ε, is what turns ln Δ into ln(Δ/μ²).
μ an arbitrary mass scale in GeV; [φ] = (d − 2)/2 because the action ∫ dᵈx L is dimensionless
Derived from topic 6.03's Γ(2 − d/2) Δ(d/2 − 2)/(4π)(d/2) by expanding every factor to first order in ε before multiplying by the pole.
lE the Euclidean loop momentum in GeV; Δ and μ² in GeV²; both sides dimensionless as ε → 0
Derived: the MS remainder at scale μ equals the MS-bar remainder at μ̄. At one loop, moving between these two schemes is a change of scale and nothing more.
MS subtracts 2/ε; MS-bar subtracts 2/ε − γE + ln 4π = 2/ε + 1.954; μ and μ̄ in GeV
Derived: the symmetry factor ½ times the master integral's 1/16π² gives 1/32π². The pole does not depend on x, so every channel brings the same 2/ε.
s the channel's p² in GeV², m the scalar mass in GeV, i0 the Feynman iε; the t and u bubbles replace s by t and u
Derived: Δ is negative on an x-interval of length β, where ln(Δ − i0) = ln|Δ| − iπ. So Im Mₛ = λ²β/32π, topic 5.08's sβσₜᵣₑₑ, whatever μ or scheme.
β = √(1 − 4m²/s), the speed of either particle in the centre-of-momentum frame, for s above 4m²
Expand every factor before multiplying by the pole
Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. Topic 6.03's master integral with n = 2 is Γ(2 − d/2) Δ(d/2 − 2)/(4π)(d/2). Put d = 4 − ε and it reads (1/16π²) Γ(ε/2) (4π)(ε/2) Δ(−ε/2). Expand each factor to first order. Γ(ε/2) = Γ(1 + ε/2)/(ε/2), and with the quoted Γ(1 + z) = 1 − γE z + O(z²) this is 2/ε − γE + O(ε). The two powers are exponentials: (4π)(ε/2) = 1 + (ε/2) ln 4π and Δ(−ε/2) = 1 − (ε/2) ln Δ. Multiplied out, their O(ε) pieces meet the 2/ε and become finite, and the bracket is 2/ε − γE + ln 4π − ln Δ, derived. Nothing here may be dropped for being close to 1: discarding (4π)(ε/2) would lose ln 4π = 2.531. The truncation can be tested at finite ε. At ε = 0.2 the exact Γ(0.1)(4π)0.1 is 12.254 against 11.954 from 2/ε − γE + ln 4π; at ε = 0.1 the gap falls from 0.300 to 0.143, the O(ε) remainder heading to zero with ε.
Why a scale μ has to appear
−ln Δ is not yet a number: Δ has mass dimension two, and the same Δ written in MeV² instead of GeV² shifts ln Δ by ln 10⁶ = 13.8. The cure is in the Lagrangian. The action ∫ dᵈx L is dimensionless, so [φ] = (d − 2)/2 = 1 − ε/2 and [λ] = d − 4[φ] = ε, derived. Write the coupling as λμε with λ dimensionless and μ any mass. A one-loop term has two vertices, λ²μ(2ε), one με more than the tree; that spare factor is (μ²)(ε/2) = 1 + (ε/2) ln μ², and against 2/ε it adds ln μ², turning −ln Δ into −ln(Δ/μ²), derived. The με still common to tree and loop sets the dimension of M in d dimensions and goes to 1 once the bracket is finite. μ is not a cutoff: no integral was cut, and μ is the same at every loop momentum. It is a reference scale, and at fixed λ moving it does change the bracket, doubling μ adding 2 ln 2 = 1.386. How λ must move to cancel that is topic 6.08's β function.
What the pole means, and what MS and MS-bar subtract
Cut the same integral off instead. With u = lE², ∫ d⁴lE/(2π)⁴ (lE² + Δ)⁻² up to |lE| = Λ is (1/16π²)[ln(1 + Λ²/Δ) − Λ²/(Λ² + Δ)], derived, which for Λ² far above Δ is (1/16π²)[ln(Λ²/Δ) − 1]. Beside the dimensional result the dictionary reads 2/ε − γE + ln 4π + ln μ² ↔ ln Λ² − 1: the pole is a logarithmic divergence in disguise, and both regulators agree on −ln Δ, which carries all the dependence on momenta and masses. The regulator lives only in a constant, which is why one constant counterterm can absorb it in topic 6.06. Minimal subtraction (MS) removes 2/ε and keeps −γE + ln 4π = 1.954; modified minimal subtraction (MS-bar) removes that as well. Neither is more correct: deciding what to subtract is deciding what the symbol λ means. Because ln μ̄² = ln μ² + 1.954, MS at μ leaves the same remainder as MS-bar at μ̄ = 2.656μ, so the bar costs nothing and keeps 1.954 out of every later logarithm.
The bubble assembled, and its real part
Topic 6.02 wrote the s-channel bubble as (−iλ)²/2 times ∫ dᵈk/(2π)ᵈ of two propagators i/(k² − m² + i0), parametrised to (l² − Δ + i0)⁻² with Δ = m² − x(1 − x)p². The Wick rotation brings a factor i, so iMₛ = (λ²/2)(i/16π²) ∫₀¹ dx [2/ε − γE + ln 4π − ln((Δ − i0)/μ²)], which is iλ²/32π² times the bracket, derived. The pole and constants do not depend on x, so each of the s, t and u bubbles carries λ²/(16π²ε), 3λ²/(16π²ε) in all, the pole topic 6.06's δλ must absorb. Call what survives subtraction B(p²), so each channel adds (λ²/32π²)B to M. Writing Δ = p²(x − x₊)(x − x₋) with x± = (1 ± β)/2, β = √(1 − 4m²/p²), and integrating each logarithm gives ∫₀¹ dx ln(Δ/m²) = −2 + β ln((β + 1)/(β − 1)) for spacelike p², where β exceeds 1, and, continued to the complex roots between p² = 0 and 4m², −2 + 2a arctan(1/a) with a = √(4m²/p² − 1), derived. Both vanish at p² = 0, so in MS-bar at μ = m the forward t-channel bubble is exactly zero.
Above threshold, the i0 writes the imaginary part
For p² above 4m² the roots x± are real and Δ is negative between them, on an interval of length x₊ − x₋ = β. Feynman's +i0 in each propagator survives in the parametrised denominator as Δ − i0, so on that interval ln(Δ − i0) = ln|Δ| − iπ, and the −ln in B picks up +iπ over a length β: Im Mₛ = (λ²/32π²)πβ = λ²β/(32π), derived. That is topic 5.08's result, where unitarity demanded Im M = sβσₜᵣₑₑ with σₜᵣₑₑ = λ²/(32πs): the loop already knows the tree cross-section. No subtraction reaches it, since every scheme subtracts real constants and μ enters only through the real ln μ². Re B is 2 − β ln((1 + β)/(1 − β)) + 2 ln(μ/m) in MS-bar, peaking with a cusp at threshold, where it equals 2 + 2 ln(μ/m). At s = 8m², β = 0.707 and Im Mₛ = 7.03 × 10⁻³ λ²; at high energy β → 1 and Im Mₛ → λ²/(32π) = 9.95 × 10⁻³ λ².
Checks: dimensions, limits and the measured analogue
Dimensions: at d = 4 the bracket is a pure number and M for two-to-two scattering is dimensionless, as the tree −λ is; in d dimensions both carry the same με. Free-field limit: the bubble is O(λ²), so beside the tree it is (λ/32π²)B, about 0.3% of the tree per unit of B at λ = 1. Massless, or high-energy, limit: as m → 0, ∫₀¹ dx ln(−x(1 − x)p²/μ²) = ln(−p²/μ²) − 2, so B → 2 − ln(−p²/μ²) and all memory of m is lost; at s = 100m² and μ = m its real part, −2.605, sits 0.113 below the exact −2.492, matching the leading correction (2m²/s)[ln(s/m²) + 1] = 0.112 that expanding β to first order in m²/s gives. Threshold limit: Im Mₛ grows as β, the velocity of either particle, as two-body phase space must. A measured number: no φ⁴ amplitude has been measured, so the external test here is unitarity. The same expansion applied to QED's vacuum polarisation in Unit 13 gives the running of α, measured as α⁻¹ falling from 137.036 at q² = 0 to about 128 at the Z mass, quoted.
Change one variable at a time
Make the relationship visible.
Raise μ/m, or switch the scheme to MS: the solid curve moves up rigidly, by 2 ln(μ/m) or by 1.954, and the dashed curve never moves. Then slide s down to 4m² and watch Im B close like β, the opening of two-particle phase space.
RE B(s)0.402
IM B(s) = πβ2.433
B(t) = B(u), REAL-0.393
IM Mₛ/λ² = β/32π0.0077
Live interpretationRE B(s): 0.402. IM B(s) = πβ: 2.433. B(t) = B(u), REAL: −0.393. IM Mₛ/λ² = β/32π: 0.0077
Catch the common trap
Explain before calculating.
In MS-bar the s-channel bubble is Mₛ = −(λ²/32π²) ∫₀¹ dx ln[(m² − x(1 − x)s − i0)/μ²]. At s = 10m², μ is doubled while λ is held fixed. What happens to Mₛ?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyIn φ⁴ theory at d = 4 − ε, the t-channel bubble in exactly forward scattering, t = 0, has Δ = m² for every Feynman parameter x. Take λ = 0.6 and μ = 2m. Write its bracket, find what MS and MS-bar each keep, the bubble's contribution Mₜ in each scheme, and the MS-bar scale μ̄ that reproduces the MS number.
- With Δ = m² the x-integral is trivial, so the bracket is 2/ε − γE + ln 4π − ln(m²/μ²) = 2/ε + 1.954 + ln 4, and ln 4 = 1.386 at μ = 2m.
- MS removes only 2/ε and keeps 1.954 + 1.386 = 3.340. MS-bar also removes −γE + ln 4π and keeps 1.386.
- The prefactor is λ²/32π² = 0.36/315.83 = 1.140 × 10⁻³, so Mₜ = 3.807 × 10⁻³ in MS and 1.580 × 10⁻³ in MS-bar.
- MS-bar at μ̄ keeps ln(μ̄²/m²). Setting that to 3.340 gives μ̄ = m e(1.670) = 5.31m, which is 2.656 × 2m, the scheme's rescaling of μ.
- Checks: Mₜ is a pure number, as M must be at d = 4, and the MS-bar term is 0.26% of the tree's magnitude λ = 0.6, the size an O(λ²) correction should have.
AnswerMS keeps 3.340 and MS-bar 1.386, so Mₜ = 3.81 × 10⁻³ (MS) or 1.58 × 10⁻³ (MS-bar); MS-bar at μ̄ = 5.31m gives the MS value.
MediumEvaluate the s-channel bubble at s = 8m² in MS-bar with μ = m and λ = 1. Find where Δ(x) is negative, the real and imaginary parts of Mₛ, and check Im Mₛ against the tree cross-section, which for a scalar of mass m = 0.50 GeV you should also give in microbarns.
- β = √(1 − 4m²/s) = √0.5 = 0.7071, and Δ = m² − x(1 − x)s vanishes at x± = (1 ± β)/2 = 0.1464 and 0.8536; it is negative between them.
- The closed form gives ∫₀¹ dx ln((Δ − i0)/m²) = −2 + β ln((1 + β)/(1 − β)) − iπβ = −2 + 0.7071 × 1.7627 − 2.2214i = −0.7535 − 2.2214i.
- In MS-bar at μ = m the subtracted bracket is minus this, 0.7535 + 2.2214i, so Mₛ = (0.7535 + 2.2214i)/315.83 = (2.386 + 7.034i) × 10⁻³.
- Unitarity check: with the identical-particle ½, σₜᵣₑₑ = λ²/(32πs), so sβσₜᵣₑₑ = λ²β/(32π) = 0.7071/100.53 = 7.034 × 10⁻³, equal to Im Mₛ.
- Units: s = 8 × 0.25 = 2.0 GeV², so σₜᵣₑₑ = 1/(32π × 2.0) GeV⁻² = 4.974 × 10⁻³ GeV⁻², and with 1 GeV⁻² = 0.3894 mb this is 1.94 × 10⁻³ mb, or 1.94 μb.
AnswerMₛ = (2.39 + 7.03i) × 10⁻³; Im Mₛ equals sβσₜᵣₑₑ exactly, with σₜᵣₑₑ = 1.94 μb for m = 0.50 GeV.
HardAt √s = 10m and a 90° scattering angle, evaluate the one-loop bubble sum over the s, t and u channels in MS-bar with λ = 0.5 at μ = m. Compare its real part with the tree, find the μ at which the summed real bracket vanishes, and test the massless approximation.
- s = 100m² and s + t + u = 4m², so at 90° t = u = −(s − 4m²)/2 = −48m².
- s channel: β = √0.96 = 0.97980 and ln((1 + β)/(1 − β)) = 4.5849, so Re Bₛ = 2 − 0.97980 × 4.5849 = −2.4922 and Im Bₛ = πβ = 3.0781.
- t and u channels: β = √(1 + 4/48) = 1.04083 and ln((β + 1)/(β − 1)) = 3.9116, so Bₜ = Bᵤ = 2 − 4.0713 = −2.0713, real because Δ is positive for every x.
- The real parts sum to −6.6349. With λ²/32π² = 0.25/315.83 = 7.916 × 10⁻⁴ the one-loop real part is −5.25 × 10⁻³, 1.05% of the tree −0.5, and Im M = 7.916 × 10⁻⁴ × 3.0781 = 2.44 × 10⁻³.
- Each bracket gains 2 ln(μ/m), the sum 6 ln(μ/m), so the real sum vanishes at μ = m e(6.6349/6) = m e1.1058 = 3.02m. The coefficient (λ²/32π²) × 6 = 3λ²/16π² is the one topic 6.08 turns into the β function.
- Massless limit B → 2 − ln(|p²|/μ²): −2.605 for s and −1.871 for t and u, a sum of −6.348 within 4.3% of the exact −6.635. The gaps, 0.11 and 0.20, are corrections of order m²/|p²| times a logarithm, larger in t because |t| is the smaller.
AnswerM = −0.5 − 5.25 × 10⁻³ + 2.44 × 10⁻³ i; the loop's real part is 1.05% of the tree and vanishes for μ ≈ 3.0m, about a third of √s.