University Quantum Field Theory I · One Loop in φ⁴: Dimensional Regularisation, Counterterms and the Running Coupling · 6.8

Minimal subtraction and the one-loop beta function

Topic 6.07 ended with an amplitude that was finite but pinned to one measured point. Here you swap that pin for an arbitrary scale μ, insist that nothing physical notices the swap, and get the course's first beta function in four lines. Then you check it against how the scattering itself changes with energy.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Minimal subtraction takes out of each divergent function only its pole in ε (MS), or the pole together with the universal −γ_E + ln 4π (MS-bar), so no measurement fixes the coupling and a scale has to take its place. In d = 4 − ε the bare coupling has mass dimension ε. Writing it λ₀ = μ^ε(λ + δ_λ) keeps the renormalised λ dimensionless, but the price is an arbitrary μ that then sits in every amplitude as ln(s/μ²), ln(−t/μ²) and ln(−u/μ²). The bare coupling existed before μ was invented, so it cannot depend on it.

That one requirement, with δ_λ = 3λ²/(16π²ε) and Z = 1 at one loop, forces β(λ) = μ dλ/dμ = −ελ + 3λ²/(16π²). The derivation looks like bookkeeping about an unphysical parameter, but at d = 4 the same coefficient is the high-energy slope of the finite, regulator-free on-shell amplitude against ln√s at fixed angle. At one loop it therefore belongs to φ⁴ scattering, not to the scheme. Only the value of the coupling at a given μ is a convention: at μ = m the MS-bar coupling exceeds the on-shell λ by λ²/(16π²).

The model stops where one loop stops. It needs 3λ/(16π²) ≪ 1 and logarithms that stay moderate, two-loop terms enter at relative order λ/(16π²), and below four dimensions the −ελ term competes with the loop and creates a fixed point that is only named here.

Simple definition: The beta function β(λ) = μ dλ/dμ is the rate at which a renormalised coupling must change with the arbitrary scale μ so that the bare coupling, and with it every physical amplitude, stays the same.
Example: In φ⁴ at d = 4, β = 3λ²/(16π²). A coupling λ̄ = 1.000 at μ = 10 GeV has β = 0.0190, so doubling μ to 20 GeV raises it by 0.0190 × ln 2 = 0.0132, to 1.013 at order λ².

Bare coupling held fixedλ₀ = μ^ε (λ + δ_λ), d = 4 − ε, δ_Z = 0 at one loop

From topic 6.04: μ enters only to keep λ dimensionless, so λ₀, which defines the theory, cannot depend on it. That single condition fixes how λ must change with μ.

λ₀ has mass dimension ε (GeV^ε); λ and δ_λ are pure numbers; μ is an arbitrary scale in GeV

Minimal-subtraction countertermδ_λ = 3λ²/(16π² ε)

Derived: each of the s, t and u bubbles carries the pole (λ²/32π²)(2/ε), and three channels give the 3. No measurement enters, so the coupling it defines is labelled by μ.

MS removes the pole alone; MS-bar also removes ln 4π − γ_E, which is MS with μ̄² = 4π e^(−γ_E) μ²

The one-loop beta functionβ(λ) ≡ μ dλ/dμ = −ελ + 3λ²/(16π²)

Derived from μ dλ₀/dμ = 0: the 1/ε of δ_λ meets the ε of μ^ε and leaves a finite coefficient. β is positive, so λ grows at short distances.

β is a pure number; at d = 4 it is 3λ²/(16π²) = 0.0190 λ²; the −ελ term is the tree-level dimension of λ₀

The MS-bar amplitudeM = −λ̄ − (λ̄²/32π²) Σ ∫₀¹ dx ln[(m² − x(1 − x)p² − i0)/μ²], p² = s, t, u

Derived: topic 6.04's three bubbles minus δ_λ. μ now sits inside every logarithm, and μ dM/dμ = 0 at order λ² gives the same β back.

M is dimensionless for 2 → 2 at d = 4; s, t, u, m² and μ² in GeV²; λ̄ ≡ λ̄(μ); i0 is the Feynman iε

Energy growth of the physical amplituded(−Re M)/d ln√s → 3λ²/(16π²) for s ≫ m² at fixed θ

Shown: in the massless limit the three logarithms add to 3 ln(s/μ²) plus angle terms, so the slope is β itself, read off a finite observable that belongs to no scheme.

M is topic 6.07's on-shell amplitude, √s in GeV; mass corrections to the slope fall roughly as (m²/s) ln(s/m²)

MS-bar coupling against the on-shell couplingλ̄(μ) = λ + (λ²/16π²)[1 + 3 ln(μ/m)]

Computed by equating the two amplitudes at threshold, where ∫₀¹ ln(1 − 2x)² dx = −2. At μ = m the excess is λ²/(16π²): 0.0063 for λ = 1.

λ is the on-shell coupling, −M at s = 4m², t = u = 0; m and μ in GeV; true to order λ²

Minimal subtraction trades a measurement for a scale

Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout. Topic 6.06 fixed δ_λ by a measurement: M = −λ at s = 4m², t = u = 0. Minimal subtraction asks for less. In d = 4 − ε the bare coupling has mass dimension ε, so it is written λ₀ = μ^ε(λ + δ_λ) with λ a pure number and μ an arbitrary scale, and topic 6.04's bubbles then carry (λ²/32π²)[2/ε − γ_E + ln 4π − ∫₀¹ dx ln(Δ/μ²)] per channel. MS chooses δ_λ to cancel the pole and nothing else, δ_λ = 3λ²/(16π²ε). MS-bar also cancels −γ_E + ln 4π, so the MS-bar coupling at μ equals the MS coupling at μ/2.656, with 2.656 = √(4π e^(−γ_E)). What is left, M = −λ̄ − (λ̄²/32π²) Σ ∫ ln(Δ/μ²), is finite and has used no data. The price shows in two places: the coupling now carries a label μ, and every logarithm reads ln(s/μ²) or ln(−t/μ²) instead of a ratio of two physical invariants.

The bare coupling cannot know about μ

λ₀ sits in the Lagrangian before any renormalisation choice, so it cannot depend on a scale introduced only to keep λ dimensionless. At one loop δ_Z = 0, so Z²λ₀ = μ^ε(λ + δ_λ) reduces to λ₀ = μ^ε(λ + δ_λ). Apply μ d/dμ and set the result to zero: ε(λ + δ_λ) + (1 + ∂δ_λ/∂λ)β = 0. Try β = −ελ + bλ². With δ_λ = Cλ²/ε and C = 3/(16π²), the order-λ² terms are Cλ² from εδ_λ, −2Cλ² from (∂δ_λ/∂λ)(−ελ), and bλ². So b = C, and the result, derived, is β(λ) = μ dλ/dμ = −ελ + 3λ²/(16π²). The 1/ε of the counterterm has met the ε of μ^ε, so the coefficient is finite even though δ_λ is not. The −ελ is classical: it is the dimension of λ₀ at work. The 3 counts the s, t and u bubbles, each worth λ²/(16π²), and the 16π² is the (4π)^(d/2) of the loop measure at d = 4. As a dimension check, β is a pure number, like λ.

The same number read off the amplitude

The bare-coupling route has a twin that never mentions λ₀. M is a physical amplitude, so μ dM/dμ = 0. Each of the three logarithms in M = −λ̄ − (λ̄²/32π²) Σ ∫ ln(Δ/μ²) contains −ln μ², so the loop's explicit μ-dependence is +6(λ̄²/32π²) ln μ. Running in −λ̄ must cancel it, which needs μ dλ̄/dμ = 3λ̄²/(16π²). Now take s ≫ m² at fixed angle. The integrals give ln(s/μ²) − 2 − iπ, ln(−t/μ²) − 2 and ln(−u/μ²) − 2, with −t = s sin²(θ/2) and −u = s cos²(θ/2), so −Re M = λ̄ + (λ̄²/32π²)[3 ln(s/μ²) + ln(sin²θ/4) − 6]. Differentiate with respect to ln√s and the slope is 6λ̄²/(32π²) = 3λ̄²/(16π²) at every angle, shown. For λ = 1 the amplitude grows by 0.0190 per e-fold of √s.

Why the one-loop coefficient belongs to no scheme

Topic 6.07's on-shell amplitude has no μ in it, yet once λ̄(μ) is written in terms of λ it is the same function as the MS-bar one, because both describe the same scattering. So at large s the on-shell −Re M also climbs at 3λ²/(16π²) per unit ln√s, a slope that could be read off a finite observable with every regulator removed. The algebra agrees. Suppose two schemes define couplings related by λ′ = λ + cλ² + O(λ³), with c a pure number. Then μ dλ′/dμ = (1 + 2cλ)β(λ) = bλ² + O(λ³), and since λ² = λ′² + O(λ′³), β′(λ′) = bλ′² + O(λ′³): the first coefficient is universal, derived. One more order of the same algebra shows the second is universal too. For this theory it is −(17/3)λ³/(16π²)², quoted, and from three loops on the coefficients depend on the scheme. What does depend on the scheme at one loop is the value of the coupling at a given scale.

Matching MS-bar to the on-shell coupling

The on-shell λ is a measurement, −M at s = 4m², t = u = 0, so evaluate the MS-bar amplitude at that point. There Δₛ = m² − 4m²x(1 − x) = m²(1 − 2x)², and ∫₀¹ ln(1 − 2x)² dx = −2, while the t and u channels give Δ = m² exactly. So Σ ∫ ln(Δ/μ²) = 3 ln(m²/μ²) − 2, and −λ = −λ̄ − (λ̄²/32π²)[3 ln(m²/μ²) − 2]. To order λ², computed, λ̄(μ) = λ + (λ²/16π²)[1 + 3 ln(μ/m)]. At μ = m the MS-bar coupling exceeds the on-shell one by λ²/(16π²). That is 0.0063 for λ = 1, a 0.6% shift that is pure convention, since both couplings describe the same amplitude. The 3 ln(μ/m) is β at work, and it grows without limit as μ moves away from the physical scale. The next topic turns that into a rule for choosing μ.

Limits: below four dimensions, weak coupling, one loop

Three checks close the derivation. Free field: as λ → 0, β → 0 and nothing runs. Dimension: away from d = 4 the tree term −ελ survives. For positive ε it gives λ = λ₀μ^(−ε) at tree level, so the dimensionless coupling falls at short distances, the super-renormalisable behaviour of topic 6.01. Against the loop term it makes β vanish at a non-zero coupling, λ* = 16π²ε/3, the Wilson-Fisher fixed point. It is named here only; its use belongs to the Wilsonian renormalisation group of a later course, and as ε → 0 only λ = 0 remains. Validity: one loop is the first term of β = 3λ²/(16π²) − (17/3)λ³/(16π²)² + …, so it holds while 3λ/(16π²) ≪ 1, that is λ ≪ 53. At λ = 1 the two-loop term changes β by 1.2%. No measured particle is a lone φ⁴ scalar, so the comparison with data waits for QED in Units 12 and 13. There α grows from 1/137.036 at low energy to about 1/128 at the Z mass, 91.19 GeV, quoted.

Change one variable at a time

Make the relationship visible.

Interactive model

On-shell coupling λ3.0

Scattering angle θ90 °

Probe energy ln(√s/m)2.0

Slide the probe from √s/m = 4.5 up to 1.3 × 10⁴ and watch slope/β climb to 1.000 as the tangent turns parallel to the dashed line. Then swing θ to 10°: the solid curve sinks and turns up late, yet its high-energy slope is still β. Raising λ steepens both lines together.

β = 3λ²/16π²0.1710

λ̄(m) − λ0.0570

GAP λ̄(√s) + Re M, s ≫ m²0.2105

PROBE √s/m7

Live interpretationβ = 3λ²/16π²: 0.1710. λ̄(m) − λ: 0.0570. GAP λ̄(√s) + Re M, s ≫ m²: 0.2105. PROBE √s/m: 7

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading is that the beta function is an artefact. It comes from a 1/ε pole and from an arbitrary scale μ, so the argument goes that another regulator or another subtraction would give another running, and that nothing physical runs at all. Both halves are wrong. The physical amplitude is μ-independent, but it is not energy-independent. Topic 6.07's on-shell M, finite and free of μ and of every regulator, grows at high energy as (3λ²/16π²) ln√s at fixed angle, and that slope is a property of φφ → φφ scattering. μ only decides how that growth is split between λ̄(μ) and the explicit logarithms. A change of scheme, λ′ = λ + cλ², shifts the coupling's value at a given μ (by λ²/(16π²) between MS-bar at μ = m and on shell) but leaves the first coefficient of β untouched. The number λ̄(μ) is a convention; the rate is physics.

In d = 4 − ε at one loop, δ_Z = 0 and the minimal-subtraction counterterm is δ_λ = 3λ²/(16π²ε). If λ₀ = μ^ε(λ + δ_λ) must not depend on μ, what is β(λ) = μ dλ/dμ at d = 4, to order λ²?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn MS-bar calculation fixes λ̄ = 0.500 at μ = 2.0 GeV. Evaluate β there, find λ̄ at μ = 6.0 GeV to order λ², and show that at fixed s, t and u the explicit logarithms in M shift by exactly what cancels the change in −λ̄. Express both scales as distances.

At d = 4, β = 3λ̄²/(16π²) = 3 × 0.250/157.91 = 4.749 × 10⁻³. It is a pure number, as μ dλ/dμ must be when λ is.
To order λ², β can be taken at its starting value across the interval, so λ̄(6.0 GeV) = 0.500 + β ln(6.0/2.0) = 0.500 + 4.749 × 10⁻³ × 1.0986 = 0.500 + 5.22 × 10⁻³ = 0.5052.
Each of the three ln(Δ/μ²) falls by 2 ln 3 = 2.197 when μ triples, so −(λ̄²/32π²) Σ ln(Δ/μ²) rises by (λ̄²/32π²) × 3 × 2.197 = (3λ̄²/16π²) ln 3 = +5.22 × 10⁻³. Meanwhile −λ̄ falls by 5.22 × 10⁻³, so M is unchanged at order λ².
As lengths, ħc/μ = 0.197327 GeV fm/2.0 GeV = 0.0987 fm and 0.197327/6.0 = 0.0329 fm. The coupling defined at the shorter distance is the larger, as a positive β requires.

Answerβ = 4.75 × 10⁻³ and λ̄(6.0 GeV) = 0.5052 to order λ². The logarithms shift M by +5.22 × 10⁻³, cancelling the −5.22 × 10⁻³ from −λ̄.

MediumA scalar of mass m = 0.50 GeV has on-shell coupling λ = 0.90, defined by M = −λ at s = 4m², t = u = 0. By evaluating the MS-bar amplitude at that point, find λ̄(μ) at μ = m and at μ = 4.0 GeV to order λ², and check the difference against β.

At s = 4m², Δₛ = m² − 4m²x(1 − x) = m²(1 − 2x)², and at t = u = 0, Δ = m². None is negative, so the amplitude is real there.
∫₀¹ ln(1 − 2x)² dx = 2∫₀¹ ln|1 − 2x| dx = −2, so the three channels give Σ∫ ln(Δ/μ²) = 3 ln(m²/μ²) − 2.
Setting M = −λ̄ − (λ̄²/32π²)[3 ln(m²/μ²) − 2] equal to −λ and solving to order λ² gives λ̄(μ) = λ + (λ²/32π²)[2 − 3 ln(m²/μ²)] = λ + (λ²/16π²)[1 + 3 ln(μ/m)].
λ²/(16π²) = 0.81/157.91 = 5.129 × 10⁻³, so at μ = m, λ̄ = 0.900 + 0.0051 = 0.9051.
At μ = 4.0 GeV, ln(μ/m) = ln 8 = 2.079, the bracket is 1 + 6.238 = 7.238, and λ̄ = 0.900 + 5.129 × 10⁻³ × 7.238 = 0.900 + 0.0371 = 0.9371. Check: the rise from μ = m is 0.0320, and β ln 8 = 0.01539 × 2.079 = 0.0320.

Answerλ̄(m) = 0.9051 and λ̄(4.0 GeV) = 0.9371, both to order λ². The 0.0051 excess at μ = m is the λ²/(16π²) of this topic.

HardFor on-shell coupling λ = 1.20, evaluate topic 6.07's −Re M at θ = 90° for √s/m = 10, 100 and 1000. Find the slope of −Re M per unit ln√s across each decade and compare it with β. Then compare −Re M at √s/m = 1000 with the MS-bar coupling λ̄ at μ = √s, and explain the gap.

On shell, −Re M = λ + (λ²/32π²)[Fₛ(σ) + Fₜ(τ) + Fᵤ(τ) + 2], with σ = s/m² and, at 90°, τ = −t/m² = −u/m² = (σ − 4)/2. Here Fₛ = −2 + v ln[(1 + v)/(1 − v)] with v = √(1 − 4/σ), and Fₜ = −2 + w ln[(w + 1)/(w − 1)] with w = √(1 + 4/τ). The +2 removes Fₛ(4) = −2, the threshold subtraction.
λ²/(32π²) = 1.44/315.83 = 4.559 × 10⁻³. At √s/m = 10 (σ = 100, τ = 48): Fₛ = 2.492 and Fₜ = Fᵤ = 2.071, so the bracket is 8.635 and −Re M = 1.2394. At 100: Fₛ = 7.208 and Fₜ = 6.521, bracket 22.249, −Re M = 1.3014. At 1000: Fₛ = 11.815 and Fₜ = 11.122, bracket 36.060, −Re M = 1.3644.
Each decade spans ln 10 = 2.3026 in ln√s. The slopes are (1.30145 − 1.23937)/2.3026 = 0.02696 and (1.36442 − 1.30145)/2.3026 = 0.02735.
β = 3λ²/(16π²) = 4.32/157.91 = 0.02736. The first decade is 1.5% low and the second agrees to 0.03%: the mass corrections fall roughly as (m²/s) ln(s/m²), and in the massless limit each channel adds 2λ²/(32π²) per unit ln√s.
λ̄(μ/m = 1000) = 1.20 + (1.44/157.91)(1 + 3 ln 1000) = 1.20 + 9.119 × 10⁻³ × 21.723 = 1.3981. It lies 0.0337 above −Re M, and (λ²/32π²)[6 − ln(sin²θ/4)] = 4.559 × 10⁻³ × 7.386 = 0.0337. The gap is a constant set by the angle, not a logarithm, so λ̄(√s) and −Re M run in parallel.

Answer−Re M = 1.2394, 1.3014 and 1.3644. The slopes are 0.02696 and 0.02735 per unit ln√s against β = 0.02736, and λ̄(μ = √s) = 1.3981 lies a constant 0.0337 above −Re M.