University Quantum Field Theory I · One Loop in φ⁴: Dimensional Regularisation, Counterterms and the Running Coupling · 6.3

Dimensional regularisation: the master integrals

Every one-loop integral in this course ends up here. After Feynman parameters and the Wick rotation one radial integral is left; evaluate it in d dimensions and you can read each divergence as a pole, reduce any numerator to it, and see which integrals vanish outright, and why that is not cheating.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Topic 6.02 reduced the bubble to i times a Euclidean integral of (l_E² + Δ)⁻², logarithmically divergent in four dimensions. A cutoff would make it finite but is not invariant under shifts of the loop momentum, harmless here and costly for worse divergences; dimensional regularisation keeps the integrand and changes the number of dimensions instead. An integrand depending only on l_E² needs from d nothing but the area of the unit sphere, and the d-th power of the one-dimensional Gaussian integral fixes that area as 2π^(d/2)/Γ(d/2) for any real d.

The radial integral is a Beta function, so ∫ d^d l_E/(2π)^d (l_E² + Δ)⁻ⁿ = Γ(n − d/2)Δ^{(d/2 − n)}/((4π)^(d/2)Γ(n)), derived for d between 0 and 2n, where it converges, and defined elsewhere by the same formula, since Γ continues analytically. A divergence becomes a pole of Γ. A logarithmic one in four dimensions is the pole at d = 4; a quadratic one is a pole at d = 2 plus one at d = 4 whose residue is proportional to Δ, and no power of a cutoff appears.

Numerators follow from l^μ l^ν → g^μν l²/d and the vanishing of odd powers, and an integral with no scale is zero, an ultraviolet pole cancelling an infrared one. The method stops short in two places: it makes the poles visible without removing them, the counterterms' job in topic 6.06, and Dirac matrices in the integrand need their algebra continued to d dimensions, in topic 12.03.

Simple definition: Dimensional regularisation evaluates a loop integral in d dimensions, where it converges, as an analytic function of d, so that a divergence of the four-dimensional integral appears as a pole at d = 4 instead of as a power or logarithm of a cutoff.
Example: With n = 3 the master integral converges at d = 4 and equals 1/(32π²Δ), or 0.0127 GeV⁻² for Δ = 0.25 GeV². With n = 2 it is Γ(2 − d/2)/(4π)^(d/2) in units of Δ^{(d/2 − 2)}: 0.140 at d = 3.9, of which the pole term (1/16π²) × 2/(4 − d) supplies 0.127.

Area of the unit sphere in d dimensionsΩ_d = 2π^(d/2)/Γ(d/2)

Derived by writing (∫ dx e^(−x²))^d = π^(d/2) in spherical coordinates. For an integrand that depends only on l_E², it is the only place the dimension enters.

d the number of Euclidean dimensions, any real number above 0; Ω_d is a pure number: Ω₂ = 2π, Ω₃ = 4π, Ω₄ = 2π² = 19.74

The Euclidean master integral∫ d^d l_E/(2π)^d (l_E² + Δ)⁻ⁿ = Γ(n − d/2) Δ^{(d/2 − n)}/((4π)^(d/2) Γ(n))

Derived where it converges, d between 0 and 2n, then defined by continuation in d. The poles of Γ at d = 2n, 2n + 2, … are the divergences.

l_E the Euclidean loop momentum in GeV, Δ above 0 in GeV², n the power of the denominator; mass dimension d − 2n

An l_E² in the numerator∫ d^d l_E/(2π)^d l_E² (l_E² + Δ)⁻ⁿ = (d/2) Γ(n − 1 − d/2) Δ^{(d/2 − n + 1)}/((4π)^(d/2) Γ(n))

Derived from l_E² = (l_E² + Δ) − Δ and Γ(z + 1) = zΓ(z). Keep the d/2 as it stands until the poles are expanded: setting it to 2 early loses a finite term.

same symbols; mass dimension d + 2 − 2n, so for n ≥ 2 the first pole sits 2 lower in d than without the numerator; for n = 1, (d/2)Γ(−d/2) = −Γ(1 − d/2) and the first pole stays at d = 2

Symmetric integration∫ d^d l l^μ l^ν f(l²) = (g^μν/d) ∫ d^d l l² f(l²)

Derived: with no vector available the tensor must be a multiple of g^μν, and contracting with g_μν fixes the 1/d. For the same reason a lone l^μ integrates to zero.

f any function of l² alone; g^μν the metric, δ^μν after the Wick rotation, with g^μ_μ = d; l^μ in GeV

Scaleless integrals vanish∫ d^d l (l²)⁻ⁿ = 0

Shown by splitting at any μ₀: the region above gives a term in μ₀^{(d − 2n)}/(2n − d), the region below its negative, so at n = 2 an ultraviolet pole at d = 4 cancels an infrared one.

no mass and no external momentum, so nothing in GeV can carry the mass dimension d − 2n; any real n

Minkowski form, as the Feynman rules give it∫ d^d l/(2π)^d (l² − Δ)⁻ⁿ = (−1)ⁿ i Γ(n − d/2) Δ^{(d/2 − n)}/((4π)^(d/2) Γ(n))

Derived from the Euclidean card in one line, for Δ above 0. The bubble enters with n = 2, the tadpole with n = 1; Δ below 0 needs the i0 of topic 6.04.

l the Minkowski loop momentum, metric (+, −, −, −); l⁰ = i l_E⁰ gives d^d l = i d^d l_E and l² = −l_E²

What an integral in d dimensions has to satisfy

Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. Topic 6.02 left each bubble as i∫ d⁴l_E/(2π)⁴ (l_E² + Δ)⁻², an integrand falling as l_E⁻⁴ against a measure growing as l_E³: a logarithmic divergence, as topic 6.01's D = 0 predicted. A cutoff on |l_E| makes it finite, but a ball of radius Λ is not carried into itself by a shift such as l = k + xp. For this logarithmic integral the damage vanishes as Λ grows; for a linear or quadratic divergence it leaves a finite surface term, and the answer depends on how the loop momentum was routed. Dimensional regularisation keeps the integrand and replaces 4 by d. It asks four things of a d-dimensional integral: linearity; shift invariance, ∫ d^d l f(l + q) = ∫ d^d l f(l); scaling, ∫ d^d l f(λl) = λ^(−d) ∫ d^d l f(l); and ∫ d^d l e^(−l²) = π^(d/2). Ordinary integrals have all four in every integer d. That they fix the integral uniquely for every real d, for the integrands of perturbation theory, is quoted (Wilson 1973; Collins 1984). Nothing with 3.9 axes is imagined: the integral is defined by these rules.

The Gaussian fixes the area of the unit sphere

Take the one-dimensional Gaussian to the d-th power: (∫ dx e^(−x²))^d = π^(d/2). The same integral in spherical coordinates is Ω_d ∫₀^∞ dr r^(d−1) e^(−r²), and with u = r² the radial factor is Γ(d/2)/2. So Ω_d = 2π^(d/2)/Γ(d/2), derived. Check it where the answer is known: Ω₁ = 2, the two ends of a line; Ω₂ = 2π; Ω₃ = 4π, since Γ(3/2) = √π/2; Ω₄ = 2π² = 19.74. Nothing in the formula needs d to be an integer. At d = 3.9, π^1.95 = 9.321 and Γ(1.95) = 0.9799 give Ω = 19.02, 3.6 per cent below Ω₄. This is all an integrand depending only on l_E² asks of the dimension: once the angles are done, d^d l_E = Ω_d l_E^(d−1) dl_E. Topic 6.02's Feynman parameters and shift bring every scalar one-loop integral of this unit to that form.

The radial integral is a Beta function

With the angles done, Iₙ = (Ω_d/(2π)^d) ∫₀^∞ dl l^(d−1) (l² + Δ)⁻ⁿ, the label E dropped. Put u = l² and then u = Δ(1 − t)/t, so that u + Δ = Δ/t: the integral becomes (1/2)Δ^{(d/2 − n)} ∫₀¹ dt t^{(n − d/2 − 1)}(1 − t)^{(d/2 − 1)}. That is Euler's Beta integral, which equals Γ(n − d/2)Γ(d/2)/Γ(n), quoted, and converges while both of its exponents exceed −1, that is for d between 0 and 2n. There Γ(d/2) cancels against Ω_d and π^(d/2)/(2π)^d = (4π)^(−d/2), so Iₙ = Γ(n − d/2)Δ^{(d/2 − n)}/((4π)^(d/2)Γ(n)), derived. Dimensions: d^d l carries mass dimension d and the denominator −2n, as Δ^{(d/2 − n)} does. Limits: n = 3 at d = 4 gives 1/(32π²Δ), which the elementary radial integral confirms; n = 1 at d = 1 gives 1/(2√Δ), and with Δ = ω² that is ∫ dω′/(2π) (ω′² + ω²)⁻¹ = 1/(2ω), the equal-time propagator of a field in one Euclidean dimension: the oscillator's ⟨x²⟩ = ħ/(2Mω) for unit mass. No measured number enters a master integral; that comparison comes when topic 6.07 assembles an amplitude.

Continuation turns divergences into poles

Γ(z) is finite except at z = −j, j = 0, 1, 2, …, where it has simple poles of residue (−1)ʲ/j!, quoted, so Iₙ continues to every d except d = 2n, 2n + 2, …. For the bubble, n = 2, the first pole is at d = 4, where the integral first diverges, and Γ(2 − d/2) ≈ 2/(4 − d) gives I₂ ≈ (1/16π²) × 2/(4 − d) plus terms finite at d = 4, which topic 6.04 expands. For the tadpole, n = 1, a cutoff in d = 4 gives (1/16π²)[Λ² − Δ ln(1 + Λ²/Δ)]. Its continuation Γ(1 − d/2) has a pole at d = 2, where the integral is logarithmic, and another at d = 4, where Γ(1 − d/2) ≈ −2/(4 − d) gives I₁ ≈ −(Δ/16π²) × 2/(4 − d). The pole carries the coefficient of ln Λ², and the Λ² term has no counterpart at all. In d = 3 the tadpole diverges linearly, yet Γ(−1/2) = −2√π is finite and I₁ = −√Δ/(4π), with no pole. Every pole is a logarithm somewhere: the leading Λ² shows up only as the pole at d = 2.

Numerators: odd terms vanish, l^μ l^ν becomes g^μν l²/d

After the shift, numerators are polynomials in l. A term odd in l integrates to zero, because l → −l leaves the measure and l² unchanged; that needs shift invariance, since the symmetric point is the shifted one. A tensor ∫ d^d l l^μ l^ν f(l²) can be built from nothing but g^μν, so it equals A g^μν, and contracting with g_μν, where g^μ_μ = d, gives A = (1/d) ∫ d^d l l² f(l²), derived. In Euclidean form, l_E² = (l_E² + Δ) − Δ turns an l_E² numerator into Iₙ₋₁ − ΔIₙ, and Γ(z + 1) = zΓ(z) collects it as (d/2)Γ(n − 1 − d/2)Δ^{(d/2 − n + 1)}/((4π)^(d/2)Γ(n)), derived; for n ≥ 2 two more powers of l move the first pole down by 2, while for n = 1 the d/2 cancels the pole of Γ(−d/2) at d = 0 and the first pole stays at d = 2. The d must stay a d. Near d = 4, 1/d = 1/4 + (4 − d)/16 + …, and multiplied by a pole 2/(4 − d) the second term leaves a finite 1/8. Replace 1/d by 1/4 too early and a finite part of the answer is lost, as the Hard example below finds.

Integrals with no scale are zero, honestly

Set Δ = 0 with n = 2: ∫ d^d l (l²)⁻² has nothing in GeV to carry its mass dimension d − 4. Rescale l by λ and the scaling rule says the integral equals λ^{(d − 4)} times itself for every λ, so it is zero unless d = 4, and continuation keeps it zero there. Split it at any μ₀ to see what the zero contains. Above μ₀, ∫ dl l^(d−5) = μ₀^(d−4)/(4 − d), convergent for d below 4: an ultraviolet pole. Below μ₀ the same integrand gives μ₀^(d−4)/(d − 4), convergent for d above 4: an infrared pole. Each continued from where it converges, the two cancel for every d, and with Ω₄/(2π)⁴ = 1/(8π²) the zero at d → 4 is (1/8π²)[1/(4 − d) − 1/(4 − d)], ultraviolet minus infrared. The ultraviolet half has exactly the massive bubble's pole, (1/16π²) × 2/(4 − d). The massless tadpole vanishes the same way, so massless φ⁴ gets no one-loop mass shift in dimensional regularisation, as topic 6.05 finds.

Change one variable at a time

Make the relationship visible.

Interactive model

Denominator power n2

Numerator power k of l_E²0

Dimension d3.5

Start on the bubble, n = 2 and k = 0: the shaded strip ends at the pole d = 4, coefficient 1. Set n = 1, the tadpole: a pole appears at d = 2 and the d = 4 coefficient turns to −1, times Δ, not Λ². Step d to 3.9 and the value nearly equals the pole term; at 4.1 it has changed sign.

DEGREE D = d + 2k − 2n-0.5

FIRST POLE AT d4

COEFFICIENT OF 2/(4 − d)1

POLE TERM AT THIS d4.000

Live interpretationDEGREE D = d + 2k − 2n: −0.5. FIRST POLE AT d: 4. COEFFICIENT OF 2/(4 − d): 1. POLE TERM AT THIS d: 4.000

Catch the common trap

Explain before calculating.

Misconception check

Because dimensional regularisation sets ∫ d^d l (l²)⁻² to zero, it is tempting to conclude that in massless φ⁴ the one-loop four-point function at zero external momentum has no ultraviolet divergence and needs no coupling counterterm. The zero is not the absence of a divergence: it is two divergences cancelling. Split at any μ₀ and the momenta above contribute (1/16π²) × 2/(4 − d), exactly the massive bubble's ultraviolet pole, while those below contribute minus the same amount, an infrared pole from l → 0. Give the integral any scale, a mass or an external p², and the infrared region is cut off while the ultraviolet pole survives, with the coefficient from which the counterterm of topic 6.06 and β(λ) in topic 6.08 are built. Setting a scaleless integral to zero is legitimate only while you remember which of its poles was ultraviolet.

With a Euclidean cutoff Λ in four dimensions, the tadpole integral ∫ d⁴l_E/(2π)⁴ (l_E² + m²)⁻¹ equals (1/16π²)[Λ² − m² ln(1 + Λ²/m²)]. What does dimensional regularisation make of the same integral?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyUse the d-th power of the Gaussian integral to find the area Ω_d of the unit sphere for d = 3, 4 and 3.9. Then evaluate the master integral with n = 3 in d = 4 for Δ = 0.25 GeV², confirm it by doing the radial integral directly, and check its mass dimension.

(∫ dx e^(−x²))^d = π^(d/2), and the same integral in spherical coordinates is Ω_d ∫₀^∞ dr r^(d−1) e^(−r²) = Ω_d Γ(d/2)/2, using u = r². So Ω_d = 2π^(d/2)/Γ(d/2).
d = 3: Γ(3/2) = √π/2, so Ω₃ = 4π = 12.57, the familiar full solid angle. d = 4: Γ(2) = 1, so Ω₄ = 2π² = 19.74. d = 3.9: π^1.95 = 9.321 and Γ(1.95) = 0.9799, so Ω = 19.02, between the two.
Master integral: Γ(3 − 2)Δ^{(2 − 3)}/((4π)² Γ(3)) = 1/(16π² × 2 × Δ) = 1/(32π²Δ). With Δ = 0.25 GeV², 32π²Δ = 78.96 GeV², so the integral is 0.01267 GeV⁻².
Directly: Ω₄/(2π)⁴ = 2π²/(16π⁴) = 1/(8π²), and ∫₀^∞ dl l³ (l² + Δ)⁻³ = (1/2)∫₀^∞ du u (u + Δ)⁻³ = (1/2) × 1/(2Δ) = 1/(4Δ). The product is 1/(32π²Δ), the same.
Dimensions: d − 2n = 4 − 6 = −2, and the result is in GeV⁻². Limit: as Δ grows the integral falls as 1/Δ, as a heavy particle in the loop should suppress it.

AnswerΩ₃ = 4π = 12.57, Ω₄ = 2π² = 19.74 and Ω = 19.02 at d = 3.9; the n = 3 master integral is 1/(32π²Δ) = 0.0127 GeV⁻².

MediumIn three Euclidean dimensions J = ∫ d³k/(2π)³ (k² + m²)⁻¹ diverges linearly. Take m = 139.57 MeV, the charged pion's mass. Evaluate J by the master formula, then with a sharp cutoff Λ = 1 GeV on |k|, and compare both with the Yukawa potential e^(−mr)/(4πr) at small r.

Master formula with d = 3, n = 1, Δ = m²: J = Γ(−1/2) m/(4π)³⁄². Γ(−1/2) = −2√π and (4π)³⁄² = 8π³⁄², so J = −m/(4π) = −11.11 MeV. Γ(1 − d/2) is finite at d = 3, so the linear divergence leaves no pole. Dimension d − 2n = 1: an energy, as found.
Cutoff: J_Λ = (4π/(2π)³) ∫₀^Λ dk k²/(k² + m²) = (1/2π²)[Λ − m arctan(Λ/m)]. With Λ/m = 7.165, arctan(Λ/m) = 1.4321, so J_Λ = 50.66 MeV − 10.13 MeV = 40.53 MeV.
Expand for Λ much larger than m, with arctan(Λ/m) = π/2 − m/Λ + m³/(3Λ³): J_Λ = Λ/(2π²) − m/(4π) + m²/(2π²Λ) − m⁴/(6π²Λ³) = 50.66 − 11.11 + 0.99 − 0.01 MeV = 40.53 MeV. A pure power of Λ, plus the dimensionally regularised value, plus terms that vanish as Λ grows.
A third regulator, point splitting: ∫ d³k/(2π)³ e^(ik⋅r) (k² + m²)⁻¹ = e^(−mr)/(4πr), the Yukawa potential of topic 5.06, quoted. At small r it is 1/(4πr) − m/(4π) + m²r/(8π) + …: a different power divergence, 1/(4πr), and the same finite −m/(4π).
Limits: as m → 0, J → 0, the scaleless integral, while the cutoff result tends to Λ/(2π²), a pure power that dimensional regularisation never produces.

AnswerJ = −m/(4π) = −11.11 MeV, with no pole; the cutoff gives 40.53 MeV = 50.66 MeV of pure Λ/(2π²) − 11.11 MeV + 0.98 MeV that vanishes as Λ grows, and point splitting gives 1/(4πr) − 11.11 MeV.

HardThe Euclidean tensor integral T^μν = ∫ d^d l_E/(2π)^d [2l^μ l^ν − δ^μν (l_E² + Δ)] (l_E² + Δ)⁻² is minus the integral of the total derivative ∂/∂l_μ [l^ν/(l_E² + Δ)]. Show that it vanishes in dimensional regularisation for every d, find what replacing 1/d by 1/4 early would leave, and evaluate T^μν with a cutoff on |l_E| at Λ² = 100Δ.

Symmetric integration replaces 2l^μ l^ν by (2/d)δ^μν l_E², so T^μν = δ^μν[(2/d)A − I₁], where A = ∫ l_E² (l_E² + Δ)⁻² and I₁ = ∫ (l_E² + Δ)⁻¹, both over d^d l_E/(2π)^d.
Master integrals: A = (d/2)Γ(1 − d/2)Δ^{(d/2 − 1)}/(4π)^(d/2) and I₁ = Γ(1 − d/2)Δ^{(d/2 − 1)}/(4π)^(d/2). So (2/d)A = I₁ exactly and T^μν = 0 for every d: the total derivative integrates to zero, as shift invariance requires.
With 1/4 in place of 1/d, T^μν = δ^μν (d/4 − 1)Γ(1 − d/2)Δ^{(d/2 − 1)}/(4π)^(d/2). Near d = 4, Γ(1 − d/2) ≈ −2/(4 − d), so (d/4 − 1)Γ(1 − d/2) → 1/2 and T^μν → δ^μν Δ/(32π²) = 3.17 × 10⁻³ Δ δ^μν, a finite error from a factor that looked harmless.
Cutoff in d = 4, with Ω₄/(2π)⁴ = 1/(8π²) and u = l_E²: A(Λ) = (1/16π²)[Λ² − 2Δ ln(1 + Λ²/Δ) + ΔΛ²/(Λ² + Δ)] and I₁(Λ) = (1/16π²)[Λ² − Δ ln(1 + Λ²/Δ)]. So T^μν = δ^μν[A(Λ)/2 − I₁(Λ)] = −δ^μν Λ⁴/(32π²(Λ² + Δ)): the logarithms cancel and a surface term survives.
At Λ² = 100Δ: A(Λ) = 0.5811Δ, I₁(Λ) = 0.6040Δ and T^μν = −δ^μν × 10⁴Δ/(32π² × 101) = −0.3135Δ δ^μν, half the size of I₁(Λ) and growing as Λ². Gauss's theorem on the sphere |l_E| = Λ gives −(1/8π²)Λ³ × (δ^μν/4) × Λ/(Λ² + Δ), the same.
A ball of radius Λ is not shift invariant, so total derivatives leave surface terms in δ^μν Λ²; in vacuum polarisation such a term would be a photon mass, which is why topic 13.01 chooses dimensional regularisation. Limit Δ → 0: the cutoff value tends to −δ^μν Λ²/(32π²), while the dimensionally regularised one stays 0.

AnswerT^μν = 0 in dimensional regularisation for every d; using 1/4 for 1/d leaves a spurious δ^μν Δ/(32π²); a cutoff leaves −δ^μν Λ⁴/(32π²(Λ² + Δ)) = −0.313Δ δ^μν at Λ² = 100Δ.