University Quantum Field Theory I · One Loop in φ⁴: Dimensional Regularisation, Counterterms and the Running Coupling · 6.1

Power counting and renormalisable couplings

Before you evaluate a single loop you can know which ones will come out infinite and how fast. That count is why renormalising φ⁴ is a finite job and renormalising a φ⁶ coupling an endless one, and it is the first thing to do with any new interaction: read off its mass dimension, and you know its fate.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A loop integral can only diverge where its loop momenta are large, so the first question to put to a diagram is how its integrand scales when all of them grow together. In four dimensions each loop brings d⁴ℓ, four powers of momentum, and each scalar propagator i/(ℓ² − m² + iε) takes two away, so a diagram with L loops and I internal lines behaves like ∫ρ^(D−1)dρ with D = 4L − 2I: under a cutoff Λ it grows as Λ^D when D is positive, as ln Λ when D = 0, and converges by the count when D is negative. Topic 3.05's relations between loops, lines, vertices and legs remove the order and leave D = 4 − E for every φ⁴ diagram: only the two-point function, quadratically, and the four-point function, logarithmically, diverge, at every order, and their divergent parts have the shapes of φ², (∂φ)² and φ⁴.

In d dimensions the same algebra gives D = d − ((d − 2)/2)E − [g]V, so a coupling's mass dimension decides everything: positive, and each vertex lowers D until nothing diverges; zero, and a fixed finite list diverges at every order; negative, and each order brings new divergent functions with new constants to measure. The count is superficial. Scaling every momentum at once misses a divergent subdiagram inside a convergent whole and is blind to cancellations a symmetry enforces; Weinberg's theorem, quoted, makes it reliable only when every subdiagram is counted too.

Simple definition: The superficial degree of divergence D of a Feynman diagram is the power of momentum its loop integral carries when every loop momentum is scaled up together: with a cutoff Λ it grows as Λ^D for positive D, as ln Λ at D = 0, and converges by the count for negative D.
Example: The two-loop sunset of φ⁴ has L = 2 loops and I = 3 internal lines, so D = 4 × 2 − 2 × 3 = 2: superficially quadratic, like the one-loop tadpole. The one-loop six-point triangle, with L = 1 and I = 3, has D = −2 and converges.

Superficial degree of divergenceD = 4L − 2I

Derived: each ∫d⁴ℓ adds four powers of momentum and each propagator i/(ℓ² − m² + iε) removes two. Negative D means convergent by the count, which is not yet convergent.

L loops and I internal scalar lines, pure counts; with a cutoff Λ in GeV the integral grows as Λ^D (ln Λ at D = 0)

φ⁴ in four dimensionsD = 4 − E

Derived: the order cancels, so E = 2 (D = 2) and E = 4 (D = 0) diverge at every order, and E ≥ 6 never diverges by the count. Odd E is excluded by φ → −φ.

E external legs, a pure count; from L = I − V + 1 and 4V = 2I + E, quoted from topic 3.05, with V the vertices

Mass dimension of a φⁿ coupling[gₙ] = d − n(d − 2)/2

Derived: in four dimensions [g₃] = 1, [λ] = 0 and [g₆] = −2. It vanishes at d = 2n/(n − 2): six dimensions for φ³, four for φ⁴, three for φ⁶.

gₙ multiplies φⁿ/n! in d dimensions; [φ] = (d − 2)/2 makes the action dimensionless; mass dimensions in powers of GeV

Power counting in d dimensionsD = d − ((d − 2)/2)E − [g]V

Derived from D = dL − 2I and nV = 2I + E. Positive [g] makes D fall with the order (super-renormalisable), zero keeps it fixed, negative makes it climb.

V vertices of a coupling g of mass dimension [g]; each further coupling adds its own −[gᵢ]Vᵢ; d, E and V pure numbers

A φ⁶ coupling in four dimensionsD = 3 + L − E/2

Derived: every function up to E = 2L + 6 diverges at L loops, so one loop already needs a φ⁸ counterterm and two loops a φ¹⁰ one, each with a new constant.

pure −(g₆/6!)φ⁶ at L loops, from D = 4 − E + 2V₆ and 6V₆ = 2I + E; g₆ in GeV⁻², E and L pure counts

Weinberg's convergence theoremD(γ) negative for G and for every subgraph γ of G ⇒ the integral of G converges

Quoted (Weinberg, 1960), not proved here. A negative D for the whole graph is not enough: a divergent subgraph inside it still diverges.

G a Feynman graph with massive lines, in Euclidean momentum space; γ any subgraph containing a loop; D counted as above

Scale every loop momentum together

Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout. A loop integral can diverge only where loop momenta are large, so scale all L of them at once, ℓᵢ → sℓᵢ. The measure grows as s^(4L), and each of the I propagators i/(ℓ² − m² + iε) falls as s⁻² once ℓ² dwarfs m² and the external momenta. In the 4L-dimensional space of loop momenta the radial integral is then ∫ρ^(D−1)dρ with D = 4L − 2I, derived: it grows as Λ^D under a cutoff Λ for positive D, as ln Λ at D = 0, and converges by the count for negative D. D is also the mass dimension of the loop integral, so a positive D has to be paid in powers of the largest scale on offer, Λ. Three one-loop φ⁴ integrals at zero external momentum, Wick-rotated as topic 6.02 will justify, show the three cases at large Λ: the tadpole, I = 1 and D = 2, goes as Λ²/16π²; the bubble, I = 2 and D = 0, as [ln(Λ²/m²) − 1]/16π²; the triangle, I = 3 and D = −2, tends to 1/(32π²m²).

Trade loops and lines for external legs

D = 4L − 2I hides the answer, because L and I both grow with the order. Topic 3.05 supplies two relations, quoted: L = I − V + 1, since momentum conservation at each vertex fixes one momentum except the overall one, and 4V = 2I + E, since each vertex has four ends, each internal line uses two and each external leg one. Then D = 4(I − V + 1) − 2I = 2I − 4V + 4, and 2I = 4V − E leaves D = 4 − E, derived. The order has dropped out, so one short table holds at every loop. E = 0: D = 4, vacuum bubbles, which cancel from normalised correlators (topic 3.03). E = 2: D = 2, the one-loop tadpole and the two-loop sunset (V = 2, I = 3) alike. E = 4: D = 0, the one-loop bubble and the two-loop double bubble (V = 3, I = 4) alike. E ≥ 6: D ≤ −2. Odd E never appears, since 4V = 2I + E makes E even and φ → −φ forbids it to all orders, though the count would give E = 1 and E = 3 the values D = 3 and D = 1. Tree check, L = 0: I = E/2 − 2, so −2I = 4 − E, the tree amplitude's own mass dimension.

What a divergence is allowed to depend on

The shape of a divergence matters as much as its size. Differentiate a one-loop diagram with respect to an external momentum p. The derivative lands on a propagator, ∂/∂p^μ [(ℓ + p)² − m²]⁻¹ = −2(ℓ + p)_μ [(ℓ + p)² − m²]⁻², which carries one power of ℓ fewer, so D drops by one. After D + 1 derivatives the integral converges, so the divergent part is a polynomial in the external momenta of degree D: derived at one loop, quoted to all orders as part of the BPHZ theorem. Lorentz invariance allows only even powers. The four-point function, D = 0, can diverge only by a constant: the bubble's ln Λ² carries no s, t or u, and their dependence sits in a finite remainder. The two-point function, D = 2, can diverge as a + bp². At one loop the tadpole supplies only a, a pure mass shift (topic 6.05), and b first appears with the two-loop sunset, quoted. The three shapes are the operators φ², (∂φ)² and φ⁴ already in the Lagrangian, which is why topic 6.06's three counterterms can absorb them.

In d dimensions the coupling's mass dimension decides

The action ∫dᵈx ℒ is dimensionless, so [φ] = (d − 2)/2 and a coupling gₙφⁿ/n! has [gₙ] = d − n(d − 2)/2, extending topic 2.01. Each loop now brings d powers, D = dL − 2I, and nV = 2I + E. Eliminating L and I gives D = d − ((d − 2)/2)E − [g]V, derived. Dimensions check it: the amputated E-point function has mass dimension d − ((d − 2)/2)E, the factor g^V carries [g]V of it, and the loop integral supplies the rest, which is D. In four dimensions φ³ has [g] = 1 and D = 4 − E − V, so every vertex lowers D and the divergent list ends: super-renormalisable. φ⁴ has [λ] = 0 and D = 4 − E: renormalisable. A φ⁶ coupling has [g₆] = −2 and D = 4 − E + 2V₆: non-renormalisable. The verdict belongs to the pair of interaction and dimension. [gₙ] vanishes at d = 2n/(n − 2), so φ³ is renormalisable in six dimensions, φ⁴ in four and φ⁶ in three, while φ⁴ is super-renormalisable in three dimensions and non-renormalisable in five and six.

A coupling of dimension −2: ever more functions diverge

Follow −(g₆/6!)φ⁶ in four dimensions through the orders. With L loops, 6V = 2I + E and I = L + V − 1 give D = 3 + L − E/2, shown, so at L loops every function up to E = 2L + 6 diverges superficially. At one loop two φ⁶ vertices joined by two lines make an eight-point function with D = 0, a logarithm multiplying φ⁸ at a point, which nothing in the Lagrangian can absorb: it needs a φ⁸ counterterm whose constant only a new measurement can fix. Two loops need φ¹⁰, and the six-point function there has D = 2, so a derivative term φ⁴(∂φ)² is needed too. Infinitely many constants mean no predictions at high energy, but not none at low: write g₆ = c/M², and against a dimensionless coupling a φ⁶ vertex at a process energy Q weighs c(Q/M)², small while Q ≪ M. Fermi's interaction is the measured case. G_F/(ħc)³ = 1.1664 × 10⁻⁵ GeV⁻², also of dimension −2, reaches G_F Q² = 1 at Q = 292.8 GeV, a distance ħc/Q = 6.7 × 10⁻⁴ fm, and W exchange, with m_W c² = 80.4 GeV, replaces it at a lower energy, before that limit is reached, quoted.

What the count cannot see

D scales every loop momentum at once, so it misses a divergence that lives in part of a diagram. Insert the one-loop bubble at any vertex of a convergent φ⁴ graph: the whole keeps D = 4 − E, as negative as E makes it, yet when only the bubble's momentum grows that subintegral has D = 0 and diverges. Weinberg's theorem of 1960, quoted, gives the test that can be trusted: in a massive theory a graph converges if D is negative for it and for every subgraph. The cure is already on the books, since the one-loop δ_λ counterterm at the same place cancels the subdivergence, and the BPHZ theorem, quoted, shows such subtractions work at every order. The count is also blind to symmetry, which can make the truth milder than D. φ → −φ removes the odd functions; in QED, Units 12 and 13 find the electron's mass shift proportional to m and only logarithmic although the count gives D = 1, and gauge invariance cuts the photon self-energy from D = 2 to a logarithm. And D sees only large momenta: infrared divergences from massless lines, met in Unit 14, are a separate matter.

Change one variable at a time

Make the relationship visible.

Interactive model

Spacetime dimension d4

Loop order marked, L2

At d = 4 the three lines are flat: the same two functions diverge at every order. Set d = 3 and they fall, so beyond two loops only vacuum graphs diverge; set d = 5 or 6 and they climb, and at L = 4 in d = 6 every function up to E = 12 lies in the band.

COUPLING DIMENSION [λ]0

D, TWO-POINT AT L2

D, FOUR-POINT AT L0

LARGEST DIVERGENT E AT L4

Live interpretationCOUPLING DIMENSION [λ]: 0. D, TWO-POINT AT L: 2. D, FOUR-POINT AT L: 0. LARGEST DIVERGENT E AT L: 4

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading is that D is the diagram's verdict: negative D means finite. It is only the verdict on the whole diagram scaled at once. Open one vertex of the one-loop φ⁴ triangle into a bubble and the six-point graph has V = 4, I = 5 and L = 2, so D = 8 − 10 = −2, yet the bubble inside it, with I = 2 and one loop, has D = 0 and diverges logarithmically when its own momentum alone grows. The graph is finite only after the one-loop δ_λ counterterm, inserted at the same vertex, cancels that subdivergence. The reliable statement, Weinberg's theorem, needs D negative for every subgraph as well as for the whole. The opposite reading fails too: D ≥ 0 does not guarantee a divergence, because a symmetry can forbid or cancel it: φ → −φ leaves the odd functions, which the count gives D = 3 and D = 1, with no diagrams at all.

A real scalar field in six spacetime dimensions has the interaction −(g/3!)φ³. What does power counting say about this theory?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyWith a Euclidean cutoff Λ, the one-loop φ⁴ tadpole, bubble and triangle at zero external momentum reduce to J_I = (1/16π²)∫₀^(Λ²) u du/(u + m²)^I with I = 1, 2 and 3. Find D for each, evaluate all three at Λ/m = 10 and Λ/m = 20, and check that doubling the cutoff does what D predicts.

Each has L = 1, so D = 4 − 2I: the tadpole has D = 2, the bubble D = 0 and the triangle D = −2, which is D = 4 − E with E = 2, 4 and 6. Dimensions agree: J_I has mass dimension 4 − 2I = D, so J₁ is a mass squared, J₂ a pure number and J₃ an inverse mass squared.
With w = u + m² each integral is elementary: 16π²J₁ = Λ² − m² ln(1 + Λ²/m²), 16π²J₂ = ln(1 + Λ²/m²) − Λ²/(Λ² + m²), and 16π²J₃ = 1/(2m²) − 1/(Λ² + m²) + m²/(2(Λ² + m²)²), each derived.
At Λ/m = 10: 16π²J₁/m² = 100 − ln 101 = 95.385, 16π²J₂ = ln 101 − 100/101 = 3.6250 and 16π²m²J₃ = 0.49015. At Λ/m = 20 the same three are 394.006, 4.9965 and 0.49751.
Tadpole: 394.006/95.385 = 4.131, close to 2^D = 4, the excess being the m² ln term. Bubble: it gains 1.3714/16π² = 0.00868, close to ln 4/16π² = 0.00878, and gains as much again at every further doubling. Triangle: it rises 1.5% and is already within 0.5% of its limit, m²J₃ → 1/(32π²) = 3.17 × 10⁻³.

AnswerD = 2, 0 and −2. Raising Λ/m from 10 to 20 multiplies the tadpole by 4.13, adds 0.0087 to the bubble, and moves the triangle up 1.5%, to within 0.5% of 1/(32π²m²).

MediumTreat a −(g/3!)φ³ interaction in four dimensions as a counting laboratory (its potential is unbounded below, so it is not a stable theory). Find [g] and D, list every superficially divergent class of diagrams with at least one loop, say which counterterms the theory needs, and with g/m = 2 find how much the one-loop self-energy changes when Λ/m rises from 100 to 200.

[φ] = 1, so [g] = 4 − 3 = 1: g is a mass. The d-dimensional count at d = 4 gives D = 4 − E − V, so every vertex lowers D by one.
Topology: 3V = 2I + E and L = I − V + 1 give V = 2L + E − 2, so D = 6 − 2E − 2L. D ≥ 0 then needs E + L ≤ 3, with L ≥ 1 and V ≥ 1.
That leaves five classes: (E, L) = (1, 1), the one-loop tadpole, D = 2; (2, 1), the self-energy bubble, D = 0; (1, 2), the two-loop tadpoles with V = 3, D = 0; and the vacuum graphs (0, 2), D = 2, and (0, 3), D = 0. Beyond three loops nothing diverges.
Counterterms: the tadpoles need a term linear in φ, the bubble only δₘ, because D = 0 allows a constant but no p², and the vacuum graphs a constant energy. No E = 3 function ever diverges, so g receives only finite corrections, and so does the field normalisation.
In magnitude the bubble at zero momentum is (g²/2)J₂, with J₂ from the Easy example: 16π²J₂ = 8.2105 at Λ/m = 100 and 9.5967 at Λ/m = 200. With g/m = 2 the self-energy moves by 2 × 1.3861/16π² = 1.76% of m², the same at every doubling: a logarithm, absorbed by δₘ.
Dimensions: Σ is a mass squared and g² already supplies it, so the loop must be dimensionless, which is D = 0. As g → 0 every entry of the list vanishes while the list itself, set by [g] alone, stays the same.

Answer[g] = 1 and D = 6 − 2E − 2L: five classes diverge, (E, L) = (1, 1), (2, 1), (1, 2), (0, 2) and (0, 3); only φ, φ² and vacuum counterterms are needed; raising Λ/m from 100 to 200 moves Σ by 1.76% of m² at g/m = 2.

HardAdd −(g/6!)φ⁶ with g = 1/M² and M = 2.0 TeV to the φ⁴ theory of a scalar of mass m = 200 GeV in four dimensions. Count the one-loop divergences and the counterterm each needs, size the φ⁶ tadpole's shift of λ with a cutoff Λ = M, find the largest divergent function at two loops, and say up to what energy the theory stays predictive.

[g] = 4 − 6 = −2, so g = 1/M² = 2.5 × 10⁻⁷ GeV⁻². At one loop I = V, and 4V₄ + 6V₆ = 2I + E gives E = 2V₄ + 4V₆, so D = 4L − 2I = 4 − 2V for any mix of vertices: only graphs with one or two vertices can diverge.
V = 1, D = 2: the φ⁴ tadpole (E = 2) needs δₘ, and the φ⁶ tadpole (E = 4), a constant, needs δ_λ. V = 2, D = 0: two φ⁴ vertices give the E = 4 bubble (δ_λ), one of each gives E = 6 (a φ⁶ counterterm), and two φ⁶ vertices give E = 8, which needs a φ⁸ counterterm the Lagrangian never had.
The φ⁶ tadpole is −i(g/2)J₁, with J₁ = [Λ² − m² ln(1 + Λ²/m²)]/16π² from the Easy example, so it shifts λ by (g/2)J₁. At Λ = M, with M/m = 10, this is (1/32π²)(1 − ln 101/100) = 3.166 × 10⁻³ × 0.9538 = 3.02 × 10⁻³: the Λ² is cancelled by g's 1/M², leaving a loop factor.
Two loops: I = V + 1, E = 2V₄ + 4V₆ − 2 and D = 6 − 2V. Three φ⁶ vertices give E = 10 with D = 0, a φ¹⁰ counterterm; two give E = 6 with D = 2, whose divergent part a + bp² needs the derivative term φ⁴(∂φ)², and one of each gives E = 4 with D = 2, which needs φ²(∂φ)². In general E = 2L + 6 diverges at L loops.
Against λ a φ⁶ vertex at a process energy Q weighs gQ² = (Q/M)²: 0.010 at Q = 200 GeV and 1 at Q = M. The new counterterms carry more powers of 1/M, so φ⁸ effects go as (Q/M)⁴: at low energy a few measured constants predict everything to a stated accuracy, and near Q = M nothing does.
Limits: as M → ∞ at fixed Λ every φ⁶ effect vanishes, the tadpole's shift as Λ²/(32π²M²), and D = 4 − E returns with φ⁴'s three counterterms. Dimensions: gΛ² and (Q/M)² are pure numbers, as a shift of the dimensionless λ must be.

AnswerOne loop needs δₘ, δ_λ, a φ⁶ counterterm and a new φ⁸ one; the φ⁶ tadpole shifts λ by 3.0 × 10⁻³ at Λ = M; two loops reach E = 10 and need φ¹⁰ and derivative terms such as φ⁴(∂φ)²; predictive while (Q/M)² ≪ 1, which is 1% at Q = 200 GeV.