University Quantum Field Theory I · One Loop in φ⁴: Dimensional Regularisation, Counterterms and the Running Coupling · 6.6

Bare and renormalised: counterterms and conditions

Every loop so far has handed back a mass and a coupling that are not the ones in the Lagrangian, plus a divergence. This lesson turns that around: rewrite the theory in the mass and coupling an experiment measures, let the difference become three counterterm vertices, and fix each one by a condition you could check in a laboratory.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The m₀ and λ₀ written in the φ⁴ Lagrangian are not the mass and coupling anyone measures. Interactions dress them: topic 6.05's tadpole moves the propagator's pole away from m₀², the field φ₀ reaches the one-particle state with amplitude √Z rather than 1 (topic 4.03), and the bubbles of topic 6.04 make the threshold amplitude differ from −λ₀, each by a regulator-dependent amount. Renormalised perturbation theory reverses the bookkeeping.

Rescale φ₀ = √Z φ and define δ_Z, δₘ and δ_λ by Z = 1 + δ_Z, Zm₀² = m² + δₘ and Z²λ₀ = λ + δ_λ: the bare Lagrangian becomes the same Lagrangian in the measured m and λ plus three terms that the Feynman rules turn into vertices. Nothing is added, since summing the counterterm insertions returns the bare propagator divided by Z. What is new is that each counterterm is fixed by a renormalisation condition, a pole at m, a residue of 1 and an amplitude of −λ at threshold, one for each divergent structure power counting found, so the regulator dependence lands in the counterterms and in the bare parameters, which no experiment sees.

At one loop δ_Z = 0, δₘ is minus the tadpole and δ_λ is the three bubbles at threshold. The scheme has limits: it needs a stable massive particle with a threshold, so a massless theory must be renormalised off shell, and three counterterms suffice only because λ is dimensionless; a φ⁶ term would demand new ones.

Simple definition: A renormalisation condition is a statement that defines a measured parameter and so fixes a counterterm: in φ⁴ theory the propagator has its pole at the measured mass m with residue 1, and the amputated four-point amplitude equals −λ at threshold.
Example: With λ = 1 and a Euclidean cutoff at Λ/m = 10, the three conditions give δ_Z = 0, δₘ = −0.302 m² and δ_λ = 0.0407, so the bare parameters are m₀² = 0.698 m² and λ₀ = 1.041 while the measured mass and coupling stay m and 1.

The bare Lagrangian rewrittenL = ½(∂φ)² − ½m²φ² − (λ/4!)φ⁴ + ½δ_Z(∂φ)² − ½δₘφ² − (δ_λ/4!)φ⁴

Derived and exact: no term is added or dropped. Perturbation theory now runs in the measured m and λ, and everything that depends on the regulator sits in the last three terms.

φ₀ = √Z φ, Z = 1 + δ_Z, Zm₀² = m² + δₘ, Z²λ₀ = λ + δ_λ; φ and m in GeV, δₘ in GeV², λ, δ_Z, δ_λ pure numbers

Counterterm vertices, read as in topic 3.07two-point: i(p²δ_Z − δₘ)four-point: −iδ_λ

Derived. Summing chains of the two-point vertex on i/(p² − m²) gives i/[Z(p² − m₀²)], the bare propagator divided by Z: nothing new was put in.

p the four-momentum through the vertex, p² in GeV²; δₘ starts at order λ, δ_λ and δ_Z at order λ²

Pole at the measured mass, residue 1Σ(m²) = 0 and dΣ/dp² = 0 at p² = m²

Stated, as the definitions of m and of the field's normalisation. The bare field φ₀ then has residue Z, topic 4.03's, and LSZ for φ needs no √Z.

−iΣ(p²) sums the one-particle-irreducible two-point diagrams, counterterm included; full propagator i/(p² − m² − Σ); m in GeV

The coupling defined at thresholdiM(s = 4m², t = 0, u = 0) = −iλ

Stated. Two particles at rest: the threshold cross-section λ²/(128πm²) then holds to one loop, so λ is the number that measurement returns.

s, t, u the Mandelstam invariants in GeV²; M dimensionless for 2 → 2 in four dimensions; Im M = 0 at this point

One-loop mass countertermδₘ = −(λ/32π²)[Λ² − m² ln(1 + Λ²/m²)]

Derived: δₘ is minus the whole tadpole, finite part included, and δ_Z = 0 because the tadpole has no p². λ = 1 and Λ/m = 10 give −0.302 m².

Λ a sharp Euclidean cutoff in GeV; dimensional regularisation, d = 4 − ε, gives (λm²/32π²)[2/ε + 1 − γ_E + ln 4π − ln(m²/μ²)]

One-loop coupling countertermδ_λ = (λ²/2)[B(4m²) + 2B(0)] → (λ²/32π²)[3 ln(Λ²/m²) − 1] for Λ ≫ m

Derived from the threshold condition; the +2 is −∫₀¹ ln(1 − 2x)² dx, and the cutoff's −1 is that +2 less 1 per bubble. It makes λ₀ grow by 3λ²/(16π²) per unit of ln Λ.

(λ²/2)B(p²) is one bubble at p²; dimensional regularisation gives (λ²/32π²)[3(2/ε − γ_E + ln 4π − ln(m²/μ²)) + 2]

Why the Lagrangian's m₀ and λ₀ are not what is measured

Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. The bare Lagrangian L = ½(∂φ₀)² − ½m₀²φ₀² − (λ₀/4!)φ₀⁴ fixes three things an experiment could test, and loops move all three. Topic 6.05's tadpole adds a constant to the inverse propagator, so the pole is no longer at m₀². The Källén–Lehmann residue of φ₀ at the pole is Z, not 1 (topic 4.03), and LSZ charges √Z per leg (topic 4.05). The s-, t- and u-channel bubbles of topic 6.04 make the threshold amplitude −λ₀ plus a regulator-dependent amount. A prediction written in m₀ and λ₀ therefore relates numbers nobody has measured. The cure is algebra, not a new principle. Rescale φ₀ = √Z φ, so that L = ½Z(∂φ)² − ½Zm₀²φ² − (Z²λ₀/4!)φ⁴, and define δ_Z, δₘ and δ_λ by Z = 1 + δ_Z, Zm₀² = m² + δₘ and Z²λ₀ = λ + δ_λ. Then L is the φ⁴ Lagrangian with m and λ plus ½δ_Z(∂φ)² − ½δₘφ² − (δ_λ/4!)φ⁴, derived and exact. Dimensions: [φ] = [m] = 1, so [δₘ] = 2 while λ, δ_Z and δ_λ are pure numbers.

Counterterms are vertices, and they add nothing new

Topic 3.07 reads a vertex as i times the coefficient of the fields in L, times the number of ways to attach the lines. In momentum space the quadratic counterterm is ½φ̃(−p)(p²δ_Z − δₘ)φ̃(p), with two ways to attach two lines, so the two-point vertex is i(p²δ_Z − δₘ); the quartic one gives −iδ_λ, its 4! attachments cancelling the 4!, both derived. Propagators now carry the measured mass, i/(p² − m² + iε), and vertices the measured λ. Each counterterm counts at its own order: δₘ starts at λ, because it must cancel the tadpole, δ_λ starts at λ², and δ_Z at λ². A check that nothing was smuggled in: sum the chain of two-point insertions on a free line. i/(p² − m²) times Σₙ [i(p²δ_Z − δₘ) i/(p² − m²)]ⁿ is i/[(1 + δ_Z)p² − m² − δₘ] = i/[Z(p² − m₀²)], derived: exactly the free bare propagator divided by Z, as φ = φ₀/√Z requires. The counterterms only re-express the old Lagrangian, and in the free-field limit λ → 0 all three vanish and bare equals renormalised.

Three conditions, one for each divergent structure

Write the full propagator as i/(p² − m² − Σ(p²)), where −iΣ sums the one-particle-irreducible two-point diagrams, counterterm vertex included. Expanding about the measured mass gives i/[(p² − m²)(1 − Σ′(m²)) − Σ(m²)], so the pole sits at m if Σ(m²) = 0 and has residue 1 if Σ′(m²) = 0. These two conditions define m and the normalisation of φ; φ₀ = √Z φ then has residue Z, topic 4.03's, and LSZ applied to φ needs no √Z. The third defines λ: the amputated four-point function equals −iλ at s = 4m², t = u = 0, two particles at rest, where Im M ∝ √(1 − 4m²/s) vanishes and M is real. Why exactly three? Topic 6.01 found D = 2 for the two-point function and D = 0 for the four-point. Each derivative with respect to an external momentum lowers D by one, so the divergent part of the first is a + bp², a scalar function of p being a function of p², and that of the second a constant: three constants, matched by δₘ, δ_Z and δ_λ. At one loop this can be checked, since B(p²) − B(0) converges. That no order needs more is the BPHZ theorem, quoted in topic 6.09.

At one loop: δ_Z = 0 and δₘ cancels the whole tadpole

With its counterterm, the one-loop self-energy is Σ(p²) = λA/2 + δₘ − p²δ_Z, where λA/2 is topic 6.05's tadpole and A = ∫d⁴l_E/(2π)⁴ (l_E² + m²)⁻¹ does not depend on p. So Σ′ = −δ_Z, the residue condition gives δ_Z = 0, and Σ(m²) = 0 gives δₘ = −λA/2, both derived. The counterterm cancels the tadpole entirely, finite part included, so the renormalised one-loop Σ vanishes at every p²: at this order the tadpole has no observable effect at all. The value of δₘ depends on the regulator. A sharp Euclidean cutoff gives δₘ = −(λ/32π²)[Λ² − m² ln(1 + Λ²/m²)], which at λ = 1 is −0.302 m² for Λ/m = 10 and −31.6 m² for Λ/m = 100. Dimensional regularisation in d = 4 − ε gives +(λm²/32π²)[2/ε + 1 − γ_E + ln 4π − ln(m²/μ²)], of the opposite sign and with no Λ² anywhere. Each is a mass squared, as a coefficient of φ² must be. Massless limit: in dimensional regularisation δₘ → 0, the tadpole being scaleless; with a cutoff it does not, which is topic 6.05's fine-tuning.

At one loop: δ_λ is the three bubbles at threshold

To order λ², iM = −iλ + (iλ²/2)[B(s) + B(t) + B(u)] − iδ_λ, with B(p²) = ∫₀¹dx ∫d⁴l_E/(2π)⁴ (l_E² + Δ)⁻² and Δ = m² − x(1 − x)p², the bubbles of topics 6.02 and 6.04. Setting M = −λ at s = 4m², t = u = 0 gives δ_λ = (λ²/2)[B(4m²) + 2B(0)], derived. In dimensional regularisation B(p²) = (1/16π²)∫dx[2/ε − γ_E + ln 4π − ln(Δ/μ²)]; at threshold Δ = m²(1 − 2x)², and ∫₀¹ ln(1 − 2x)² dx = −2, so δ_λ = (λ²/32π²)[3(2/ε − γ_E + ln 4π − ln(m²/μ²)) + 2]. With a cutoff on l_E, δ_λ = (λ²/32π²)[3 ln(1 + Λ²/m²) + (Λ/m) atan(m/Λ) − 2Λ²/(Λ² + m²)], tending to (λ²/32π²)[3 ln(Λ²/m²) − 1]; both derived, and 0.0407 at λ = 1, Λ/m = 10. The 3 counts the channels, and the coefficient 3λ²/(16π²) of ln Λ returns in topic 6.08. δ_λ is real because the condition sits exactly where the bubble's imaginary part switches on.

The bare parameters are never observed

At one loop Z = 1, so m₀² = m² + δₘ and λ₀ = λ + δ_λ. With λ = 1 and a cutoff, m₀² is 0.698 m² at Λ/m = 10, passes through zero near Λ/m = 18 and is −30.6 m² at Λ/m = 100, while λ₀ rises from 1.041 to 1.084; in dimensional regularisation both carry poles in ε. A negative bare mass squared would be alarming if it described anything. It does not, because no measurement returns m₀ or λ₀: every observable is a pole, a residue or an amplitude, and the conditions write all three in m and λ, which experiment fixes once. The threshold cross-section λ²/(128πm²) does it here; in QED the same roles are played by the electron's pole mass, 0.51099895 MeV, and α = 1/137.036 defined at q² = 0, quoted for Units 12 and 13. Change the regulator and the bare parameters move so that m and λ do not: that is what being unphysical means. Bethe's m₀ in UAQM-13.04 was never weighed either. Where the scheme strains: as m → 0, B(0) diverges in the infrared, so the coupling must be defined at a spacelike point or by topic 6.08's minimal subtraction.

Change one variable at a time

Make the relationship visible.

Interactive model

Measured coupling at threshold1.0

Cutoff in units of the mass10.0

Hold the coupling at 1 and drag the cutoff from 1 to 20: m₀² sinks through zero near Λ/m = 18 while λ₀ gains only about 5 per cent, quadratic and logarithmic divergence side by side. Then halve the coupling: both departures halve, and the measured m and λ never move.

MASS COUNTERTERM-0.302 m²

BARE MASS SQUARED0.698 m²

COUPLING COUNTERTERM0.0407

BARE COUPLING1.0407

Live interpretationMASS COUNTERTERM: −0.302 m². BARE MASS SQUARED: 0.698 m². COUPLING COUNTERTERM: 0.0407. BARE COUPLING: 1.0407

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading is that counterterms are extra interactions bolted on to cancel infinities, so that the renormalised theory is a doctored version of the bare one and each counterterm is just the divergent part of a loop. Neither holds. The counterterm Lagrangian is the bare Lagrangian term for term, with its coefficients re-expressed through m, λ and the field normalisation: summing the two-point insertions returns the bare propagator divided by Z, and no parameter is added, since the three conditions trade m₀, λ₀ and Z for m, λ and a unit residue. Nor is an on-shell counterterm only a divergence. δₘ cancels the whole tadpole, finite part included, and δ_λ carries the finite +2 from the threshold channel; removing only the pole is a different scheme, minimal subtraction, with a different λ (topic 6.08). What is a choice is where the conditions are imposed, never whether.

A one-loop φ⁴ calculation with a sharp Euclidean cutoff at Λ/m = 10 imposes the three on-shell conditions with measured mass m and coupling λ = 1. The cutoff is then raised to Λ/m = 100 and the conditions imposed again. What changes?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA scalar has measured pole mass m = 0.50 GeV and threshold coupling λ = 0.50. With a sharp Euclidean cutoff Λ = 5.0 GeV on the loop momentum, impose the on-shell conditions at one loop to find δ_Z, δₘ and the bare mass m₀. Then repeat for Λ = 20 GeV and say what the result means.

The one-loop two-point function is the tadpole, which does not depend on p, plus the counterterm: Σ(p²) = λA/2 + δₘ − p²δ_Z. So Σ′(m²) = −δ_Z, and the residue condition Σ′(m²) = 0 gives δ_Z = 0.
The pole condition Σ(m²) = 0 gives δₘ = −λA/2 = −(λ/32π²)[Λ² − m² ln(1 + Λ²/m²)]. Here Λ²/m² = 100 and ln 101 = 4.615, so the bracket is 25.00 − 0.25 × 4.615 = 23.846 GeV².
With λ/32π² = 0.50/315.83 = 1.5831 × 10⁻³, δₘ = −0.03775 GeV². Since Z = 1 at this order, m₀² = m² + δₘ = 0.2500 − 0.0378 = 0.2122 GeV², so m₀ = 0.461 GeV.
At Λ = 20 GeV, Λ²/m² = 1600 and ln 1601 = 7.378, so the bracket is 400 − 1.845 = 398.16 GeV², δₘ = −0.6303 GeV² and m₀² = −0.380 GeV²: no real bare mass at all.
Checks: δₘ is a mass squared, as the coefficient of φ² must be, and it vanishes as λ → 0, where m₀ → m. The negative m₀² harms nothing: the pole stays at the measured 0.50 GeV at either cutoff, and no experiment reads m₀.

Answerδ_Z = 0, δₘ = −0.0378 GeV² and m₀ = 0.461 GeV at Λ = 5.0 GeV; at Λ = 20 GeV, δₘ = −0.630 GeV² and m₀² = −0.380 GeV², while the measured mass stays 0.50 GeV.

MediumFor a scalar of pole mass m = 0.50 GeV, the φφ → φφ cross-section extrapolated to threshold is measured to be 1.00 μb. Use the coupling condition to extract λ, then find δ_λ and the bare coupling λ₀ for a sharp Euclidean cutoff at Λ/m = 10 and at Λ/m = 100. By how much does the one-loop prediction for the threshold cross-section change between the two?

The condition sets M = −λ at s = 4m², t = u = 0, where Im M = 0, so to one loop |M|² = λ² at threshold. For identical particles σ = ½ × 4π × |M|²/(64π²s) = λ²/(32πs), which at s = 4m² is λ²/(128πm²).
Convert: 1.00 μb = 1.00 × 10⁻³ mb = (1.00 × 10⁻³/0.3894) GeV⁻² = 2.568 × 10⁻³ GeV⁻². Then λ² = 128πm²σ = 402.12 × 0.2500 GeV² × 2.568 × 10⁻³ GeV⁻² = 0.2582, a pure number as it must be, and λ = 0.508.
δ_λ = (λ²/32π²)[3 ln(1 + L) + √L atan(1/√L) − 2L/(1 + L)] with L = Λ²/m². At L = 100 the bracket is 13.845 + 0.997 − 1.980 = 12.862, so δ_λ = 0.2582 × 12.862/315.83 = 0.01051 and λ₀ = λ + δ_λ = 0.5186.
At L = 10⁴ the bracket is 27.631 + 1.000 − 2.000 = 26.631, so δ_λ = 0.02177 and λ₀ = 0.5299. The rise, 0.01126, is close to (3λ²/16π²) ln 10 = 0.01129; the small gap is the m²/Λ² terms at Λ/m = 10.
The threshold amplitude is −λ at both cutoffs by construction, so σ = λ²/(128πm²) = 1.00 μb at both. The prediction changes by zero; only the unobservable λ₀ moved.

Answerλ = 0.508; at Λ/m = 10, δ_λ = 0.0105 and λ₀ = 0.519; at Λ/m = 100, δ_λ = 0.0218 and λ₀ = 0.530; the threshold cross-section stays 1.00 μb.

HardTwo groups renormalise the same φ⁴ theory with the same pole mass m. Group A imposes M = −λ_A at threshold, s = 4m², t = u = 0, and quotes λ_A = 0.500. Group B imposes M = −λ_B at the symmetric point s = t = u = 4m²/3, an unphysical point where M is real. Find λ_B at one loop, show that the answer contains no regulator, and say whose coupling is the physical one.

Both describe one bare theory and Z = 1 at this order, so λ₀ = λ_A + δ_A = λ_B + δ_B and λ_B − λ_A = δ_A − δ_B. Each condition fixes its own counterterm: δ_A = (λ²/2)[B(4m²) + 2B(0)] and δ_B = (λ²/2) × 3B(4m²/3), where λ_A and λ_B may be swapped in the λ² prefactor at this order.
Every B(p²) carries the same divergent constant, 2/ε − γ_E + ln 4π + ln μ² or ln Λ², which cancels in differences: B(p²) − B(p′²) = −(1/16π²)∫₀¹dx ln[Δ(p²)/Δ(p′²)]. So λ_B − λ_A = (λ²/32π²)(∫dx ln[Δ(4m²/3)/Δ(4m²)] + 2∫dx ln[Δ(4m²/3)/m²]).
With Δ(4m²) = m²(1 − 2x)², ∫₀¹ ln(1 − 2x)² dx = −2. With Δ(4m²/3) = m²[1 − (4/3)x(1 − x)], positive for every x, I₁ = ∫₀¹ ln[1 − (4/3)x(1 − x)] dx = −2 + 2√2 atan(1/√2) = −2 + 1.7408 = −0.2592.
The braces give (I₁ + 2) + 2I₁ = 3I₁ + 2 = 1.2225. With λ²/32π² = 0.250/315.83 = 7.916 × 10⁻⁴, λ_B − λ_A = 9.68 × 10⁻⁴, so λ_B = 0.50097. A sharp cutoff gives the same 1.2225, to within 10⁻⁴ already at Λ/m = 10, because no ε, μ or Λ survives the difference.
Checks: the shift is dimensionless, second order and vanishes as λ → 0, as a change of conditions at one loop must. Neither coupling is more physical: A's gives the threshold cross-section directly, B's needs this same relation to do so, and both predict one amplitude at every s, t, u to order λ².

Answerλ_B − λ_A = (λ²/32π²)(3I₁ + 2) = 9.7 × 10⁻⁴, so λ_B = 0.501. The shift is finite and regulator-free, and the two couplings are equally valid labels for one theory.