University Quantum Field Theory I · One Loop in φ⁴: Dimensional Regularisation, Counterterms and the Running Coupling · 6.9
The running coupling, leading logs and the Landau pole
Topic 6.08 ended with a differential equation; this topic solves it and puts the answer to work. The solution tells you which coupling to use at the energy you probe, sums a logarithm at every order for free, and then marks, with a pole, the place where it stops being believable.
Build the model
Connect the measurement to the mechanism.
Integrating β(λ) = 3λ²/(16π²) from μ₀ to μ shows that 1/λ falls linearly in ln μ, by 3/(16π²) per e-fold whatever λ(μ₀) is, so λ(μ) = λ(μ₀)/[1 − (3λ(μ₀)/(16π²)) ln(μ/μ₀)], derived. Expanded in λ(μ₀) it is a geometric series with one term λ(μ₀)(n+1) lnⁿ(μ/μ₀) at every order: the leading logarithms, the most an n-loop diagram can carry, and the μ-independence of physical amplitudes fixes every one of their coefficients from the one-loop β alone, shown. That is what the running coupling is for.
A calculation truncated in λ(μ₀) degrades at energy √s even when λ(μ₀) is small, because what must be small is (3λ(μ₀)/(16π²)) ln(√s/μ₀); evaluating the coupling at μ ≈ √s moves every such logarithm into λ(√s), which becomes the expansion parameter at that energy. The same solution says where it stops. λ grows at short distance, and the formula diverges at the Landau pole μ₀ exp[16π²/(3λ(μ₀))], 7.3 × 10²² μ₀ for λ(μ₀) = 1, computed. But the formula is a one-loop result, and it carries λ past 8.4, where the quoted two-loop term of β is already a tenth of the one-loop term, 6.3 e-folds before the pole whatever λ(μ₀) is.
The pole marks the end of perturbation theory's reach; whether φ⁴ exists at all distances is the non-perturbative question of triviality, named here for later.
- Simple definition
- The running coupling λ(μ) is the φ⁴ coupling renormalised at scale μ, its μ-dependence fixed so that physical amplitudes do not depend on μ; at one loop it grows with μ as λ(μ₀)/[1 − (3λ(μ₀)/(16π²)) ln(μ/μ₀)].
- Example
- With λ = 1.00 at μ₀ = 100 GeV, 1/λ falls by 0.0437 per decade, so eight decades up, at 10¹⁰ GeV, 1/λ = 0.650 and λ = 1.54; the fixed-order one-loop value 1 + 0.350 = 1.35 is 12 per cent low.
Derived by separating μ dλ/dμ = 3λ²/(16π²). Equivalently 1/λ falls by 3/(16π²) = 0.0190 per e-fold, 0.0437 per decade, whatever λ(μ₀) is.
λ dimensionless in four dimensions; μ, μ₀ renormalisation scales in GeV (1 GeV = 1.602 × 10⁻¹⁰ J); ln(μ/μ₀) a pure number
Shown from the μ-independence of M: every coefficient follows from the one-loop β. Stopping after term n leaves out a fraction r(n+1) of λ(μ).
term n is (3/(16π²))ⁿ λ₀(n+1) lnⁿ(μ/μ₀), the most logarithms n loops can carry; the series converges for |r| below 1
Quoted from topics 6.07 and 6.08. Setting μ = Q leaves −M = λ(Q) + Cλ(Q)²: the coupling at the energy probed is the expansion parameter.
s, |t|, |u| all of order Q² ≫ m²; M dimensionless for 2 → 2 in four dimensions; C a pure number with no large log
Derived: where 1/λ reaches zero, and where r = 1, the edge of the leading-log series. Exponential in 1/λ(μ₀), so no finite order in λ(μ₀) can see it.
μL, μ₀ in GeV; 16π²/3 = 52.64, so λ(μ₀) = 1 gives μL = 7.3 × 10²² μ₀ and λ(μ₀) = 0.1 gives 4.0 × 10²²⁸ μ₀
Derived from the solution. Wherever perturbation theory is judged to end, the pole lies only 16π²/(3λb) e-folds further on, inside the region it cannot describe.
μb the scale where λ(μ) reaches a chosen breakdown value λb, independent of λ(μ₀); λb = 8.36 gives 6.30 e-folds, 2.73 decades
Quoted. The ratio of the terms is 17λ/(144π²) = 0.0120λ: 1.2 per cent at λ = 1 and 10 per cent at λ = 8.36; its zero at λ = 83.6 lies far past where it holds.
for (λ/4!)φ⁴ with one real scalar; the first two coefficients are the same in every mass-independent scheme
Integrate the flow: 1/λ falls linearly in ln μ
Natural units ħ = c = 1 and the metric (+, −, −, −) throughout, with ℒ = ½(∂φ)² − ½m²φ² − (λ/4!)φ⁴, in which Unit 1 showed λ to be dimensionless. Topic 6.08 derived β(λ) = μ dλ/dμ = 3λ²/(16π²) from the μ-independence of the bare coupling; it is used here, not re-derived. The equation separates, dλ/λ² = (3/(16π²)) d ln μ, and integrating from μ₀, where λ = λ₀, gives 1/λ₀ − 1/λ(μ) = (3/(16π²)) ln(μ/μ₀), so λ(μ) = λ₀/[1 − (3λ₀/(16π²)) ln(μ/μ₀)], derived. The first form is the clearer one: 1/λ is a straight line in ln μ with universal slope −3/(16π²) = −0.0190 per e-fold, or −0.0437 per decade, and λ₀ sets only its intercept. For λ₀ = 0.5, 1/λ starts at 2 and λ has doubled when 1/λ reaches 1, after 1/0.0437 = 22.9 decades, half the distance to where 1/λ reaches zero. Checks: every term is a pure number; μ = μ₀ returns λ₀; λ₀ → 0 gives no running, the free field; and to first order the formula is λ₀ + (3λ₀²/(16π²)) ln(μ/μ₀), the one-loop shift of topic 6.08.
One leading logarithm at every order
Expand in λ₀ with t = ln(μ/μ₀): λ(μ) = λ₀ + (3/(16π²))λ₀²t + (3/(16π²))²λ₀³t² + …, one term λ₀(n+1)tⁿ at each order, and since each loop integration supplies at most one logarithm, that is the most an n-loop term can carry. Why should loop diagrams obey this series? Write the high-energy amplitude as −M = λ + aλ²L + c₂λ³L² + …, with a = 3/(16π²), L = ln(Q/μ) and λ = λ(μ). It cannot depend on μ, and dλ/d ln μ = aλ² while dL/d ln μ = −1. At order λ³L the second term contributes 2aλ(dλ/d ln μ)L = 2a²λ³L and the third −2c₂λ³L, so c₂ = a², shown, and the same step at each order gives cₙ = aⁿ. The two-loop β and the non-log constants first enter at λ(n+2)Lⁿ, the next-to-leading tower. The tower is not a bubble chain: the s-channel chain alone gives (1/(16π²))ⁿ, a third of the one-loop log and a ninth of the two-loop one. Other diagrams supply the rest, and the running coupling sums them without drawing them.
Choose μ near √s, and λ(√s) becomes the expansion parameter
Topics 6.07 and 6.08 give the renormalised one-loop amplitude. In MS-bar, with s, |t| and |u| all of order Q² ≫ m², it reads −M = λ(μ) + (3λ(μ)²/(16π²)) ln(Q/μ) + Cλ(μ)², quoted, where C carries the angles and the s-channel cut but no large logarithm: at 90° with Q² = s, Re C = −7.39/(32π²) = −0.0234. Differentiate with respect to ln μ and the log term cancels the running, so −M is μ-independent through order λ², a check on β. Now take λ₀ = 0.30 at μ₀ = 10 GeV and probe Q = 10¹⁵ GeV. Kept at μ₀, the log term is r = 0.184 of the tree term and the fixed-order answer is 0.355. Run instead: λ(Q) = 0.30/(1 − 0.184) = 0.368, and with μ = Q the log term vanishes. The difference, 0.012, is the leading logs from two loops up, four times the constant 0.0234λ² = 0.003 that the one-loop calculation took care to keep. No single μ is required: any μ within a factor of a few of √s leaves the omitted terms free of large logarithms, and λ(√s) is what must be small.
The Landau pole: λ grows at short distance
1/λ reaches zero, and λ diverges, at ln(μL/μ₀) = 16π²/(3λ₀), derived. For λ₀ = 1 that is 52.6 e-folds or 22.9 decades, μL = 7.3 × 10²² μ₀, computed; for λ₀ = 0.1 it is 4.0 × 10²²⁸ μ₀. The dependence on λ₀ is exp[16π²/(3λ₀)], an essential singularity at λ₀ = 0 that no finite order in λ₀ can reproduce, and in the leading-log series the pole is r = 1, its radius of convergence. Restore units to see the scale: with μ₀ = 100 GeV and λ₀ = 1, μL = 7.3 × 10²⁴ GeV, more than five decades above the Planck mass of 1.22 × 10¹⁹ GeV, and ħc/μL = 0.197 GeV fm/(7.3 × 10²⁴ GeV) = 2.7 × 10⁻⁴¹ m. The coupling grows towards short distance. Towards long distance it falls, approaching 16π²/(3 ln(μ₀/μ)) once that logarithm is large, but the fall is physical only down to μ ≈ m: for momenta well below m the bubble's logarithm becomes ln(m²/μ²), independent of the momenta, and the effective coupling stops changing.
Perturbation theory fails before the pole
The pole was found by solving a one-loop equation, so it is only as reliable as that equation where λ is large. Quote the next term: β = 3λ²/(16π²) − (17/3)λ³/(16π²)², with first two coefficients that are the same in every mass-independent scheme. The second term is 17λ/(144π²) = 0.0120λ times the first: 1.2 per cent at λ = 1 and 10 per cent at λ = 8.36. Take that as the edge, λb = 8.36. From λ₀ = 1 the one-loop solution reaches it at 1.3 × 10²⁰ μ₀, 2.7 decades below the pole. In general the solution puts ln(μL/μb) = 16π²/(3λb), derived and independent of λ₀: 6.3 e-folds for λb = 8.36, and fewer for any looser criterion. So the last stretch before the pole lies where the equation that predicts it is not valid. The two-loop β sharpens the point: it vanishes at λ = 83.6, where the running would stop instead of diverging, but that zero lies even deeper in the region where neither truncation can be trusted. The pole is the end of the approximation, not a derived property of φ⁴.
What is quoted, and what is named for later
Three results frame this one, and none is derived here. The BPHZ theorem, quoted, states that the three counterterms of topic 6.06, δZ, δₘ and δλ, make every Green's function of φ⁴ finite at every order, so β exists to all orders and the leading-log argument above holds to all orders. Triviality, named, is the statement that the only φ⁴ theory valid down to arbitrarily short distances is the free one: long supported by lattice studies, and proved by Aizenman and Duminil-Copin in 2021 for the continuum limit of lattice-regularised φ⁴ in four dimensions, quoted; read that way, the Landau pole says φ⁴ is an effective theory below some cutoff. Asymptotic freedom, named, is the opposite sign: non-abelian gauge theories with few enough fermion species have β below zero, and the strong coupling, measured as αₛ = 0.118 at the Z mass, falls at short distance. No elementary φ⁴ coupling has been measured to run, but QED runs the same way, and its running is measured: 1/α is 137.036 at low energy and about 129 at the Z mass, 91.19 GeV, quoted; the QED units of this course compute the leptons' share of that fall and quote the hadrons'. The Wilsonian renormalisation group, which reads running as the integrating-out of short-distance modes, belongs to a later course.
Change one variable at a time
Make the relationship visible.
At n = 1 the dashed line is the fixed-order one-loop result, linear in the logarithm. Raise n: it closes on the solid curve fast where r is small and slowly near the pole, where r → 1. Halve λ(μ₀) and the pole moves twice as many decades away, yet λ still enters the shaded band first.
LANDAU POLE, log₁₀(μL/μ₀)22.9
LEADING-LOG RATIO r AT √s0.700
1/λ(√s), ZERO AT THE POLE0.300
SERIES TO ORDER n AT √s1.700
Live interpretationLANDAU POLE, log₁₀(μL/μ₀): 22.9. LEADING-LOG RATIO r AT √s: 0.700. 1/λ(√s), ZERO AT THE POLE: 0.300. SERIES TO ORDER n AT √s: 1.700
Catch the common trap
Explain before calculating.
In MS-bar, λ(μ₀) = 0.80 at μ₀ = 100 GeV. You want φφ → φφ at √s = 10¹⁴ GeV, with every scattering angle large, at leading order with the logarithms under control. Which coupling should replace λ in −M ≈ λ, and why?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyIn MS-bar, λ(μ₀) = 0.25 at μ₀ = 1.0 TeV. Use one-loop running to find λ at μ = 1.0 × 10¹⁶ GeV, compare it with the fixed-order value λ₀ + (3λ₀²/(16π²)) ln(μ/μ₀), check the limits, and give the distance that scale probes, in metres.
- Dimensions: λ is a pure number in four dimensions, and ln(μ/μ₀) = ln(10¹⁶ GeV/10³ GeV) = 13 ln 10 = 29.934 is one too, so every term below is dimensionless.
- Run: 1/λ(μ) = 1/λ₀ − (3/(16π²)) ln(μ/μ₀) = 4.0000 − 0.018998 × 29.934 = 4.0000 − 0.5687 = 3.4313, so λ(μ) = 0.2914.
- Fixed order: r = 3 × 0.25 × 29.934/157.91 = 0.1422 and λ₀(1 + r) = 0.2855. The gap, 0.0059, equals λ₀r²/(1 − r) = 0.25 × 0.02021/0.8578, the leading logs from two loops up.
- Limits: as λ₀ → 0 both results tend to λ₀ with no running, the free field; as μ → μ₀, r → 0 and they agree; and the gap starts at order λ₀³, as a two-loop term must.
- Distance: ħc/μ = 0.197327 GeV fm/(1.0 × 10¹⁶ GeV) = 1.97 × 10⁻¹⁷ fm = 2.0 × 10⁻³² m.
Answerλ(10¹⁶ GeV) = 0.291, 2.1 per cent above the fixed-order 0.286, at a distance scale of 2.0 × 10⁻³² m.
MediumIn MS-bar, λ(μ₀) = 0.60 at μ₀ = 200 GeV. For φφ → φφ at √s = 2.0 × 10¹¹ GeV, with s, |t| and |u| all of order s, take −M = λ(μ) + (3λ(μ)²/(16π²)) ln(√s/μ) and drop the constant term. Evaluate −M with μ = μ₀ and with μ = √s, find the two-loop leading log the first choice omits, and see how far −M moves as μ goes from √s/2 to 2√s.
- L₀ = ln(√s/μ₀) = ln(10⁹) = 20.723, so r = 3 × 0.60 × 20.723/157.91 = 0.2362. With μ = μ₀: −M = 0.60 × (1 + 0.2362) = 0.7417.
- With μ = √s the log vanishes and −M = λ(√s) = 0.60/(1 − 0.2362) = 0.7856. The two choices differ by 0.0438, 5.6 per cent of the answer.
- The first choice omits λ₀r² = 0.60 × 0.055798 = 0.03348 at two loops and λ₀r³/(1 − r) = 0.01035 from three loops up; together 0.04383, the whole difference.
- Scale variation: λ(2√s) = 1/(1/0.78556 − 0.018998 ln 2) = 0.79377, so −M = 0.79377 − 0.018998 × 0.79377² × ln 2 = 0.78548. At √s/2, λ = 0.77752 and −M = 0.77752 + 0.018998 × 0.77752² × ln 2 = 0.78548.
- Both lie about 8 × 10⁻⁵ below the value at μ = √s, because d(−M)/d ln μ = 2a²λ³ ln(√s/μ) at this order, which vanishes at μ = √s. A spread of 10⁻⁴ against 0.044 is the case for μ near √s, though it cannot see the constants of order λ³/(16π²) a two-loop calculation would add.
Answer−M = 0.742 with μ = μ₀ and 0.786 with μ = √s; the first omits 0.034 at two loops and 0.044 in all. For μ between √s/2 and 2√s, −M stays within 9 × 10⁻⁵ of 0.7856.
HardIn MS-bar, λ(μ₀) = 2.0 at μ₀ = 1.0 TeV. Find the one-loop Landau pole in GeV and as a distance. Then find where λ reaches 8.36, the value at which the quoted two-loop term of β = 3λ²/(16π²) − (17/3)λ³/(16π²)² is 10 per cent of the one-loop term, first with one-loop running and then by integrating the two-loop β exactly, and say what the comparison implies for the pole.
- Pole: ln(μL/μ₀) = 16π²/(3λ₀) = 157.91/6.0 = 26.32, so μL = 10³ GeV × e26.32 = 2.7 × 10¹⁴ GeV, below the Planck mass of 1.22 × 10¹⁹ GeV. As a distance, ħc/μL = 0.197327 GeV fm/(2.7 × 10¹⁴ GeV) = 7.3 × 10⁻¹⁶ fm = 7.3 × 10⁻³¹ m.
- One loop: with a = 3/(16π²) = 0.018998, ln(μb/μ₀) = (1/λ₀ − 1/λb)/a = (0.5000 − 0.1196)/0.018998 = 20.02, so μb = 5.0 × 10¹¹ GeV. The remaining 26.32 − 20.02 = 6.30 e-folds is 16π²/(3 × 8.36), the same for every λ₀.
- Two loops: with b = (17/3)/(16π²)² = 2.2724 × 10⁻⁴, write dλ/(aλ² − bλ³) = d ln μ and split 1/(λ²(a − bλ)) = 1/(aλ²) + b/(a²λ) + b²/(a²(a − bλ)). Integrating, ln(μ/μ₀) = (1/a)(1/λ₀ − 1/λ) + (b/a²) ln[λ(a − bλ₀)/(λ₀(a − bλ))].
- Numbers: b/a = 0.011962 and b/a² = 0.6296. λb/λ₀ = 4.180 and (1 − 0.02392)/(1 − 0.1000) = 1.0845, so the logarithm is ln 4.533 = 1.5115 and the correction is +0.9517. Then ln(μb/μ₀) = 20.97 and μb = 1.3 × 10¹² GeV, reached 0.95 e-folds, a factor 2.6 in μ, later than at one loop.
- Meaning: up to λ = 8.36 the two truncations differ by 4.8 per cent of the 20-e-fold run, so the running there is under control. Beyond it they part: one loop reaches its pole 6.3 e-folds on, while the two-loop β, which vanishes at λ = 83.6, makes λ approach 83.6 and never diverge. Neither holds for λ of order 10 and above, so the pole is a property of the one-loop truncation, not a prediction.
AnswerμL ≈ 2.7 × 10¹⁴ GeV, a distance of 7.3 × 10⁻³¹ m; λ = 8.36 is reached at 5.0 × 10¹¹ GeV at one loop and 1.3 × 10¹² GeV at two. Past λ ≈ 8 neither truncation holds, so the pole is not a prediction.