research29 min studyLesson 683 in Physics

Subtract short-distance subgraphs before the overall divergence

Explain lower-order counterterm insertions and the locality of the remaining ultraviolet subtraction.

On this page

What you’ll understand

  • Identify nested, disjoint, and overlapping ultraviolet subgraphs.
  • Insert lower-order counterterms inside higher-order graphs.
  • Test whether the remaining overall divergence is local.

At two loops, subtracting the pole of the final integrated expression is not always the correct first operation. A divergent subgraph can sit inside a larger momentum-dependent structure, producing a pole multiplied by a nonlocal function. Lower-order counterterms must be inserted in that same larger structure. After the required subtractions, locality constrains the remaining overall ultraviolet divergence. We will isolate this logic with a simple product model and a second example that tests cancellation of a logarithm times a pole.

A subgraph carries its divergence into a larger graph

An ultraviolet-divergent subgraph arises when some loop momenta become large while the remainder is held fixed. Nested divergences have one such subgraph inside another; disjoint ones involve separate subgraphs; overlapping ones share lines without either containing the other. Each describes a different potentially singular region. A counterterm for a proper subgraph replaces that subgraph by its local ultraviolet approximation, while its external legs remain connected to the rest of the graph. Consequently the inserted contribution can depend nonlocally on the full graph’s external momenta.

This nonlocal dependence of an inserted graph is compatible with a local counterterm vertex. Locality constrains the replacement itself, not the complete surrounding propagation. A pole multiplied by ln(Q/μ) in an unsubtracted two-loop expression is therefore not permission to add a logarithmic momentum-dependent vertex to the Lagrangian. It is often evidence that a subdivergence has not yet been removed. One must also separate ultraviolet from infrared poles; dimensional regularization can display both as poles in the same ε, and a scaleless zero can hide their cancellation.

RΓ=(1−K)R′Γ,R′Γ=Γ+∑∅≠S(∏γ∈SCγ)Γ/S,Cγ=−KR′γR\Gamma=(1-K)R'\Gamma,\qquad R'\Gamma=\Gamma+\sum_{\varnothing\ne S}\left(\prod_{\gamma\in S}C_\gamma\right)\Gamma/S,\qquad C_\gamma=-K R'\gamma
Γ is a regulated graph, K extracts its chosen ultraviolet subtraction part, and S runs over sets of mutually disjoint proper divergent subgraphs; nested subtractions are already included recursively in each Cγ. Γ/S denotes their contraction. This recursive subtraction can equivalently be organized as a forest formula, with compatible regulators and infrared-safe UV extraction.↔ Scroll wide equations horizontally.

Forests prevent omissions and double counting

For disjoint subgraphs, subtracting either one alone is insufficient because their common ultraviolet region would be counted incorrectly. The forest expansion includes the product of their counterterms. Nested graphs require the inner subtraction before the outer remaining ultraviolet part is identified. Overlapping subgraphs cannot simply be put simultaneously into one forest, since their contractions are incompatible; the forest organization sums the permitted alternatives. This combinatorial structure is what replaces an informal instruction to subtract every visible loop independently.

After proper subdivergences are subtracted, the remaining overall ultraviolet divergence is a polynomial in external momenta and masses of bounded degree, under the assumptions of perturbative locality and power counting. It can be cancelled by an allowed local interaction. This need not mean there is only a simple 1/ε pole: higher poles can occur, with consistency relations tying them to lower-order data. Nonrenormalizable effective theories remain systematic when the required local operators are organized at a specified order in the low-energy expansion.

B(ϵ)=ϵ−1+L,B2−2ϵ−1B+ϵ−2=L2,L=ln⁡(Q/μ)B(\epsilon)=\epsilon^{-1}+L,\qquad B^2-2\epsilon^{-1}B+\epsilon^{-2}=L^2,\qquad L=\ln(Q/\mu)
This dimensionless product is an illustrative model with two independent logarithmic subgraph factors, not a universal two-loop Feynman integral. The last term is the double-counterterm forest contribution. Q and μ are positive energy scales.↔ Scroll wide equations horizontally.

The diagnostic is consistency across all embeddings

A lower-order counterterm is fixed once and then reused wherever that interaction appears. Its coefficient cannot be retuned separately for each larger diagram. This reuse is a strong check on signs, symmetry factors, and field-normalization conventions. Gauge identities further correlate counterterms in gauge theories. If a nonlocal ultraviolet pole survives after the complete set of insertions, likely causes include a missing graph, a mistaken combinatorial factor, a mixed convention, or infrared contamination; a new nonlocal subtraction is not the routine remedy.

The waves lab illustrates cancellation of a shared functional shape. Reset and set both controls to phase difference zero and amplitude ratio one, predicting summed amplitude two. Reset and set phase difference 180 degrees and ratio one, predicting cancellation of the entire profile. This resembles an insertion cancelling a nonlocal surrounding factor because both contributions contain it. These equal-frequency classical waves contain no subgraphs, momentum integrals, or ultraviolet regulator. Their exact cancellation illustrates an algebraic requirement, while locality and the forest rules supply the field-theory content.

Keep the double-counterterm contribution

Use the product model B=1/ε+L with ε=0.10 and L=2. Evaluate the bare product B², the two single-counterterm contributions together, and the double-counterterm contribution. Find the fully subtracted value.

  1. The one-loop factor is B=10+2=12, so the bare product is B²=144. It includes both a double pole and a logarithm times a pole.

  2. The two single insertions sum to −2×10×12=−240. Adding only these gives −96, which is not the finite answer.

  3. The double insertion contributes +10²=100. The total is 144−240+100=4, equal to L² and independent of ε in this model.

The result

The complete forest combination gives 4. Omitting the double insertion would fail to cancel the subdivergent regions consistently.

Explore the model

Interactive lab

Interactive lab · waves

Cancel a common nonlocal shape

Reset and set BOTH controls to phase difference 0° and amplitude ratio 1, predicting amplitude 2. Reset and set phase difference 180° and ratio 1, predicting cancellation at every plotted position. This is an algebraic analogy, with no loop or forest calculation.

Cancel a common nonlocal shapeTwo waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Current readings: Sum at x = 0: 0.866; Resultant amplitude: 1.732; Phase difference: 60°. All parameters are adjustable with the labeled controls below.−3.14−1.5701.573.14−3.4−1.701.73.4x · radiansAmplitude
SumFirst componentSecond component
Sum at x = 0
0.866
Resultant amplitude
1.732
Phase difference
60°
t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA local replacement inside a larger graph can cancel the same surrounding momentum-dependent structure.

Check your understanding

What is the appropriate response to a surviving ultraviolet pole proportional to ln(Q/μ) before subdivergence subtraction?

Try it yourself

A dimensionless regulated graph has pole part 2/ε²+(3L+5)/ε, with L=ln(Q/μ). The complete proper-subgraph insertions contribute −2/ε²−3L/ε−1/ε. Find the remaining pole and the overall minimal counterterm. Is the remainder local?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Combine double-pole and logarithmic-pole coefficients before examining the constant pole.

Compare with the solution

The double poles cancel, and the 3L/ε terms cancel. The remaining divergence is (5−1)/ε=4/ε, so the overall counterterm is −4/ε. This residue is independent of the external logarithm and is local for the assumed dimensionless vertex. A leftover L/ε term would instead signal an incomplete or inconsistent subtraction in this setup.

Take it with you

  • Subgraph counterterms must appear inside the same larger structures.
  • Forest terms handle nested and disjoint subtractions without double counting.
  • The remaining overall UV divergence is local, but need not contain only a simple regulator pole.

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