One-loop evolution sums an infinite tower of logarithms
Derive a leading-log solution and distinguish resummation from adding unknown fixed-order information.
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The question for today
What you’ll understand
- Identify the combined expansion parameter involving a large logarithm.
- Recover a leading-log tower from the one-loop RG solution.
- Separate matching accuracy, logarithmic accuracy, and residual scale dependence.
A small coupling does not guarantee a useful fixed-order prediction when it multiplies a large logarithm of separated scales. The renormalization group can determine an infinite set of those logarithms from a few low-order coefficients. This is resummation: a controlled reorganization of known logarithmic information, not a substitute for missing matching constants or nonperturbative physics. We will derive an elementary tower and then evolve a coefficient whose anomalous dimension generates another set of leading logarithms.
A large scale ratio can invalidate naive fixed-order counting
Take an illustrative asymptotically free coupling with β(g)=−bg² and b positive. Integrating from μ to a hard scale Q gives g(Q)=g(μ)/(1+bg(μ)L), where L=ln(Q/μ). Expanding this solution gives g−bg²L+b²g³L² and higher terms. At the order g to the power n+1, the highest logarithm L to the power n is fixed by repeated one-loop running. These terms can compete when bgL is order one even though both endpoint couplings remain perturbatively small.
The logarithm appears because widely separated momentum scales contribute to the same fixed-order expression. Choosing μ close to Q removes that particular explicit large logarithm from a hard matching function, while evolving from a reference scale relates the input coupling to g(Q). Exact μ independence is the reason these two operations fit together. It does not imply the observable is independent of Q. If a process has several independent large ratios, a single choice of μ cannot remove all logarithms, and an appropriate factorization with multiple evolution equations may be needed.
An anomalous dimension produces a multiplicative evolution factor
Let a Wilson coefficient obey dlnC/dlnμ=cg in the same coupling convention. Dividing by the beta equation gives dlnC/dg=−c/(bg), so the coefficient ratio is a power of the running-coupling ratio with exponent −c/b. Expanding that power generates logarithms determined by the one-loop anomalous dimension and beta function. The sign follows the specified coefficient equation; an operator whose product with C is scale independent evolves with the opposite anomalous dimension.
The initial coefficient must still be found by matching at a suitable high scale. One-loop running alone does not determine its nonlogarithmic one-loop constant, and solving an evolution equation exactly does not make the whole calculation exact. Leading-log, next-to-leading-log, and fixed-order accuracy describe related but distinct information. Threshold matching must be applied when active fields change. Near an infrared strong-coupling scale, the same formal one-loop expressions cease to be perturbatively controlled even if a compact formula continues to return a number.
Resummation organizes known terms without certifying every error
A useful check expands the resummed answer back to the fixed order that was explicitly calculated. The logarithmic coefficients must agree. One can then add finite matching terms without double counting the already-resummed expansion, using a consistent matched prescription. Residual renormalization-scale dependence starts beyond the retained accuracy, but an accidentally flat scale curve is not proof that every omitted constant or power correction is small. Perturbative, logarithmic, and EFT power-counting errors should be assessed in their respective domains.
The function lab displays the local negative beta function behind this example. Reset and set both controls to a=−1 and x=0.20, predicting y=−0.040. Reset and set a=−1 and x=0.10, predicting y=−0.010. Assign x to g and y to β=−g²: the flow rate is negative and its magnitude decreases as the coupling weakens. The lab draws a quadratic rate rather than integrating over scale. It has no logarithm or matching control, so the resummed coupling must be obtained from the differential equation above.
Work it through
Compare a logarithmic truncation with its RG sum
Take b=2, g(μ)=0.10, and L=ln(Q/μ)=3. Find the one-loop RG-resummed coupling and the expansion through g²L. Also calculate the next leading-log term to diagnose the truncation.
The combined parameter is bgL=2×0.10×3=0.60. The resummed one-loop value is 0.10/(1+0.60)=0.0625.
The expansion through the first logarithm gives g−bg²L=0.10−2×0.01×3=0.0400. This differs substantially even though g itself is small.
The next leading-log term is b²g³L²=4×0.001×9=0.0360, moving the partial sum to 0.0760. The alternating sizeable terms show why this logarithmic tower should be summed.
The retained RG solution is 0.0625; the first and second logarithmic truncations give 0.0400 and 0.0760. Resummation improves the organization of this known tower.
The living notebook
Explore the model
Interactive lab · function
Check the local asymptotically free rate
Reset and set BOTH controls to a=−1 and x=0.20, predicting y=−0.040. Reset and set a=−1 and x=0.10, predicting y=−0.010. Assign y=−g²; this displays a rate and does not integrate running or resum a series.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. 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 forThe resummed logarithms come from integrating the RG equation, while the local rate weakens toward small coupling.
Pause & reason
Check your understanding
Your turn
Try it yourself
For the coefficient equation above, take c/b=1/2, g(Q)/g(μ)=1/4, and C(μ)=3 after factoring out any engineering dimension. Find C(Q), and explain how the associated operator matrix element must change if their product is RG invariant.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
The coefficient exponent is negative. Its paired matrix element must supply the reciprocal evolution factor.
Compare with the solution
C(Q)/C(μ)=(1/4)⁻¹ᐟ²=2, so C(Q)=6. For a single RG-invariant product C⟨O⟩, the operator matrix element changes by the reciprocal factor 1/2 between the same scales. Running one factor while leaving the other fixed would create spurious scale dependence in the physical prediction.
Keep the idea
Take it with you
- The relevant fixed-order parameter can be g times a large logarithm.
- RG evolution sums specified logarithmic towers while leaving matching information to be calculated.
- A resummed expression still has perturbative and EFT validity limits.
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