research33 min studyLesson 92 in Physics

Wilsonian matching and power counting

Integrate out a Gaussian heavy scalar, determine the effective interaction's sign, and measure the accuracy of its derivative expansion.

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What you’ll understand

  • Eliminate a heavy field by completing a Euclidean square and identify its nonlocal kernel.
  • Expand that kernel in external momentum and check operator dimensions.
  • Keep derivative truncation, loop corrections, and infrared matching distinct in an error budget.

A low-energy theory earns predictive power by reproducing specified measurements. Matching determines coefficients by comparing a more resolved description with its effective replacement. We will use a Gaussian heavy scalar, where completing a square exposes both the sign and the derivative expansion before any loop calculation is needed.

Integrate out a field with the sign visible

Work in four-dimensional Euclidean spacetime with a real light field ϕ\phi, a heavy field HH, and interaction gHϕ2/2gH\phi^2/2. Both fields have mass dimension one, so gg has dimension one. The heavy quadratic operator is A=M2−∂2A=M^2-\partial^2, whose momentum eigenvalue is positive M2+Q2M^2+Q^2. Complete the square in HH with source J=gϕ2/2J=g\phi^2/2. Gaussian integration produces a light-field term with a minus sign, because the completed square subtracts half of JA−1JJA^{-1}J.

The determinant is independent of ϕ\phi in this deliberately simple Gaussian model and only changes the vacuum normalization. Extra interactions involving HH would make further terms possible. The negative quartic contribution here does not assert that an isolated light potential is globally stable: the full model must contain appropriate stabilizing interactions and be used around its stated vacuum. Euclidean action coefficients also should not be copied into Minkowski scattering rules without translating conventions.

12HAH+JH=12(H+A−1J)A(H+A−1J)−12JA−1J,ΔSE=−g28∫d4x ϕ21M2−∂2ϕ2\frac12 HAH+JH=\frac12(H+A^{-1}J)A(H+A^{-1}J)-\frac12JA^{-1}J,\qquad \Delta S_E=-\frac{g^2}{8}\int d^4x\,\phi^2\frac{1}{M^2-\partial^2}\phi^2
The integral notation suppresses the second coordinate of the inverse differential operator. The sign is the Euclidean effective-action sign for the stated interaction, before choosing an operator basis.↔ Scroll wide equations horizontally.

A local series needs a small kinematic ratio

In a channel with Euclidean transfer QQ, expand the positive kernel g2/(M2+Q2)g^2/(M^2+Q^2) as a geometric series. This isolates a dimensionless coefficient g2/M2g^2/M^2 followed by corrections in Q2/M2Q^2/M^2. In position space the first term multiplies ϕ4\phi^4, a dimension-four operator. The next can be written using derivatives on ϕ2\phi^2 and has dimension six, so its coefficient has mass dimension minus two. Integration by parts can change its appearance but cannot change a matched observable.

Every relevant external invariant must be small compared with the heavy scale. Low total energy alone is insufficient if a different channel carries a large transfer or if another nearby threshold was omitted. A geometric expansion at small spacelike momentum has an explicitly calculable remainder in this example. More general effective theories estimate the next allowed order using power counting, symmetries, and coefficient sizes. An unexpectedly enhanced coefficient is information to investigate, rather than permission to claim a universal tiny error.

K(Q2)=g2M2+Q2=g2M2[1−Q2M2+O ⁣(Q4M4)]K(Q^2)=\frac{g^2}{M^2+Q^2}=\frac{g^2}{M^2}\left[1-\frac{Q^2}{M^2}+O\!\left(\frac{Q^4}{M^4}\right)\right]
K is the positive heavy-exchange kernel; its contribution to the Euclidean action retains the separate overall minus sign above. This expansion applies for absolute Q squared below M squared.↔ Scroll wide equations horizontally.

Matching is not the same operation as running

Matching fixes coefficients at a chosen scale by equating the same low-energy quantities in two descriptions. Renormalization-group running then transports those coefficients between scales within the effective theory. When light loops occur on both sides, common infrared contributions must cancel in the matching difference; otherwise long-distance physics would be incorrectly counted as a new short-distance coefficient. Operators related by field redefinitions can trade coefficients, so the chosen basis and renormalization prescription belong in a reproducible result.

Our Gaussian elimination is exact for the heavy field in the stated action, but evaluating the remaining light-field theory at tree level is an additional approximation. Report separately which derivative terms are retained, which loop orders are evaluated, and how measured inputs were fixed. This bookkeeping makes the effective theory useful for precision work: increasing one order has a stated purpose and test. It also exposes the boundary where a heavy degree of freedom must be restored rather than approximated by more local terms.

Compare two local approximations with the exact kernel

Use M = 10, g = 2, and Euclidean Q = 1 in consistent mass units. Compare the constant and two-term approximations to K.

  1. The exact positive kernel is 4/(100 + 1) = 4/101, approximately 0.03960396. This is a kernel value, not the signed quartic coefficient of the action.

  2. The constant approximation is 4/100 = 0.04. Its difference from the exact value divided by the exact value is 0.01, or 1%.

  3. Retain the first derivative correction: 0.04 times (1 − 0.01) = 0.0396. The relative error is 0.0001, or 0.01%, as predicted by the next power of the small ratio.

The result

A channel ratio of Q squared/M squared = 0.01 controls the local expansion here. The known remainder checks the power counting, while the Euclidean action still carries its derived negative coefficient.

Explore the model

Interactive lab

Interactive lab · spectral

Compare resolved detail with a truncated representation

Increase or decrease N in the finite sine sum Σ sin(nx−t)/nᵖ while holding p fixed. It provides an inspectable example of retaining a limited set of spatial scales. Removing Fourier terms is not the same calculation as integrating out H, and this lab neither matches Wilson coefficients nor estimates an effective-theory loop error.

Compare resolved detail with a truncated representationA finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Current readings: Sum at x = π/2: 0.867; Smallest wavelength: 2π / 5; Highest-mode coefficient: 0.2. All parameters are adjustable with the labeled controls below.01.573.144.716.28−2.56−1.2801.282.56x · radiansFinite sine sum
Finite sumHighest included mode
Sum at x = π/2
0.867
Smallest wavelength
2π / 5
Highest-mode coefficient
0.2
t = 0 s

A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. 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 truncated representation needs a stated domain. Effective matching additionally determines the interactions that replace the omitted degrees of freedom.

Check your understanding

What is the correct mass dimension of the coefficient multiplying a dimension-six operator in four dimensions?

Try it yourself

Keep M = 10 and g = 2 but raise Q to 5. Compute the constant and two-term kernels and compare their relative errors with the exact value.

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 expansion parameter is now 25/100. Evaluate the denominator before computing percentage differences.

Compare with the solution

The exact kernel is 4/125 = 0.032. The constant approximation 0.04 is 25% too large. The two-term value is 0.04 times 0.75 = 0.03, which is 6.25% too small. The errors follow the first and second powers of 0.25 in this geometric example; they do not justify using the same local expansion through a heavy-particle threshold.

Take it with you

  • Completing the heavy-field square fixes a sign that guessing local operators cannot determine.
  • Power counting requires operator dimensions, coefficient sizes, and all relevant small kinematic ratios.
  • Matching, running, derivative truncation, and loop truncation solve distinct parts of the prediction problem.

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