Tree-level matching by heavy-field elimination
Eliminate a specified heavy scalar and match local operators, keeping sign, factorial, and crossed-channel cancellations explicit.
On this page
The question for today
What you’ll understand
- Solve a heavy scalar equation perturbatively in derivatives.
- Translate a generated quartic term into a chosen coupling convention.
- Test an apparent derivative correction against an on-shell crossed-channel amplitude.
Tree-level matching determines local low-energy interactions by requiring agreement with a specified theory containing heavy fields. Solving the heavy field equation is often efficient, provided its inverse differential operator is expanded in the correct regime and its homogeneous high-energy solutions are excluded. A scalar example is enough to expose the main hazards: a sign error in completing the square, a missed factorial in a quartic coupling, and an operator that appears in the local action but cancels in a particular on-shell amplitude.
Specify the full toy theory and its boundary prescription
In Minkowski signature (+,−,−,−), take a real light scalar phi and a real heavy scalar H of mass M, with interaction . The parameter g has mass dimension one. Include any light-sector interactions in a separate Llight. After integration by parts, the H-dependent action is a quadratic form plus a linear source. Its classical equation fixes H in terms of the source using the inverse of Box plus M squared, with the scattering boundary prescription understood.
The relevant solution describes virtual heavy response to light fields. An independent oscillating solution of the homogeneous heavy equation is not retained because the EFT is not meant to describe an external on-shell H particle. Expanding the inverse differential operator is justified when derivatives acting on the source are small compared with M. Substitution back into the action gives half the source contracted with its inverse quadratic operator, not the full linear-source contribution by itself.
Translate the contact term before comparing amplitudes
If the light quartic convention is , the generated positive coefficient of phi to the fourth shifts lambda downward by three g squared divided by M squared. This is a threshold correction to a dimension-four coupling, not necessarily a higher-dimensional correction. Decoupling does not mean that all heavy dependence is accompanied by higher powers of external energy: renormalizable parameters are also matched between theories.
For identical light scalars, the heavy exchange amplitude contains s, t, and u channels. With the stated interaction, its contribution is a sum of three positive fractions with denominators M squared minus the corresponding invariant. At lowest order it equals three times g squared over M squared, matching the contact contribution. Leaving out two channels would miss a factor of three and would misidentify the quartic threshold correction even if the single-channel propagator expansion were algebraically correct.
An operator need not contribute to every matrix element
Expanding the crossed amplitude gives a two-derivative term proportional to s plus t plus u. It vanishes for the massless on-shell four-point process, although the derivative operator appears in the intermediate action. Integration by parts and the lower-order equations of motion account for this cancellation in a reduced basis. If leading light interactions are present, removing that operator can induce higher-multiplicity interactions; the cancellation is not a statement that the entire operator has no physical role in every process.
The first nonzero energy correction to this particular massless four-point amplitude is proportional to the sum of the squared invariants. Approaching any heavy-channel pole invalidates the local description. All coefficients in this lesson follow from the declared scalar toy theory and should not be promoted to universal signs or size relations for arbitrary ultraviolet completions. A different heavy spin, representation, or interaction changes the matching, while equality of the low-energy observables remains the defining test.
Work it through
A generated quartic and its coupling convention
Take M=100 GeV and g=20 GeV in the specified scalar model, with a light quartic coefficient λfull=1 in the convention −λφ⁴/4!. Find the generated coefficient of φ⁴, the quartic shift, and the constant heavy contribution to the low-energy massless four-point amplitude.
The positive Lagrangian coefficient is . Its sign follows from substituting the quadratic heavy-field solution.
Matching to gives . Thus the matched light quartic is .
Each heavy channel contributes at leading order. Summing all three gives +0.120 in the amplitude, consistent with the contact shift .
The generated φ⁴ coefficient is +0.0050, the matched quartic is 0.88, and the constant heavy-exchange amplitude contribution is +0.120.
The living notebook
Explore the model
Interactive lab · function
A heavy field responds quadratically to a light-field source
For each fresh comparison, Reset and set BOTH Coefficient a=0.50 and Position x. Compare x=0.20 and x=0.40, predicting y=0.020 and y=0.080. Up to an overall sign and units, this illustrates the leading heavy response proportional to φ² at fixed g/M². The instrument is y=ax², with no differential inverse, crossed channels, threshold pole, or matching procedure.
- 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 source dependence is simple only after the mass and derivative hierarchy have been specified.
Pause & reason
Check your understanding
Your turn
Try it yourself
For the same toy theory, let s/M²=0.040 and t/M²=u/M²=−0.020, with g²/M²=0.040. Compute the massless heavy contribution through terms quadratic in these invariant ratios. Identify the constant, linear, and quadratic pieces separately.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Expand each channel as 1+x+x². Use the on-shell sum before multiplying by the common factor g²/M².
Compare with the solution
The constants sum to three. The linear terms sum to . The quadratic terms give . Hence . Its leading constant is 0.120 and the retained energy-dependent correction is 0.000096. These are small invariant ratios in all channels. A single-channel expansion would contain a nonzero linear correction and would not reproduce this identical-particle amplitude or the reduced-basis on-shell cancellation.
Keep the idea
Take it with you
- Heavy-field elimination must retain the quadratic-action factor and its sign.
- Dimension-four threshold shifts require the same matching care as higher-dimensional operators.
- Crossing and equations of motion can remove an apparent correction from a specific on-shell amplitude.
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