research38 min studyLesson 692 in Physics

Operator dimensions and EFT power counting

Build dimensionally consistent interactions and translate canonical dimensions into a qualified amplitude expansion.

Before you begin

On this page

The question for today

What you’ll understand

Derive canonical field dimensions from kinetic terms.
Normalize Wilson coefficients with explicit powers of a heavy scale.
Keep amplitude, rate, coupling, and derivative expansions consistent.

An EFT may contain infinitely many symmetry-allowed local interactions, but a useful prediction requires only finitely many at a specified order. Power counting supplies that ordering. Canonical mass dimension is the first tool in a relativistic theory near a weakly coupled fixed point, although it is not the final answer in every regime. Operator normalization, external states, couplings, symmetry suppressions, and loop factors all determine how a term contributes to an actual observable.

Derive dimensions from the action

In four spacetime dimensions with natural units, the action is dimensionless and a Lagrangian density has mass dimension four. A derivative has dimension one. Requiring the scalar and fermion kinetic terms to have that dimension gives scalar dimension one and fermion dimension three halves. A gauge field strength has dimension two. These are canonical dimensions; anomalous scaling near an interacting fixed point is a separate issue and does not change how the Lagrangian's engineering units are balanced.

An operator of dimension d needs a coefficient of mass dimension four minus d. It is conventional to factor a heavy scale Lambda and write a dimensionless coefficient cd. This convention does not assert that cd equals one, or even that all coefficients have similar size. Couplings, loop suppressions, symmetry-breaking spurions, and normalization factors may be displayed explicitly. Comparing two coefficient tables requires checking which of these factors their authors absorbed into the operator definitions.

[\mathcal L]=4,\quad [\partial_\mu]=1,\quad [\phi]=1,\quad [\psi]=\frac32,\quad [F_{\mu\nu}]=2,\qquad \mathcal L=\mathcal L_{\leq4}+\sum_{d>4,i}\frac{c_i^{(d)}}{\Lambda^{d-4}}\mathcal O_i^{(d)}

Brackets denote canonical mass dimension in four spacetime dimensions. The ci(d) are dimensionless in this normalization, and Λ is an energy scale. Symmetry, integration-by-parts identities, and equation-of-motion redundancy determine the independent operator index i.↔ Scroll wide equations horizontally.

Power count a specified matrix element

A four-fermion operator has dimension six, so a typical relativistic two-to-two amplitude receives a correction proportional to $c_6E^2/\Lambda^2$ after external spinors supply the appropriate energy dependence. That statement is about a specified process with nonsingular kinematics and chosen normalization. A scalar six-field operator also has dimension six, but contributes first to different external multiplicities. The same canonical dimension does not give an identical correction to every observable.

Derivative power counting must also include masses treated as soft scales, possible background field amplitudes, and perturbative loop orders. In nonrelativistic systems energy and spatial momentum can scale differently; in chiral theories quark masses count as two derivatives. These are organized alternatives rather than failures of dimensional analysis. The central requirement is a declared small parameter and a consistent rule assigning every relevant diagram an order before deciding which terms to discard.

\mathcal A=\mathcal A_0+\delta\mathcal A_6+\delta\mathcal A_8+\cdots,\qquad |\mathcal A|^2=|\mathcal A_0|^2+2\operatorname{Re}(\mathcal A_0^*\delta\mathcal A_6)+|\delta\mathcal A_6|^2+2\operatorname{Re}(\mathcal A_0^*\delta\mathcal A_8)+\cdots

For the illustrated counting, δA6 is order Λ−2 and δA8 is order Λ−4. The square of the dimension-six contribution and interference with dimension eight enter at the same inverse-scale order. Coupling or helicity assumptions may further distinguish their sizes.↔ Scroll wide equations horizontally.

A rate needs its own consistent truncation

Squaring a truncated amplitude automatically generates higher-order terms. Keeping those terms can be a deliberate phenomenological choice, but it is not a complete calculation at that higher order unless the other contributions of equal size are controlled. In particular, a squared dimension-six contribution is not automatically more trustworthy than unknown dimension-eight interference. If leading interference vanishes by helicity or a symmetry, the bookkeeping should be reorganized and the assumption stated rather than hidden.

A very small leading amplitude can make a modest absolute correction into a large fractional rate change. The expansion parameter should therefore be assessed at the amplitude level together with the relevant observable normalization. A large measured deviation does not alone locate the heavy scale because the coefficient is also unknown. Conversely, a small fitted coefficient can reflect weak coupling or symmetry rather than a much higher mass. Dimensional analysis constrains units and scaling, while physical assumptions determine quantitative coefficient estimates.

Work it through

A small amplitude correction and a larger rate fraction

For an explicitly normalized toy amplitude, take real A0=0.20 and δA6=c6(E/Λ)² with c6=0.80, E=0.20 TeV, and Λ=2.0 TeV. Compute the dimension-six amplitude correction, its interference contribution to the squared amplitude, and the relative interference correction to the leading rate.

The ratio is E/Λ=0.10, so $\delta\mathcal A_6=0.80(0.10)^2=0.0080$ . All amplitudes in this example are dimensionless.
The leading squared amplitude is $0.20^2=0.0400$ . The interference is $2(0.20)(0.0080)=0.0032$ , using the stated real relative phase.
The fractional interference correction is $0.0032/0.0400=0.080$ , or eight percent. The additional square is 0.000064, but keeping it as a complete next-order prediction would require information about other terms of order Λ−4.

The result

The amplitude shift is 0.0080 and its interference adds 0.0032 to a leading squared amplitude of 0.0400, an eight-percent rate correction.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

Energy scaling for a dimension-six amplitude

For each fresh comparison, Reset and set BOTH Coefficient a=1 and Position x. Compare x=0.10 and x=0.20, predicting y=0.010 and y=0.040. Interpret x as E/Λ and y as a normalized dimension-six amplitude correction. The actual graph is y=ax²; it has no spinor states, loop factors, interference phases, or complete scattering rate.

y = ax²TangentSigned integral

y at chosen x: 0.608
Tangent slope · 2ax: 1.35
Signed area · ax³/3: 0.182

Coefficient a0.75Position x0.9

t = 0 s

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 forCanonical dimension gives a conditional energy scaling of a matrix element, not a universal correction to every process.

Pause & reason

Check your understanding

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Read the checkpoint explanation

Both |δA6|² and 2Re(A0*δA8) scale as Λ−4 in the stated expansion. Their relative sizes need additional physical assumptions.

Your turn

Try it yourself

A four-fermion interaction has coefficient $C=10^{-6}\,\mathrm{GeV^{-2}}$ . Verify its coefficient dimension, infer the scale Λ if the associated dimensionless coefficient is assumed to be one, and compute the normalized amplitude CE² at E=50 GeV and E=100 GeV.

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

Give me a hint

Four fermion fields contribute four times three halves to the operator dimension. The scale extraction uses an explicit unit-coefficient assumption, not a model-independent mass measurement.

Compare with the solution

The operator has dimension six, so C must have dimension minus two, as given. With $C=1/\Lambda^2$ , one gets $\Lambda=1000\,\mathrm{GeV}$ . The amplitudes are $10^{-6}(50)^2=0.0025$ and $10^{-6}(100)^2=0.0100$ . Doubling the energy increases this correction fourfold. If the dimensionless coefficient were different from one, the same measured C would correspond to a different inferred scale; the coefficient and heavy mass cannot be separated by this dimensional relation alone.

Keep the idea

Take it with you

Canonical dimensions fix coefficient units but not their numerical natural size.
Power counting must refer to specified external states and all relevant small parameters.
Consistent rate truncation includes every contribution at the claimed inverse-scale order.

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