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Gauge theory and EFT

L3 · Chapter 4

How do you calculate the influence of a particle you cannot produce?

Sandbox Physics ·

A heavy particle can change light-particle scattering through virtual exchange even when the collision cannot produce it. Expanding its propagator at low energy replaces that exchange with local interactions. Each additional term retains more of the energy dependence.

A scattering process with one exchange channel

Take distinguishable light real scalars ϕ,χ\phi,\chi and a heavy real scalar SS of mass MM. The light fields are massless in this model. Set Lint=−S(g1ϕ2+g2χ2)/2\mathcal L_{\rm int}=-S(g_1\phi^2+g_2\chi^2)/2. Each gig_i has one power of mass, expressed in TeV in natural units.

Study ϕϕ→χχ\phi\phi\to\chi\chi at center-of-mass energy EE, with Mandelstam invariant s=E2s=E^2. These species and couplings give a single tree-level ss channel. Define the real amplitude below. “Full” means full within this tree model; loops and a resonance width are absent.

Afull(s)=g1g2M2−s=α1−x,α=g1g2M2,x=sM2\mathcal A_{\rm full}(s)=\frac{g_1g_2}{M^2-s}=\frac{\alpha}{1-x},\qquad \alpha=\frac{g_1g_2}{M^2},\quad x=\frac{s}{M^2}
The amplitude and both ratios are dimensionless. The scan stays below the pole, where additional physics would be needed.

Replace propagation with local interactions

For ∣x∣<1|x|<1, expand the denominator as a geometric series. The leading term retains a constant, the next term retains energy squared, and the next adds energy to the fourth power. Matching requires agreement with the full amplitude through the retained order.

Writing A=C0+C2s+⋯\mathcal A=C_0+C_2s+\cdots gives C0=g1g2/M2C_0=g_1g_2/M^2 and C2=g1g2/M4C_2=g_1g_2/M^4. The first is dimensionless and the second has inverse-mass-squared units. The first four-scalar operator here has dimension four; two additional derivatives raise the dimension to six. Count fields and derivatives before assigning an operator dimension.

AN(s)=α∑n=0Nxn,A0=α,A1=α(1+x)\mathcal A_N(s)=\alpha\sum_{n=0}^{N}x^n,\qquad \mathcal A_0=\alpha,\quad \mathcal A_1=\alpha(1+x)
The order labels count the energy expansion. Every displayed amplitude remains at tree level.
Afull−AN=αxN+11−x,Afull−ANAfull=xN+1\mathcal A_{\rm full}-\mathcal A_N=\frac{\alpha x^{N+1}}{1-x},\qquad \frac{\mathcal A_{\rm full}-\mathcal A_N}{\mathcal A_{\rm full}}=x^{N+1}
Multiply the finite series by its denominator: intermediate terms cancel, leaving this exact remainder as an independent analytic check.

Source: Aneesh V. Manohar

Eliminate the heavy field in the action

Group the light fields into J=(g1ϕ2+g2χ2)/2J=(g_1\phi^2+g_2\chi^2)/2. After integration by parts, the heavy-field terms are −S(M2+□)S/2−SJ-S(M^2+\Box)S/2-SJ. Solve for the classical stationary field and substitute it back. Expanding the resulting inverse differential operator gives local terms.

The heavy field follows the slowly varying light source. Every pair of derivatives keeps finer variations. Our convention is □=∂t2−∇2\Box=\partial_t^2-\nabla^2, which gives −s-s on a plane wave of total momentum squared ss. The alternating action expansion therefore agrees with the same-sign amplitude expansion.

Scl=−(M2+□)−1J,ΔLeff=12J1M2+□J=J22M2−J□J2M4+⋯S_{\rm cl}=-(M^2+\Box)^{-1}J,\qquad \Delta\mathcal L_{\rm eff}=\frac12J\frac{1}{M^2+\Box}J=\frac{J^2}{2M^2}-\frac{J\Box J}{2M^4}+\cdots
This is tree-level elimination. In this quadratic-heavy, linear-source model, exact Gaussian integration gives a light-field-independent determinant that only changes vacuum normalization. Light and mixed-loop matching need separate calculations.

A complete numerical example

Choose M=2 TeVM=2\,\mathrm{TeV}, α=0.4\alpha=0.4 and E=1 TeVE=1\,\mathrm{TeV}. Then x=0.25x=0.25 and the full amplitude is 0.533333…0.533333\ldots. Successive approximations give 0.40.4, 0.50.5 and 0.5250.525, with relative errors 25%25\%, 6.25%6.25\% and 1.5625%1.5625\%.

At lower energies, another term brings a large improvement. Near the heavy scale it helps less. Specify an acceptable error before choosing the order and maximum energy.

0.40.81.20.00.20.40.60.8
Vertical: amplitude A\mathcal AHorizontal: energy ratio E/ME/M
  • Full tree model
  • Leading order
  • Next order
  • Two corrections
Original ideal-reference plot from the chapter’s scalar model, with coupling ratio 0.4. The horizontal axis stops below the heavy pole. Curves are analytic predictions; observations appear in the linked experiment.
Analytic ideal-amplitude references for a 2 TeV mediator and coupling ratio 0.4. These are not acquired samples.
Energy ratio E/ME/MAfull\mathcal A_{\rm full}A0\mathcal A_0A1\mathcal A_1A2\mathcal A_2
0.250.250.4266670.4266670.4000000.4000000.4250000.4250000.4265630.426563
0.500.500.5333330.5333330.4000000.4000000.5000000.5000000.5250000.525000
0.750.750.9142860.9142860.4000000.4000000.6250000.6250000.7515630.751563

Matching scale and collision energy have different roles

Matching commonly uses a renormalization scale μ\mu near the heavy mass to align the full and effective theories. Collision energy EE specifies the physical process. These scales answer different questions.

The tree-level coefficients here contain no renormalization logarithms and have no running to vary. A “matching scale” control should not invent such a dependence. Chapter 6 calculates a separate published example with genuine quantum running. This model tests energy range, expansion order and omitted terms.

Try the reasoning in an experiment

Find where the approximation loses accuracy

  1. Use the exact remainder to predict the maximum energy ratios for a one-percent relative-amplitude tolerance at the first two orders.
  2. Drag the energy probe and compare the full model with the truncated series. Record where the tolerance is crossed.
  3. Hold mass and seed fixed and save synthetic amplitude observations. Fit first the lower-energy subset and then include the high-energy samples. Compare the fit shift with the sign of the omitted terms. The observations are real amplitudes with Gaussian noise.

Exercises and explanations

  1. How do the first two orders meet a one-percent relative-amplitude tolerance?

    Read the explanation

    Leading order needs x<0.01x<0.01, hence E/M<0.1E/M<0.1. The next order needs x2<0.01x^2<0.01, hence E/M<0.1≈0.316E/M<\sqrt{0.1}\approx0.316. Equality reaches exactly one percent.

  2. Double the mediator mass while holding the two dimensionful couplings fixed. What happens to the leading amplitude?

    Read the explanation

    It falls by a factor of four because the denominator quadruples. Holding the dimensionless coupling ratio fixed would give a different result. Specifying what remains fixed is essential.

  3. Does squaring a next-order amplitude give its squared magnitude consistently truncated at the same energy order?

    Read the explanation

    The square contains higher powers. Through order xx, expand α2(1+x)2\alpha^2(1+x)^2 as α2(1+2x)\alpha^2(1+2x). Keeping the unexpanded square is a different approximation that should be stated.

Sources and reproduced scope

Derivations, tables and the plot are authored for this course. The reading locations below separate source results from our teaching extensions. Numerical examples are model references or labeled synthetic observations. No experimental dataset or source figure is copied.

  1. David Tong, Gauge Theory (2018 lecture notes)

    Sections 2.1–2.1.3 (connections and Wilson loops), 2.7.3 (Higgs phase), 3.1.1–3.2.2 (two-dimensional anomaly and Ward identities), 4.2.1 (link endpoints) and 4.3.1–4.3.3 (lattice fermions). Printed pages 26–37, 105–108, 122–139, 207–210, 220–228.

  2. Aneesh V. Manohar, Introduction to Effective Field Theories, arXiv:1804.05863v1 (2018)

    Sections 3.5 and 4 (integrating out and power counting), 5.5–5.7 (matching), 5.10.1, PDF equations (5.51)–(5.60) (QCD running and operator mixing), and 6.2–6.3 (field redefinitions). The scalar-channel examples are derived for this course. The known-log boundary expansion checks truncation of the published running factors; it does not calculate finite hard matching.

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