research31 min studyLesson 725 in Physics

Tree-amplitude factorization tests

Check pole residues, internal-state phases and ordered channels while recognizing the contact information factorization cannot fix.

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What you’ll understand

  • Extract a channel residue without confusing it with a value near the pole.
  • Track the propagator i and opposite internal momentum assignments.
  • Diagnose contact ambiguities and incorrect residues in finite candidate functions.

A tree amplitude develops a pole when an intermediate particle can propagate on shell. Locality and unitarity relate that pole's residue to lower-point amplitudes with a sum over physical intermediate states. This provides a stringent test of a proposed expression, independent of how it was obtained. Yet matching every pole does not usually determine regular contact terms. A reliable test therefore distinguishes admissible singularities, their normalized residues, and analytic freedom invisible to those residues.

Residues isolate long-distance propagation

Let P be the sum of a consecutive subset of incoming momenta in an ordered tree. Near an isolated simple massless pole, multiply the amplitude by P squared and then take the limit. Terms regular in that channel vanish in the product. The remaining residue describes two lower-point processes joined by an on-shell state. This limit is more informative than substituting a small arbitrary number for P squared and comparing large values.

Use a convention in which the analytic partial amplitude retains its Feynman i factors and the internal scalar propagator is i divided by P squared. The internal momentum entering the left amplitude is minus P, while the right amplitude receives plus P. Helicity labels are opposite across the sewing, with crossing phases chosen consistently. Omitting the propagator i or independently rephasing only one lower amplitude can change a residue's sign or complex phase.

RP=lim⁡P2→0P2An=∑hAL(…,(−P)−h) i AR(Ph,…),An=RPP2+O((P2)0)R_P=\lim_{P^2\to0}P^2A_n=\sum_h A_L(\ldots,(-P)^{-h})\,i\,A_R(P^h,\ldots),\qquad A_n=\frac{R_P}{P^2}+O((P^2)^0)
This fixes a propagator-phase convention for A with its overall Feynman phases retained; compare all subamplitudes in that same convention. For massive exchange use P²−m² and the appropriate physical-state sum. A four-dimensional n-point amplitude has energy dimension 4−n, so RP has dimension 6−n.↔ Scroll wide equations horizontally.

Kinematics and helicity control whether a residue survives

An ordered channel must first be allowed by cyclic adjacency. Next, each lower-point amplitude must exist for the indicated species and helicities. A nominal pole can have zero residue because a required subamplitude vanishes. For complex three-point corners, the relevant angle or square branch must match the channel solution. These checks prevent counting every denominator in a possible diagram as a nonzero singularity of every helicity amplitude.

Spurious denominators can appear in intermediate representations, especially after solving momentum constraints or reducing a basis. Their residues must cancel in the complete expression if no corresponding physical channel exists. Tests should be made at generic points of the chosen pole surface, away from simultaneous additional singularities. Otherwise a special zero in another factor can conceal a wrong residue or make a legitimate pole appear absent.

Atest(s,t)=Rss+Rtt+C(s,t),lim⁡s→0sAtest=Rs,lim⁡t→0tAtest=RtA_{\mathrm{test}}(s,t)=\frac{R_s}{s}+\frac{R_t}{t}+C(s,t),\qquad \lim_{s\to0}sA_{\mathrm{test}}=R_s,\qquad \lim_{t\to0}tA_{\mathrm{test}}=R_t
The displayed model has independent nonzero-channel variables s,t with energy-squared units and a function C regular at both tested poles. It illustrates residue extraction, not a claimed arbitrary-helicity gauge amplitude. Limits hold with the other invariant generic; contact terms can remain undetermined.↔ Scroll wide equations horizontally.

Contact terms require information beyond pole matching

A local four-field interaction contributes a term regular in all exchange channels. Adding it changes scattering at generic kinematics while preserving every exchange residue. Factorization alone consequently cannot determine its Wilson coefficient. Derivative power counting, symmetry, large-complex-momentum behavior, matching to heavier physics, or independent measurements may supply that missing information. Which input is appropriate depends on whether the target is a renormalizable theory or an effective expansion.

Gauge invariance supplies another useful diagnostic: replacing an external polarization by its momentum should annihilate a properly assembled physical amplitude. A candidate can pass one numerical test and fail this structural one. At loops, poles coexist with cuts, ultraviolet renormalization, and infrared singularities; the simple meromorphic-tree analysis becomes only part of the problem. Passing all intended tree tests establishes consistency within their assumptions, rather than a general proof of a complete quantum theory.

A complex residue and an invisible contact constant

A four-point scalar channel has lower-point values AL=2i GeV and AR=3i GeV in the stated convention. Another channel has Rt=−4i GeV². For A=Rs/s+Rt/t+5i, find Rs and evaluate A at s=0.10 GeV², t=−2.0 GeV². Does the constant change either residue?

  1. Sew the first channel with the propagator numerator i: Rs=(2i)i(3i)=−6i GeV2R_s=(2i)i(3i)=-6i\,\mathrm{GeV^2}. Three factors of i give minus i; using only the product of the two subamplitudes would miss the phase.

  2. The exchange pieces at the chosen finite point are −60i and +2i. Adding the regular 5i gives −53i, a dimensionless four-point amplitude as required by the units of the residues and invariants.

  3. Multiplying by s before taking s to zero eliminates both the t-channel term and the constant, leaving −6i GeV². The analogous t limit gives −4i GeV². The finite value changes with the contact constant even though these residues do not.

The result

The first residue is −6i GeV² and the finite-point value is −53i. The 5i contact contribution is genuine additional information invisible to both simple-pole tests.

Explore the model

Interactive lab

Interactive lab · vectors

Track the phase of a sewn contribution

Treat a complex contribution of magnitude 2 as a planar phasor. Reset and set both controls to Angle −90° and Magnitude 2; predict (0,−2), representing −2i. For a fresh comparison, Reset and set both controls to Angle 90° and Magnitude 2; predict (0,+2). The sign reversal illustrates why sewing phases must be consistent.

Track the phase of a sewn contributionPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. 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 forComplex phases matter before coherent addition. This vector lab does not extract a pole, sum physical intermediate states, or evaluate the scalar example or a gauge amplitude.

Check your understanding

Two tree candidates have identical residues in every physical channel but differ by a local polynomial. Which conclusion is justified?

Try it yourself

The desired residues are Rs=−6i GeV² and Rt=−4i GeV². A candidate is Ac=(−6i GeV2/s)(1+s/t)A_c=(-6i\,\mathrm{GeV^2}/s)(1+s/t). Does it pass both channel tests? Can adding a constant contact term repair a failed residue?

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 the candidate before taking either limit, keeping the other invariant generic. A regular constant disappears after multiplication by the vanishing channel invariant.

Compare with the solution

The candidate expands to −6i GeV²/s−6i GeV²/t. Its s residue is correct, but its t residue is −6i GeV² rather than −4i GeV². Adding any constant changes neither residue, so it cannot repair this failure. A term +2i GeV²/t would repair the t residue but is another pole contribution, not a local contact term. The distinction identifies which part of the reconstruction is actually wrong.

Take it with you

  • An isolated pole residue is a channel limit with normalized internal-state sewing.
  • Allowed channels can vanish through helicity selection, while spurious poles must cancel.
  • Contact terms survive beyond factorization tests and need independent physical assumptions or input.

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