Skip to main content
Sandbox Physics
Scattering amplitudes

L3 · Chapter 6

Reconstruct a tree amplitude from its poles

Sandbox Physics ·

A tree amplitude is a rational function of complex momenta. Along a path preserving mass shells and total momentum, its poles mark on-shell intermediate particles. If the contribution at infinity vanishes, residues supplied by lower-point amplitudes reconstruct the original answer.

Complex momenta make three-point amplitudes useful

Three real massless momenta summing to zero must be parallel. All pair invariants vanish. Complex momenta allow angle brackets to vanish with square brackets nonzero, or the reverse, preserving nontrivial three-point information.

A three-point amplitude with two negative helicities uses angle brackets; two positive helicities use square brackets. These are the recursion seeds. At a pole, a larger tree splits into two smaller trees, which can be reduced until only three-point objects remain.

A3(1−,2−,3+)=i⟨12⟩3⟨23⟩⟨31⟩,A3(1+,2+,3−)=−i[12]3[23][31]A_3(1^-,2^-,3^+)=i\frac{\langle12\rangle^3}{\langle23\rangle\langle31\rangle},\qquad A_3(1^+,2^+,3^-)=-i\frac{[12]^3}{[23][31]}
The first expression uses the complex branch with vanishing square brackets; the second uses vanishing angle brackets. The real three-point limit needs separate care.

Source: Lance J. Dixon

Move two legs while preserving every constraint

For the four-point (−−++)(--++) example, shift the dotted spinor of leg one and the undotted spinor of leg four by a complex parameter zz. Each momentum remains a spinor outer product, so its determinant stays zero. The same four-vector is added to one momentum and removed from the other.

The complex path is a calculation tool. Its complex energies are not detector events. Returning to zero parameter recovers the amplitude at the original real kinematics.

λ~^1=λ~1−zλ~4,λ^4=λ4+zλ1\widehat{\widetilde\lambda}_1=\widetilde\lambda_1-z\widetilde\lambda_4,\qquad\widehat\lambda_4=\lambda_4+z\lambda_1
All other spinors remain fixed. This is a good shift for negative-helicity leg one and positive-helicity leg four.
k^1=k1−zq,k^4=k4+zq,qαα˙=λ1αλ~4α˙,q2=q⋅k1=q⋅k4=0\widehat k_1=k_1-zq,\quad\widehat k_4=k_4+zq,\quad q_{\alpha\dot\alpha}=\lambda_{1\alpha}\widetilde\lambda_{4\dot\alpha},\qquad q^2=q\cdot k_1=q\cdot k_4=0
The shifts cancel. Orthogonality preserves each squared mass for every complex parameter.

Source: Lance J. Dixon

Worked example with one finite pole

Insert the shift into the four-point expression. Its numerator stays fixed, as does ⟨41⟩\langle41\rangle. Only ⟨34⟩\langle34\rangle becomes linear in the parameter. The calculation reduces to the residue of a simple rational function.

Collect the fixed factors in CC. Solve the denominator for its zero. The residue of A4(z)/zA_4(z)/z there follows by multiplying by distance from the pole and taking the limit, giving C/(z∗⟨31⟩)=−C/⟨34⟩C/(z_*\langle31\rangle)=-C/\langle34\rangle.

Cauchy’s theorem adds the residue at zero, this finite-pole residue and the contribution at infinity. Here A4(z)A_4(z) falls inversely with the parameter, so infinity contributes zero. The residue at zero is A4(0)A_4(0), giving the reconstruction below.

C=i⟨12⟩3⟨23⟩⟨41⟩,A4(z)=C⟨34⟩+z⟨31⟩,z∗=−⟨34⟩⟨31⟩C=\frac{i\langle12\rangle^3}{\langle23\rangle\langle41\rangle},\qquad A_4(z)=\frac{C}{\langle34\rangle+z\langle31\rangle},\qquad z_*=-\frac{\langle34\rangle}{\langle31\rangle}
Assume the displayed brackets are nonzero and the original configuration avoids a physical pole. Special configurations follow by continuity.
A4(0)=−Res⁡z=z∗A4(z)z=C⟨34⟩=i⟨12⟩4⟨12⟩⟨23⟩⟨34⟩⟨41⟩A_4(0)=-\operatorname*{Res}_{z=z_*}\frac{A_4(z)}{z}=\frac{C}{\langle34\rangle}=i\frac{\langle12\rangle^4}{\langle12\rangle\langle23\rangle\langle34\rangle\langle41\rangle}
This is a complete residue check of the known four-point function. General recursion replaces the residue by the product of two three-point subamplitudes.

Obtain residues from smaller amplitudes

The pole also satisfies P^122=0\widehat P_{12}^2=0. The amplitude factorizes there into a left three-point (−−+)(--+) object and a right three-point (−++)(-++) object. Their internal helicities are opposite. Apply the preceding chapter’s pole relation to each channel and divide by the shifted propagator’s slope to obtain BCFW recursion.

Each pair of subamplitudes is evaluated at its own channel’s complex pole. The connecting propagator uses the original unshifted channel momentum. Multiplicity decreases in each subproblem. Four points has one planar partition; higher points require all allowed partitions and intermediate helicities.

An(0)=∑allowed channels P∑h=±AL(zP,P^h)iP2AR(zP,−P^−h)+B∞A_n(0)=\sum_{\text{allowed channels }P}\sum_{h=\pm} A_L(z_P,\widehat P^h)\frac{i}{P^2}A_R(z_P,-\widehat P^{-h})+B_\infty
The infinity contribution can be set to zero only after establishing decay for the chosen shift.

Source: Lance J. Dixon · R. Britto

A bad shift loses a boundary contribution

Reverse the choice: shift the undotted spinor of leg one and the dotted spinor of leg four. The numerator ⟨12⟩4\langle12\rangle^4 generally grows as the fourth power, with only one growing denominator factor, so the amplitude behaves as z3z^3. Finite-pole residues alone now miss a boundary term.

A stated theory, correct factorization and controlled infinity behavior support recursion together. Higher-derivative contact interactions can preserve pole positions while changing a polynomial part of the amplitude. Their information appears in the boundary behavior.

Try the reasoning in an experiment

Compare a four-point residue with the direct result

  1. Choose a nonsingular angle and record the diagram sum as a complex number. Insert spinor products for that same configuration into this chapter’s one-pole expression.
  2. Calculate the pole, residue and zero-parameter value step by step. Compare the complete complex amplitudes.
  3. Repeat at another angle. Explain why the complex pole moves but the reconstruction procedure stays the same.

Exercises and explanations

  1. Add a constant dd to a rational function. Do its finite-pole residues change? Can those residues uniquely reconstruct the new function?

    Read the explanation

    The residues are unchanged but the zero-parameter value increases by the constant. Infinity behavior has changed. Finite-pole data cannot determine the answer without its boundary information.

  2. Why must the shifted external legs stay on shell as well as conserving momentum?

    Read the explanation

    The recursion uses on-shell external-state amplitudes and their factorization. Off-shell legs introduce additional gauge-dependent structures and require corresponding off-shell objects and residue relations.

Sources and reproduced scope

Worked examples and schematic figures are authored for this course. Numerical values are calculated model references. The published comparison recalculates a specified analytic CDD curve; it does not copy experimental data, source artwork or the paper’s numerical optimization points. Source locations and reproduced scope are listed below.

  1. Lance J. Dixon, A brief introduction to modern amplitude methods, arXiv:1310.5353v1 (2013)

    Section 2, equation (5), color decomposition; section 3.1, equations (14), (19), (21)–(27), spinors and all-outgoing conventions; section 3.3, equations (41)–(46), polarizations; section 4.1, equations (55)–(57), soft factors; section 4.3, equations (68)–(70), MHV and collinear factors; sections 4.5 and 5.1, equations (74)–(84), complex three-point seeds and recursion. Our conserving scan and four-point residue example are derived extensions.

  2. R. Britto, F. Cachazo, B. Feng and E. Witten, Direct Proof Of Tree-Level Recursion Relation In Yang-Mills Theory, hep-th/0501052v2; Physical Review Letters 94, 181602 (2005)

    Section 2 introduces null-momentum spinors, constructs the shift in equation (2.1), and derives recursion in equation (2.7); section 3 controls large-parameter behavior; section 4 proves the MHV construction. The course keeps Dixon’s overall amplitude phase and specializes the original shift to legs one and four. Its soft scan uses the MHV function as input and derives the conserving real-momentum path separately.

FeedbackSomething broken, confusing, or missing on this page?Tell us
What kind of feedback?

Sent with this page’s address, its Lab ID, the site language and your device type. Please leave out personal details. Feedback is kept for up to 12 months.Privacy policy