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research32 min studyLesson 724 in Physics

BCFW complex shifts and boundary terms

Preserve nullness and momentum conservation under a complex shift and retain the contour contribution at infinity unless it is justified to vanish.

Before you begin
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The question for today

What you’ll understand

  • Check the on-shell conditions of a two-leg spinor shift.
  • Locate shifted factorization poles.
  • Use the residue theorem with a correctly signed boundary contribution.

BCFW recursion reconstructs suitable tree amplitudes from simpler on-shell amplitudes by exploring a one-complex-parameter family of momenta. The shift is designed to preserve every external mass and total momentum. Finite poles then represent internal factorization channels. The remaining question is the behavior at infinity: residues determine the amplitude only when the large-contour contribution vanishes or is supplied independently. This condition belongs to the theory and chosen shift, rather than following from momentum conservation alone.

Shift two spinors without leaving the mass shell

Choose two distinct legs i and j, add z times the undotted spinor of j to that of i, and subtract z times the dotted spinor of i from that of j. Each shifted momentum is still an outer product and hence null. Their changes are equal and opposite, so the total incoming sum is preserved. The complex parameter is dimensionless, and the shift vector has momentum units.

This deformation is analytic kinematics, not a laboratory acceleration of two particles. Dotted and undotted spinors must remain independent. Requiring them to stay conjugate would usually destroy the construction. A channel containing exactly one shifted leg has an invariant linear in z because the shift vector is null; a channel containing both or neither remains unchanged. The moving-channel root identifies where an internal line becomes on shell.

λ^i=λi+zλj,λ~^j=λ~j−zλ~i,q=λjλ~i,p^i=pi+zq,p^j=pj−zq,q2=q⋅pi=q⋅pj=0\widehat\lambda_i=\lambda_i+z\lambda_j,\qquad \widehat{\widetilde\lambda}_j=\widetilde\lambda_j-z\widetilde\lambda_i,\qquad q=\lambda_j\widetilde\lambda_i,\qquad \widehat p_i=p_i+zq,\quad \widehat p_j=p_j-zq,\quad q^2=q\cdot p_i=q\cdot p_j=0
Unwritten spinors are fixed. The equations use the previous null-matrix convention and all-incoming arrows. z is dimensionless and q has energy units. Distinct generic shifted legs avoid degenerate special configurations.↔ Scroll wide equations horizontally.

Cauchy's theorem keeps a boundary term

At a fixed tree multiplicity, a local perturbative amplitude continued under this shift is meromorphic in z. Integrate the amplitude divided by z around a large counterclockwise circle. The pole at zero contributes the unshifted amplitude; finite nonzero poles contribute their residues. Defining the large-circle integral as the boundary term gives a reconstruction formula with a minus sign in front of the finite residue sum.

When the amplitude tends to zero sufficiently rapidly, that boundary integral vanishes and factorization residues give ordinary recursion. A constant contact term has no finite pole yet contributes at infinity, so it is an immediate counterexample to residue-only reconstruction without a growth assumption. Individual diagrams can grow faster than their gauge-invariant sum, making diagram counting useful but insufficient unless the required cancellations are controlled.

P^2(z)=P2+2zP⋅q,z∗=−P22P⋅q,A(0)=B∞−∑z∗≠0Res⁡z=z∗A(z)z,B∞=lim⁡R→∞12πi∮∣z∣=RA(z)z dz\widehat P^2(z)=P^2+2zP\cdot q,\quad z_*=-\frac{P^2}{2P\cdot q},\qquad A(0)=B_\infty-\sum_{z_*\ne0}\operatorname*{Res}_{z=z_*}\frac{A(z)}{z},\quad B_\infty=\lim_{R\to\infty}\frac{1}{2\pi i}\oint_{|z|=R}\frac{A(z)}{z}\,dz
The channel formula is for the +zq orientation; a channel carrying −zq has the corresponding minus sign. The contour is counterclockwise. The boundary definition presumes the indicated limit exists; the ordinary recursion requires B∞=0, with finite residues evaluated in one consistent factorization convention.↔ Scroll wide equations horizontally.

Constructibility depends on the interaction and the shift

For Yang–Mills trees, suitable helicity choices and shift orientations have favorable large-z behavior, while an oppositely oriented mixed-helicity choice can fail. Higher-derivative interactions often worsen growth. One must specify which spinors are shifted before quoting a good-helicity rule, since exchanging dotted and undotted shifts changes that rule. A vanishing boundary result proved for one theory cannot simply be transferred to an effective theory with additional contact vertices.

Finite residues sew lower-point amplitudes with an internal-state sum, compatible complex three-point branches, and the prescribed propagator phase. Color ordering restricts which partitions appear. At loop level integrated amplitudes have branch cuts and infrared divergences, so this elementary rational-tree contour argument does not apply unchanged. Loop-integrand recursion and rational-part recursion require additional hypotheses. Establishing a useful recursion therefore means checking its analytic object, allowed poles, large-parameter behavior, and normalization together.

Work it through

A contact term survives the large circle

As a dimensionless meromorphic model of a shifted amplitude, take F(z)=3/(z−2)+1. Compute the finite residue of F(z)/z, its boundary B∞, and F(0). What would a boundary-free reconstruction incorrectly predict?

  1. The only nonzero finite pole is z=2. Its residue in F/z is the numerator 3 divided by the remaining factor z evaluated there, giving 3/2. The regular constant contributes no residue at that pole.

  2. At large z the rational pole term decays and F tends to one. The large-circle integral of F/z therefore gives B∞=1. This is precisely the information omitted by examining finite factorization poles alone.

  3. The contour formula gives F(0)=1−3/2=−1/2F(0)=1-3/2=-1/2, matching direct substitution. Setting the boundary to zero would give −3/2, an error of one from the unseen contact term.

The result

The finite residue is 3/2, the boundary is 1, and the correct value is −1/2. The example isolates the contour mechanism without pretending that an arbitrary rational function is a complete gauge amplitude.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

Polynomial growth has no finite poles

Inspect the real slice of the polynomial z². Reset and set both controls to Coefficient a=1 and Position x=1; predict y=1. For a fresh comparison, Reset and set both controls to a=1 and x=2; predict y=4. Such growth is a reminder that finite poles alone cannot specify all analytic functions.

Polynomial growth has no finite polesThis 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. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
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 forThe quadratic lab is neither a complex contour plot nor a BCFW evaluator. It illustrates polynomial freedom; the actual shifted amplitude's boundary must be calculated or constrained analytically.

Pause & reason

Check your understanding

What additional fact permits ordinary boundary-free BCFW recursion?

Your turn

Try it yourself

Let G(z)=2/(z+1)−4/(z−2), with dimensionless z. Reconstruct G(0) from its nonzero finite residues, checking the boundary instead of assuming it. Separately locate the zero of a shifted channel P²(z)=(6+3z) GeV².

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

Give me a hint

For each simple pole of G/z, divide the original numerator by the pole location. Inspect G at large z, and solve the channel's linear equation.

Compare with the solution

The residues are −2 at z=−1 and −2 at z=2. G decays as 1/z, so B∞=0 and G(0)=−(−2−2)=4, also obtained directly. The channel becomes null at z=−2. Its root location is a kinematic condition, whereas the residues and boundary determine a function's reconstruction; the two kinds of information should not be confused.

Keep the idea

Take it with you

  • A reciprocal two-leg spinor shift preserves nullness and total incoming momentum.
  • Finite residues reconstruct a tree amplitude only after accounting for the large contour.
  • Theory, helicity orientation, contact terms and loop analytic structure limit constructibility.

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