# Some more questions about the BCFW reduction

This question is a continuation of this previous question of mine and I am continuing with the same notation.

One claims that one can actually split this $n$-gluon amplitude such that there is just a single gluon propagating between two $n-$point amplitudes and the $p_{n-1}(z)^{-}$ and $p_n(z)$ are on two sides. Define $q_{i,n-1}(z) = p_i + p_{i+1} + ...+ p_{n-1}(z)$ and define $h$ to be the helicity of the gluon when propagating out of the left amplitude. This is summarizied in saying that the following expression holds,

$A(1,2,..,n,z) = \sum _{i=1} ^{n-3} \sum _ {h = \pm 1} A^L(p_i,p_{i+1},..,p_{n-1}(z),q^h_{i,n-1}(z)) \frac{1}{q_{i,n-1}(z)^2}A^R(p_n(z),p_1,p_2,...,p_{i-1},q^{-h}_{i,n-1}(z))$

• Is there a "quick" explanation for the above split and why the propagating gluon has to flip helicity? (..it seems to be way of putting in the helicity conservation at high-energies but I can't make it very precise..)

• In the above split shouldn't the sum be from $i=2$ since one can't get lower than $3$-gluon vertices on either side?

Now one can apparently write the momentum squared of the propagator in the following way, $q_{i,n-1}(z)^2 = q^2_{i,n-1} - z[p_{n-1}|\gamma_\mu q^\mu_{i,n-1}|p_n>$, where $q_{i,n-1}(0) = q_{i,n-1}$ and then apparently using the previous expression of $A(1,2,..,n,z) = \sum _{i} \frac{R_i}{(z-z_i)}$ one can re-write the amplitude as,

$A(1,2,..,n) = \sum _{i=1} ^{n-3} \sum _ {h = \pm 1} A^L(p_i,p_{i+1},..,p_{n-1}(z_i),q_{i,n-1}^h(z_i)) \frac{1}{q_{i,n-1}^2}A^R(p_n(z_i),p_1,p_2,...,p_{i-1},-q_{i,n-1}^{-h}(z_i))$

where $z_i$ is such that $q_{i,n-1}(z_i)^2 = 0$

• I would like to know how the above expression for $A(1,2,..,n)$ was obtained. (..it looks like Cauchy's residue theorem but I can't make it completely precise..)
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First question: Yes. Second question: No. Third question: try to integrate $A(z)/z$ in two ways by contour deformation. –  Sidious Lord Mar 13 '12 at 7:39
@Sidious Lord Can you kindly add some more explanation to your comment above? –  user6818 Mar 20 '12 at 19:33

1.) The helicities flip because one assumes all particles incoming (or outgoing depending on your conventions). I.e. if a particle propagates it will have "the wrong direction" for the vertex it propagtes to. Hence one has to reverse its direction which corresponds to flipping its helicity.

2.) Don't worry about the indicies there -- the important thing is to remember that one cannot go below three-point indices. Even if the sum runs over wrong indices the "wrong" contributions would vanish because there is not 2-point amplitude.

3.) Yes, it is indeed Cauchy's theorem. The physical amplitude is obtained from the complex amplitude by a contour integral around $z=0$. Pushing the boundary to infinity one finds that the contour integral is just a sum over residues for finite $z$ plus a residue at $z=\infty$. This is standard complex calculus. If the amplitude falls off fast enough as a function of $z$ the latter can be neglected and the physical amplitude is just

$A(0)=\sum_{residues}(\text{finite }z)$.

But by the usual factorization properties we know that the amplitude has to factorize at the poles of $z$ into two subamplitudes with fewer legs connected by a propagator.

I hope that helps you :)

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