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  Some more questions about the BCFW reduction

+ 3 like - 0 dislike

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..)
This post has been migrated from (A51.SE)
asked Mar 6, 2012 in Theoretical Physics by Anirbit (585 points) [ no revision ]
retagged Apr 19, 2014 by dimension10
First question: Yes. Second question: No. Third question: try to integrate $A(z)/z$ in two ways by contour deformation.

This post has been migrated from (A51.SE)
@Sidious Lord Can you kindly add some more explanation to your comment above?

This post has been migrated from (A51.SE)

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