# 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..)
• 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

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.