A certain gluon scattering amplitude

Question

I am stuck with this process of calculating the tree-level scattering amplitude of two positive helicity (+) gluons of momentum say $p_1$ and $p_2$ scattering into two gluons of negative (-) helicity with momentum $p_3$ and $p_4$.

This is apparently $0$ for the diagram where one sees this process as two 3 gluon amplitudes with a propagating gluon (of say momentum $p$) and $p_1$ and $p_2$ are attached one each to the two 3 gluon amplitudes. I want to be able to prove this vanishing.

So let $p_2^+$ be with $p$ and $p_3^-$ and rest on the other 3 gluon vertex.

I am working in the colour stripped formalism. Let the Lorentz indices be $\rho$, $\sigma$ for the propagating gluon. And for the external gluons $p_1^+$, $p_2^+$, $p_3^-$, $p_4^-$ let $\nu, \lambda, \beta, \mu$ respectively be their Lorentz indices. Let the auxiliary vectors chosen to specify the polarizations of these external gluons be $p_4, p_4, p_1, p_1$ respectively. So the "wave-functions" of these four gluons be denoted as, $\epsilon^{+/-}(p,n)$ where $p$ stands for its momentum and $n$ its auxiliary vector and in the spinor-helicity formalism one would write,

1. $ \epsilon^{+}_\mu(p,n) = \frac{<n|\gamma_\mu|p]}{\sqrt{2}<n|p>} $

2. $\epsilon^{-}_\mu(p,n) = \frac{[n|\gamma_\mu|p>}{\sqrt{2}[p|n]} $

Hence I would think that this amplitude is given by,

$\epsilon^{-}_{\mu}(p_4,p_1)\epsilon_{\nu}^{+}(p_1,p_4)\epsilon_\lambda^+(p_2,p_4)\epsilon_\beta^-(p_3,p_1)\left( \frac{ig}{\sqrt{2}} \right)^2 \times \{ \eta^{\mu \nu}(p_4-p_1)^\rho + \eta^{\nu \rho}(p_1-p)^\mu + \eta^{\rho \mu}(p - p_4)^\nu\} \left ( \frac{-i\eta_{\rho \sigma}}{p^2}\right)\{ \eta^{\lambda \beta}(p_2-p_3)^\sigma + \eta^{\beta \sigma}(p_3-p)^\lambda + \eta^{\sigma \lambda}(p - p_2)^\beta \} $

One observes the following,

1. $\epsilon^{-}_\mu(k_1,n). \epsilon^{- \mu}(k_2,n) = \epsilon^{+}_\mu(k_1,n).\epsilon^{+\mu} (k_2,n) = 0$

2. $\epsilon^{+}_\mu(k_1,n_1).\epsilon^{-\mu}(k_2,n_2) \propto (1-\delta_{k_2 n_1})(1-\delta_{k_1,n_2})$

Using the above one sees that in the given amplitude the only non-vanishing term that remains is (upto some prefactors),

$\epsilon^{-}_{\mu} (p_4,p_1) \epsilon_{\nu}^{+}(p_1,p_4) \epsilon{_\lambda}^{+}(p_2,p_4)\epsilon_{\beta}^{-}(p_3,p_1) \left\{ \eta^{\nu}_{\sigma}(p_1-p)^\mu + \eta_\sigma^\mu(p - p_4)^\nu\right\}\times \{ \eta^{\lambda \beta}(p_2-p_3)^\sigma\}$

(..the one that is the product of the last two terms of the first vertex factor (contracted with the index of the propagator) and the first term from the second vertex factor..}

• Why is this above term zero? (..the only way the whole diagram can vanish..)

Answer 1

I was trying to nudge you in the right direction, but here's the explicit calculation. Focus on the vertex where gluons 1 and 4 meet. There you have a factor $\epsilon(p_1)_\nu \epsilon(p_4)_\mu \left(\eta^{\mu \nu} (p_4 - p_1)^\rho + \eta^{\nu\rho} (p_1 - p)^\mu + \eta^{\rho\mu} (p - p_4)^\nu\right)$. But, by construction, $\epsilon(p_1) \cdot \epsilon(p_4) = 0$, so the $\eta^{\mu \nu}$ term vanishes. In the second term, we use $p = -p_1 - p_4$ to note that we have a factor $(2 p_1 - p_4) \cdot \epsilon(p_4)$. But $\epsilon(p_4) \cdot p_4 = 0$ because gauge bosons are transverse, whereas $p_1 \cdot \epsilon(p_4) = 0$ by your choice of the reference spinor for gluon 4. So the second term is zero. The last term is zero in analogous fashion. So this vertex is zero and you don't need to think about the rest of the diagram.