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A certain gluon scattering amplitude

+ 2 like - 0 dislike
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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..)
This post has been migrated from (A51.SE)
asked Mar 20, 2012 in Theoretical Physics by user6818 (955 points) [ no revision ]
retagged Apr 19, 2014 by dimension10
@Matt Reece Thanks for your comment. I guess the point you are raising is now clarified with me specifying the polarization vectors ($\epsilon^{+/-}(p,n)$) that I had in mind. I have added their explicit expressions. There are $3$ colour ordered diagrams contributing to this scattering process, the one whose expression I have written above is one of the two of those 3 which vanish. It would be very helpful if you can explain as to why is the final expression of mine zero (or if there is something wrong about the expression itself!)

This post has been migrated from (A51.SE)
@Matt Reece Thats exactly what I have clarified in the third paragraph. My choice for the reference spinors correspond to momenta $p_4, p_4, p_1, p_1$ for particles with momentum-and-helicity $p_1^+$, $p_2^+$, $p_3^-$, $p_4^-$ respectively. I guess that clarifies the complete meaning of the amplitude that I have written down.

This post has been migrated from (A51.SE)
Choosing a gauge means choosing a reference spinor $\left|n\right>$ (or $\left|n\right]$, depending on helicity). Often you want to choose different particles to have the same reference spinor, or maybe the reference spinor to come from another momentum in the process, to make as many terms as possible zero.

This post has been migrated from (A51.SE)
It's not physical to talk about a diagram vanishing. Diagrams aren't gauge invariant, amplitudes are. It only makes sense when you specify a gauge, i.e. a choice of polarization vectors. (I haven't read any of your details, but maybe thinking about the general fact will be useful to you.)

This post has been migrated from (A51.SE)
I can't see why my last equation has gotten garbled! It would be great if someone can rectify that.

This post has been migrated from (A51.SE)

1 Answer

+ 0 like - 0 dislike

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.

This post has been migrated from (A51.SE)
answered Mar 24, 2012 by Matt Reece (1,630 points) [ no revision ]

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