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I am calculating the cross-section for quark gluon to quark gluon scattering in the spinor helicity formalism. This process has contributions from the Mandelstam channels $s$, $t$ and $u$.

Using the usual brute-force Feynman calculus, I obtain for the average over initial and sum over final states for the $u$-channel

\begin{equation}\tag{1} \frac{1}{4 N_{C}\left(N_{C}^{2}-1\right)} \sum_{\text{color}}\sum_{\text{spin}}\left|\mathcal{M}_{u}\right|^{2}=-g^{4} \frac{4}{9} \frac{s}{u}, \end{equation}

where I take $N_C=3$. I believe that this is correct.

Performing the same calculation in the spinor-helicity formalism, I find that the average over initial and sum over final states gives

\begin{align} &\frac{1}{4 N_{C}\left(N_{C}^{2}-1\right)} \sum_{\text{color}}\sum_{h_{1}, h_{2}, h_{3}, h_{4}=+,-}\left|\mathcal{M}_{u}\left(1^{h_{1}} 2^{h_{2}} 3^{h_{3}} 4^{h_{4}}\right)\right|^{2}\\\tag{2} &\qquad=-g^{4} \frac{4}{9}\bigg( \ \frac{us}{t^2}+\frac{s^3}{ut^2}\bigg), \end{align}

where $h_i$ denotes the helicity of particle $i$.

Now, one should find the same squared amplitude for the process at hand, using either the usual Feynman calculus or spinor-helicity techniques.

My question is the following. Since Eq. (1) $\neq$ (2), have I already made a mistake in my calculation, or is there a chance that the contributions from the other channels may conspire to yield the correct squared amplitude for the entire process, even though Eq. (1) $\neq$ (2)?

In other words, should I expect to find that the squared amplitude for each channel, calculated using Feynman calculus, is identical to the squared amplitude for the same channel calculated in the spinor helicity formalism? Or is the equality only true for the squared amplitude for the full process, which is the sum of the squared amplitudes for each channel, plus cross-terms?

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  • $\begingroup$ what exactly do u mean by spinor helicity formalism calculation? Are you using the BCFW relations? If this is the case, then you would Not expect the amplitude to agree in a channel. This is in fact the whole point of these calculation. what would agree is the u channel residue, and in this case, it does. So maybe, you are right. $\endgroup$
    – Anonjohn
    Oct 28 '20 at 20:54
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I have since resolved my question, after calculating the correct final result using the usual Feynman techniques as well as the spinor helicity formalism.

It is indeed possible that one can obtain a particular result for the squared amplitude for one Mandelstam channel using Feynman calculus, and a different result for that same channel in the spinor helicity formalism while still obtaining the correct average over initial and sum over final states of the squared amplitude for the full process. For example, there exists a choice of reference spinors associated with each polarization vector such that the $t$-channel amplitude (and hence it's squared amplitude) for $qg \rightarrow qg$ vanishes identically. However, using Feynman techniques, the $t$-channel's average over initial and sum over final states does not vanish.

In short:

My question is the following. Since Eq. (1) ≠ (2), have I already made a mistake in my calculation, or is there a chance that the contributions from the other channels may conspire to yield the correct squared amplitude for the entire process, even though Eq. (1) ≠ (2)?

You did not make a mistake! You were correct a month ago! Indeed, the contributions from the other channels do conspire to yield the correct squared amplitude for the entire process, even with Eq. (1) ≠ (2).

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