- #1

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When I calculate the matrix element ##i\mathcal{M}##, I get the following result:

$$i\mathcal{M}=\left[\bar{q}^\prime_i(k^\prime)igt^a_{ij}\gamma_\mu q^\prime_j(k)\right]

\frac{-ig^{\mu\nu}\delta_{ab}}{t}

\left[\bar{q}_k(p^\prime)igt^b_{kl}\gamma_\nu q_l(p)\right]

=ig^2t^a_{ij}t^a_{kl}\frac{1}{t}\left[\bar{q}^\prime_i(k^\prime)\gamma_\mu q^\prime_j(k)\right]\left[\bar{q}_k(p^\prime)\gamma^\mu q_l(p)\right],$$

where ##t=q^2=(p-p^\prime)^2##.

Now, upon squaring the matrix element, I am not sure what happens with the colour indices ##i,j,k,l##. Since they are summed over, one should assume that I would have to introduce another set of colour indices ##m,n,o,p## and one new adjoint index ##b## since ##a## is also summed over, as one does for the spacetime indices ##\mu,\nu## on the Dirac matrices. However, the standard result in most of the text books is that somehow we end up with the traces ##tr[t^at^{a\ast}]=C_F(N^2-1)## and the same for ##b##, while the spinors ##q,q^\prime## somehow lose their colour indices, i.e. the squared matrix element (not averaged over spin and colour) looks like

$$|\mathcal{M}|^2=\frac{g^4}{t^2}\times C^2_F(N^2-1)^2

\times tr\left[\gamma_\mu q^\prime(k)\bar{q}^\prime(k)\gamma_\nu q^\prime(k^\prime)\bar{q}^\prime(k^\prime)\right]

\times tr\left[\gamma^\mu q(p)\bar{q}(p)\gamma^\nu q(p^\prime)\bar{q}(p^\prime)\right].$$

It is precisely the mathematics by which the spinors lose their colour indices in-between these steps, that interests me!

Now, if I take the matrix element from above and introduce said new set of colour indices after squaring the matrix element, I get this result:

$$

|\mathcal{M}|^2=\frac{g^4}{t^2}\times t^a_{ij}t^{a\dagger}_{mn}t^b_{kl}t^{b\dagger}_{op}

\times tr\left[\gamma_\mu q^\prime_j(k)\bar{q}^\prime_n(k)\gamma_\nu q^\prime_i(k^\prime)\bar{q}^\prime_m(k^\prime)\right]

\times tr\left[\gamma^\mu q_l(p)\bar{q}_p(p)\gamma^\nu q_k(p^\prime)\bar{q}_o(p^\prime)\right].$$

So, how do I get from there to the textbook result above? I see that somehow we must get factors like ##\delta_{im},\delta_{jn},\delta_{ko},\delta_{lp}##, such that

$$\delta_{im}\delta_{jn}\delta_{ko}\delta_{lp}t^a_{ij}t^{a\dagger}_{mn}t^b_{kl}t^{b\dagger}_{op}

=t^a_{ij}t^{a\dagger}_{ij}t^b_{kl}t^{b\dagger}_{kl}

=t^a_{ij}t^{a\ast}_{ji}t^b_{kl}t^{b\ast}_{lk}\\

=tr[t^at^{a\ast}]tr[t^bt^{b\ast}]

=C^2_F(N^2-1)^2.$$