# How to deal with colour indices on spinors

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## Main Question or Discussion Point

I want to calculate transition amplitudes in QCD for processes like $q(k)q^\prime(p)\rightarrow q(k^\prime)q^\prime(p^\prime)$, where $q,q^\prime$ are quarks. However, I am unsure what to do with the colour indices of the quark spinors upon squaring the matrix element. For the sake of illustration let us consider the scattering mentioned above. The Feynman diagram including the colour indices $i,j,k,l$ and the adjoint indices of the gluon $a,b$ is shown below. The spin indices are suppressed.

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
$$\delta_{im}\delta_{jn}\delta_{ko}\delta_{lp}t^a_{ij}t^{a\dagger}_{mn}t^b_{kl}t^{b\dagger}_{op} =t^a_{ij}t^{a\dagger}_{ij}t^b_{kl}t^{b\dagger}_{kl} =t^a_{ij}t^{a\ast}_{ji}t^b_{kl}t^{b\ast}_{lk}\\ =tr[t^at^{a\ast}]tr[t^bt^{b\ast}] =C^2_F(N^2-1)^2.$$