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Consider the $s$-channel mediated top quark production process
$$ d + \overline d \rightarrow t + \overline t$$

Using the Feynman rules for QCD, the amplitude contains a color factor $$[c^\dagger _{\overline d} ~t^\mu ~c_d][c^\dagger _{t} ~t_\mu ~c_{\overline t}] $$

where $t^\mu$ are the generators of the $SU(N)$ color group and summation over $\mu$ is implicit. To evaluate the cross section $\sigma$, one has to sum over final colors and average over initial colors. One gets

$$ \sigma = {1\over N^2} \sum_{initial} \sum_{final} [c^\dagger _{\overline d} ~t^\mu ~c_d][c^\dagger _{t} ~t_\mu ~c_{\overline t}][c^\dagger _{\overline d} ~t^\nu ~c_d]^*[c^\dagger _{t} ~t_\nu ~c_{\overline t}]^* $$

My question is, how does one proceed from here? The answer has a term proportional to ${N^2 -1 \over N^2}$, and I can only account for the $N^2$ in the denominator.

PS: My understanding is limited to what is discussed in Griffiths' book. I have no background in QFT/QED/QCD. Please mention sources if possible.

Consider the $s$-channel mediated top quark production process
$$ d + \overline d \rightarrow t + \overline t$$

Using the Feynman rules for QCD, the amplitude contains a color factor $$[c^\dagger _{\overline d} ~t^a ~c_d][c^\dagger _{t} ~t^a ~c_{\overline t}] $$

where $t^a$ are the generators of the $SU(N)$ color group and summation over a is implicit. To evaluate the cross section $\sigma$, one has to sum over final colors and average over initial colors. One gets

$$ \sigma \propto {1\over N^2} \sum_{initial} \sum_{final} [c^\dagger _{\overline d} ~t^a ~c_d][c^\dagger _{t} ~t^a ~c_{\overline t}][c^\dagger _{\overline d} ~t^b ~c_d]^*[c^\dagger _{t} ~t^b ~c_{\overline t}]^* $$

My question is, how does one proceed from here? The answer has a term proportional to ${N^2 -1 \over N^2}$, and I can only account for the $N^2$ in the denominator.

PS: My understanding is limited to what is discussed in Griffiths' book. I have no background in QFT/QED/QCD. Please mention sources if possible.

Edit: Many have suggested that I use $[c^\dagger _{\overline d} ~t^a ~c_d][c^\dagger _{t} ~t_a ~c_{\overline t}] $ (note Einstein's convention) but I have not seen this in Griffith' book. He has used superscripts for both indices. Also, I have correctly changed the color index to latin.

Consider the s-channel mediated top quark production
$$ d + \overline d \rightarrow t + \overline t$$

Using Feynman rules for QCD, the invariant amplitude (M) is contains a color factor given as : $$ f = [c^\dagger _{\overline d} ~t^\mu ~c_d][c^\dagger _{t} ~t^\mu ~c_{\overline t}] $$

where $t^\mu$ are generators of SU(N) colour group. The summation over $\mu$ is implied. To evaluate cross section ($\sigma$), one has to 'sum over final, and average over initial color'.

$$ \sum f f^* = {1\over N^2} \sum_{initial} \sum_{final} [[c^\dagger _{\overline d} ~t^\mu ~c_d][c^\dagger _{t} ~t^\mu ~c_{\overline t}][c^\dagger _{\overline d} ~t^\nu ~c_d]^*[c^\dagger _{t} ~t^\nu ~c_{\overline t}]^* $$

My question is, how does one proceed from here? The answer has a term proportional to ${N^2 -1 \over N^2}$, and I can only account for $N^2$ in the denominator.

PS: My understanding is limited to what is discussed in Grifiths book. I have no background in QFT/QED/QCD. Please mention source if possible.

Consider the $s$-channel mediated top quark production process
$$ d + \overline d \rightarrow t + \overline t$$

Using the Feynman rules for QCD, the amplitude contains a color factor $$[c^\dagger _{\overline d} ~t^\mu ~c_d][c^\dagger _{t} ~t_\mu ~c_{\overline t}] $$

where $t^\mu$ are the generators of the $SU(N)$ color group and summation over $\mu$ is implicit. To evaluate the cross section $\sigma$, one has to sum over final colors and average over initial colors. One gets

$$ \sigma = {1\over N^2} \sum_{initial} \sum_{final} [c^\dagger _{\overline d} ~t^\mu ~c_d][c^\dagger _{t} ~t_\mu ~c_{\overline t}][c^\dagger _{\overline d} ~t^\nu ~c_d]^*[c^\dagger _{t} ~t_\nu ~c_{\overline t}]^* $$

My question is, how does one proceed from here? The answer has a term proportional to ${N^2 -1 \over N^2}$, and I can only account for the $N^2$ in the denominator.

PS: My understanding is limited to what is discussed in Griffiths' book. I have no background in QFT/QED/QCD. Please mention sources if possible.

Consider the s-channel mediated top quark production
$$ d + \overline d \rightarrow t + \overline t$$

Using Feynman rules for QCD, the invariant amplitude (M) is contains a color factor given as : $$ f = [c^\dagger _{\overline d} ~t^\mu ~c_d][c^\dagger _{t} ~t^\mu ~c_{\overline t}] $$

where $t^\mu$ are generators of SU(N) colour group. The summation over $\mu$ is implied. To evaluate cross section ($\sigma$), one has to 'sum over final, and average over initial color'.

$$ \sum f f^* = {1\over N^2} \sum_{initial} \sum_{final} [[c^\dagger _{\overline d} ~t^\mu ~c_d][c^\dagger _{t} ~t^\mu ~c_{\overline t}][c^\dagger _{\overline d} ~t^\nu ~c_d]^*[c^\dagger _{t} ~t^\nu ~c_{\overline t}]^* $$

My question is, how does one proceed from here? The answer has a term proportional to ${N^2 -1 \over N^2}$, and I can only account for $N^2$ in the denominator.

PS: My understanding is limited to what is discussed in Grifiths book. I have no background in OFT/QED/QCD. Please mention source if possible.

Consider the s-channel mediated top quark production
$$ d + \overline d \rightarrow t + \overline t$$

Using Feynman rules for QCD, the invariant amplitude (M) is contains a color factor given as : $$ f = [c^\dagger _{\overline d} ~t^\mu ~c_d][c^\dagger _{t} ~t^\mu ~c_{\overline t}] $$

where $t^\mu$ are generators of SU(N) colour group. The summation over $\mu$ is implied. To evaluate cross section ($\sigma$), one has to 'sum over final, and average over initial color'.

$$ \sum f f^* = {1\over N^2} \sum_{initial} \sum_{final} [[c^\dagger _{\overline d} ~t^\mu ~c_d][c^\dagger _{t} ~t^\mu ~c_{\overline t}][c^\dagger _{\overline d} ~t^\nu ~c_d]^*[c^\dagger _{t} ~t^\nu ~c_{\overline t}]^* $$

My question is, how does one proceed from here? The answer has a term proportional to ${N^2 -1 \over N^2}$, and I can only account for $N^2$ in the denominator.

PS: My understanding is limited to what is discussed in Grifiths book. I have no background in QFT/QED/QCD. Please mention source if possible.