7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Are the $t^\mu$ defined somewhere in this book? As far as I know the generators of $SU(N)$ have no Lorentz indices. $\endgroup$ – Photon Apr 14 '14 at 11:38
  • $\begingroup$ @Photon My bad! I should not have used greek indices for color index. I have corrected them to latin ones after the edit. $t^a$ are generators of SU(N) color group. In case of SU(3) you can relate then to Gell-Mann matrices as $t^a = {\lambda^a \over 2}$. $\endgroup$ – negligible_singularity Apr 14 '14 at 15:50
3
$\begingroup$

Part of your confusion here probably comes from your notation. Usually we reserve the index $\mu$ for spacetime. The generators of $SU(N)$ are more commonly labelled with Latin indices $t^a$. See for example here.

We can split the amplitude into two parts, according to whether they concern color or kinematics. You are just interested in the color part. Each quark comes with a color "polarization", i.e. a normalized basis vector $c$ in the fundamental representation of $SU(N)$. The quark-gluon vertices just give color factors of $t^a$.

Now the cross section is given by the modulus squared of the amplitude so we have

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

Now using that the color polarizations are normalized we can reinterpret the sum as a double trace term

$$ \sigma \propto {1\over N^2}Tr(t^a t^b)Tr(t^a t^b) $$

Now we can use the color algebra relation

$$Tr(t^a t^b) = \frac{1}{2}\delta^{ab}$$

to conclude that

$$\sigma \propto \frac{(N^2 - 1)}{N^2}$$

As for a reference, you could do worse than read Part III of Peskin and Schroeder's book. It's quite a big step up from Griffiths though. But if you want to understand what's really going on then you'll need to do QFT at some stage!

Where Does the Trace Come From?

Observe that all the square bracket factors in my first expression for $\sigma$ are scalars, so we may commute them. We can thus write

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

Now expanding out the complex conjugation and using that the $t^a$ are Hermitian we find

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

Now the color polarizations are normalized, so the central terms in each factor just give $1$. This leaves

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

But now the sum over initial colors gives exactly the definition of $Tr(t^at^b)$. Similarly for the final colors.

$\endgroup$
  • $\begingroup$ Can you elaborate on the part where you say: "using that the color polarizations are normalized we can reinterpret the sum as a double trace term" $\endgroup$ – negligible_singularity Apr 14 '14 at 15:55
  • $\begingroup$ Certainly - I've added an explanation to my answer. $\endgroup$ – Edward Hughes Apr 15 '14 at 11:22
  • 1
    $\begingroup$ The trick I missed the first time was $\delta^{ab}\delta_{ab} = N^2 -1$. I kept thinking $\delta^{ab}\delta_{ab} = N$. $\endgroup$ – negligible_singularity Apr 19 '14 at 5:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.