We are given the Breit-Wigner formula for the process $ud\rightarrow W\rightarrow e\nu$ as $$\sigma=\frac{1}{N_c^2}\frac{2J_W+1}{(2J_u+1)(2J_d+1)}\frac{4\pi}{s}\frac{\Gamma_{ud}\Gamma_{e\nu}}{(\sqrt{s}-m_W)^2+\Gamma^2/4}.$$ where $N_c$ is the number of possible initial color states. However, I am unsure why it appears as $N_c^2$ rather than simply $N_c$. I would think that the initial $ud$ must have equal and opposite color in order that color is conserved, giving $N_c$ possibilities, whereas $N_c^2$ implies that the two can have different colors which seems incorrect. Is the formula wrong, or am I misunderstanding something?


1 Answer 1


I think the extra $1/N_c$ might come from the definition of $\Gamma_{ud}$.

If $\Gamma_{ud}$ is the decay width of the $W$ into $u$ and $d$ then it is proportional to $N_c$. In your computation you want to "move" this amplitude to the initial state, so you have to remove this $N_c$ to get the $ud\to W$ interaction and divide by an additional $N_c$ when averaging over the initial color.

EDIT for clarity:

The amplitude can be factorized as $$|\mathcal M|^2 = |W\to u d|^2 |W \to l\nu| BW(s)$$

where $BW(s)$ is the usual Breit-Wigner denominator. Now to get the cross section you want to average over the initial spin and color, so schematically something like this

$$\sigma \propto \frac{1}{N_c}\frac{1}{(2 J_u +1)(2 J_d +1)} \sum |\mathcal M|^2 $$

while your formula is written in terms of the decay width, that is obtained by summing over the final spin/color

$$\Gamma \propto N_c \sum |W\to ud|^2 $$

And this is why I think you have an additional $N_c$ factor in your formula. If you go through the full derivation of the formula in detail

  • $\begingroup$ What do you mean by move the amplitude to the initial state? $\endgroup$ Jun 1, 2021 at 12:39
  • 1
    $\begingroup$ @AlexGhorbal yeah sorry I tried explaining it just by words, but maybe I should be more detailed. I'll edit the question as soon as I have time! $\endgroup$
    – FrodCube
    Jun 1, 2021 at 12:40
  • $\begingroup$ @AlexGhorbal I added a bit more details $\endgroup$
    – FrodCube
    Jun 1, 2021 at 21:56
  • $\begingroup$ My confusion is in the last step I think. If you sum over $|W\to ud|^2$, isn't the $N_c$ already included in this? $\endgroup$ Jun 2, 2021 at 10:54

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