# Color factor in Breit-Wigner formula

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?

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

• What do you mean by move the amplitude to the initial state? Jun 1, 2021 at 12:39
• @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! Jun 1, 2021 at 12:40
• @AlexGhorbal I added a bit more details Jun 1, 2021 at 21:56
• 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? Jun 2, 2021 at 10:54