I am trying to answer a question that asks to find the branching fraction of $\tau^{+}$ decays.

In the answers, it gives that the widths are as follows:

$$\Gamma(\tau^- \rightarrow ν_{\tau} d \overline{u}) = 3\cos^2 \theta_c\Gamma(\tau^- \rightarrow e^- \nu_{\tau^{+}} \overline{\nu_{e^-}})$$

$$\Gamma(\tau^- \rightarrow \nu_{\tau} s \overline{u}) = 3\sin^2 \theta_c\Gamma(\tau^- \rightarrow e^- \nu_{\tau^{+}} \overline{\nu_{e^-}})$$

I understand that the factors of sin and cos come from the CKM matrix elements, but I am confused as to where the factor of $3$ comes from and why the additional factor of $\theta_c\Gamma(\tau \rightarrow e \nu \overline{\nu})$ must be included.In the answers,it says that the extra factor of $3$, is the color factor for the quark case, but I am confused as to what this means, so any explanation would be appreciated.

  • $\begingroup$ in case you get no good answer from an expert see this lecture nikhef.nl/~h24/qcdcourse/section-5.pdf $\endgroup$ – anna v Apr 26 '20 at 13:06
  • $\begingroup$ What is the $\bar v$ in your $\Gamma(\tau \to e \nu \bar v)$ supposed to be? I would have though it should be $\nu_\tau$ or $\nu_e$ so one is comparing a purely leptonic process to one involving quarks. Where did you get your quoted formulae from? $\endgroup$ – mike stone Apr 26 '20 at 14:16
  • $\begingroup$ I think it should be $\Gamma(\tau \to \nu_\tau e \bar \nu_e)$ as that is a purely weak process (and so exactly calculable) making it natural to compare it to the semi-hadronic decay so as to extract information about the hadronic part $\endgroup$ – mike stone Apr 26 '20 at 14:25
  • $\begingroup$ Yes, sorry it should be that, the formula I quoted from is just off the answer sheet, so probably wasn't included as an error. $\endgroup$ – Xenophilius Apr 26 '20 at 14:28
  • $\begingroup$ So do you now understand how you should answer your homeowrk question? I don't want to do your homework for you! $\endgroup$ – mike stone Apr 26 '20 at 14:31

The decay rate is given by computing a tree digaram and squaring and summing over the phase-space for the decay products. This sum can be calulated much more efficiently from from the imaginary part of a loop diagram in which each decay-product particle is connected to itself and then back to the $\tau^-$ so as to reverse the decay process and give a diagrm in which the $\tau$ does not decay. The loop diagram for all the $d\bar u$ and $s\bar u$ processes or the $ e \nu_e$ are basically the same and, because the $u,d,s$ quarks and the $e^-, \nu_e$ leptons are much lighter than the $\tau$, the masses in the propagators can be ignored so the loop momentum integrations are essentialy the same. In all three processes there is a closed loop with either the quarks running round it, or the $\nu_e, e^-$. The only significant difference between the three processes is the way the $W^-$ couples (see the Wikipedia anticle on the $\tau$ for picture of the decay process) to the quarks. This gives the $\sin^2 \theta_{\rm Cabibbo}$ and $\cos^2 \theta_{\rm Cabibbo}$ factors. Next, consider the fact that quarks come in three colours while the leptons have no colour. This means that the quark loop is repeated three times, once for each colour. From this we get the factor of three.


The reason for the extra factor $3$ is: Quarks come in 3 different colors (red, green, blue). For example, the $d$ quark can occur in 3 color states: $d_r$, $d_g$, and $d_b$. And like there is a conservation law for electrical charge, there are also conservation laws for red, green and blue color-charge.

Therefore the reaction $$\tau^- \to \nu_\tau d \bar{u}$$ is actually a short-hand notation for the 3 possible reactions

$$\tau^- \to \nu_\tau d_r \bar{u}_r$$ $$\tau^- \to \nu_\tau d_g \bar{u}_g$$ $$\tau^- \to \nu_\tau d_b \bar{u}_b$$

Each of these 3 reactions has the same statistical weight as the single reaction $$\tau^- \to e^-\nu_\tau \bar{\nu}_e$$


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