# Leptonic Decay of Muon and Tau

(a) Draw the Feynman diagrams showing the dominant leptonic decay mode of the $\mu^-$ and the $\tau^-$.

(b) Assuming $m_\tau \gg m_\mu \gg m_e$, estimate the ratio of the rate of the two decay modes.

Part (a) is simple.

Part (b), I'm not sure what I am meant to do. I know the decay widths for the decay modes are $$\Gamma(\mu^- \to e^-+\bar{\nu}_e+\nu_\mu) = \frac{G_F^eG_F^\mu m_\mu^5}{192 \pi^3},$$ $$\Gamma(\tau^- \to e^-+\bar{\nu}_e+\nu_\tau) = \frac{G_F^eG_F^\tau m_\tau^5}{192 \pi^3},$$ $$\Gamma(\tau^- \to \mu^-+\bar{\nu}_\mu+\nu_\tau) = \frac{G_F^\mu G_F^\tau m_\tau^5}{192 \pi^3}$$

and by lepton universality, $$G_F^e=G_F^\mu = G_F^\tau.$$

I'm not sure how to proceed from here. Is it simply finding the branching ratios and dividing one by the other?

• Note that the question asks for a single ratio. Do you know what ratio is intended and why? Question that might help out: "Is the decay of the tau to a muon distinguishable from that to an electron?" – dmckee --- ex-moderator kitten Dec 10 '15 at 3:19
• I do not know what ratio is intended. As for the question, surely you can distinguish the decay of the tau to a muon or to an electron? The muon is heavier than the electron so the energy loss by Bremsstrahlung for a muon is far smaller than the electron. For example in the CMS experiment muons will travel further than electrons. – John Sweeney Dec 10 '15 at 3:53
• Good. Now what does quantum mechanics tell you about combining distinguishable final states vis a vis combining indistinguishable final states? – dmckee --- ex-moderator kitten Dec 10 '15 at 4:28
• I know about distinguishable and indistinguishable particles. If two particles are distinguishable then the quantum state of the system is simply a tensor product of the two states the particles are in. If they are indistinguishable then there is a linear combination of tensor products, e.g $(|a\rangle |b \rangle + |b\rangle|a \rangle )/ sqrt{2}$. Is this what you are asking for? – John Sweeney Dec 10 '15 at 5:44