As we know that particles having structure (quarks) decays to lighter products. We can show the decay of them by quark level Feynman diagram. But what about muon decays to electron, neutrino (muon type) and anti neutrino (electron type). Is there a way to make Feynman diagram of it?
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2 Answers
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The Feynman diagram is completely analogous to the Feynman diagram of the decay of neutron (beta-decay):
The muon $\mu^-$ splits to a muon neutrino (there is a mistake on top, it should say $\nu\mu$, not $\nu_e$, sorry) and the $W^-$-boson, and the latter splits to $e^-$ and $\bar \nu_e$. The word "decay" doesn't mean that the original decaying particle had to be composite. Instead, "decay" means that the number of final particles is higher than the number of initial particles and the initial particle ceases to exist.
Quantum field theory – and Nature – guarantees that the number of particles may change and it is routinely changing. Energy may be converted to mass via $E=mc^2$ and all processes compatible with the conservation laws (which don't include the "number of particles" conservation law!) occur with a nonzero probability.
answered Aug 14, 2014 at 7:45
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Sorry, but I don't understand what you mean when you say that quarks have a structure. What we know is that quarks are point-like particle, like was confirmed from SLAC years ago. Quarks aren't so much different from Leptons, unless they are "colored".
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2$\begingroup$ This should be a comment and not an answer. $\endgroup$ Commented Aug 14, 2014 at 9:45
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$\begingroup$ It seems that @Chetanverma meant that the structure of some composite particles is quarks, not that quarks have structure. $\endgroup$ Commented Aug 14, 2014 at 12:46
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$\begingroup$ But I don't understand why is plausible a diagram of : $t\rightarrow b ~ W^+$ but is mysterious ${\mu}^- \rightarrow \nu_{\mu} ~ {\bar{\nu}}_e ~ e^- $ $\endgroup$– KarozoCommented Aug 14, 2014 at 14:09