My question is, can a physical process be deemed valid/invalid, such as muon decay, purely from the allowed standard model vertices? Or do we need to take into account things like lepton numbers and energy conservation?

For example, consider muon decay, $ \ \mu^- \to \nu_\mu + e^- + \bar{\nu_e}$, described by the Feynman diagram,

But it seems that another valid Feynman diagram according to the standard model would be the following,

Which represents $e^- \to \nu_e + \mu^- + \bar{\nu_\mu}$. The reason I thought to put the $\bar{\nu_\mu}$ was because of lepton numbers, not because of valid standard model vertices, as $\nu_\mu$ would have been a valid choice as well.

Since electrons have smaller rest mass than muons, it seems that the second diagram would violate conservation of energy in the rest frame of the "decaying" electron, so this is not a valid physical process.

Edit: I got the arrow on the outgoing $\bar{\nu_\mu}$ wrong. Oops.

  • $\begingroup$ You claim the SM lagrangian violates lepton number? $\endgroup$ – Cosmas Zachos Sep 2 '18 at 6:24
  • $\begingroup$ @Cosmas Zachos I am yet to come to that in my particle physics course, haha. By the sounds of it, the thing works. However, would an outgoing $\nu_e$ rather than $\bar{\nu_e}$ in muon decay not be a valid vertex? What am I missing here? $\endgroup$ – user154080 Sep 2 '18 at 6:28
  • $\begingroup$ You might have to specify what "works" means in your question... $\bar{\nu}_e$ is defined as what accompanies the electron in a resolution of $W^-$. $\endgroup$ – Cosmas Zachos Sep 2 '18 at 6:41

Assuming that the electron decay you mentioned is indeed a valid diagram, then the amplitude for this process does not necessarily vanish. However, the decay rate (or any other measurable physical quantity) will vanish due to kinematic factors that relate the decay rate to these Feynman amplitudes. In particular, the delta function corresponding to energy conservation in e.g. Fermi's golden rule will always vanish for any configuration of external particles, so long as the electron is at rest. Thus, the decay rate vanishes and the process doesn't happen.

I hope this helps!


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