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While studying SM, I was taught that weak force bosons $V=\{W^\pm,Z^0\}$ do not interact with right/left-chiral fermions/antifermions. For this reason, we cannot observe right-handed neutrinos (if they do exist). However, we do observe right-chiral charged leptons as $e^-_R, \mu^-_R\dots$ through EM interactions (thanks to their charge).

But the decay of heavily charged leptons as $\mu$ and $\tau$ must involve weak force otherwise lepton family number would be violated,

\begin{equation}\label{eq1} \mu^-_L\rightarrow W^-+\nu_{\mu L} \rightarrow e^-_L+\bar{\nu}_{eR}+\nu_{\mu L} \quad \text{Involving weak bosons} \end{equation}

$$\mu\rightarrow e+\gamma \quad \text{Cannot happen since LFN is violated}$$

If right-chiral heavy fermions as $\mu_R^-$ and $\tau_R^-$ cannot decay through weak force? then how do they decay?

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A physical muon is a mass eigenstate. This is a superposition of the left and right chirality components. Because of this, the physical muon can always interact via the weak currents through its left-chiral component, and as such it (and a physical $\tau$) will always decay.

Directly addressing your question - the right-chiral muon $\mu_R$ is stable (because it cannot decay via weak currents or EM, as you point out), but this is not not very physically relevant. You cannot stabilize a $\mu$ by isolating a right-chiral $\mu_R$, because the muon's mass causes mixing between the components. That is, the coefficients of the above-mentioned superposition are in general not constant in time - instead they have some phase factor with a frequency related to the muon's mass.

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  • $\begingroup$ I can understand from your reasoning that a free Dirac muon is a superposition of a left-chiral muon and a right-chiral muon. But this raised another question in my mind, when a Dirac muon undergoes a weak decay, you say only the left-chiral component changes, and the right-chiral component remains unaffected, does that mean after interaction we get a superposition of a left-chiral electron and right-chiral muon? $\endgroup$ Commented Jun 13 at 15:19
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    $\begingroup$ @A.M.MElsayed马克 I did not say anything about how the components are affected by the interaction - only that the left-chiral component must not be zero in order for the interaction to occur. The $W$ interaction occurring basically constitutes a chirality measurement, taking the right-chiral component to zero, if you want to think about it that way. In the decay frame immediately after the interaction you must have a purely left-chiral electron (although it will be subject to the same chiral oscillations in its future) $\endgroup$ Commented Jun 13 at 15:40
  • $\begingroup$ Thank you, it is much more clear now $\endgroup$ Commented Jun 14 at 7:36

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