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## Q. 14.6

Examine the lepton conservation laws in the decay of the $\pi^{-}$ and $\mu^{-}$.

## Verified Solution

The decays are

\begin{aligned}&\pi^{-} \rightarrow \mu^{-}+\bar{\nu}_{\mu} \\&\mu^{-} \rightarrow e^{-}+\bar{\nu}_{e}+\nu_{\mu}\end{aligned}

The $\pi^{-}$ first decays to $\mu^{-}$, which in turn decays into $e^{-}$. However, we must also have neutrinos in order to conserve lepton number. The first reaction has $L_{\mu}=0$ on the left and $L_{\mu}=+1-1=0$ on the right. For the second reaction, we must conserve both $L_{e} \text { and } L_{\mu}$. On the left side we have $L_{e}=0 \text { and } L_{\mu}=1$. and on the right side we have $L_{e}=1-1+0=0$ and $L_{\mu}=0+0+1=1 \text {, so both } L_{e} \text { and } L_{\mu}$ are conserved.