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I am aware that the following decay mode may happen: $$\pi^- → e^-\overline\nu $$

My question is why is the following decay mode not possible

$$\pi^- → e^-\nu $$

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It is called conservation of lepton number . The pion has lepton number zero. The names: neutrino and antineutrino have been assigned to the neutral leptons carrying the opposite lepton number of the electron (+1) electron antineutrino (-1), positron (-1) positron antineutrino (+1), so that the original particle and the final state have both lepton number zero.

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  • $\begingroup$ Anna - is this simply a "trick" of mathematics (and don't read anything into my use of the word trick) or does the conservation have a deeper meaning? $\endgroup$ – Rick Mar 25 at 12:59
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    $\begingroup$ The deeper meaning is that that is what fitted the experimental data, to start with. By now it is a part of the standard model of partilcle physics, the mainstream theory en.wikipedia.org/wiki/Standard_Model $\endgroup$ – anna v Mar 25 at 13:14
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    $\begingroup$ @Rick It very much has a deeper meaning: it tells you that if you used your decaying pions to make a neutrino beam, and you pointed that at a lump of lead (say), then further interactions would be able to turn that electron antineutrino into a positron but not an electron. $\endgroup$ – Emilio Pisanty Mar 25 at 15:03
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What anna v said is correct, but as far as I can tell, we still don't know the nature of the neutrino (maybe it's a Majorana particle?). We pair ( $l~\overline{\nu}_l$ ) and ( $\overline{l}~\nu_l$ ) to conserve lepton number, but it may turn out that the neutrino is its own anti-particle - see, e.g. neutrinoless double beta decay. So, maybe the answer to your question is that you wrote the same allowed decay twice.

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