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Why we consider Hadrons as the eigenstate of parity operator? In doing so we assign each hadron a quantum no. +1 or -1. What about the parity of leptons, and more specifically parity of neutrino? The whole thing boils down to asking what will be the parity of
$$\pi^{-}+e^{+}+\nu_{e}.$$

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  • $\begingroup$ Can't you look that up in a table in Wikipedia? $\endgroup$ – my2cts Nov 10 '19 at 10:49
  • $\begingroup$ I do not get parity of neutrino.The above reaction I think may not have definite parity.To clear my doubt i have just asked. $\endgroup$ – p rakesh kumar dora Nov 10 '19 at 11:09
  • $\begingroup$ I believe neutrinos have the same parity as the leptons. $\endgroup$ – my2cts Nov 10 '19 at 15:13
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A question whose answer would only moot the real answer to the real question underlying it.

The strong interactions preserve parity, so it is a good quantum number for the strong interaction of hadrons; and charge leptons, when considering their electromagnetic interactions, also preserving parity.

While the parity of the pion and the electron are well defined, it would be self-defeating, and confusing, to insist on a parity for neutrinos, as you'd be unlikely to use it.

The reason is neutrinos only interact weakly (and gravitationally, irrelevant here) and the weak interactions violate parity (maximally in the charged current sector!), in sharp contrast to the strong and electromagnetic ones; and so parity would not be a good quantum number in everything involving neutrinos.

I cannot fathom the context of the collection of particles you proffer, but you are effectively asking for the parity of $W^+ \pi^-$, which should be ringing alarm bells.

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