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In nuclear physics lessons I've been taught that pions are pseudoscalar particles and thus their intrinsic parity is odd. The professor said that an experimental proof of this can be derived observing the process: $$ \pi^- + D \quad \longrightarrow \quad n + n $$

where the pion is considered as an s-wave ($\ell = 0$ ) and the deuteron is in triplet state ($s=1$).

I can't understand why this decay can prove pions have a negative intrinsic parity.

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The answer is based on a comment by Cosmas Zachos.

The pion has no spin, and so can't flip the spin of either nucleon in the deuteron. The nucleons in the deuteron have their spins aligned ($S=1$), so the two neutrons in the final state must be in a spin triplet as well. In order for the final two neutrons to be antisymmetric under exchange, their wavefunction must have $L=\text{odd}$, which means the final state has odd parity.

If the final state has odd parity, and the $\pi^-\rm D$ are initially in an $s$-wave state, then the parities of the $\pi^-$ and the $\rm D$ must be opposite. The negative parity of the $\pi$ follows from the positive parity of the deuteron.

This seems to have been shown originally by Steinberger.

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