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If $\rm p\bar p$ (proton-antiproton) annihilation at rest proceeds via the $S$ state ($L=0$), why is it that the reaction: $\rm p\bar p\to \pi^0 +\pi^0$ is forbidden as strong interaction (i.e. parity conserved).

The initial state has odd parity as $L=0$, the final state must have the same: $L=$ odd for the final state. How do I determine the value of $L$ for pions? (Do I use the fact that total angular momentum $J$ is conserved and the known values of spin $S$?)

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    $\begingroup$ Pions are spin-0, so you can't have odd parity. $\endgroup$ – probably_someone Mar 2 '18 at 20:31
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You've concluded that, to conserve parity, $L$ must be odd. By the two pions are identical bosons, and so the wavefunction must by symmetric under exchange. If $L$ is odd, the wavefunction is anti-symmetric under exchange, and so this is forbidden.

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  • $\begingroup$ As I understands it, L is the only contribution to the wavefunction for a boson because it's spin-0. So L has to be even for the wavefunction to be symmetric. Is that right? $\endgroup$ – Ret Mar 3 '18 at 14:15
  • $\begingroup$ @Z.Yang For a spin-0 boson, yes. $\endgroup$ – Chris Mar 3 '18 at 20:11

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