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I understand that it is due to C conservation, but I'd like some more information - perhaps it's simpler than I'm imagining.

Similarly for the decay $\omega \rightarrow \rho \gamma$.

Thanks!

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  • $\begingroup$ I think the answer lies in $G$-parity, although Non-conservation of a property doesn't mean violation under all circumstances. $\endgroup$ – Kevin Aug 15 '19 at 12:07
  • $\begingroup$ look at isospin conservation hyperphysics.phy-astr.gsu.edu/hbase/Particles/parint.html, it does not have decaying quark, but the isospin balance left and right is wrong, pions have isospin one (for the title decay). $\endgroup$ – anna v Aug 15 '19 at 13:04
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It is not clear to me what is troubling you.

You must have ascertained by now from your PDG listings that $J^{PC}$ for the particles involved is $$ \omega, \rho^0, \gamma : \qquad 1^{--} \\ \eta, \pi^0 :\qquad 0^{-+} ~~. $$ Both strong and EM interactions preserve C, here clearly mismatched (multiplicatively) in the two decays you are proposing.

What qualms would you have?

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