Here it says this decay mode is a violation of C-parity. I don't understand how that works.

So $\omega$ has $C=-1$ and $J=1$ and $\pi^0$ has $C=1$ and $J=0$, that means the orbital angular momentum of final state is $L=1-0=1$. Consider C-parity of final state we get $C=(-1)^L=-1$. This seems to agree with C-parity conservation?

I've also seen an argument saying that our final state with $L=1$ is anti-symmetric if we exchange two $\pi^0$, which is forbidden since they are bosons. But I don't understand how that's related to C-parity.

  • $\begingroup$ what does the angular momentum have to do with it. C conservation has to do with charge conjugation and in isospin space, not in space. en.wikipedia.org/wiki/C_parity . The generalization is G parity, also displayed in the table you show en.wikipedia.org/wiki/G-parity where the isospin vectors appear and for neutral particles only I appears. page 3 of pdg.lbl.gov/rpp-archive/files/RevModPhys.52.S1.pdf $\endgroup$ – anna v Dec 2 '17 at 5:13
  • $\begingroup$ @annav Thanks for your reply. I understand the definition of c parity and g parity, I just don’t know why this decay mode is listed under “charge conjugation violating modes” in the picture I posted. $\endgroup$ – Salmonella mayonnaise Dec 7 '17 at 19:17
  • $\begingroup$ I thinnk it is because G and C are the same for neutral states $\endgroup$ – anna v Dec 7 '17 at 19:19

@anna_v 's comment is your answer. The ω has C=- and G=- , while the C of each π0 is +, and thus of their aggregate +. (Your involving the angular momentum here in connection to C is fundamentally unsound--You are probably confusing the aggregate C of a fermion-antifermion pair or π+ π- which go into each other. Stay away from it here.)

The G of each pion is -, so their aggregate's is +, so G-parity also forbids the decay. That's what G parity is: evenness versus oddness of the number of pions. The ω is thus destined to only decay to an odd number of π s. However, G parity is violated more than C, since it involves isospin in its definition, which is not as perfect a quantum number as C in the strong interactions.

The PDG entry is thus spot on.


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