I'm reading Martin and Shaw's "Particle Physics."

In Section 5.4.1 they show how $C$-parity restricts the number of photons in pion decay to an even number of photons:

$$ \pi^0=u\bar u\textrm{ (or }d\bar d) \rightarrow 2n\gamma,~n\in\mathcal I$$

And in Section 1.3.3, they mention that positronium decay can give both even and odd number of photons:

$$ e^-+e^+\rightarrow n\gamma,~n\in\mathcal I $$

In both cases a particle-antiparticle pair annihilates, so why do we not get odd suppression in both cases?

Is there perhaps an unstated assumption that for the pion we look at the ground state, while for positronium we also allow excited states?

  • 1
    $\begingroup$ This might not affect the question, but excited states of the pion usually aren't called pions. Figure 2.1 in hep-ph/0001312 shows the mass and angular momentum of one of the "excited states," which is called $b_1(1235)$ in the Particle Data Group summary table $\endgroup$ Nov 13, 2021 at 15:40

1 Answer 1


For a fermion-antifermion pair, you can easily see that $$ C= (-)^{L+S}, $$ so, e.g., $$ C(^1S_0)=+ , ~~~\leadsto ~~~~ \to 2\gamma,\\ C(^3S_1)=- , ~~~\leadsto ~~~~ \to 3\gamma, $$ the photons being odd under C.

The first case covers the lowest-lying pseudoscalars, like the pion, and parapositronium; while the second covers the ρ, ψ, ..., and orthopositronium.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.