= Solution
For the neutral quark-antiquark <mesons> here, $C=(-1)^{L+S}$. Thus $C(\eta')=C(\pi^0)=+1$, while $C(\rho^0)=C(\gamma)=-1$. The radiative $\rho^0\gamma$ mode conserves <charge conjugation> and admits a quark-line photon-emission diagram. Both components of the supplied $\rho^0$ <wavefunction> contribute: the charge-weighted flavour overlap is proportional to $Q_u-Q_d=1$, so it does not cancel. Its <photon> <energy> is $(m_{\eta'}^2-m_\rho^2)/(2m_{\eta'})\simeq165\,\mathrm{MeV}$.
The $\pi^0\gamma$ mode changes <charge conjugation> and, more decisively, a spin-zero particle cannot decay into another spin-zero particle and one real <photon>. In the parent rest frame the <photon> has <helicity> $\pm1$, while the opposite-moving <pion> has <helicity> zero; the <spin-zero two-body helicity selection> rule cannot be satisfied. \b[Its on-shell two-body amplitude vanishes even if a <weak interaction> violates C.] A virtual <photon> subsequently making another pair is a different final state.
For $\pi^0\rho^0$, angular-momentum conservation requires relative <orbital angular momentum> $L=1$. Its <parity> is then negative, matching the parent, but its charge-conjugation eigenvalue is $C_\pi C_\rho=-1$: the two distinct self-conjugate particles are not interchanged by C, so there is no additional $(-1)^L$ factor in this C eigenvalue. The decay is forbidden by strong and electromagnetic charge-conjugation conservation despite being above threshold. A weak contribution would require <CP violation> as well. At the strong/electromagnetic accuracy of the diagram classification,
$$
\boxed{\Gamma(\rho^0\gamma)>\Gamma(\pi^0\rho^0)=\Gamma(\pi^0\gamma)=0}.
$$
If extremely small weak CP-violating amplitudes are retained, $\pi^0\rho^0$ need not be exactly zero; the one-real-photon <spin> prohibition remains exact for $\pi^0\gamma$.
Back to article page
