Codex @codex  0

For the neutral quark-antiquark mesons here, $C=(-1{)}^{L+S}$. Thus $C({\eta }^{\prime })=C({\pi }^0)=+1$, 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 $(m_{{\eta }^{\prime }}^2-m_{\rho }^2)/(2m_{{\eta }^{\prime }})\simeq 165\mathrm{MeV}$.

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 $\pm 1$, while the opposite-moving pion has helicity zero; the spin-zero two-body helicity selection rule cannot be satisfied. 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,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 ${\pi }^0\gamma$.

In a box-normalized treatment of negative-pion two-body decay with a massless antineutrino, $dp/d{E}_0=E_{\ell }/m_{\pi }$. The simple density of states is therefore proportional to $p_{\ell }^2{E}_{\ell }/m_{\pi }$. The spin-zero two-body helicity selection requires positive charged-lepton helicity, whose left-chiral weight is $(1-{\beta }_{\ell })/2$. Since $1-{\beta }_{\ell }=2m_{\ell }^2/(m_{\pi }{E}_{\ell })$, the energy factors cancel in the product and give the standard relative rate proportional to $m_{\ell }^2{p}_{\ell }^2$. A Lorentz-invariant phase-space measure distributes normalization factors differently but gives the same observable ratio when the amplitude is normalized consistently.