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I have to determine whether the neutral rho meson could decay into a neutral pion and eta. I have checked the quantum numbers and all of them seem to be conserved adding an orbital angular momentum to the pion-eta system.

Nevertheless, this reaction seems not to be allowed and I was wondering if this could be because the rho meson doesn't have enough energy to provide angular momentum to the system. Is this correct?

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Looks like this decay via the strong interaction is not allowed due to violation of G-parity, as @Cosmas Zachos hinted in a comment, but it is probably allowed via the electromagnetic interaction, but such decay may be difficult to observe as rho decays very fast via the strong interaction.

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Yes,I have done the following:

Initial information: rho_0: I^G(J^{PC}) = 1+(1--) eta: I^G(J^{PC}) = 0+(0-+) pi_0: I^G(J^{PC}) = 1-(1--)

*Angular momentum: J_rho = 1 J_eta_pi = 0 then we need additional angular orbital momentum to conserve the momentum. I assume L = 1.

  • Parity: J_rho = (-) J_eta_pi = (-)(-)(-)^L = (-)

  • Charge conjugation: C_rho = (-) C_eta_pi = (+)(+)(-)^L = (-)

  • I tried even G-parity too: G_rho = (+) G_eta_pi = (+)(-)(-)^L+S+I = (+)

My guess is that maybe, the mass of rho (~775MeV) is not enough to give angular momentum to the system (m_pion=140MeV; m_eta~550MeV), but I don´t know how to check it.

Cheers, Ateniger

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    $\begingroup$ G of eta is + G of pion is - , G of rho is + . See books.google.gr/… . "the G parity of a system of particles each of which has a definite G parity is equal to the product of the G parities of the separate particles". G parity is conserved in strong decay, so this could only be electromagnetic, which will be down by powers of 1/137 $\endgroup$ – anna v Dec 1 '16 at 13:57
  • $\begingroup$ The above point suggests that your additional mysterious antisymmetrizer $(-)^{L+S+I}$ is distinctly unsound: G just counts pions and it is as good as isospin, violated by electromagnetism, but not the strong interactions. @dukwon alerts you to the equivalent 4π mode, electromagnetic, except there you do not need to reconstruct the η. $\endgroup$ – Cosmas Zachos Dec 1 '16 at 16:18

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