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?


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.


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

  • 1
    $\begingroup$ 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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