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Homework Help: Rho0 decays

  1. Jul 2, 2012 #1
    1. The problem statement, all variables and given/known data
    I saw here, but since the thread is closed I need to ask my question in a new thread.

    I have two decays: [itex]\rho^0\to \pi^+\pi^-[/itex] and [itex]\rho^0\to \pi^0\pi^0[/itex]. The text says that both are allowed with respect to laws of conservation of energy and impulse. Why is the second one forbidden?

    Both rho meson and pi meson, are a triplet in isospin space, so I don't see a problem there. But how are both allowed with respect to conservation of impulse, if the impulse of rho is 1, and of pions is 0 (impulse as in J=L+S)?

    Is it just because the Clebsch-Gordan coefficient in the second decay is 0 and in the first it's [itex]\frac{1}{\sqrt{2}}[/itex]?

    So the coupling of two pi0 is forrbidden in the isospin coupling? Is that the answer?
  2. jcsd
  3. Jul 3, 2012 #2


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    The final state is a two-particle state, so it is possible to have nonzero orbital angular momentum (##L\neq 0##). However, once you identify the possible orbital state, you must also consider the symmetry of the wavefunction under exchange of the ##\pi##s.

    This is not an issue, since that coupling exists. In fact, ##\rho^0\to \pi^0\pi^0\gamma## is an allowed decay channel.
  4. Jul 3, 2012 #3
    Thank you :)
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