1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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 :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Rho0 decays
  1. Higgs decay (Replies: 1)

  2. Meson decay (Replies: 0)

  3. Pion Decay Angle (Replies: 25)

  4. Meson decay (Replies: 0)

  5. Radioactive decay (Replies: 7)

Loading...