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Why can't rho^o decay into two pi^o?

  1. Dec 9, 2008 #1
    As title, why does the following decay is forbidden?
    [tex]\rho0\rightarrow\pi0\pi0[/tex]
     
    Last edited: Dec 9, 2008
  2. jcsd
  3. Dec 9, 2008 #2

    Vanadium 50

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    Isospin. Take a look at the Clebsch-Gordon coefficient table.
     
  4. Dec 9, 2008 #3
    Thank you. You are correct. I look at the CG 1x1 table.
    |10> = (1/2)|11>|1-1>-(1/2)|-11>|11>. The coefficient of |10>|10> is zero!

    But why is that physically? The isospin is still conserved for |10>|10> case.
     
    Last edited: Dec 10, 2008
  5. Dec 10, 2008 #4

    Vanadium 50

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    What kind of answer are you looking for? Why are there Clebsch-Gordon coefficients at all? Why does isospin use the same Clebsch-Gordon coefficients as angular momentum? Why is there isospin at all? I don't know how to answer you.
     
  6. Dec 10, 2008 #5

    clem

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    You can also use Bose statistics and conservation of angular momentum.
    The rho has spin one. There is no pi0-pi0 state with J=1.
     
  7. Dec 10, 2008 #6
    Hello,
    I don't understand why pi0-pi0 can't have J=1.
    I thought (probably wrong) that they could have some relative angular momentum L for example L=1 so that then pi0-pi0 would have J=1.

    Maybe it's trivial but could someone please explain me why that isn't possible :)

    Thank you!
     
  8. Dec 10, 2008 #7
    Isn't isospin conservation just a manifestation of nucleon SU(2) invariance of the Hamiltonian?
     
  9. Dec 10, 2008 #8

    Vanadium 50

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    If you put it in a J=1 state the total wavefunction is antisymmetric, and it has to be symmetric because of Bose statistics.

    JDStokes, yes, isospin is SU(2). Which is why you use the Clebsch-Gordon coefficients as opposed to something else.
     
  10. Dec 11, 2008 #9

    clem

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    It is important to note that Ispin symmetry is broken, while Bose statistics and conservation of angular momentum are not.
     
  11. Dec 11, 2008 #10
    It can't decay into pi0 pi0 because of conservation of angular momentum.

    It can decay into pi0 pi0 gamma. In a world with unbroken isospin symmetry, that decay would also be prohibited (because rho0 is symmetric under u-d interchange, and pi0's are antisymmetric). In our world, it's merely suppressed. Relative probability of this decay is 4.5 * 10^-5 %.
     
  12. Dec 11, 2008 #11

    clem

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    1. You also need Bose statistics for the pi0s.
    2. Once there is a gamma, Ispin is irrelevant.
    2. The rho is u-ubar. It has C=-, but no symmetry property.
     
  13. Dec 11, 2008 #12
    1. Gamma is isospin invariant. u and d quarks couple to the gamma with different strength, because isospin is not a perfect symmetry.

    2. Rho is [tex](u\bar{u} - d\bar{d})/\sqrt{2}[/tex].
     
  14. Dec 11, 2008 #13
    If there is a difference between [itex](u\bar{u} - d\bar{d})/\sqrt{2}[/itex] and [itex]u\bar{u}[/itex], it must break isospin ! This is why (I think) you miss clem's point.
     
  15. Dec 12, 2008 #14
    :confused:

    We're located in different points on the surface of the Earth, but that does not mean that rotational symmetry of the Earth is broken. [itex](u\bar{u} - d\bar{d})/\sqrt{2}[/itex] and [itex]u\bar{u}[/itex] are degenerate and there is a symmetry transformation that takes one into the other, but that does not mean that they are the same state.
     
  16. Dec 13, 2008 #15

    clem

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    I know the rho is not simply u-ubar. I just gave a simple example to show that symmetry didn't enter.
     
  17. Dec 13, 2008 #16
    Sure, I think we agree. The states would be impossible to distinguish IF isospin were exact. Of course, the proton and the neutron have different interactions with the photon :smile:
     
  18. Dec 13, 2008 #17

    clem

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    I am confused by the last few posts. u-ubar is not an Ispin eigenstate.
    The combination [tex](u\bar{u} - d\bar{d})/\sqrt{2}[/tex] is.
     
    Last edited: Dec 13, 2008
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