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Decays possible? Parity conservation, bosons, fermions

  1. Dec 21, 2014 #1
    1. The problem statement, all variables and given/known data

    The question is to determine which decays are possible for:

    i) ##P^0 ->\prod^+ \prod^-##
    ii)##P^0 ->\prod^0 \prod^0##


    2. Relevant equations

    where ##J^p = 0^-, 1^- ## respectively for ##\prod^+, \prod^- , \prod^0## and ##P^0## respectively.

    3. The attempt at a solution

    For part i, the LHS has odd parity. ##P=(-1)^l##, so on the RHS we require ##l## to be odd.
    Also need to conserve total angular momentum ##J=(l+s)+(l+s-1)+....+ | l-s |## *
    On LHS ##J=1.##
    ##s=0##, so conservation gives ##l=-1## , which is consistent with an odd parity , so the decay is allowed.


    part ii) We have the same J and P arguments, so I would have concluded the decay is possible.
    The solution however is that is not because the RHS now has 2 identical bosons so the final wavefunction must be symmetric under the exchange of the two neutral pions. However this requires that the orbital angular momentum is even, so we have inconsistency.

    So here's what I know :
    If you swap 2 bosons the wave function has to be unchanged, but if you swap 2 fermions the wave function changes sign.
    So , with this, I now don't see why we cant apply the argument to the decay in part i) - unless this property is only true for a system of identical particles??

    More importantly, I don't follow the argument completely: The angular momentum being odd or even, i.e- as far as I can see the only way for ##l## to creep in, comes from the parity being odd or even- ##P=(-1)^l##, but parity is describing how the wave function behaves under a change from ##\vec r ## to ##\vec -r## So what has this got to do with swapping 2 bosons? The only possible arguement I can think of would be along the lines of considering the particular case were one of the particles is situated at ##\vec r ## and the other at ##\vec -r## when we swap the bosons position??


    Thanks in advance !
     
  2. jcsd
  3. Dec 21, 2014 #2

    mfb

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    Staff: Mentor

    That is the point.
    Swapping two non-identical particles leads to a completely different system, where no symmetry is relevant. Only swapping two identical particles has an interesting result.

    What is P, by the way? ##\rho##?
     
  4. Dec 22, 2014 #3
    ##P## is the parity.
     
  5. Dec 22, 2014 #4

    mfb

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    2016 Award

    Staff: Mentor

    I mean the particle that decays. I do not recognize capital P as a particle name, only p for protons but that does not make sense here.
     
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