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Proton-Antiproton Annihilation

  1. Apr 8, 2009 #1
    1. The problem statement, all variables and given/known data

    A proton and antiproton at rest in an S-state annihilate to produce [tex]\pi[/tex]0[tex]\pi[/tex]0 pairs. Show
    that this reaction cannot be a strong interaction.

    2. Relevant equations

    I interpret this problem as:

    p + p_bar -> [tex]\pi[/tex]0 + [tex]\pi[/tex]0

    3. The attempt at a solution

    If this were a strong interaction, Parity should be conserved, yes? I think that in the annihilation
    described above, parity is not being conserved.

    But really, my first question is, am I interpreting it correctly? Is the annihiliation producing [tex]\pi[/tex]0 + [tex]\pi[/tex]0?
    Or is it producing [tex]\pi[/tex]0 + [tex]\pi[/tex]0 + [tex]\pi[/tex]0 + [tex]\pi[/tex]0? Not sure what they mean by "[tex]\pi[/tex]0[tex]\pi[/tex]0 pairs".

    Then my next question is, what is the Parity of p_bar?
    Last edited: Apr 8, 2009
  2. jcsd
  3. Apr 8, 2009 #2
    Okay, so I think I've come to the conclusion that the Parity of the antiproton is -1.
    I read that, in general, P(fermion * antifermion) = -1, and since proton's parity is +1, I'm
    going to conclude that parity of the antiproton is -1.

    Okay, so total parity on the left hand side of the equation is 0.

    But I still don't know if I'm interpreting the problem correctly. Is the annihilation producing:

    [tex]\pi[/tex]0 + [tex]\pi[/tex]0 ?

    [tex]\pi[/tex]0 + [tex]\pi[/tex]0 + [tex]\pi[/tex]0 + [tex]\pi[/tex]0 ?

    [tex]\pi[/tex]0[tex]\pi[/tex]0 ?

    [tex]\pi[/tex]0[tex]\pi[/tex]0 + [tex]\pi[/tex]0[tex]\pi[/tex]0 ?

    Pardon the ignorance here. I'm taking this Nuclear Physics class despite not meeting some
    of the pre-reqs. I'm a math major that was just interested in the material, so I'm a bit

    Thanks for any help.
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