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[noob] simple parity violation

  1. Jul 3, 2012 #1
    hi,
    I'm studying parity violation in particle physics..

    I have this decay:

    1+ ---> 0+ + 0+

    in J^P notation.

    Why this process violate parity? All terms have positive parity.
    The parity of the products is just (+1)*(+1)*(-1)^l,
    so this means that the two scalar particles form an odd orbital angular momentum system?
     
  2. jcsd
  3. Jul 3, 2012 #2

    Bill_K

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    There certainly must be orbital angular momentum. Or else you cannot conserve J.
     
  4. Jul 3, 2012 #3
    well, J shouldn't be conserved in any ways in this situation, right?
    the composition of the 2 particles should give J_tot = 0 in any case
     
  5. Jul 3, 2012 #4

    Bill_K

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    :confused: J is always conserved.
     
  6. Jul 3, 2012 #5
    The conservation of angular momentum should give J_inital=J_final. In this case J_in=1 so this lead to J_fin=1. The final particles are spin 0 particle so J=L=1. This also bring to:

    $$P_{f}=(+1)(+1)(-1)^L=-1$$

    which is in contrast with the initial parity.
     
  7. Jul 3, 2012 #6
    yes, but we have a composition of 2 J=0 states, so there shouldn't be any ways of conserving angular momentum, and this reaction shouldn't happen for this reason.
    However, this book: http://books.google.it/books?id=HNcQ_EiuTxcC&lpg=PP1&hl=it&pg=PA88#v=onepage&q&f=false

    says that the reaction has parity conservation problems, but in principle can happen if strong interaction would violate parity, which sounds wrong to me, because it would violate angular momentum, not parity...

    EDIT: wait, I'm reading the last reply :P
    EDIT2: Einj, you're right, but we have to assume that the final particles have S=0, in the book above it is written that J=0, so that J_f = 0
     
  8. Jul 3, 2012 #7
    Can you please tell me the page? However we are not "assuming" that final particle have 0 spin, we know it because the decay is a:

    $$1^+ \rightarrow 0^+ + 0^+$$

    I'm still convinced of what I wrote above. :tongue:
     
  9. Jul 3, 2012 #8
    page 88, just below eq 3.7. It clearly says that J=0 for both final particles
     
  10. Jul 3, 2012 #9
    it says also that they are 2 scalar states, so S=0. Maybe it's just a notation error
     
  11. Jul 3, 2012 #10
    Yes, but it says that J=0 for the single particle. It just means, as you said that S=0. But this does not exclude that the two body system could have an orbital momentum. Probably you have been confused by the notation, as when you say that a particle is a J^P state, with J you obviously mean its spin as a single particle can't have an orbital momentum:tongue:
     
  12. Jul 3, 2012 #11
    I can't understand, sorry :)

    we have 2 single particles, both with J=S=0. So L must be 0.
    if we compose the 2 system, we must have all quantum number=0
    So J_f = L_f = S_f = 0.

    where is my error?
     
  13. Jul 3, 2012 #12

    Bill_K

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    Einj's last post said it perfectly.

    When you quote the JP for a single particle, the assumption is that the particle is located at the origin. Orbital angular momentum L = r x p depends on where you put the particle.

    When you combine two particles into a single system, you must also include their relative orbital angular momentum: J = S1 ⊗ S2 ⊗ L. So even if S1 = S2 = 0, the combined system can have nonzero L (and therefore nonzero J) if the two particles are "orbiting" each other.
     
    Last edited: Jul 3, 2012
  14. Jul 4, 2012 #13
    ok, I didn't know this fact. So the rules of composition of angular momentum work exactly with the spin hilbert space, but not for the orbital momentum, where we can add a relative momentum?
    Is there a formalization. maybe in some basic quantum mechanics book?
     
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