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Neutral pion decay: JPC conservation

  1. May 27, 2013 #1
    [itex]\pi^0[/itex]s decay to two photons via the EM interaction. The [itex]J^{PC}[/itex] of the pion is [itex]0^{-+}[/itex] and of a [itex]\gamma[/itex] is [itex]1^{--}[/itex].
    [itex]\gamma\gamma[/itex] therefore has [itex]J^{PC} = 0^{++}, 1^{-+}, 2^{++}[/itex].
    This does not match the pion, so how can this decay occur?
     
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  3. May 27, 2013 #2

    Vanadium 50

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    The polarization state is E1 x E2, and as you can see, this is odd under parity.
     
  4. May 27, 2013 #3
    Sorry, I don't quite understand your answer. Are you referring to the polarization of a single photon? If so, then yes I agree: that's why I put the Parity eigenvalue of the photon as -1, making the parity of a two photon system [itex](-1) \times (-1)^L = (-1)^{L+1}[/itex]
     
  5. May 27, 2013 #4

    Bill_K

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    The parity of a two-photon state depends on their relative polarizations. They are both transverse, of course, and in the case of a pion decay the photon polarization vectors are also perpendicular to each other. This is what V50 means by E1 x E2, and such a state has odd space parity.
     
  6. May 27, 2013 #5
    Hmm ok. Is it possible to think of this in terms of combining quantum numbers? So adding two [itex]1^{--}[/itex] systems and obtaining a [itex]0^{-+}[/itex]? If not, what's special about photon which makes this possible?

    Also, does this imply that a two photon system can have either parity depending on what decayed?
     
  7. May 27, 2013 #6

    Vanadium 50

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    The states you wrote down have the photons polarization vectors all pointing in the same direction. In this case, they are perpendicular to each other.
     
  8. May 28, 2013 #7

    Bill_K

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    Consider the case where the pion decays at rest, and the two photons are emitted along the ± z-axis. A photon can either have helicity + (spin parallel to momentum) or helicity - (spin antiparallel to momentum) Since the total Jz of the two photons must be zero, they are either both helicity + or both helicity -. Call these states |++> and |-->.

    Now the parity operation reverses helicity, so neither of these states is an eigenstate of parity. Rather the eigenstates are (|++> + |-->)/√2 (even parity) and (|++> - |-->)/√2 (odd parity). The odd parity state is the one we want.

    So yes.

    If you write the helicity states in terms of the states of transverse polarization, i.e. |±> = (|x> ± i |y>)/√2, you'll see that in the odd parity state the polarizations of the two photons come out perpendicular.
     
    Last edited: May 28, 2013
  9. May 28, 2013 #8
    Brilliant, thanks for both your help in understanding this!
     
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