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QFT anomaly in electromagnetic, neutral Pion decays

  1. Apr 30, 2005 #1
    The electromagnetic neutral pion decay is a three-point interaction: it decays into two virtual and charged Kaons or Protons, of which one then radiates a photon and then annihalates with the other to produce a second photon. (Obviously, a neutral particle cannot radiate photons directly)


    1.) Let's go into the CMS of the neutral Pion. There, it neither has spin nor orbital momentum. The total angular momentum is therefore J=L+S=0 (where + denotes the spin addition). The Pion is a P=-1 eigenstate of parity and a C=+1 eigenstate of the charge conjugation operator. The C value of the two photons is C=(-1)(-1)(-1)^(l+s). Because of momentum conservation, it is L=0 and therefore l=0. Conservation of J then requires that the two photon spins are antiparallel, which requires S=0 and therefore s=0, where s is the spin-3-component. This shows that the two-Photon decay agrees with C-conservation. Using a similar argumentation, the one-Photon decay contradicts C-conservation.

    Let's have a look at parity conservation. The two photons have intrinsic parity -1 each, which "adds" up to +1. Orbital angular momentum must be zero according to classical conservation laws. So we have a violation of parity. Is that correct?

    2.) Has that to do with the (a?) QFT anomaly, namely Ward-Identities as a substitution for classical conservation laws which get broken by the process of quantization? If yes, could someone please give an overview of what is actually happening here?
  2. jcsd
  3. Apr 30, 2005 #2


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    You may want to consider posting your series of questions in the Nuclei and Particles section, which I think is more relevant.

  4. Apr 30, 2005 #3
    You are right. Did not see it at first. Seems like I'm not allowed to move it. Who is?
  5. Apr 30, 2005 #4


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    Oh, don't worry about it. When one of the moderators sees this, it'll get moved. Till then, you should just continue as usual with the existing threads.

  6. Apr 30, 2005 #5
  7. May 3, 2005 #6

    Meir Achuz

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    There are some errors in your analysis.
    1. C for two photons is just (-1)(-1)=+1. There is no (-1)^{l+s) because the photon is its own antiparticle.
    2. I don't understand "Because of momentum conservation, it is L=0 and therefore l=0." Angular momentum conservation applies to J and not to L
    3. Conservation of parity from a pseudoscalar \pi^0 requires L=1 for the two photon state, and then the two photons must be in total spin S=1+1=1. This spin state of the photons was verified by looking at their relative polarization. This argument, in reverse, was originally proposed by Yang in the 50's to measure the parity of the pi0. The experiment found crossed polarizations for the photons, showing they were in a spin one state.
    4. The above results just depends on conservation laws, and have nothing to do with more detailed theory.
  8. May 4, 2005 #7
    Thx. But where is the anomaly? :-)
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