Pion decay in QED

  • Thread starter Rudipoo
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  • #1
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Hi,

I was wondering whether the decay of the Pi-0 meson in QED to an electron positron pair can occur as follows:

Pi-0 -> virtual photon -> e+e-

or does it have to go via

Pi-0 -> two virtual photons -> e+e- (the Feynman diagram has a 'square' of virtual electrons/photons)?

I have a feeling it is forbidden via the first route but I'm unsure why, unless angular momentum has to be conserved within the diagram as well as for the beginning and end states?

Many thanks.
 

Answers and Replies

  • #2
Vanadium 50
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More fundamental than that. What is the coupling between the pi0 and the photon?
 
  • #3
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The coupling strength is the charge of the up quark multiplied by e, the electronic charge, so (2/3)*e?
 
  • #4
Vanadium 50
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No, that's the coupling to an up quark. What is the coupling to the entire pi0?
 
  • #5
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For the diagram that just has the up/anti-up annihilating to a photon I don't know what else the coupling could be...

For the two photon diagram you'd get (2/3*e)+(-2/3*e)=0?!
 
  • #6
Bill_K
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Rudipoo, as you say, the π0 is composed of a quark and an antiquark. It's like positronium, only with quarks instead of electrons. And we know the answer: that positronium only decays into two photons.

For π0 the quarks in the ground state have their spins anti-aligned. That is, it's a singlet S state. It has space parity P = -1 (a pseudoscalar) and charge parity C = +1. An n-photon state on the other hand has C = (-1)n. Therefore you can only have an even number of photons, and this holds even in the intermediate state.
 

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