Why 2 photons are formed from positron-electron annihilation at rest

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SUMMARY

Positron-electron annihilation at rest predominantly results in the formation of two photons due to the conservation of momentum and the nature of interaction points in Feynman diagrams. While it is theoretically possible to produce more than two photons, such occurrences are significantly suppressed, with a 3-photon annihilation being suppressed by a factor of 370. The decay of positronium, which can occur in short-lived and long-lived states, further illustrates that only two or three photons are typically produced, with two photons being the most likely outcome due to the simplicity of the interaction. This understanding is crucial for particle physics and radiation detection.

PREREQUISITES
  • Understanding of Feynman diagrams and interaction points
  • Knowledge of particle physics, specifically positronium states
  • Familiarity with conservation laws in physics, particularly momentum conservation
  • Basic principles of photon behavior in particle interactions
NEXT STEPS
  • Research the theoretical framework of Feynman diagrams in particle interactions
  • Study the properties and decay mechanisms of positronium
  • Explore experimental methods for detecting photon emissions in particle annihilation
  • Investigate the statistical mechanics behind photon production ratios in annihilation events
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Particle physicists, researchers in quantum mechanics, and students studying radiation detection and particle interactions will benefit from this discussion.

phosgene
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This isn't a homework question, though it kind of relates to a practical I did recently. Sorry if it's posted in the wrong section!

So, why are only 2 photons formed from positron-electron annihilation at rest? I understand why you can't have just one, as then you won't get conservation of momentum. But why do you only get 2 photons instead of say, 10 or 1000? I've looked up various particle physics textbooks and one on radiation detection, but they only explain that you can't get a single photon.
 
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As far as I know, you can get any number of photons except 1, but the reaction to two photons is really dominating, as it just needs [strike]one[/strike] two vertices ("interaction points") (edit: fixed) in the Feynman graph. In addition, the photons always have the same energy, so it is easy to detect.

3-photon annihilation, suppressed by a factor of 370.

Positronium is interesting, as it has one short-living state (which decays to two photons) and one long-living state (which cannot do this, and has to decay to three photons)
 
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Thanks for the reply. I don't quite understand, though. I thought that positron-electron annihilation always occurs when a positron captures an electron to form positronium, which then decays. If positronium can only decay into 2 and 3 photons, doesn't this rule out the decay into other numbers of photons?
 
Those numbers for positronium are just the most likely numbers (as all lower numbers are forbidden), and a positron/electron pair does not have to form positronium to annihilate.
 
but the reaction to two photons is really dominating, as it just needs 1 vertex ("interaction point") in the Feynman graph
there are two vertex in annihilation.There is a electron or/and positron internal line between the two photons.
 
Oops, thanks. I imagined two and wrote one.
 
Ah, I see. I guess the other mechanism is a plan old collision? The wording of Das and Ferbel's intro to nuclear and particle physics seemed to imply that positronium always formed before an annihilation, so I just assumed that it formed in a collision.
 
This:
http://arxiv.org/pdf/hep-ph/0310099v1.pdf
gives the theoretic ratios of 4 photon annihilation to 2, and 5 photon annihilation to 3, in the region of 1/1 000 000. Experiments are similar, though they have measurement errors.
 

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