A question about Feynman diagrams

  1. jcsd
  2. If time is going from left to right, then the diagram on the left means an electron and a positron annihilate and become a photon, which then decays into an electron-positron pair. The diagram on the right shows an electron and positron exchanging a photon.
     
  3. yes, but this photon exchange (for the diagram on the right) - does it happen instantaneously?
     
  4. No that does not mean it happens instantaneously. The "axes" on the Feynman diagrams should not be taken too literally. For example, many diagrams are drawn with exchanged particles following curved paths, but this does not mean that it actually does this.
     
  5. Bill_K

    Bill_K 4,160
    Science Advisor

    In calculating the Feynman amplitude, both ends of the photon line are integrated over. That will include all cases: t1 < t2, t1 > t2, and t1 = t2.
     
  6. The cliche thing that answers this question is that Feynman diagrams don't show the real physical process that takes place, but instead give us a nice insight in order to calculate amplitudes of different processes.
    In that sense, the above diagrams do give you information on annihilation and scattering... However what happens in real is what you really measure, and that's the amplitudes (or better the cross section), which the diagrams allow you to easily calculate.
     
  7. This is a virtual photon. It can travel faster than light.
     
  8. I see. Thank you all
     
  9. Actually, it looks like the diagrams would be the same if one of them is rotated by 90 degrees. Since timing (which event precedes which) doesn't matter, shouldn't they be identical?
     
  10. formally they are identical (the term describing the process in the Lagrangian is the same, the coupling strengths are the same).
    Physically they describe something different. The one is annihilation and the other is electromagnetic scattering.
     
  11. but then isn't it mathematically counting twice? if topologically the diagrams are indistinguishable (since rotating does not change the topology) and we are counting both diagrams
     
  12. you have to count them in if you don't know what process is taking place... For example if someone told you to calculate the cross section of an electron and positron, in that case you'd have to take both diagrams in consideration...
    The same thing you'd do if you had 2 identical particles (two possible configurations)
     
  13. ok, but what makes them distinguishable topologically? is it the fact that when we rotate them it's not exactly the same diagram but rather electron is switched with a positron? otherwise, if it is exactly the same, I could for example rotate the diagram in 45 degrees. the process would be the same (only timing of the events are different) and we would count the diagram again. for each rotation we would have a different diagram...
    I'm sure it doesn't work this way, I just want to understand
     
  14. When deciding whether two diagrams are "distinguishable", you can't move the external legs, only the internal lines.
     
  15. No. you can do the rotation on your paper for an artistic reason, nothing would change the process in the end.
     
  16. The two diagrams represents two possible channels for the e- e+ interaction. In both, the photon represents a virtual particle (mediator of electromagnetic interaction), then, due to the photon is virtual it can't satisfy the energy-momentum relation. Indeed, remember that into the QFT the propagators are two point functions.
     
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