A question about Feynman diagrams

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Discussion Overview

The discussion revolves around the interpretation and implications of Feynman diagrams in the context of second-order electron-positron scattering. Participants explore the meaning of specific diagrams, the nature of virtual particles, and the mathematical treatment of these diagrams in quantum field theory.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants question whether the photon exchange in the right diagram occurs instantaneously, leading to discussions about the interpretation of time in Feynman diagrams.
  • Others clarify that the axes in Feynman diagrams should not be taken literally, as the paths of exchanged particles do not represent actual trajectories.
  • A participant mentions that in calculating the Feynman amplitude, all time orderings of events are considered, including cases where events occur simultaneously.
  • There is a suggestion that Feynman diagrams do not depict real physical processes but rather serve as tools for calculating amplitudes and cross sections.
  • Some participants note that the diagrams can represent virtual photons, which may not adhere to the speed of light constraints.
  • One participant raises a question about the topological indistinguishability of diagrams when rotated, leading to a discussion about how diagrams are counted in calculations.
  • Another participant emphasizes that while diagrams may appear identical under rotation, they represent different physical processes (annihilation vs. scattering).
  • There is a discussion about the criteria for distinguishing diagrams, focusing on the movement of external legs versus internal lines.
  • One participant asserts that the diagrams represent different channels for electron-positron interactions, with virtual photons as mediators that do not satisfy the energy-momentum relation.

Areas of Agreement / Disagreement

Participants express a range of views on the interpretation of Feynman diagrams, the nature of virtual particles, and the implications of diagram rotation. No consensus is reached on whether the photon exchange is instantaneous or how to treat topological indistinguishability in calculations.

Contextual Notes

Limitations include the dependence on interpretations of time in diagrams, the nature of virtual particles, and the mathematical treatment of indistinguishable diagrams in quantum field theory.

spookyfish
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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.
 
yes, but this photon exchange (for the diagram on the right) - does it happen instantaneously?
 
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.
 
spookyfish said:
For a 2nd order electron-positron scattering, the following diagrams are possible:
http://www.quantumdiaries.org/wp-content/uploads/2010/02/eepp-1024x228.png
My question is: What is the meaning of the right diagram - is a photon created and annihilated instantaneously? how is this possible?
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.
 
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.
 
This is a virtual photon. It can travel faster than light.
 
I see. Thank you all
 
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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