What is the theoretical difference between the physical elementary interaction that causes an e+ to attract an e− when they exchange a virtual photon? Why is this exchange different from an e-e- scattering (which produces repulsion)?
This question is related to but more specific than the recently posted The exchange of photons gives rise to the electromagnetic force. It has not been answered there. It would be good to see the answer somewhere.
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$\begingroup$ Isn't this basically the same thing as physics.stackexchange.com/questions/2244/…? If not, it'd help to edit the question to specify exactly how this one differs from #2244 and why the answers to that question don't cut it for you. (And in that case I will certainly reopen the question) $\endgroup$– David ZDec 25, 2010 at 0:08
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$\begingroup$ @David: It looks like this one might be contained within the question you linked, but latter is so vague that it's hard to say (and I don't think this one will be answered on the linked one). Anyway, this one should be at least edited to make the text longer than the title. =P $\endgroup$– MalabarbaDec 25, 2010 at 0:12
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2$\begingroup$ I think this is a great question that shows appreciations for QED going a little beyond qualitative hand-waving. In particular, one has to understand the difference between $e^- e^-$ scattering and $e^- e^+$ scattering which are great elementary processes in their own right. So I'd like it to be opened. At the same time though, I suggest OP to formulate the body of the question clearer and remove the speculative statements that just decrease the value of this great question, in my opinion. $\endgroup$– MarekDec 25, 2010 at 0:57
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2$\begingroup$ @sigoldberg1: Edit your question to mention the one linked by David. Explain that that question is not likely to yield the specific answer you are looking for. Then specify a little better what you are looking for: "Is there an understanding of the physical interaction the causes an $e^+$ to attract an $e^-$ when they exchange a virtual photon? Why is this exchange different from an $e^-" $e^-$ scattering (which produces repulsion)? $\endgroup$– MalabarbaDec 25, 2010 at 4:08
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2$\begingroup$ @David this is a great question and a very specific one, contrary to the one you are comparing it to. And it has a very specific answer, which has to do with Fermi statistics and Wick contraction. You can find the details on pages 121 and 122 of Peskin and Schroeder. I realize how this question is thematically similar to the one you cite. However, the similarity ends there! $\endgroup$– user346Dec 25, 2010 at 8:59
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1 Answer
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Short answer: Calculate the scattering amplitude (matrix element) for electron-electron or electron-positron scattering. They will differ by a minus sign. This means that the electron electron interaction will have a positive sign in front of the 1/r Coulomb potential which corresponds to a positive (repulsive) force. The same procedure for electron-positron scattering will lead to a negative 1/r Coulomb potential and hence a negative (attractive) force.
Longer answer: To decide whether any scattering process is attractive or repulsive you calculate the scattering amplitude (matrix element). Then compare your answer to the Born approximation for nonrelativistic QM scattering to see what classical potential your QFT result corresponds to. If the potential increases as you separate the particles the corresponding classical force will be attractive and vice versa. The minus signs in the scattering calculation can arise from many sources (anticommuting fermion fields, vertex rules, propagator terms, etc.). When you choose different incoming or outgoing states, like electron-electron or electron-positron, the calculation results in an extra negative sign. Interestingly, for the Coulomb force, one of the deciding minus signs comes from the photon vector propagator. Any vector propagator will cause like fermion charges to repel and unlike fermion charges to attract. Fermions don't always have to repel other fermions of the same type. If fermions exchange a scalar particle (instead of a photon) they can attract another fermion of the same charge! Something else to consider is that the Coulomb potential 1/r you arrive at is only for first order Feynman diagrams. If you include higher order terms you get corrections to the usual Coulomb potential/force.
Note: Using a Born approximation from nonrelativistic QM is an admittedly weird way of going about answering this question. It seems strange that we would take a result from our more fundamental theory (QFT) and then have to translate it into a less fundamental theory like nonrelativistic QM. But by asking questions posed in classical or nonrelativistic language (e.g. is this interaction force attractive or repulsive), maybe we are forcing ourselves to use a classical or nonrelativistic approximation/theory.
Check out Peskin pg 125-126 for more on deriving the Coulomb potential/force from QED.
answered Dec 31, 2010 at 8:46