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QED as a gauge invariant theory 
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#1
Mar2311, 04:46 PM

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I'm just beginning to learn about Feynman diagrams and wanted to make sure I've got the correct basic understanding of QED. This is what I believe to be true right now:
QED allows us to describe the interaction between an EM field and light/matter. The QED vertex is composed of a photon and one particle before and after the interaction. If for example an electron interacts with an EM field we can describe this interaction by saying that a virtual photon is transferred from the electron. This photon adds grad(a) to the vector potential of the field and da/dt to the scalar potential, thus the field is unchanged. If this is true, can we reverse the argument and say that a photon can interact with an EM field via the exchange of virtual electrons? I may be completely wrong, just finding my lecture notes a bit difficult to comprehend. Thanks. 


#2
Mar2311, 05:49 PM

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PF Gold
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No. A photon carries no electric charge and as such does not interact with the electric field.



#3
Mar2311, 06:22 PM

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PF Gold
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There is indeed a 1loop contribution to the process [tex]\gamma\gamma \rightarrow\gamma\gamma[/tex]. Any charged particle can run in the loop, but the electron dominates because of its small mass.



#4
Mar2311, 07:16 PM

P: 29

QED as a gauge invariant theory
So is what I said about the electron interacting with the EM field correct? 


#5
Mar2311, 07:23 PM

P: 29

I think I see the answer now. If we know the direction of the arrows for the horizontal propagators then we must know those of the vertical propagators. 


#6
Mar2311, 08:03 PM

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PF Gold
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#7
Mar2311, 08:05 PM

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#8
Mar2411, 12:21 PM

P: 29

Quick question, if there are n Feynman diagrams possible at first order say, is the rate proportional to n^2? I would have thought so since the matrix element is given by the expression M=M1+M2+M3+.... where 1,2 and 3 are the consecutive orders. I've been looking at a few examples that say the rate is proportional to n which is why I'm confused.



#9
Mar2411, 01:14 PM

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PF Gold
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For instance [tex]e^+e^\rightarrow e^+e^[/tex] gets contributions from either an intermediate photon or [tex]Z^0[/tex] particle. When the energy of the initial particles is very small, the contribution from the intermediate [tex]Z^0[/tex] is extremely small. 


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