# QED as a gauge invariant theory

by jmz34
Tags: gauge, invariant, theory
 P: 29 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.
 Mentor P: 14,634 No. A photon carries no electric charge and as such does not interact with the electric field.
 PF Patron HW Helper Sci Advisor P: 2,595 There is indeed a 1-loop contribution to the process $$\gamma\gamma \rightarrow\gamma\gamma$$. Any charged particle can run in the loop, but the electron dominates because of its small mass. Attached Thumbnails
P: 29

## QED as a gauge invariant theory

 Quote by Vanadium 50 No. A photon carries no electric charge and as such does not interact with the electric field.
But a photon has an EM field?

So is what I said about the electron interacting with the EM field correct?
P: 29
 Quote by fzero There is indeed a 1-loop contribution to the process $$\gamma\gamma \rightarrow\gamma\gamma$$. Any charged particle can run in the loop, but the electron dominates because of its small mass.
For the process (e+)(e-)->photon + photon the simplest Feynman diagram is just half of what you've drawn, but I'm confused about the propagator. I initially joined the two vertices for the above process with a line propagator (with no arrow) but in your diagram you have included arrows. I initially thought that vertices that coincide at the same time cannot have any arrows because it could mean the transfer of a particle of anti particle. On the other hand vertices that do not correspond to the same time can have an arrow on the propagator joining them since we know which way the particle is going (forwards in time). My time axis is horizontal by the way.

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.
PF Patron
HW Helper
P: 2,595
 Quote by jmz34 For the process (e+)(e-)->photon + photon the simplest Feynman diagram is just half of what you've drawn, but I'm confused about the propagator. I initially joined the two vertices for the above process with a line propagator (with no arrow) but in your diagram you have included arrows. I initially thought that vertices that coincide at the same time cannot have any arrows because it could mean the transfer of a particle of anti particle. On the other hand vertices that do not correspond to the same time can have an arrow on the propagator joining them since we know which way the particle is going (forwards in time). My time axis is horizontal by the way. 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.
I'm not sure what you mean about vertices coinciding in time. If we were to compute this amplitude in spacetime rather than momentum space, we would integrate over the location of the vertices. In any case, the arrows are conventional. Matching the directions of the arrows when forming diagrams helps to ensure that charge is conserved, momenta match up, etc. In the case of the loop, they correspond to the direction of the momentum going around the loop.
P: 29
 Quote by fzero I'm not sure what you mean about vertices coinciding in time. If we were to compute this amplitude in spacetime rather than momentum space, we would integrate over the location of the vertices. In any case, the arrows are conventional. Matching the directions of the arrows when forming diagrams helps to ensure that charge is conserved, momenta match up, etc. In the case of the loop, they correspond to the direction of the momentum going around the loop.
OK that makes sense, thanks.
 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.
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HW Helper
I wouldn't say that the question is one that can be answered in general, since to get the rate, we have to compute $$|M|^2$$ and integrate over phase space. Generally some diagrams will contribute more or less to the computation.
For instance $$e^+e^-\rightarrow e^+e^-$$ gets contributions from either an intermediate photon or $$Z^0$$ particle. When the energy of the initial particles is very small, the contribution from the intermediate $$Z^0$$ is extremely small.