
#1
Mar2311, 04:46 PM

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. 



#2
Mar2311, 05:49 PM

Mentor
P: 15,569

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




#3
Mar2311, 06:22 PM

Sci Advisor
HW Helper
PF Gold
P: 2,606

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 theorySo 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

Sci Advisor
HW Helper
PF Gold
P: 2,606





#7
Mar2311, 08:05 PM

P: 29





#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

Sci Advisor
HW Helper
PF Gold
P: 2,606

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. 


Register to reply 
Related Discussions  
Gravity as a diffeomorphism invariant gauge theory (new Krasnov paper)  Beyond the Standard Model  1  
electromagnetic energy is not Gauge invariant?  Classical Physics  4  
gaugeinvariant measure in LQG  Beyond the Standard Model  6  
How to quantize a gauge theory which is invariant onshell?  General Physics  3 