eoghan
- 201
- 7
Hi!
I'm studying an introduction to QED and I don't understand the bosonic propagators.
Consider the electron/positron scattering with the exchange of a virtual photon. According to perturbation theory, the propagator is:
T=<f|H|i>+\sum<f|H|n>\frac{1}{E_i-E_n}<n|H|i>+...
where f is the final state and i the initial state.
Now, I guess both the final and the initial states are the states of an electron and a positron, but with different momenta.
Moreover there is just one intermediate state |n> which is the state of the virtual photon.
Now, looking at the final formula for T and doing a sort of back-engineering I guess that the perturbation term H is e, so the first term in the approximation becomes
e<f|i>
and the second term gives:
\frac{e^2}{E_i-E_{\gamma}}<f|i><i|f>
Again, looking at the final formula I see that <i|f>=1 so that the second term becomes:
\frac{e^2}{E_i-E_{\gamma}}
Now, my questions are:
1) the first term in the approximation vanishes in the final formula so <i|f>=0. But then also the second term vanishes! So, what is <f|i>?
2)Why the perturbation hamiltonian is constant and equal to the charge of the electron?
P.s. the final formula of the propagator is obtained by considering two different diagrams. Here I considered only one of them.
I'm studying an introduction to QED and I don't understand the bosonic propagators.
Consider the electron/positron scattering with the exchange of a virtual photon. According to perturbation theory, the propagator is:
T=<f|H|i>+\sum<f|H|n>\frac{1}{E_i-E_n}<n|H|i>+...
where f is the final state and i the initial state.
Now, I guess both the final and the initial states are the states of an electron and a positron, but with different momenta.
Moreover there is just one intermediate state |n> which is the state of the virtual photon.
Now, looking at the final formula for T and doing a sort of back-engineering I guess that the perturbation term H is e, so the first term in the approximation becomes
e<f|i>
and the second term gives:
\frac{e^2}{E_i-E_{\gamma}}<f|i><i|f>
Again, looking at the final formula I see that <i|f>=1 so that the second term becomes:
\frac{e^2}{E_i-E_{\gamma}}
Now, my questions are:
1) the first term in the approximation vanishes in the final formula so <i|f>=0. But then also the second term vanishes! So, what is <f|i>?
2)Why the perturbation hamiltonian is constant and equal to the charge of the electron?
P.s. the final formula of the propagator is obtained by considering two different diagrams. Here I considered only one of them.
Last edited: