QED: Electron/Positron Scattering & Bosonic Propagators

[eoghan]

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:
[tex]T=<f|H|i>+\sum<f|H|n>\frac{1}{E_i-E_n}<n|H|i>+...[/tex]
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 [tex]e[/tex], so the first term in the approximation becomes
[tex]e<f|i>[/tex]
and the second term gives:
[tex]\frac{e^2}{E_i-E_{\gamma}}<f|i><i|f>[/tex]
Again, looking at the final formula I see that <i|f>=1 so that the second term becomes:
[tex]\frac{e^2}{E_i-E_{\gamma}}[/tex]

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.

 

[fzero]

You seem to be missing a lot of background. Rather than backtrack from the answer to try to guess at how to solve the problem, you should go back and figure out the very basics again.

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:
[tex]T=<f|H|i>+\sum<f|H|n>\frac{1}{E_i-E_n}<n|H|i>+...[/tex]
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.

This is true.

Moreover there is just one intermediate state |n> which is the state of the virtual photon.

Not quite. Each intermediate state has an electron, a positron and a photon. We sum over all energies of the virtual photon, so a continuum of states actually contributes to the sum above.

Now, looking at the final formula for T and doing a sort of back-engineering I guess that the perturbation term H is [tex]e[/tex], so the first term in the approximation becomes
[tex]e<f|i>[/tex]

You should really look through your notes to figure out what Hamiltonian [tex]H[/tex] that you're using. Written in terms of quantum fields, it should look like [tex] H\sim e \bar{\psi}\gamma^\mu \psi A_\mu[/tex] where [tex]\psi[/tex] is the quantum field for the electron and [tex]A_\mu[/tex] is the photon field. Since neither the initial nor final state has a photon in it, [tex] \langle f| H| i\rangle =0[/tex].

and the second term gives:
[tex]\frac{e^2}{E_i-E_{\gamma}}<f|i><i|f>[/tex]
Again, looking at the final formula I see that <i|f>=1 so that the second term becomes:
[tex]\frac{e^2}{E_i-E_{\gamma}}[/tex]

It's a bit more complicated than this, since we're summing over all intermediate states. Furthermore, because of momentum conservation, the electron and positron have different momenta in the intermediate state as compared to the initial state, so [tex]|n\rangle \neq | \gamma\rangle |i\langle[/tex].

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>?

[tex]\langle i|f\rangle \neq 0[/tex], only [tex]\langle i|H|f\rangle =0 [/tex], see above.

2)Why the perturbation hamiltonian is constant and equal to the charge of the electron?

It isn't, check your notes or text to figure out what Hamiltonian they're using.

 

[eoghan]

You seem to be missing a lot of background.

Yes, it is. The course is just an introduction to particle physics. I've studied quantum mechanics and relativistic quantum mechanics, but I haven't studied yet QFT. The notes just says that the propagator is found by applying the second order perturbation theory. I thought it was simpler, but it's too complicated! I think I'm going to learn the propagator by heart without trying to justify it.
Thank you!