QFT- Compton Scattering Crossection

[G01]

Homework Statement

I'm working through the derivation of the Compton scattering crossection and I'm getting stuck partway through. More details below:

Homework Equations

The Attempt at a Solution

So, I'm following along with Zee section II.8.

We have two diagrams that contribute to order e^2, one that corresponds to the following amplitude:

[tex]A(\epsilon',k',\epsilon,k)=ie^2\bar{u}'\frac{{\not}\epsilon' {\not} \epsilon {\not} k}{2pk}u[/tex]

The other diagram corresponds to [itex]A(\epsilon,-k,\epsilon',k')[/itex]

Now, I want to find the crossection, which means I have to square the total amplitude, so I end up with a term like:

[tex]|A(\epsilon',k',\epsilon,k)|^2=\frac{e^4}{(2pk)^2} Tr(u'\bar{u}' {\not}\epsilon' {\not} \epsilon{\not}k u\bar{u} {\not}k {\not} \epsilon {\not}\epsilon') [/tex]

[tex]=\frac{e^4}{(2pk)^2} Tr(({\not}p'+m) {\not}\epsilon' {\not} \epsilon{\not}k ({\not}p +m ){\not}k {\not} \epsilon {\not}\epsilon') [/tex]From here, we simplify, and multiply everything out. The m^2 term drops out because [itex]{\not}k{\not}k=k^2=\mu^2=0[/itex] where mu is the photon mass.

The [itex]{\not}p m[/itex] terms involve odd numbers of gamma matrices and thus drop out when we take the trace. So, we're left with:

[tex]=\frac{e^4}{(2pk)^2} Tr({\not}p' {\not}\epsilon' {\not} \epsilon{\not}k {\not}p{\not}k {\not} \epsilon {\not}\epsilon') [/tex]

Now we simplify using the normalization for the polarization vectors and the anti commutation relations. We get down to:

[tex]=\frac{e^4}{(2pk)} Tr({\not}p' {\not}\epsilon' {\not} {\not}k {\not} \epsilon) =4\frac{e^4}{(2pk)}(2(k\epsilon')^2 +k'p))[/tex]

This last step is where I am getting stuck. I can fill in all the gaps up to this point. This is what I do:

I anticommute the k and epsilon' and then use the normalization [itex]{\not}\epsilon'^2=-1[/itex]

Thus, I am left with:

[tex]=4\frac{e^4}{(2pk)}Tr({\not}p'(2k\epsilon') {\not} \epsilon'+{\not}p'{\not}k)[/tex]

I don't see how I can get to the result given in the line above. The factor of four comes from the trace over the gamma matrices, but specifically, how do I get a term proportional to [itex](k\epsilon')^2[/itex]

I think I have to express p' in terms of p and k', but I still don't see how I can get to the answer given.

Thanks alot. Any help will be greatly appreciated.

 

[mark726]

Hello! It looks like you're on the right track with your derivation. To get to the result given in the line above, you can use the fact that the trace of an odd number of gamma matrices is always zero. This means that the only term that contributes to the trace is the one with all even numbers of gamma matrices, which is the term proportional to (k\epsilon')^2.

To get to this term, you can use the anti-commutation relations for the gamma matrices and the fact that {\not}p'={\not}p - {\not}k + {\not}k'. This will allow you to rearrange the terms in the trace and get the desired result.

I hope this helps! Let me know if you have any further questions or if you'd like me to go into more detail. Good luck with your derivation!