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QFT- Compton Scattering Crossection

  1. Jun 26, 2011 #1

    G01

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    1. The problem statement, all variables and given/known data

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

    2. Relevant equations



    3. 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.
     
  2. jcsd
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