Assume we have the abelian gauge theory with single fermion. Suppose the following diagram:
Here the initial "photon" γ is in the same state as the final one, so this is the diagram of self-energy. I want to check whether the optical theorem holds for it.
The optical theorem for the total amplitude reads 2ImMγ→γ=(2π)4∑n|Mγ→n|2δ(pn−pγ)≃Γγ
The question: have I also take into account the diagram which is also 6th order of perturbation theory?
An edit
It seems that the formal answer is indeed yes. Really, the rhs of (1) contains the half of the squared diagram, and the diagram above is needed in the lhs.
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