Sunday, March 18, 2018

quantum field theory - Diagrams involved in 1-loop electron self-energy in QED


I'm following the derivation of electron self-energy at 1-loop in QED on Peskin-Schroeder, page 216. To second order in the coupling the considered diagram (7.15) is


1-loop diagram, different vertices



The 2-point correlator at second order in the coupling contains, beyond the 2 external fields, 2 interactions: up to integrals and γ-matrices Ω|Tψ(x)ˉψ(y)Aμ(z)ˉψ(z)ψ(z)Aν(w)ˉψ(w)ψ(w)|Ω.


Since every ψ can in principle be contracted with every ˉψ this would provide 3!=6 diagrams, factorizing in three pairs of identical diagrams (so that the factor of 2 cancels against the 1/2 from the second order expansion of the exponential): the one above plus


more 1-loop diagrams


Diagram B is 2-loop and contains a vacuum bubble, so is to be discarded upon the usual vacuum bubble factorization argument.


But why is diagram A not considered in Peskin-Schroeder derivation?



Answer



As suggested in the comments the diagram evaluates to 0, introducing a photon mass μ \begin{equation} \begin{split} \text{Fourier amputated diagram} &= (-ie)^2(-1)\int \frac{d^4k}{2\pi^4} \gamma^{\mu} \text{Tr} \left[ \gamma^{\nu} \frac{\require{cancel}\cancel{k}+m}{k^2-m^2} \right] \frac{-i\eta_{\mu\nu}}{-\mu^2} \\ & \propto \int d^4k \, \gamma^\mu k_\alpha \frac{\text{Tr}\left[\gamma^\mu \gamma^\alpha \right]}{k^2-m^2} \\ & \propto \int d^4k \, \frac{\cancel{k}}{k^2-m^2} = 0 \end{split} \end{equation}


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