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 $\gamma$-matrices $$ \langle \Omega | T \, \psi(x) \, \bar{\psi}(y) \, \, A_{\mu}(z) \, \bar{\psi}(z) \, \psi(z) \, \, A_{\nu}(w) \, \bar{\psi}(w) \, \psi(w) | \Omega \rangle . $$

Since every $\psi$ can in principle be contracted with every $\bar{\psi}$ 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?

  • $\begingroup$ I don't have that book at hand, but: 1) That diagram is the Hartree part of the interaction, so in general has to be included, of course. 2) Probably the authors are talking about some specific situation. For example, if you are studying the Homogeneous Electron Gas, in which the classical part (Hartree) of the coulombian interaction cancels out with the positive background, the only thing remaining to first order is the exchange, which is precisely the first diagram you showed above. $\endgroup$
    – Qwertuy
    Commented Oct 1, 2018 at 12:21
  • 3
    $\begingroup$ Diagram A is zero in dimensional regularisation. Diagram B is disconnected. $\endgroup$ Commented Oct 4, 2018 at 11:58
  • $\begingroup$ The first one is tadpole diagram, essentially it contributes to vacuum. $\endgroup$
    – Turgon
    Commented Oct 6, 2018 at 5:39

1 Answer 1


As suggested in the comments the diagram evaluates to $0$, introducing a photon mass $\mu$ \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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