The electron self-energy at one-loop is given by the one-particle irreducible graph

enter image description here

I know how to calculate it using the Feynman rules but I was wondering how this diagram appears in the QED effective action (which it in principal should as a 1PI graph). With \begin{equation} \Gamma = S + i \,{\rm Str}(\ln(S^{(2)})) \end{equation} I do not see how I get a loop with two different particles in it.

So how does this diagram appear in the QED effective action?


1 Answer 1

  1. The 1-loop quantum correction to the 1PI effective action is given by a functional superdeterminant$^1$ $$ \begin{align}\Gamma_{\text{1-loop}}[\phi_{\rm cl}] ~=~&\frac{i\hbar}{2} \ln{\rm sDet}\left(S^{-1}_2 \frac{\delta^2 S[\phi_{\rm cl}]}{\delta \phi_{\rm cl} \delta \phi_{\rm cl}}\right)\cr ~=~&\frac{i\hbar}{2} {\rm sTr}\ln\left(S^{-1}_2 \frac{\delta^2 S[\phi_{\rm cl}]}{\delta \phi_{\rm cl} \delta \phi_{\rm cl}}\right)\cr ~=~&\frac{i\hbar}{2} {\rm sTr}\ln\left({\bf 1}+S^{-1}_2 \frac{\delta^2 S_{\neq 2}[\phi_{\rm cl}]}{\delta \phi_{\rm cl} \delta \phi_{\rm cl}}\right)\cr ~=~&-\frac{i\hbar}{2}\sum_{n\in\mathbb{N}}\frac{1}{n} {\rm sTr}\left(-S^{-1}_2 \frac{\delta^2 S_{\neq 2}[\phi_{\rm cl}]}{\delta \phi_{\rm cl} \delta \phi_{\rm cl}}\right)^n,\end{align}$$ cf. e.g. my Phys.SE answer here.

  2. OP's amputated 1-loop 1PI QED self-energy diagram with 1 fermion propagator and 1 gauge field propagator is included in the term with $n=2$ containing 2 propagators $G^{k\ell}_0=-(S^{-1})^{k\ell}$. The 2 propagators do not need to correspond to the same field.


$^1$For notation, see my Phys.SE answer here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.