In QED, the Lagrangian with gauge-fixing terms is $$\tag{7.2}L=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-\frac{1}{2}\xi(\partial^\sigma A_\sigma)^2 $$ (See Greiner field quantization), from which we can obtain the propagator to be (in momentum space)

$$\tag{3}D_F^{\mu\nu}(k)=\frac{-g^{\mu\nu}}{k^2+i\epsilon}+\frac{\xi-1}{\xi}\frac{k^\mu k^\nu}{(k^2+i\epsilon)^2}$$ (see Greiner field quantization page 190).

I understand that $\xi$ drops out at the end of the calculation of any QED scattering cross sections. My question is: how to see it explicitly and directly at the level of perturbation theory, using Feynman diagrams? (No path integrals.)

Greiner did such an explicit example in chapter 8 on top of page 247 for the simplest case of electron electron scattering up to 2nd order perturbation. In general I don't immediately see how to do it because for individual Feynman diagrams $\xi$ actually gives different values, it is only when we sum over different Feynman diagrams does the contribution of $\xi$ drop out.

$\textbf{EDIT}:$ I now realized this is more or less the Ward identity. Similar to how Peskin and Schroeder argues that for the purposes of computing S matrix elements we can make the abbreviation (7.75) on page 246. The proof of the Ward identity is diagrammatic.


1 Answer 1


Greiner is only considering a few specific Feynman diagrams at tree-level. For a systematic approach OP's question is best asked within the framework of BRST quantization. The scattering cross sections and $S$-matrix elements can be expressed in terms of correlation functions, which in turn represents Feynman diagrams. Then one can show that BRST-invariant correlation functions are independent of the gauge-fixing condition, cf. my Phys.SE answer here.

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    $\begingroup$ This is not what OP is asking. They are asking for a diagrammatic/combinatorial proof. Your answer is still based on the path integral, at least implicitly. Also, in your other answer you say "one may show that..." but give no proof, so again this is not what OP is asking... $\endgroup$ Oct 17, 2022 at 13:01
  • $\begingroup$ I updated the answers. $\endgroup$
    – Qmechanic
    Oct 17, 2022 at 13:46

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