Consider the Lagrangian of the Gupta-Bleuer formalism given by: $$\mathcal{L} =-\frac{1}{4}F^{\mu\nu}F_{\mu\nu} -\frac{1}{2\xi}(\partial A)^2.$$ I understand the necessity of the gauge fixing term: Without it, we would have $\Pi^\kappa=F^{\kappa 0}$ and therefore $\Pi^0=F^{00}=0$ due to the antisymmetry of the electromagnetic tensor, resulting in a contradiction with the commutator $[A^0(t,x),\Pi^0(t,x')]=i\delta^3(x-x')$ given by canonical quantization.

The Wikipedia page of gauge fixing claims, that the Feynman gauge $\xi=1$ and the Yennie gauge $\xi=3$ are often the most useful ones for calculations. I understand this for the Feynman gauge, as for example the Gupta-Bleuer propagator is (compare with equation 4.95 here for example) given by: $$\widetilde{\Delta}_{\mathrm{GB},\mu\nu}(k) =-\frac{i}{k^2+i\epsilon}\left(\eta_{\mu\nu}+(\xi-1)\frac{k_\mu k_\nu}{k^2}\right).$$ But where is the Yennie gauge useful? So far, I have not seen an example or found one. ("Yennie" is only mentioned in eight posts in total on PSE.) What is an example or a sketch of a derivation in Gupta-Bleuer formalism or QED, where the Yennie gauge is useful, or an explanation why it is special enough to deserve its own name?

  • $\begingroup$ Did you do your research before posting this question? Literally the first result upon Google searching “Yennie gauge” is the article sciencedirect.com/science/article/pii/0003491680903309”. The very first paragraph of this paper answers your question. $\endgroup$
    – Prahar
    Commented Jun 7, 2023 at 14:32
  • $\begingroup$ @Prahar You can be assured I did do my research. (Given a seemingly simple question like this, it would be rather inappropriate not to do so.) But as my question indicates, I have searched mostly with "Gupta-Bleuer formalism" added or through scripts about it, that have only mentioned, but not explained the existence of the Yennie gauge - leaving me to assume, that there must be a deeper reason. Thank you for the link, the paper is not too long to not give it a full read. $\endgroup$ Commented Jun 7, 2023 at 15:30

1 Answer 1


Well, the answer to your question is probably found the easiest in the original article by Yennie, Frautschi and Suura They found that the charge renormalization constant in QED is free of infrared divergencies in Yennie gauge (in the article it is shown in eq.(5.9)) to all orders.

Actually I am not aware of any other use of this specific gauge choice, but maybe another answer will give more insights on that.

  • $\begingroup$ Thank you! It seems like footnote 13 on page 430 about "the gauge, in which the renormalized photon propagator has the form $\delta_\nu^{\;\mu}/k^2$ for small $k^2$" is a simple answer, that doesn't even require the need to understand most of or all of the paper. $\endgroup$ Commented Jun 7, 2023 at 17:37

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