Let $\psi(x)$ be the field of the electron. Its Fourier transformed two-point function reads $$ \langle\psi\bar\psi\rangle=\frac{1}{\not p-m-\Sigma(\not p)}. $$

If we calculate $\Sigma(\not p)$, we observe that it depends on the gauge parameter $\xi$, which in principle is not a problem because $\Sigma(\not p)$ is not observable by itself.

But if we think of a gauge transformation as taking $\psi\to\mathrm e^{i\alpha(x)}\psi(x)$, then the two-point function should satisfy $$ \langle\psi\bar\psi\rangle\to \langle\psi\mathrm e^{i\alpha(x)}\mathrm e^{-i\alpha(x)}\bar\psi\rangle=\langle\psi\bar\psi\rangle $$

Therefore, one would naïvely expect $\Sigma(\not p)$ to be gauge invariant, and therefore it shouldn't depend on $\xi$. What is the solution to this contradiction? Why do our expectations fail?

  • $\begingroup$ The catch, I suspect, lies in the calculation of $\Sigma$. Strictly speaking it is the summation of an infinite number of orders, but in calculations one necessarily have to truncate it at some finite order. This truncation introduces the gauge dependence. If one were able to compute it to all orders, it would be gauge independent, I think. $\endgroup$ Commented Dec 7, 2016 at 11:37
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    $\begingroup$ @flippiefanus if that works it would certainly be satisfying. But Im not 100% sure that that works: after all, when computing $S$ matrix elements, we ask for $\xi$-independence order by order in perturbation theory, right? $\endgroup$ Commented Dec 7, 2016 at 11:41
  • $\begingroup$ How does the gauge transformation $\mathrm{e}^{\mathrm{i}\alpha(x)}$ change $\xi$? In a theory with $\xi$, you have abandoned gauge invariance because you fixed the gauge, so I don't see how the requirement that $\langle\psi\bar\psi\rangle$ be "gauge invariant" is consistent with having fixed the gauge by introducing $\xi$. $\endgroup$
    – ACuriousMind
    Commented Dec 7, 2016 at 14:51
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    $\begingroup$ In any case, the two point function is non-local $\psi(0)\bar\psi(x)$, so it is not gauge invariant. $\endgroup$
    – Thomas
    Commented Dec 7, 2016 at 16:05
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    $\begingroup$ You need $S(x,y)=\langle\psi(x)\bar\psi(y)\rangle$. By Lorentz invariance $S(x,y)$ only depends on $x-y$, and $S(p)$ is the FT in $x-y$. $\endgroup$
    – Thomas
    Commented Dec 7, 2016 at 16:21

3 Answers 3


The propagator $S(p)$ is the Fourier transform of the two-point function $S(x,y)=\langle\psi(x)\bar\psi(y)\rangle$, $$ S(p) = \int \frac{d^p}{(2\pi)^4} \, \exp(-ip\cdot(x-y)) \, S(x,y)\, . $$ Note that because of Lorentz invariance $S(x,y)$ does not depend on $x+y$. Clearly, the two-point function is non-local and not gauge invariant.


As an alternative to Thomas answer, we note that if we write the transformation law explicitly, we get $$ \langle\psi(x)\bar\psi(y)\rangle\to \langle\psi(x)\mathrm e^{i\alpha({\color{red}x})}\mathrm e^{-i\alpha({\color{red}y})}\bar\psi(y)\rangle=\mathrm e^{i(\alpha(x)-\alpha(y))}\langle\psi(x)\bar\psi(y)\rangle $$

We see that the two-point function fails to be gauge invariant because the fields are evaluated at different points and thus the local phases don't cancel off each other. This wasn't evident in the OP because I didn't write the space-time labels explicitly. Silly me.

  • $\begingroup$ In this sense the Dirac field $\psi(x)$ is also gauge non-invariant since it changes. The more important thing is that the variable change is not obliged to preserve the form of equations, if it helps solve the equations. $\endgroup$ Commented Dec 7, 2016 at 17:48
  • $\begingroup$ It is incorrect to take the exponential outside of the expectation value. The gauge parameter $\alpha$ should be considered as an independent (scalar) degree of freedom after quantisation and as such has a non trivial expectation value. See answer below. $\endgroup$
    – nox
    Commented Jan 17, 2021 at 11:28
  • $\begingroup$ @lux I disagree. It is simply not true that $\alpha$ should be taken as an operator. You can introduce a Stückelberg field by performing a gauge transformation whose gauge parameter is an operator. But you most certainly do not have to. In standard QED, without Stückelberg fields, the gauge transformation is definitely a $c$-number and can be pulled out of correlation functions. $\endgroup$ Commented Jan 17, 2021 at 23:25
  • $\begingroup$ OK. Just look at Landau's treatment $\endgroup$
    – nox
    Commented Jan 17, 2021 at 23:26
  • $\begingroup$ @lux Sorry, but no need to: Landau may very well regard $\alpha$ as an operator. I am not saying you cannot do that; you definitely can. I am saying you do not have to if you don't want to. In fact, in the most standard formulation of QED you do not regard $\alpha$ as an operator. Your approach, while indeed correct, is the exception rather than the rule. It is a fine approach, very interesting, but you should keep in mind that is non canonical and you should not be making such categorical claims. $\endgroup$ Commented Jan 17, 2021 at 23:30

The propagator - or any, arbitrary correlation function - depends strongly on the gauge of internal photons (the Ward identity deals with the variations of external photons' gauge).

This was first noted by Landau and Khalatnikov (and around the same time by Fradkin) who basically analyse the quantised version of the gauge transformation field called $\alpha(x)$ by OP:

  • L. Landau, I. Khalatnikov, Sov. Phys. JETP2,69 (1956).
  • E. S. Fradkin, Zh. Eksp. Teor. Fiz.29, 258261 (1955).

The treatment of $\alpha$ as a Stueckelberg type field is clearer in

  • M. A. L. Capri, D. Fiorentini, M. S. Guimaraes, B. W. Mintz, L. F. Palhares, S. P. Sorella, Phys. Rev. D 94, 065009 (2016)
  • T. De Meerleer, D. Dudal, S. P. Sorella, P. Dall'Olio, A. Bashir, Phys. Rev. D 97, 074017 (2018)

For the generalisation to arbitrary Green functions (involving simple products of the fermion field - see comments) see

  • T De Meerleer, D. Dudal, S. P. Sorella, P. Dall'Olio, A. Bashir, Phys. Rev. D 101, 085005 (2020)
  • N. Ahmadiniaz, J. P. Edwards, J. Nicasio, C. Schubert, arXiv:2012.10536 [hep-th]
  • $\begingroup$ It is not true that an arbitrary correlation function depends on the choice of gauge for internal photons. Correlation functions of gauge invariant operators (such as $F^{\mu\nu}$ or $\bar\psi(x)W(x,y)\psi(y)$, with $W$ a Wilson line) are gauge invariant and do not depend on $\xi$. $\endgroup$ Commented Jan 17, 2021 at 23:28
  • $\begingroup$ The original question is talking about correlators of pure products of $\psi$ fields. A new topic should be discussed in a separate question. $\endgroup$
    – nox
    Commented Jan 17, 2021 at 23:30
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    $\begingroup$ Why? You are making a wrong statement, whether relevant to the question or not. It is simply not true that arbitrary correlation functions depend on the gauge. This phrase is false. Whether this answers the question in the OP is entirely irrelevant. If someone asks about black holes, and you claim that 1+2=7, you would still be making a false claim, even if unrelated to black holes. $\endgroup$ Commented Jan 17, 2021 at 23:34

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