I am a bit puzzled by a statement in this question here. In particular, the claim that the residual gauge symmetry in Yang-Mills theory disappears upon Wick rotation to the Euclidean theory.

For concreteness, lets consider a gauge group $U(1)$, so our action is simply $$ S=-\frac{1}{4}\int d^{4}x F_{\mu\nu}F^{\mu\nu} $$ Where $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$ is the field strength tensor, and for now we are in Minkowski space with signature $(-1,+1,+1,+1)$.

It is clear that this action has a gauge symmetry, $A_{\mu}\to A_{\mu}':=A_{\mu}+\partial_{\mu}\Lambda$, for an arbitrary scalar function $\Lambda$. Furthermore, by choosing $\Lambda$ such that $\Box\Lambda=-\partial^{\mu}A_{\mu}$, we find that we can set $\partial^{\mu}A_{\mu}'=0$, this is called the Lorenz gauge.

There is a residual gauge freedom given by choosing $\Lambda'$ such that $\Box\Lambda'=0$, since under this gauge transformation $\partial^{\mu}A_{\mu}'\mapsto \partial^{\mu}A_{\mu}'':=\partial^{\mu}A_{\mu}'+\Box\Lambda'=0$. So to fully fix the gauge, we also have to choose a function satisfying $\Box\Lambda'=(-\partial_{t}^{2}+\partial_{x}^{2}+\partial_{y}^{2}+\partial_{z}^{2})\Lambda'=0$.

The claim is that after a Wick rotation to Euclidean theory, this residual gauge freedom disappears. Definging $t=-i\tau$, the Wick rotated action is: $$ S^{E}=\frac{1}{4}\int d^{4}xF_{\mu\nu}F^{\mu\nu} $$ where the signature is now $(+1,+1,+1,+1)$.

This action again has a gauge symmetry $A_{\mu}\to A_{\mu}+\partial_{\mu}\Lambda$, and we can again choose Lorenz gauge by picking a $\Lambda$ such that $\Box\Lambda=-\partial^{\mu}A_{\mu}$, where now $\Box=\partial^{\mu}\partial_{\mu}=\nabla^{2}$, the Laplacian.

It appears to me that we once again have residual gauge freedom. Namely, that we can transform by a function $\Lambda$ such that $\nabla^{2}\Lambda=0$, so it really seems like the residual gauge symmetry has not disappeared in the Wick rotated theory!

My guess is that I am either misinterpreting what it means for the residual gauge freedom to disappear, or that I am missing an issue with the existence of solutions to the equations defining $\Lambda$ in Euclidean space. I would really appreciate some help resolving this issue.


1 Answer 1


In chapter 74 of Srednicki's QFT book (which the previous Phys.SE post refers to) it is implicitly assumed that in Euclidean 4D $x$-space the gauge field $A_{\mu}$ vanishes sufficiently fast as $|x|\to \infty$. In particular, large gauge transformations are implicitly excluded. In this case, the only harmonic function $\Lambda$ (i.e. such that $\Box\Lambda=0$) is the zero-function $\Lambda\equiv 0$, cf. the maximum principle, so that the gauge is completely fixed.

  • $\begingroup$ Thank you, this answer is definitely starting to clear things up. However, since (in Lorenz gauge), we have the equation of motion $\Box A_{\mu}=0$ for the gauge fields, would this not also imply that the $A_{\mu}$ themselves are rapidly decaying harmonic functions and thus zero? $\endgroup$
    – CoffeeCrow
    Nov 29, 2021 at 0:18

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