# Why not regard all large gauge transformations as genuine ones?

A large gauge transformation is a gauge transformation that is not connected to the identity. When quantizing a gauge theory, we must take configurations related by ordinary gauge transformations to represent the same physical state, but it is ambiguous whether large gauge transformations should be considered as true gauge transformations.

For example, the typical treatment of Yang-Mills compactifies space to $$S^3$$, and finds several vacua $$|n \rangle$$ which are related only by large gauge transformations. Since instantons allow tunneling between them, the physical vacuum is a $$\theta$$-vacuum of the form $$|\theta \rangle = \sum_n e^{i n \theta} |n \rangle.$$ However, E. Weinberg presents a different view in his book Classical Solutions in Quantum Field Theory, with more detail in this paper. Suppose one works in a gauge where there is a one-to-one correspondence between $$F_{\mu\nu}$$ and $$A_\mu$$, e.g. $$A_3 = 0, \quad A_2|_{z = 0} = 0, \quad A_1|_{y = z = 0}, \quad A_0|_{x = y = z = 0} = 0.$$ There is no gauge freedom here because the first condition leaves only $$z$$-independent gauge transformations, then the second leaves only $$z$$-independent and $$y$$-independent gauge transformations, and so on. Hence there is a unique vacuum, corresponding to $$A_\mu = 0$$, and there is no such thing as a $$\theta$$-vacuum. The key here is that establishing this gauge requires large gauge transformations, so Weinberg has implicitly taken them to be do-nothing operations.

Though this formulation is different from the usual one, it seems to give all the same physical predictions. For example, instantons still exist, but they are tunneling events from one vacuum to itself, analogous to a pendulum rotating by a full turn. The observable effects of instantons, such as baryon number violation, hold just as well. The $$\theta$$-term of QCD need not be induced by the $$\theta$$ vacuum, but can simply be put into the Lagrangian since it is allowed by symmetries.

Hence for Yang-Mills, we seem to lose nothing by taking all large gauge transformations to be do-nothing operations, and we gain simplicity and clarity. Are there any downsides? Specifically, is there any measurable quantity that Weinberg's formalism would get wrong, and the more common one would get right? More generally, why don't we always mod out by large gauge transformations?

• If you treat large gauge transformations as redundencies you gauge out the whole Aharonov-Bohm effect. – David Bar Moshe Jul 31 '18 at 8:19
• @DavidBarMoshe I don't see why that would be. In the bundle picture ($U(1)$ bundle over, say, $S^1$) I thought the Aharanov-Bohm phase was built into the bundle's transition functions, and hence cannot be changed by any gauge transformation, large or small. – knzhou Jul 31 '18 at 10:15
• @DavidBarMoshe I wouldn't call what you propose a large gauge transformation. My impression is that large gauge transformations would have $\phi(2 \pi) = \phi(0)$ but with the map $\phi(\theta)$ not homotopic to the identity, as it can nontrivial winding. Such maps only change the Aharanov-Bohm phase by multiples of $2\pi$, so there's no problem. – knzhou Jul 31 '18 at 11:04
• @DavidBarMoshe I define a gauge transformation to be a fiber-preserving bundle automorphism $P \to P$, and a large gauge transformation to be one not connected smoothly to the identity. Your proposed gauge transformation is not a map to begin with, since it's not single valued. – knzhou Jul 31 '18 at 13:41
• Another argument: the Aharahov-Bohm effect relates a phase to a magnetic flux. Gauge transformations, large or small, can never change the magnetic flux, so they can't change the Aharahov-Bohm phase. – knzhou Jul 31 '18 at 13:42

## 1 Answer

You cannot generally use large gauge transformations as "do nothing" operations because while they connect classically equivalent systems, the quantizations of these systems may be inequivalent, cf. this answer by David Bar Moshe and references therein.

In short, the example given there is the Witten effect (named after Witten's description of it in "Dyons of charge $eθ/2π$") producing dyons with fractional electric charge when $\theta$ is non-zero. The dyon appears as "the monopole state" when quantizing the theory modulo small gauge transformations.

On a more abstract level, it is the usual quantization process itself that means that quantumly only the small gauge transformations are guaranteed to "do nothing": Both the Dirac-Bergmann recipe and the BRST formalism focus on the algebra of gauge transformations and enforcing its "do nothing" character on the quantum physical space of states. Nothing in these recipes enforces that large gauge transformations would be "do nothing" operations in the quantum gauge theory, since the algebra can only ever exponentiate to transformations connected to the identity. You might get away with claiming they do, but the standard quantization of gauge systems gives you no solid footing to do so, not even by the laxer standards of physicists.

• I don't think the Witten effect is an example. My impression is that the physical effects of the 'inequivalent quantizations' you talk about are already accounted for by the $\theta$-term in Weinberg's formalism. (Skimming a derivation of the Witten effect, it seems to depend only on the presence of this term, not on the $\theta$-vacuum structure itself.) Is there really a direct dependence on the fact that there are multiple distinct vacua? – knzhou Jul 30 '18 at 17:48
• Indeed I agree that the standard quantization schemes don't say anything either way, but from my perspective the "Weinberg scheme" which makes all large gauge transformations do nothing doesn't have any disadvantages. You might consider it an additional postulate, if you like. Is there anything wrong with doing that? – knzhou Jul 30 '18 at 17:50
• @knzhou I cannot access Weinberg's paper so I cannot tell exactly what his scheme is. However, it almost sounds as if he fixes the gauge prior to quantization. When this is possible - i.e. in the absence of Gribov ambiguities - then this is a perfectly valid choice, but the theory with completely fixed gauge is not a gauge theory anymore. You're not implementing any gauge transformations as "do nothing" operations, there just are no gauge transformations. – ACuriousMind Jul 30 '18 at 17:54