The gauge symmetry in classical pure Yang-Mills theory with a gauge field $A_{\mu}$ requires an action $S$ to be invariant under continuous transformations $$ A_{\mu}(g) \to g(A_{\mu} + i\partial_{\mu})g^{-1} $$ When we talk about quantized theory, we're dealing with the Hilbert space of rays $|\Psi (A_{\mu})\rangle$, which must be invariant under unitary transformation $U(g)$: $$ \tag 0 |\Psi(A_{\mu})\rangle \to U(g)|\Psi (A_{\mu})\rangle = |\Psi(A_{\mu})\rangle $$ Equivalently, for infinitesimal transformation with generator $G(x)$ one has to require $$ G(x)|\Psi (A_{\mu})\rangle = 0 $$ This reduces the Hilbert space by projecting it on a space with only physical gauge field polarizations. That's why the gauge symmetry is called a do-nothing transformation.

Next, suppose the "large" gauge transformation whose element $g_{(n)}$ carries a non-zero winding number $n$. We have that for the vacuum state on zero winding number configuration $|0\rangle$ $$ \tag 1 U(g_{(n)})|0\rangle = |n\rangle $$ One can introduce a $\theta$-vacuum defined as $$ |\theta\rangle =\sum_{n}e^{in\theta}|n\rangle, $$ so $$ \tag 2 U(g_{(n)})|\theta\rangle = e^{in\theta}|\theta\rangle $$ So we have that "large" gauge transformations are not do-nothing transformations; moreover, even after introducing the new vacuum it still acts non-trivially!

My questions are:

  • $(0)$ corresponds to invariance of classical gauge theory action under local gauge transformations. To which corresponds $(1)$? Naively I think that large gauge transformations change the gauge field strength tensor, but I would like to formalize this.

  • Finally, whether a classical analog of $(2)$ exists? Is this correspondence completely determined by classical gauge fields topology?


2 Answers 2


That large gauge transformations are not true gauge transformations (i.e. yield physically distinct states) is a purely quantum phenomenon due to a choice of quantization procedure that is present in the cases where there are large gauge transformations. Classically, large gauge transformations are always gauge transformations, i.e. trivial on the physical state space. See also this answer by David Bar Moshe.

Essentially, the special status of large gauge transformations arises from the fact that the quantization procedure for a gauge theory only imposes that applying the generators of gauge transformations to physical states must yield zero, and hence the physical states are invariant under gauge transformations generated by them. But, rather by definition, the transformations generated by the generators only yield the gauge transformations connected to the identity (the exponential map of a Lie algebra maps to the connected components of the corresponding group). Therefore, the quantization procedure by design only imposes invariance of the quantum theory under small gauge transformations.

There is no good reason to demand that the quantum theory be invariant under large gauge transformations because it is well-known that the same classical system can have different inequivalent quantizations, and the large gauge transformations simply become the transformations between these inequivalent quantizations, which seems physically reasonable - given a classical theory, its full quantum theory should be the "sum" of all possible quantizations.

  • 1
    $\begingroup$ Can you specify or give a reference for how "large gauge transformations simply become the transformations between these inequivalent quantizations"? In addition, for the usual gauge groups (SU(2), SU(3)), how can there be elements that are not connected to the identity? For example, SU(2) is the three-sphere $S^3$ and therefore it is far from obvious how a point on this sphere could not be connected to the identity element... $\endgroup$
    – jak
    May 13, 2017 at 9:32
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    $\begingroup$ @JakobH See the paper by Landsman linked in David Bar Moshe's answer. The group I'm talking about here is not the gauge group, but the group of gauge transformations, which is an infinite-dimensional group that can be not connected even if the gauge group is. $\endgroup$
    – ACuriousMind
    May 13, 2017 at 18:40

Some of the original papers on $\theta$ vacua already pointed out that they have no classical analogues. The physical processes they vacuum mixing describes are pure tunneling, and tunneling through a barrier does not exist in classical dynamics.

  • $\begingroup$ The tunneling interpretation is just one of arguments clarifying of importance of $\theta$-vacuum in pure gauge theories from physical point of view. One can argue this in another way, by requiring the state ray to be unchanged under acting a large gauge transformation operator on it (up to a phase, which is the origin of my question). But when doing this we operate with the gauge field topology only, which is classical. And therefore there may be classical analog of this variance. For example, smooth deformation of pure gauge belonging to a class $n$ to the one belonging to $m$. $\endgroup$
    – Name YYY
    Jan 25, 2017 at 14:25
  • $\begingroup$ @NameYYY The fact that there can be a residual phase difference between the original and the transformed state is intrinsically quantum mechanical. $\endgroup$
    – Buzz
    Jan 25, 2017 at 14:27
  • $\begingroup$ But what does it say about the gauge symmetry as "do-nothing" symmetry? And what about my first question (about explicit classical analog of gauge variance in $(1)$)? $\endgroup$
    – Name YYY
    Jan 25, 2017 at 14:30
  • $\begingroup$ @NameYYY Well, your first question's idea that large gauge transformations change the field strength is not correct. They don't; otherwise they wouldn't be pure gauge transformations. Beyond that, there isn't anything to say, because classically the state of the gauge field is defined solely by its field strength; only quantum mechanically (because states with the same field strength but different underlying gauge potentials can have different phases and thus interfere) do the pure gauge modes matter at all. $\endgroup$
    – Buzz
    Jan 25, 2017 at 14:33
  • $\begingroup$ but one can still define configurations of gauge potential on which the strength tensor vanishes. Suppose the gauge transformation $A_{\mu} = 0 \to A_{\mu} = g\partial_{\mu}g{-1}$, where $g$ carries non-trivial winding number. This is large gauge transformation. Is the strength tensor $F_{\mu\nu}$ changed under this transformation; i.e., is this transformation continuous (having zero commutator $[\partial_{\mu},\partial_{\nu}]g$ for simplicity)? $\endgroup$
    – Name YYY
    Jan 25, 2017 at 14:42

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