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In my answer at What, in simplest terms, is gauge invariance?, I mentioned that in certain contexts there can be a "gauge theory" with a local symmetry that leave the Lagrangian/Hamiltonian invariant and whose degrees of freedom are physically observable. My follow-up got very long, so I decided to address the Lagrangian and Hamiltonian cases separately. Here are my thoughts on the Lagrangian case - anyone else is of course welcome to contribute their own. The case of lattice gauge Hamiltonian is addressed at Hamiltonian lattice gauge theory with physically observable local degrees of freedom.

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  • $\begingroup$ So you mean a gauge degree of freedom, but not a Lagrangian symmetry? $\endgroup$ – Bob Bee Jul 12 '16 at 3:42
  • $\begingroup$ @BobBee Possibly an (emergent) symmetry of an effective Lagrangian. See my answer for the precise statement of the question :-) $\endgroup$ – tparker Jul 12 '16 at 3:50
  • $\begingroup$ Can you give a reference for "A gauge theory is usually defined as containing d.o.f. that are both local and physically unobservables"? E.g. in "Quantization of gauge systems" by Henneaux/Teitelboim, a gauge theory is defined by the Legendre transformation being non-invertible, which reflects in a local symmetry of the Lagrangian and constraints in the Hamiltonian theory. For "nice" systems, this is a bijection - every local symmetry belongs to a Hamiltonian constraint and vice versa. Essentially, the definition of a "gauge theory" is then a constrained Hamiltonian system that is "nice". $\endgroup$ – ACuriousMind Jul 12 '16 at 11:11
  • $\begingroup$ Furthermore, on the mathematical side, "gauge theory" is much narrower and takes only those systems that are described by a global gauge group and the prinicipal bundle formalism, which also encompasses almost all applications of "gauge theory" in physics that I know. $\endgroup$ – ACuriousMind Jul 12 '16 at 11:13
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    $\begingroup$ @ACuriousMind From Wikipedia: "a gauge theory is a type of field theory in which the Lagrangian is invariant under a continuous group of local transformations ... Two such mathematical configurations are equivalent (describe the same physical situation) if they are related by a transformation of this abstract coordinate basis" $\endgroup$ – tparker Jul 12 '16 at 17:03
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(In my experience this tends to be a rather controversial subject, so I think this answer might start some arguments!)

First of all, given a specific action, it is a purely mathematical result whether or not there exists a local transformation that depends on an arbitrary smooth function $\lambda(x)$ on spacetime and leaves the action invariant. The equations don't know whether these "gauge degrees of freedom" $\lambda(x)$ are "physically observable" or not. But the existence of such a local freedom leads to nontrivial mathematical results, which hold whether or not the local degrees of freedom are observable. The hard part is assigning the right physical interpretation to those degrees of freedom.

The precise mathematical consequence of a local symmetry is simply that the gauge degrees of freedom are completely decoupled from the gauge-invariant degrees of freedom. The point isn't that these degrees of freedom are never physical; it's that if you only care about gauge-invariant quantities, then you can get away with completely ignoring the gauge degrees of freedom by working in whichever gauge you want, and you know that whatever choice you make won't affect your results for gauge-invariant quantities.

There may be certain sets of local degrees of freedom that are physically measurable in some situations, but not in others. For example, let's say you discovered a new fundamental scalar field $\varphi(x)$ that coupled directly to the electromagnetic gauge field via a coupling $g\, \varphi\, A_\mu A^\mu$ and therefore explicitly broke the electromagnetic $U(1)$ gauge symmetry. (I'll assume all the fields are classical for now - things get really subtle once you start considering gauge quantum field theories.) You could then use this field as a probe that could directly measure the value of the gauge field $A_\mu$ itself, which would therefore become physically observable. What physical implications would this have?

Well, the most direct physical consequence of gauge symmetry is the conservation of the current to which the gauge field couples linearly. If we consider the modified Maxwell Lagrangian $$\mathcal{L} = -\frac{1}{4} F_{\mu \nu} F^{\mu \nu} + A_\mu J^\mu - g \varphi\, A_\mu A^\mu,$$ then the Euler-Lagrange equation becomes $$ \partial_\nu F^{\mu \nu} = J^\mu - 2g \varphi A^\mu$$ (which is manifestly no longer gauge-symmetric) and $$ \partial_\mu \partial_\nu F^{\mu \nu} = 0 = \partial_\mu J^\mu - 2g \partial_\mu (\varphi A^\mu) \implies \partial_\mu J^\mu = 2g \partial_\mu (\varphi A^\mu),$$ and indeed we see that electric charge is no longer conserved in general.

But here's the key point: in a region of space where the gauge-symmetry-breaking field $\varphi(x) \equiv 0$, charge is still conserved, even if $\varphi(x) \neq 0$ somewhere else. In this hypothetical, $A_\mu$ is a physically measurable quantity, because there exists a $\varphi$ field that "measures" it. So even if there's some region of space that happens to not have any $\varphi$ field in it, then the $A_\mu$ field still a unique configuration in that region. But gauge symmetry is effectively restored in that region, because you aren't measuring $A_\mu$ even though you could. (This is kind of like how particles only have mass in regions of spacetime where the Higgs field's $U(1)$ symmetry is spontaneously broken.) We've demonstrated the sensible result that the gauge symmetry's enforcement of charge conservation is "robust," in the sense that the violation of charge conservation is proportional to the violation of the gauge symmetry - so if we discovered tomorrow that the photon actually had a mass of $10^{-10}$ eV, it's not like all of a sudden charges would suddenly start appearing and disappearing out of nowhere. This is why in condensed-matter systems with emergent gauge symmetries that only hold approximately and at low temperature, the gauge symmetry is still useful even though it isn't exact - there are gauge-symmetry-breaking fields, but their effects are small.

To make this even more concrete, let's imagine you build a magic "$A_\mu$-meter" which shines a beam of $\varphi$ field on any system, and you use it to directly measure $A_\mu$. You would find that when you're pointing the beam toward a system, charge will not be conserved in that system (i.e. charge could appear or disappear). But as soon as you turn the beam off, charge suddenly becomes conserved! It's as though the beam were adding or subtracting charge from the system, even though $\varphi$ and $J_\mu$ are not coupled directly, but only indirectly via $A_\mu$'s gauge degrees of freedom.

Taking this idea further, if an entire Lagrangian describing a system has a gauge symmetry, than that means that the gauge degrees of freedom cannot be detected within the system described by that Lagrangian, by a detector also within that system. So when you have a condensed-matter system with an emergent gauge symmetry that isn't a true symmetry of the Standard Model, that system can't "see" the gauge degrees of freedom within itself, but only the gauge-invariant ones. But we "outsiders" know that those gauge degrees of freedom still have unique physical values, and could use an external probe to measure them. Or put another way, there could well be fields beyond the Standard Model that couple directly to the electromagnetic gauge degrees of freedom and can therefore directly measure the $A_\mu$ field - but as long as those fields aren't also coupled to us, and don't substantially affect the dynamics of the gauge-invariant degrees of freedom (the $F_{\mu \nu}$ tensor), we have no way of detecting their existence.

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    $\begingroup$ Sorry, it feels like you are really twisting things around, but I'll have to think about it more. Still, it seems you posit an external field that interacts with your gauge field and breaks the gauge symmetry, there is then no mystery why that gauge symmetry is not one any more, and you can measure its other degrees of freedom. Maybe that makes sense in some approximation in condensed matter, since one is always having to add another layer to the approximation, not sure how it enters in QFT, again unless you discover that new field. $\endgroup$ – Bob Bee Jul 12 '16 at 4:11
  • $\begingroup$ @BobBee Point taken, but even the Standard Model is probably just "another layer to the approximation." There are many theories suggesting that the SM gauge symmetries are only approximate and are broken at the Planck scale (or lower). The way of thinking explained above gives you a systematic way of thinking about how these possible effects would look at different scales. arxiv.org/abs/cond-mat/0404617 $\endgroup$ – tparker Jul 12 '16 at 4:19
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    $\begingroup$ I've understood it better as effective QFT. I don't doubt there will be other physics at higher energies, so if you mean that at those higher energies some gauge symmetries could break, ok. But I've understood at at the higher energies more than likely some additional symmetries emerge, or maybe they mean others get absorbed. For electroweak, U(1) survived. Is there a pubLished reputable paper, not in condensed matter, that has your points, within a plausible model? And I don't mean string theory. Edited: I'll look at your reference. Yes, hopefully this will cause controversy. $\endgroup$ – Bob Bee Jul 12 '16 at 4:24
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    $\begingroup$ @BobBee It could go either way - at low temperatures, symmetries can break, but they can also emerge. Also, I hope that you meant "a published reputable paper, not in condensed matter" as two separate conditions, not two phrasings of the same condition ... :) $\endgroup$ – tparker Jul 12 '16 at 4:40
  • $\begingroup$ Yes. I have no problem with what I called solid state. Just prefer to see in this context the QFT/high energy physics relation $\endgroup$ – Bob Bee Jul 12 '16 at 18:42

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