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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 $\alpha(... 0 Instead of the current conservation only is valid if the Maxwell-equations are fulfilled, the logic sould be the other way around, like this: the Maxwell-equations are fulfilled only if the current conservation is valid. Gauge-invariance does not depend on the behavior of$A$. It only depends on the behavior of$J$. The behavior of$J$is constrained by ... 1 For what it's worth, the off-shell Maxwell action (54.21) is gauge invariant if the background sources$J^{\mu}$satisfy the continuity equation. OP's question seems to be spurred by the ambiguity coming from the fact that one should specify which fields (among the gauge fields and matter fields) are treated quantum mechanically (off-shell) and which fields ... 1 We can identify two cases here. Case 1: The gauge potential has a singularity which produces a singularity in the physical field too. Given that the physical field is invariant under gauge transformations, it follows that you won't be able to cure this by making a gauge transformation. We can make a trivial example of this by considering a$U(1)$theory in ... -1 QFT books often jump straight to QED since this theory was one of the first figured out, and is also the first (so far as I'm aware) to produce new results matching experiment. This, however, leads to some mystery surrounding the procedure of gauge fixing because the machinery to gauge fix properly usually comes later on when non-Abelian gauge theories are ... 0 The material you've read is just talking about the fact that general relativity can be easily expressed in a coordinate-independent way. However, other theories of physics can also be expressed in a coordinate-independent way. It's just that they weren't originally/traditionally expressed that way. So the answer is no, you can't derive all of general ... 2 I'll focus on the scalar$O(n)$gauge theory that was mentioned in the question. The set of observables in the gauged version of the model is not a subset of the observables in the ungauged version. Gauging the$O(n)$symmetry does eliminate some observables, but it also introduces others. Observables in the gauged version are required to be invariant under ... 3 Doesn't the standard non abelian Lagrangian just reduce to the correct thing in the abelian case? The Lie algebra$\mathfrak{u}(1)$is just generated by the$1 \times 1$matrix known as "$1$", so here the only generator is$T^a$where$a = 1$, and$T^1 = 1$. Then$\mathrm{Tr}(T^1) = 1, \mathrm{Tr}(T^1 T^1) = 1$, so the whole thing just reduces to$...
Since most of the answers in the post mentioned by Qmechanic are either too technical or come to the wrong conclusion, I will try to give a simple yet precise answer. I will consider Maxwell theory as an example. Let's try to apply the Noether's theorem to gauge symmetries of the Maxwell theory, i.e. $A_\mu\to A_\mu +\partial_\mu\lambda$ for arbitrary ...