What exactly does it mean by gauge-invariant "operators"?

Question

For simplicity, let us consider $U(1)$ gauge theory without matter fields.

At classical level, the gauge field $A^\mu$ has the gauge transformation law \begin{equation} A^\mu \to A^\mu +\partial^\mu \chi \end{equation} for any smooth function $\chi$. And the field strength tensor $F^{\mu \nu}=\partial^\mu A^\nu-\partial^\nu A^\mu$ is invariant under the transformation.

After a quantization like the Gupta-Bleuler one, $A^\mu$ become operator-valued distributions on some "physical" Hilbert space satisfying Lorenz condition $\partial_\mu A^\mu=0$ in that Hilbert space.

Now, my question is: How is the gauge transformation defined for the quantized fields? What does it mean by gauge-invariant "operators"?

My guess is that the gauge transformation is simply $A^\mu \to A^\mu +\partial^\mu \chi$ as operators on the Hilbert space and $F^{\mu \nu}$ is a gauge-invariant "operator".

Could anyone please clarify?

Answer 1

The whole point of the quantization of gauge theories is that we don't really want a realization of the gauge symmetry in the quantum theory because the gauge symmetry is unphysical (see also this answer of mine) - we want all the quantum states and operators to be gauge-invariant by construction. Note that the Gupta-Bleuler condition is a gauge-fixing condition and the point of gauge fixing is that it gets rid of the gauge symmetry - you're fixing a gauge because you don't want to bother with gauge invariance.

Gupta-Bleuler quantization is an ad-hoc implementation of this idea, the more general procedure is BRST quantization - see this answer of mine for an explanation why Gupta-Bleuler and BRST yield the same result.

In any case, the result of the quantization procedure is a physical Hilbert space in which all states are "gauge-invariant" because by construction there is no gauge symmetry left acting on this space. Operators that act on this space properly (i.e. map physical states to physical states) are trivally gauge-invariant, operators that don't (i.e. exist on some pre-BRST or pre-gauge-fixing Hilbert space but map at least one state that ends up being physical to one state that is unphysical) are not.