A general Noether theorem in fields theory says that an infinitesimal symmetry of the action leads to a conserved current $j^\mu$, i.e. $\partial_\mu j^\mu=0$.
Below I would like to consider a minor generalization of the following well known situation in QED. The Lagrangian density in QED is $$\mathcal{L}=\bar\psi (i\gamma^\mu(\partial_\mu+ieA_\mu)-m)\psi)-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}.$$ $\mathcal{L}$ is invariant under the global symmetry
\begin{eqnarray} \psi\to (1+ie\epsilon)\psi,\\ \bar\psi\to (1-ie\epsilon)\bar\psi\\ A_\mu\to A_\mu, \end{eqnarray} where $\epsilon$ is an infinitesimal parameter. It is well known that application of the Noether theorem leads to conserved current $$j^\mu=-e\bar\psi\gamma^\mu\psi.$$ The operator $Q=\int d^3x j^0(x)$ is interpreted as electric charge operator and commutes with the Hamiltonian.
However it is equally well known that $\mathcal{L}$ is invariant under more general local transformations. Hence one could try to construct a Noether current for each such transformation. More precisely fix an arbitrary function $f(x)$. Then $\mathcal{L}$ is invariant under \begin{eqnarray} \psi\to (1+ief\epsilon)\psi,\\ \bar\psi\to (1-ief\epsilon)\bar\psi,\\ A_\mu\to A_\mu-(\partial_\mu f)\epsilon, \end{eqnarray} where again $\epsilon$ is an infinitesimal parameter. (Notice that the previous transformation corresponds to $f\equiv 1$.) It is easy to compute the Noether current: $$j^\mu=-fe(\bar\psi\gamma^\mu\psi)+\partial_\nu f(\partial^\mu A^\nu-\partial^\nu A^\mu).$$
Does the latter current has any physical interpretation? Does it play any role in the theory? More generally in any gauge theory (e.g. QCD) one can get similar currents for any gauge transformation. Are they useful?
ADDED. Let me add a comment to make my question more precise. When one has a Lie group preserving the action, the application of the Noether theorem and construction of charge operators using these currents lead presumably to a representation of the Lie algebra of this group (e.g. this is the case with the Poincare group, inner symmetries and more generally with supersymmetries). Hence if the gauge group preserves the action, one would expect that its Lie algebra acts on the Hilbert space of the theory. In the case of QED one would get a representation in the Hilbert space of the theory of the commutative (infinite dimensional) Lie algebra of real valued functions on $\mathbb{R}^{3+1}$. I am wondering if this is indeed the case. What I have seen in the literature is only one charge operator $Q=-e\int d^3x \cdot \bar\psi(x)\gamma^0\psi(x)$ corresponding to the current $-e\bar\psi\gamma^\mu\psi$ which corresponds to the rotation with the constant phase I mentioned first in the post.