The Lagrangian of QED is
\begin{equation} \mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\bar{\psi}\big(i\not{D}-m\big)\psi \end{equation}
where $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$ and $\not{D}=\gamma^\mu\big(\partial_\mu+iqA_\mu)$. Here $A_\mu$ is the EM gauge field and $q$ is the charge of the electron. This lagrangian has a $U(1)$ gauge symmetry
\begin{equation} \begin{cases} \psi\rightarrow\psi'=e^{iq\alpha(x)}\psi \\ \bar{\psi}\rightarrow\bar{\psi}'=e^{-iq\alpha(x)}\bar{\psi} \\ A_\mu\rightarrow A_\mu'=A_\mu-\partial_\mu\alpha(x) \end{cases} \end{equation}
where $\alpha(x)$ is a smooth arbitrary function of spacetime coordinate $x^\mu$. Now, we apply the definition of the Noether Current
\begin{equation} J^\mu=\frac{\delta \mathcal{L}}{\delta (\partial_\mu\phi_a)}\delta \phi_a \end{equation} (where I'm not adding the extra term due to the lagrangian changing in a total derivative since $\delta\mathcal{L}=0$). So we get
\begin{equation} J^\mu=-q\alpha(x)\bar{\psi}\gamma^\mu\psi+F^{\mu\nu}\partial_\nu\alpha(x) \end{equation}
When $\alpha(x)=constant$ we recover the usual electric charge. Now, I want to address two questions regarding the complete current $J^\mu$ for arbitrary $\alpha(x)$.
First Question: It is said in many textbooks that global symmetries give conservation laws that are satisfied on-shell while local symmetries give conservation laws satisfied off-shell. It is clear for the global part that this is true: The current is just (setting $\alpha=1$)
\begin{equation} J^\mu|_{\alpha=1}=-q\bar{\psi}\gamma^\mu \psi \end{equation}
Since the equation of motion for the gauge fields is
\begin{equation} \partial_\nu F^{\mu\nu} =-q\bar{\psi}\gamma^\mu \psi \end{equation}
we inmediately get that $\partial_\mu J^\mu=\partial_\mu\partial_\nu F^{\mu\nu}=0$. So the current get's conserved when the EoM are satisfied. Now, the question is: How do we check that either $J^\mu$ is identically zero off-shell or that $\partial_\mu J^\mu$ is zero off-shell?
Second Question: If we keep working on the full current we can use Leibniz rule to get
\begin{equation} J^\mu=-q\alpha(x)\bar{\psi}\gamma^\mu\psi+\partial_\nu\big[F^{\mu\nu}\alpha(x)\big]-\partial_\nu F^{\mu \nu}\alpha(x) \\ \ \ \ \ \ \ \ \ \ \ =\alpha(x)\Big[-q\bar{\psi}\gamma^\mu\psi-\partial_\nu F^{\mu \nu}\Big]+\partial_\nu\big[F^{\mu\nu}\alpha(x)\big] \end{equation}
Now, if we apply the EoM, the first term vanishes. We are left with a total derivative. If we find the total charge in a given volume $V$ we get
\begin{equation} Q=\int_V J^0 d^3x = \int_V \partial_\nu\big[F^{0 \nu}\alpha(x)\big] d^3x= \int_S \vec{E}\cdot \hat{n} \ \alpha(x) dS \end{equation}
Now, if the volume we choose is the whole universe and if the electric field and the gauge condition $\alpha(x)$ decay nicely to infinity (see my other question where I ask why does $\alpha(x)$ need to decay at infinity [1]) then the total charge in the universe is zero. However, that doesn't mean that this symmetry is not real or not useful. I could also choose any other volume and get a perfectly functional conservation law for a new quantity whose charge is the electric field at the boundary of the desired volume weighted by an arbitrary function $\alpha(x)$. So the question is: Why is this conserved quantity not physical?
[1] Why do we require that the gauge condition $\alpha(x)$ falls off at infinity?
