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I have realised that I am unsure about how I can apply Noether's theorem to a Lagrangian with local invariance. For instance, the following Lagrangian has a local $U(1)$ invariance:

$$\mathcal{L}=(D_{\mu}\psi)^{\dagger}(D^{\mu}\psi)-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-V(\psi^{\dagger}\psi)$$

Where $V$ is the scalar potential, $F^{\mu\nu} = \partial^{\mu}A^{\nu} - \partial^{\nu}A^{\mu}$ is the field strength tensor and $D_{\mu} = \partial_{\mu} + ieA_{\mu}$ is the covariant derivative.

Under the $U(1)$ transformation our fields transform infintesimally as:

$$\psi\mapsto \psi + i\alpha(x)\psi, \quad \psi^{\dagger} \mapsto \psi^{\dagger} - i\alpha(x)\psi^{\dagger},\quad A^{\mu}\mapsto A^{\mu} - \frac{1}{e}\partial^{\mu}\alpha(x)$$

The Noether current can be found:

$$j^{\mu} = \Pi^{\mu}_{\psi}D\psi + \Pi_{\psi^{\dagger}}^{\mu}D\psi^{\dagger} + \Pi_{A_{\nu}}^{\mu}DA_{\nu}+W^{\mu}$$

Where $D\phi \equiv \left.\frac{\partial \phi}{\partial \alpha}\right|_{\alpha = 0}$. And where $D\mathcal{L} = \partial_{\mu}W^{\mu}$. So we can see that $W^{\mu} = 0$. Using our Lagrangian and expanding the covariant derivatives, we have:

$$\mathcal{L}=(\partial_{\mu}\psi^{\dagger})(\partial^{\mu}\psi) + ieA_{\mu}(\partial^{\mu}\psi^{\dagger} - \partial^{\mu}\psi) + e^{2}A_{\mu}A^{\mu}\psi^{\dagger}\psi - \frac{1}{4}F_{\mu\nu}F^{\mu\nu} - V(\psi^{\dagger}\psi)$$

Thus:

$$\Pi_{\psi}^{\mu} = \frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\psi)}=\partial^{\mu}\psi^{\dagger} - ieA^{\mu} \\ \Pi_{\psi^{\dagger}}^{\mu} = \frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\psi^{\dagger})}=\partial^{\mu}\psi+ieA^{\mu} \\ \Pi^{\mu}_{A_{\nu}}=\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}A_{\nu})}=-F^{\mu\nu}$$

And:

$$D\psi = i\psi,\quad D\psi^{\dagger} = -i\psi^{\dagger},\quad DA_{\mu} = -\frac{1}{e}\partial_{\mu}\alpha$$

This would give us a Noether current of:

$$j^{\mu} = i(\partial^{\mu}\psi^{\dagger} - ieA^{\mu})\psi - i(\partial^{\mu}\psi + ieA^{\mu})\psi^{\dagger} + \frac{1}{e}F^{\mu\nu}\partial_{\nu}\alpha$$

But this seems strange to me, not least because it still contains $\alpha$.

What have I misunderstood here?

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