# Noether current and continuity equation in classical scalar QED

Consider the following scalar QED model \begin{align} S = \int \mathrm{d}^{d+1} x\, \left\{-\left(\mathrm{D}_{\mu}\phi\right)^{\dagger} \left(\mathrm{D}^{\mu}\phi\right) -m^2 \phi^{\dagger}\phi - \frac{1}{4} F_{\mu\nu} F^{\mu\nu}\right\}\,, \end{align} where $$\eta_{\mu\nu} = \mathrm{diag}(-,+,+,+)$$, and $$\mathrm{D}_{\mu}\phi = \left(\partial_{\mu}-\mathrm{i} e A_{\mu}\right)\phi$$.

Under the following infinitesimal gauge transformation \begin{align} \delta \phi = -\mathrm{i} e \Lambda(x)\,,\qquad \delta \phi^{\dagger} = +\mathrm{i} e \Lambda(x)\,,\qquad \delta A_{\mu} = -\partial_{\mu} \Lambda(x)\,, \end{align} the theory is expected to be invariant, leading to Noether identities, which obey continuity equations.

Under the infinitesimal gauge transformation above, calculation gives \begin{align} \delta S = \int\mathrm{d}^{d+1}x\,\bigg\{ &\mathrm{i} e \Lambda(x) \left(\mathrm{D}^{\mu}\mathrm{D}_{\mu}-m^2\right)\phi - \mathrm{i} e \Lambda(x) \left[\left(\mathrm{D}^{\mu}\mathrm{D}_{\mu}-m^2\right)\phi\right]^{\dagger} \\ & -\partial_{\mu}\Lambda(x)\left[- \mathrm{i} e \left(\phi^{\dagger}\mathrm{D}^{\mu}\phi- \left(\mathrm{D}^{\mu}\phi\right)^{\dagger}\phi\right) +\partial_{\nu} F^{\nu\mu}\right] \\ & -\partial_{\mu}\left[ \mathrm{i} e \Lambda(x) \left(\phi^{\dagger}\mathrm{D}^{\mu}\phi- \left(\mathrm{D}^{\mu}\phi\right)^{\dagger}\phi\right) -\left(\partial_{\nu} \Lambda(x)\right)F^{\mu\nu}\right]\bigg\}\,, \end{align} where the first two lines would give equations of motion, and the third line might be related to a Noether current.

One sees that since $$\Lambda(x)$$ now is a space-time function, the derivation for Noether's first theorem does not immediately apply. How can I extract a Noether current, and how does the corresponding continuity equation look like? Do I have a conservation law here?

Edit 28.04.2019:

If I fix the transformation to be rigid $$\Lambda(x) \equiv \Lambda$$, I will get the current \begin{align} j^{\mu} = \mathrm{i} e \left(\phi^{\dagger}\mathrm{D}^{\mu}\phi- \left(\mathrm{D}^{\mu}\phi\right)^{\dagger}\phi\right). \end{align} Equipped with \begin{align} \partial_{\mu}\left(\mathrm{D}^{\mu}\phi\right) &= \left(\mathrm{i} e A_{\mu}\mathrm{D}^{\mu} +m^2\right)\phi\,, \\ \partial_{\mu}\left(\mathrm{D}^{\mu}\phi\right)^{\dagger} &= \left[\left(\mathrm{i} e A_{\mu}\mathrm{D}^{\mu} +m^2\right)\phi\right]^{\dagger}\,, \end{align} which come from the equations of motion, one can derive \begin{align} \partial_{\mu}j^{\mu} = \mathrm{i} e \big\{&+ \left(\mathrm{D}_{\mu}\phi\right)^{\dagger}\left(\mathrm{D}^{\mu}\phi\right) +m^2\phi^{\dagger}\phi \\ &- \left(\mathrm{D}_{\mu}\phi\right)^{\dagger}\left(\mathrm{D}^{\mu}\phi\right) -m^2\phi^{\dagger}\phi\big\} = 0. \end{align}

Therefore,

1. From Noether's first theorem (rigid transformation), one does get a current, in the usual form.
2. The continuity equation contains $$\partial_{\mu}$$, instead $$\mathrm{D}_{\mu}$$. As a result, one does have a conserved charge. If it were $$\mathrm{D}_{\mu} j^{\mu} = 0$$, one would in general not have a conserved charge.

The question concerning what we can get from Noether's second theorem (gauge transformation) remains not answered.