= Solution
Hold $\phi,\phi^*$ fixed while varying $A_\nu$. With the specified convention $D_\mu=\partial_\mu+iqA_\mu$, the two matter <variations> are $\delta(D_\mu\phi)=iq\phi\,\delta A_\mu$ and $\delta(D_\mu\phi)^*=-iq\phi^*\delta A_\mu$. Thus
$$
\delta\mathcal L_{\rm matter}=iq\left[\phi(D^\nu\phi)^*-\phi^*D^\nu\phi\right]\delta A_\nu=-J^\nu\delta A_\nu.
$$
Adding the Maxwell <variation> from part (a) gives $\delta S=\int d^4x\,[\partial_\mu F^{\mu\nu}-J^\nu]\delta A_\nu$. Consequently
$$
\boxed{\partial_\mu F^{\mu\nu}=J^\nu,\qquad J^\mu=iq\left[\phi^*D^\mu\phi-\phi(D^\mu\phi)^*\right].}
$$
Expanding the <gauge covariant derivatives> supplies a useful convention check:
$$
J^\mu=iq(\phi^*\partial^\mu\phi-\phi\partial^\mu\phi^*)-2q^2A^\mu|\phi|^2.
$$
In particular, the last sign is negative for the specified plus sign in $D_\mu$. Reversing that convention requires consistently changing the phase transformation and current conventions as well. This current is the matter <source> in <scalar electrodynamics>; its $A$ dependence must not be dropped.
Back to article page
