Use $D_\mu=\partial_\mu+ieA_\mu$. The two Lagrangians are
$$
\mathcal L_{\mathrm{SQED}}=(D_\mu\phi)^*D^\mu\phi-M^2\phi^*\phi-\frac14F_{\mu\nu}F^{\mu\nu}
$$
and
$$
\mathcal L_{\mathrm{QED}}=\bar\psi(i\gamma^\mu D_\mu-m)\psi-\frac14F_{\mu\nu}F^{\mu\nu}.
$$
Both are invariant under the <U(1) gauge symmetry>
$$
\phi\mapsto e^{-ie\alpha(x)}\phi,
\qquad
\psi\mapsto e^{-ie\alpha(x)}\psi,
\qquad
A_\mu\mapsto A_\mu+\partial_\mu\alpha.
$$
For the constant phase subgroup, normalized to unit matter charge, the <Noether currents> are
$$
j^\mu_{\mathrm{SQED}}=i\{\phi^*D^\mu\phi-(D^\mu\phi)^*\phi\},
\qquad
j^\mu_{\mathrm{QED}}=\bar\psi\gamma^\mu\psi.
$$
Their electric currents are $e j^\mu$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 301 / 3 / a / Solution

The local <U(1) gauge symmetry> is
$$
\psi\mapsto e^{ie\chi(x)}\psi,
\qquad
\bar\psi\mapsto\bar\psi e^{-ie\chi(x)},
\qquad
A_\mu\mapsto A_\mu-\partial_\mu\chi.
$$
The global phase subgroup has <Noether current> and charge
$$
j^\mu=\bar\psi\gamma^\mu\psi,
\qquad
Q=\int d^3x\,j^0=\int d^3x\,\psi^\dagger\psi.
$$
Multiplying the fermion equation by $\bar\psi$, its adjoint by $\psi$, and subtracting shows $\partial_\mu j^\mu=0$ on shell. Therefore $\dot Q=-\int d^3x\,\boldsymbol\nabla\cdot\mathbf j=0$ under vanishing boundary flux.

The source in Maxwell's equation is
$$
J^\nu=e j^\nu-2i\lambda\partial_\mu
(\bar\psi[\gamma^\mu,\gamma^\nu]\psi).
$$
The second term is an identically conserved <magnetization current>, because it is the divergence of an antisymmetric tensor. It changes the local source but contributes only a boundary term to the total charge.

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 301 / 1 / d / Solution

{c}

For $D_\mu=\partial_\mu+ieA_\mu$, the local U(1) transformation is
$$
\psi\mapsto e^{ie\chi}\psi,
\qquad
A_\mu\mapsto A_\mu-\partial_\mu\chi.
$$


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