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.

Solution

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

{title2=$J_{\mathrm{mag}}^\mu$}

A relativistic magnetization current has the form $J_{\mathrm{mag}}^\mu=\partial_\nu M^{\nu\mu}$ for an antisymmetric polarization-magnetization tensor. It is identically conserved because commuting derivatives contract an antisymmetric tensor.


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 Incoming links: Magnetization current