Under the global <U(1) gauge symmetry>, the phases in $\bar\psi\psi$ cancel and $\alpha$ is constant, so
$$
D_\mu(e^{i\alpha}\psi)=e^{i\alpha}D_\mu\psi.
$$
Both the fermion term and $F_{\mu\nu}F^{\mu\nu}$ are therefore invariant. To extract the <Noether current>, temporarily promote $\alpha$ to a function. The variation of the action is
$$
\delta S=i\int d^4x\,(\partial_\mu\alpha)\bar\psi\gamma^\mu\psi
=-i\int d^4x\,\alpha\,\partial_\mu j^\mu,
\qquad j^\mu=\bar\psi\gamma^\mu\psi.
$$
The <Euler-Lagrange equations> then imply the <current conservation> law $\partial_\mu j^\mu=0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 304 / 3 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-304/3/a/solution