Varying $\bar\psi$ and $\psi$ independently gives
$$
\left(i\not D-i\lambda[\gamma^\mu,\gamma^\nu]F_{\mu\nu}\right)\psi=0
$$
and its adjoint
$$
i(D_\mu\bar\psi)\gamma^\mu
+i\lambda\bar\psi[\gamma^\mu,\gamma^\nu]F_{\mu\nu}=0,
$$
where $D_\mu\bar\psi=\partial_\mu\bar\psi-ieA_\mu\bar\psi$. Varying $A_\nu$ and integrating the Pauli term by parts gives the modified <Maxwell equations>
$$
\partial_\mu F^{\mu\nu}
=e\bar\psi\gamma^\nu\psi
-2i\lambda\partial_\mu\!\left(
\bar\psi[\gamma^\mu,\gamma^\nu]\psi\right).
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-301/1/c/solution