Because $n^\mu\partial_\mu\phi=-2\Pi$,
$$
Q=-n_\mu J^\mu
=i(\bar\phi\Pi-\phi\bar\Pi).
$$
Writing $\phi=\phi_R+i\phi_I$ and $\Pi=\Pi_R+i\Pi_I$ gives
$$
\boxed{Q=2(\phi_I\Pi_R-\phi_R\Pi_I)}.
$$
Every quantity on the right is real, so the Noether charge density is explicitly real valued.

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 357 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-357/3/c/solution