= Solution
Under $\psi\mapsto e^{i\alpha}\psi$ with constant real $\alpha$, the field and its <complex conjugate> acquire opposite phases. These cancel in $|\psi|^2$, $|\psi|^4$, $\nabla\psi^*\cdot\nabla\psi$, and $\psi^*\dot\psi-\psi\dot\psi^*$. The <Lagrangian density> therefore has the <particle-number phase symmetry>, a global <circle group> $U(1)$.
For a spacetime-dependent real $\alpha(\mathbf r,t)$,
$$
\dot\psi'=e^{i\alpha}(\dot\psi+i\dot\alpha\psi),\qquad
\nabla\psi'=e^{i\alpha}(\nabla\psi+i\psi\nabla\alpha).
$$
Writing $n=|\psi|^2$, the change in the <Lagrangian density> is
$$
\Delta\mathcal L=-\hbar n\dot\alpha-\frac{\hbar^2}{2M}\left[
i(\psi\nabla\psi^*-\psi^*\nabla\psi)\cdot\nabla\alpha+n|\nabla\alpha|^2\right].
$$
This is not generally zero or a <boundary term>. Thus \b[the symmetry is global, with no arbitrary local phase symmetry]. A compensating <gauge field> and <gauge-covariant derivative of a charged scalar field> would be required to promote it to a local redundancy.
Back to article page
