= Solution
Treat $\psi,\psi^*$ independently in the <Legendre transform in mechanics>. Their <canonical momenta> are
$$
\pi_\psi=\frac{i\hbar}{2}\psi^*,\qquad \pi_{\psi^*}=-\frac{i\hbar}{2}\psi.
$$
The time-derivative terms cancel in $\pi_\psi\dot\psi+\pi_{\psi^*}\dot\psi^*-\mathcal L$, giving the <Hamiltonian density>
$$
\boxed{\mathcal H=\frac{\hbar^2}{2M}|\nabla\psi|^2-\mu|\psi|^2+\frac g2|\psi|^4.}
$$
The first-order time dependence makes the momenta constrained; this direct calculation still gives the canonical density. In physical terms this is the <grand-canonical Hamiltonian of a Bose field>, $K=H-\mu N$. The physical energy density is obtained by adding $\mu|\psi|^2$ back. A stationary field in this chemical-potential frame has the usual factor $e^{-i\mu t/\hbar}$ in the physical frame.
Complete the square:
$$
\mathcal H=\frac{\hbar^2}{2M}|\nabla\psi|^2+\frac g2\left(|\psi|^2-\frac\mu g\right)^2-\frac{\mu^2}{2g}.
$$
For uniform or periodic boundary conditions, or in the homogeneous bulk, both nonconstant terms are nonnegative and vanish simultaneously at
$$
\boxed{\psi_0=\sqrt{\frac\mu g}\,e^{i\theta_0},\qquad \mathcal H_{\min}=-\frac{\mu^2}{2g}.}
$$
The <ground states> form a circle. A particular nonzero $\psi_0$ is changed by every nontrivial phase rotation, although the full family is invariant. Selecting one such <mean-field theory> state therefore exhibits <spontaneous symmetry breaking> of $U(1)$. This describes phase selection in the thermodynamic limit; an exact finite-system particle-number <eigenstate> need not itself have a nonzero field expectation.
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