= Solution
The $-\mu\psi$ term means that the generator is the <grand potential>, rather than just the physical <energy>. With boundary variations vanishing, the <Hamiltonian> <functional> is
$$
\boxed{H[\psi,\psi^*]=\int\left[\frac{\hbar^2}{2m}|\nabla\psi|^2+\frac g2|\psi|^4-\mu|\psi|^2\right]d^d x.}
$$
Its <functional derivative> is
$$
\frac{\delta H}{\delta\psi^*}=-\frac{\hbar^2}{2m}\nabla^2\psi+(g|\psi|^2-\mu)\psi,
$$
so the <Gross–Pitaevskii equation> is $i\hbar\psi_t=\delta H/\delta\psi^*$. Treating $\psi$ and $\psi^*$ as independent variables in this variation gives the factor $g$, not $2g$.
For repulsive interactions $g>0$ and $\mu>0$, complete the square in the local <energy>:
$$
\frac g2 n^2-\mu n=\frac g2\left(n-\frac\mu g\right)^2-\frac{\mu^2}{2g}.
$$
The <gradient> contribution is nonnegative, so the homogeneous <ground state> has arbitrary constant <complex argument> and
$$
\boxed{n_0=\frac\mu g,\qquad \psi_0=\sqrt{n_0}e^{i\theta_0}.}
$$
The infinite-volume constant may be subtracted to define a relative <grand potential>; it does not affect the <functional derivative>. At fixed particle number one instead minimizes the physical <energy> $H+\mu N$, with $\mu$ as its <Lagrange multiplier>. For $g>0$, $\mu\leq0$ the grand-potential minimum over $n\geq0$ is the vacuum $n=0$. For $g<0$ the displayed unconstrained <functional> is unbounded below; the positive-density stable uniform state assumed in the following parts requires repulsion.
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