 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 isIts functional derivative isso the Gross–Pitaevskii equation is $i\hslash {\psi }_t=\delta H/\delta {\psi }^{\ast }$. Treating $\psi$ and ${\psi }^{\ast }$ 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:The gradient contribution is nonnegative, so the homogeneous ground state has arbitrary constant complex argument andThe 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 $g>0$, $\mu \le 0$ the grand-potential minimum over $n\ge 0$ 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.