= Solution
A homogeneous static saddle minimizes $f(\rho)=-\mu\rho+g\rho^2/2$ for $\rho=|\psi|^2\ge0$. With $g>0$ and $\mu>0$,
$$
\boxed{\rho_0=\mu/g,\qquad\psi_0=\sqrt{\rho_0}e^{i\phi_0},\qquad\phi_0\in\mathbb R/(2\pi\mathbb Z).}
$$
The mean-field <Bose-Einstein condensate> has a macroscopic coherent occupation of the uniform mode. Its continuous <particle-number phase symmetry> is $\psi\mapsto e^{i\chi}\psi$, the <circle group> $U(1)$ generated by particle number. Choosing one phase breaks it, and the saddle manifold is a circle. For $\mu\le0$, the constrained minimum is the vacuum $\rho_0=0$, so the condensed saddle requires the stated positive-density regime.
<Spontaneous symmetry breaking> is understood through a phase-selected thermodynamic description: an exact finite-volume number <eigenstate> has vanishing field expectation. The resulting <Goldstone boson> is a gapless phase/sound mode, while phase <gradients> carry the <superfluid velocity> $\mathbf v=\nabla\phi/m$. This is the saddle-point conclusion requested, not a proof of true condensate order in every dimension or temperature. Long-wavelength phase fluctuations can invalidate the assumed order, as quantified below.
Back to article page
