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A homogeneous static saddle minimizes $f(\rho )=-\mu \rho +g{\rho }^2/2$ for $\rho =|\psi {|}^2\ge 0$. With $g>0$ and $\mu >0$,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 $\mu \le 0$, 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 $\mathbf{v}=\nabla \varphi /m$. 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.

Under $\psi \mapsto e^{i\alpha }\psi$ with constant real $\alpha$, the field and its complex conjugate acquire opposite phases. These cancel in $|\psi {|}^2$, $|\psi {|}^4$, $\nabla {\psi }^{\ast }\cdot \nabla \psi$, and ${\psi }^{\ast }\dot{\psi }-\psi {\dot{\psi }}^{\ast }$. The Lagrangian density therefore has the particle-number phase symmetry, a global circle group $U(1)$.

For a spacetime-dependent real $\alpha (\mathbf{r},t)$,Writing $n=|\psi {|}^2$, the change in the Lagrangian density isThis is not generally zero or a boundary term. Thus 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.