 Solution

Put $\rho ={\rho }_0+\delta \rho$ and retain quadratic fluctuations in the smooth zero-winding sector. The quadratic density-phase action isAt wavelengths long compared with the healing length, drop the density-gradient term. Completing the square givesThe real Gaussian density integral, or its equivalent contour shift, contributes only a field-independent determinant. Absorb that normalization and the uniform saddle action into $S_0$. To this quadratic long-wave accuracy,The Gaussian extension of $\delta \rho$ to the full real line is a fluctuation approximation around positive ${\rho }_0$; it is not an exact replacement of the global density constraint. The compact field's vortex/winding sectors also lie beyond this smooth-phonon integral.

The Bose-gas phase-only action is a continuum harmonic chain with Euclidean inverse propagator ${\omega }_n^2/g+{\rho }_0{k}^2/m$. Continuing to real frequency givesRestoring $\hslash$ gives $E_k\sim \hslash c_s|k|$ for a wavenumber $k$. This is the phonon branch of the Bogoliubov spectrum. Keeping the omitted density-gradient term replaces $g$ by $g+k^2/(4m{\rho }_0)$ and yields $E_k^2=c_s^2{k}^2+k^4/(4m^2)$ in the adopted units, so the full quadratic spectrum is $\sqrt{{ϵ}_k({ϵ}_k+2g{\rho }_0)}$ with ${ϵ}_k=k^2/(2m)$.

The phase-only action also tests the condensate assumption. At zero temperature its equal-time phase variance has an infrared contribution $\int d^{d}k/|k|$, logarithmically divergent in one dimension; at positive temperature the zero Matsubara mode gives $\int d^{d}k/k^2$, divergent in dimensions at or below two. Thus this same low-energy theory exposes the regimes where the mean-field condensate cannot describe true thermodynamic long-range order. It permits a finite infrared fluctuation at zero temperature for $d\ge 2$ and at positive temperature for $d>2$, within the remaining weak-coupling assumptions.