Quadratic density-phase action ($S_2=\int [g(\delta \rho {)}^2/2+i\delta \rho {\partial }_{\tau }\varphi +|\nabla \delta \rho {|}^2/(8m{\rho }_0)+{\rho }_0|\nabla \varphi {|}^2/(2m)]$)

Expand the density-phase action of a Bose gas around ${\rho }_0=\mu /g>0$ and a constant phase. Density and phase are coupled by $i\delta \rho {\partial }_{\tau }\varphi$, and retaining density gradients gives the full quadratic Bogoliubov spectrum. Dropping those gradients is a long-wavelength approximation. A small-fluctuation Gaussian extension around ${\rho }_0$ does not exactly remove the original nonnegative-density constraint.

A positive density curvature $g$ makes uniform static density fluctuations costly. It does not imply an additional independent gapped excitation: the first-order term makes density and phase conjugate. For a simple nonrelativistic weakly interacting Bose gas, their coupled quadratic dynamics yields one positive Bogoliubov spectrum, which is gapless at zero momentum. The word massive for the density field describes its static curvature in this setting.

In the smooth zero-winding long-wave sector, integrate the density Gaussian integral by completing $g(\delta \rho +i{\partial }_{\tau }\varphi /g{)}^2/2$. Its field-independent normalization leaves the displayed action. The real-frequency pole is ${\omega }^2=g{\rho }_0{k}^2/m$, so the phonon speed is $\sqrt{g{\rho }_0/m}$. Keeping density gradients yields ${\omega }^2=g{\rho }_0{k}^2/m+k^4/(4m^2)$, matching the Bogoliubov spectrum.

The Bose-gas phase-only action gives a thermal infrared integral $\int d^{d}q/q^2$ and a zero-temperature equal-time integral $\int d^{d}q/|q|$. Thus an infinite uniform short-range interacting gas has no true finite-temperature condensate in $d\le 2$, and the one-dimensional ground state also lacks true condensate long-range order. This does not exclude local density coherence, a finite-size condensate description or two-dimensional algebraic order and superfluidity.^[ref]