 Solution

On the physical density-phase contour put $\bar{\psi }=\sqrt{\rho }{e}^{-i\varphi }$. Direct differentiation givesIntegrating the Laplacian by parts with the stated bulk boundary convention, and dropping the periodic total derivative of $\rho$, produces the density-phase action of a Bose gas,Write $\rho ={\rho }_0+\delta \rho$ and $\varphi ={\varphi }_0+\phi$, where ${\varphi }_0$ is constant. To quadratic order,The uniform static density fluctuation has positive curvature $g$. Its inverse static susceptibility at wavevector $q$ is $A_q=g+q^2/(4m{\rho }_0)$, which remains nonzero at $q=0$. This is the sense in which the amplitude or density fluctuation is massive. The phase has no term proportional to ${\phi }^2$ without derivatives; a uniform phase rotation costs no action. It is the massless direction required by global $U(1)$ symmetry.

The term $i\delta \rho {\partial }_{\tau }\phi$ shows density-phase conjugacy. It is important to distinguish static density stiffness versus an excitation gap: the density and phase do not form two independent normal-mode branches with one gapped and one gapless particle. In this nonrelativistic first-order theory they supply the conjugate variables of one positive Bogoliubov branch, derived below. The omitted cubic terms include $\delta \rho |\nabla \phi {|}^2/(2m)$ and corrections from the density-gradient denominator.

There is also a background term $i{\rho }_0\int d\tau d^{d}r{\partial }_{\tau }\phi$. It vanishes for smooth fluctuations with zero temporal winding. Periodicity of $\psi$, however, permits $\phi (\beta )-\phi (0)=2\pi w$. The compact phase winding term must be retained when such sectors are included; it is not generally the same constant in all sectors. The local Gaussian phonon calculation concerns the smooth zero-winding sector and does not describe vortices or points where the density vanishes.