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Choose the real representative $v={\psi }_0=\sqrt{\mu /g}$. For $\psi =v+\chi$,and the fourth power through quadratic order isIn $\mu |\psi {|}^2-g|\psi {|}^4/2$, the linear terms and the $|\chi {|}^2$ terms cancel because $g{v}^2=\mu$. The remaining potential contribution is ${\mu }^2/(2g)-\mu (\chi +{\chi }^{\ast }{)}^2/2$.

The time-derivative contribution isIts first term is a total derivative and contributes only a temporal boundary term for fixed endpoint fields. Since $v$ is uniform, the gradient contribution is simply $-{\hslash }^2|\nabla \chi {|}^2/(2M)$. Dropping that boundary term and the constant, the quadratic condensate fluctuation Lagrangian isThe omitted cubic and quartic interactions describe scattering between these small fluctuations; they do not enter the linear normal mode calculation.

At long wavelength,This is the sound-like Goldstone mode of the broken global $U(1)$ continuous symmetry. A spatially uniform infinitesimal phase change is $\delta \psi =i{\psi }_0\delta \theta$, entirely in the ${\chi }_2$ direction, and costs no potential energy. Slow spatial variations acquire gradient energy and propagate as sound. Thus the dispersion relation realizes Goldstone's theorem through a gapless excitation; it is not a massive phase oscillation.

The amplitude and phase are canonically paired by the first-order time derivative in the quadratic condensate fluctuation Lagrangian. There is one positive-frequency branch, rather than an additional independent amplitude branch of a relativistic second-order-time theory. At high momentum,so the same branch becomes particle-like. The healing length $\xi =\hslash /\sqrt{2M\mu }$ sets the crossover scale.