= Solution
At long wavelength,
$$
\boxed{\omega(k)=c_s k+O(k^3),\qquad c_s=\sqrt{\frac\mu M},\qquad \omega(0)=0.}
$$
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,
$$
\hbar\omega=\epsilon_k+\mu+O(\mu^2/\epsilon_k),
$$
so the same branch becomes particle-like. The <healing length> $\xi=\hbar/\sqrt{2M\mu}$ sets the crossover scale.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2019-q3-condensate-dispersion.png]
{title=Gapless Bogoliubov dispersion and its sound and particle limits}
{description=Original dimensionless curves with $q=k\xi$ and $E/\mu=\hbar\omega/\mu$. The exact curve is $q\sqrt{q^2+2}$, the sound asymptote is $\sqrt2q$, and the large-momentum asymptote is $q^2+1$.}
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