 Solution

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.