= Solution
Choose the real representative $v=\psi_0=\sqrt{\mu/g}$. For $\psi=v+\chi$,
$$
|\psi|^2=v^2+v(\chi+\chi^*)+|\chi|^2,
$$
and the fourth power through quadratic order is
$$
|\psi|^4=v^4+2v^3(\chi+\chi^*)+2v^2|\chi|^2+v^2(\chi+\chi^*)^2+O(\chi^3).
$$
In $\mu|\psi|^2-g|\psi|^4/2$, the linear terms and the $|\chi|^2$ terms cancel because $gv^2=\mu$. The remaining potential contribution is $\mu^2/(2g)-\mu(\chi+\chi^*)^2/2$.
The time-derivative contribution is
$$
\frac{i\hbar}{2}(\psi^*\dot\psi-\psi\dot\psi^*)
=\frac{i\hbar v}{2}\partial_t(\chi-\chi^*)+\frac{i\hbar}{2}(\chi^*\dot\chi-\chi\dot\chi^*).
$$
Its 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 $-\hbar^2|\nabla\chi|^2/(2M)$. Dropping that boundary term and the constant, the <quadratic condensate fluctuation Lagrangian> is
$$
\boxed{\mathcal L_2=\frac{i\hbar}{2}(\chi^*\dot\chi-\chi\dot\chi^*)-\frac{\hbar^2}{2M}\nabla\chi^*\cdot\nabla\chi-\frac\mu2(\chi+\chi^*)^2.}
$$
The omitted cubic and quartic interactions describe scattering between these small fluctuations; they do not enter the linear <normal mode> calculation.
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