Stability requires $\lambda>0$. The classical potential $r\rho+\lambda\rho^2$ has a symmetry-breaking minimum when $r<0$, at
$$
\bar\rho=-\frac r{2\lambda}>0.
$$
Writing $\rho=\bar\rho+\delta\rho$ and discarding constants and <total derivatives> gives the <quadratic Lagrangian>
$$
\mathcal L_2=\delta\rho\,\dot\theta-
\frac{(\nabla\delta\rho)^2}{8m\bar\rho}-
\lambda(\delta\rho)^2-
\frac{\bar\rho}{2m}(\nabla\theta)^2.
$$
The density fluctuation is a gapped <amplitude mode>, while the phase is the prospective <Goldstone boson>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 337 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-337/1/b/solution