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

Put $q=\delta\theta$, $p=\delta\phi$, so $\theta=\pi/2-q$ and $\phi=p$. The <quadratic Lagrangian> is obtained by taking
$$
\mathcal L_2=sq\dot p-
\frac{\rho_s}{2}\left[(\nabla q)^2+(\nabla p)^2\right].
$$
The two real fluctuations form a coordinate and its <canonical momentum> rather than two independent modes. Their <Euler-Lagrange equations> combine to give
$$
\ddot q+\left(\frac{\rho_s}{s}\right)^2\nabla^4q=0,
$$
and similarly for $p$. The <ferromagnetic magnon> therefore has the quadratic <dispersion relation>
$$
\omega_k=\frac{\rho_s}{s}k^2.
$$

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 337 / 2 / d / Solution

A quadratic Lagrangian contains terms of second order in field fluctuations and determines their linearized propagators and dispersion relations.


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