The quadratic <path integral> is
$$
Z=\int\mathcal D\theta\,\mathcal D\delta\rho\,
\exp\left(i\int dt\,d^dx\,\mathcal L_2\right).
$$
Introduce the positive spatial operator
$$
A=2\lambda-\frac{\nabla^2}{4m\bar\rho}.
$$
Completing the square in the <Gaussian functional integral> over $\delta\rho$ yields
$$
\mathcal L_{\rm eff}=
\frac12\dot\theta A^{-1}\dot\theta-
\frac{\bar\rho}{2m}(\nabla\theta)^2.
$$
The <derivative expansion> $A^{-1}=(2\lambda)^{-1}+O(\nabla^2)$ therefore gives
$$
\mathcal L_{\rm G}=\frac1{4\lambda}
\left[\dot\theta^2-v^2(\nabla\theta)^2\right],
\qquad
v^2=\frac{2\lambda\bar\rho}{m}=-\frac rm.
$$
This is the long-wavelength <Goldstone boson> action. Keeping the spatial derivative in $A$ gives $\omega^2=v^2k^2+k^4/(4m^2)$.

Solution

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

 Solution

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