The <Euler-Lagrange equation> is
$$
\ddot\varphi+3H\dot\varphi-\frac{c^2}{a^2}\nabla^2\varphi+2H^2\varphi=0.
$$
After a spatial <Fourier transform> and the change to <conformal time>, $\partial_t=a^{-1}\partial_\tau$, this becomes
$$
\varphi_{\mathbf k}''+2\mathcal H\varphi_{\mathbf k}'
+(c^2k^2+2a^2H^2)\varphi_{\mathbf k}=0.
$$
For the exact <de Sitter spacetime> scale factor $a=-1/(H\tau)$, one has $a''/a=2a^2H^2$. Setting $u_{\mathbf k}=a\varphi_{\mathbf k}$ therefore cancels both the friction term and the effective mass term, leaving
$$
\boxed{(\partial_\tau^2+c^2k^2)(a\varphi_{\mathbf k})=0}.
$$
This is the special cancellation for a <conformally coupled scalar field>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 312 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-312/1/a/solution