Linearizing the <curvaton> equation and Fourier transforming gives
$$
\delta\sigma_k''+2\frac{a'}a\delta\sigma_k'
+(k^2+a^2m_\sigma^2)\delta\sigma_k=0.
$$
Writing $f_k^\sigma=a\delta\sigma_k$ yields
$$
(f_k^\sigma)''+\left(k^2+a^2m_\sigma^2-\frac{a''}{a}\right)f_k^\sigma=0.
$$
Since $m_\sigma\ll H$, the mass term is negligible and this is the equation stated. It has the same Bunch-Davies mode as the inflaton perturbation, so after horizon exit
$$
\boxed{\Delta_{\delta\sigma}^2(k)=\left(\frac H{2\pi}\right)^2}
$$
up to a small mass-induced spectral tilt.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 310 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-310/4/b/solution