Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 62 / 4 / a / Solution

A perturbation of the inflaton shifts its local clock: $\delta \varphi \simeq \dot{\varphi }\delta t$. Regions with different clock shifts reach the end of inflation at different times and hence undergo different amounts of expansion. The inflaton clock-shift origin of curvature perturbations gives a local expansion difference of magnitude $H\delta t=H\delta \varphi /\dot{\varphi }$. The sign depends on whether $\delta t$ denotes a clock advance or a delay to a fixed end value.

The comoving curvature perturbation has amplitude $\mathcal{R}\simeq H\delta \varphi /\dot{\varphi }$. For adiabatic evolution, superhorizon conservation of single-field comoving curvature carries this amplitude into the later growing density contrast, with the numerical conversion depending on the epoch and the density slice. Thus the approximate density-amplitude statement in the question follows; it does not identify the density contrast and $\mathcal{R}$ in every gauge.

The frozen scalar power per $d\log k$ is $(H/2\pi {)}^2$, so the slow-roll curvature power spectrum isConsequently $⟨{\mathcal{R}}^2⟩=\int (dk/k){\Delta }_{\mathcal{R}}^2(k)$, giving the requested approximate growing-density spectrum. Both $H$ and $\dot{\varphi }$ are evaluated separately when each Fourier mode undergoes cosmological horizon exit.