Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 55 / 5 / c / Solution

The slow-roll equations give $H^2\simeq V/(3M_{\mathrm{Pl}}^2)$ and $\dot{\Phi }\simeq -V^{\prime }/(3H)$. Substituting in the stated slow-roll curvature power spectrum yieldsWith the potential slow-roll parameter ${ϵ}_V=M_{\mathrm{Pl}}^2{V}^{\prime 2}/(2V^2)$,All quantities are evaluated at horizon exit. The two forms have the same normalization because $(\sqrt{8\pi }{)}^4=64{\pi }^2$.