= Solution
A perturbation $\delta\phi$ shifts the time at which the rolling inflaton reaches a fixed value by $\delta t\simeq\delta\phi/\dot\phi$. The corresponding perturbation in local expansion is $\delta N=H\delta t$, which is the <comoving curvature perturbation> up to the stated sign convention:
$$
\mathcal R=\frac H{\dot\phi}\delta\phi.
$$
Therefore
$$
\Delta_{\mathcal R}^2
=\frac{H^2}{\dot\phi^2}\Delta_{\delta\phi}^2.
$$
Using $\dot\phi^2=2\epsilon H^2M_{\rm Pl}^2$ gives the <slow-roll curvature power spectrum>
$$
\boxed{\Delta_{\mathcal R}^2(k)
=\frac1{2\epsilon M_{\rm Pl}^2}
\left(\frac H{2\pi}\right)^2}.
$$
For each $k$, the slowly varying quantities on the right are evaluated at <cosmological horizon exit>, $k=aH$.
Back to article page
