 Solution

A light inflaton behaves approximately as a scalar field in de Sitter space while a wavelength is well inside the Hubble radius. Its Bunch-Davies vacuum has quantum fluctuations. Cosmic inflation stretches each Fourier mode until $k=aH$, after which its physical wavelength exceeds the Hubble radius. The nearly constant growing field mode has a typical fluctuation per logarithmic wavenumber interval $\delta \varphi \sim H/(2\pi )$.

A field fluctuation changes the local position on the rolling background trajectory. Neighboring regions therefore reach the same field value, and the end of inflation, at slightly different times. A clock displacement of magnitude $|\delta \varphi /\dot{\varphi }|$ becomes a difference in local expansion of order $H|\delta \varphi /\dot{\varphi }|$. Equivalently, in a conventional sign choice the comoving curvature perturbation is related to the field fluctuation on spatially flat slices byChanging the sign convention for spatial curvature changes the sign of $\mathcal{R}$, but not its spectrum. This is the inflaton clock-shift origin of curvature perturbations. For a single-field slow-roll attractor there is no independent entropy mode. On a super-Hubble scale, gradient terms are negligible and the superhorizon conservation of single-field comoving curvature preserves the growing adiabatic mode, so fluctuations generated near exit persist as primordial curvature perturbations. This conservation requires the attractor and adiabatic assumptions; a freely chosen non-attractor background would not have the same conclusion.