Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 48 / 4 / c / Solution

The annihilation operators kill the vacuum, and their commutator givesTherefore the scale-invariant inflationary power spectrum has the superhorizon limitThe canonical mode grows as $1/|\tau |$ outside the Hubble radius, while the physical field perturbation $f/a$ freezes. A slowly varying $H$ produces an almost scale-invariant spectrum rather than exact scale invariance.

For a slowly rolling single field, fluctuations in the inflaton clock become the comoving curvature perturbation, $\mathcal{R}\simeq -H\delta \varphi /\dot{\bar{\varphi }}$ up to sign convention. Restoring the reduced Planck mass and using $ϵ={\dot{\bar{\varphi }}}^2/(2M_{\mathrm{pl}}^2{H}^2)$ givesThis conversion presumes nonzero background roll; a strictly constant test scalar in exact de Sitter spacetime does not by itself define a finite comoving curvature perturbation through this formula. The curvature perturbations seed the later density fluctuations.

When metric fluctuations and their Einstein-gravity normalization are restored, the two graviton polarizations have the same massless mode equation. The usual total primordial tensor power spectrum is ${\Delta }_t^2=2H^2/({\pi }^2{M}_{\mathrm{pl}}^2)$, so the single-field leading ratio is $r=16ϵ$. Tensor amplitudes therefore probe the inflationary energy scale, whereas scalar amplitudes also depend on the roll rate. These tensor statements explain the relevance of the mode result; they are not derived from a scalar action with metric perturbations omitted.