Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 312 / 2

The massless field equation isFor a homogeneous solution this becomesThereforeSince $\dot{\bar{\varphi }}={\bar{\varphi }}^{\prime }/a$, the cosmic-time velocity decays as $a^{-3}$.

The second Friedmann equation gives the first Hubble slow-roll parameterBecause $\dot{\bar{\varphi }}\propto a^{-3}$,ThusAs $a\to \infty$, the kinetic energy decays as $a^{-6}$ and the constant $\Lambda$ dominates. Hence $H$ approaches a constant,This non-attractor background is ultra-slow-roll inflation.

Varying the quadratic action givesFor a Fourier mode,At late times $ϵ\propto a^{-6}$, soThe equation becomes ${\mathcal{R}}_k^{\prime \prime }+4{\mathcal{R}}_k^{\prime }/\eta +k^2{\mathcal{R}}_k=0$. At $k=0$, its independent solutions are a constant and ${\eta }^{-3}$. The leading late-time solution in the form given in the question is thereforeThis is the growing curvature perturbation in ultra-slow-roll inflation.

Expand the quantized field using the normalized modeThe canonical commutation relation fixes the Wronskian $M_{\mathrm{Pl}}^2{z}^2(u_k{u}_k^{\ast \prime }-u_k^{\ast }{u}_k^{\prime })=i$. Writing $x=k\eta$, the primed correlators areEach unprimed expression includes $(2\pi {)}^3{\delta }^{(3)}(\mathbf{k}+{\mathbf{k}}^{\prime })$. The imaginary part of the mixed correlator givesConsequently the classicality parameter of a cosmological perturbation isIt tends to zero on superhorizon scales, so the perturbation becomes effectively classical even while its amplitude grows as $a^3$. In ordinary slow-roll inflation, $\mathcal{R}$ instead freezes to a constant; its analogous normalized commutator also vanishes, but behaves as $|k\eta |$ for the leading de Sitter mode rather than $|k\eta {|}^3$.