Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 310 / 4

With the symmetric Fourier normalization displayed in the question, the creation and annihilation operators obey the canonical commutation relationSince $\delta \hat{\varphi }=\hat{f}/a$, its vacuum two-point correlation function isComparison with the definition in the question givesUsing $|f_k{|}^2=(1+1/(k^2{\tau }^2))/(2k)$ and $a=-1/(H\tau )$,On superhorizon scales, $k\ll aH$ or $|k\tau |\ll 1$, so the scale-invariant inflationary power spectrum freezes to

A perturbation $\delta \varphi$ shifts the time at which the rolling inflaton reaches a fixed value by $\delta t\simeq \delta \varphi /\dot{\varphi }$. The corresponding perturbation in local expansion is $\delta N=H\delta t$, which is the comoving curvature perturbation up to the stated sign convention:ThereforeUsing ${\dot{\varphi }}^2=2ϵH^2{M}_{\mathrm{Pl}}^2$ gives the slow-roll curvature power spectrumFor each $k$, the slowly varying quantities on the right are evaluated at cosmological horizon exit, $k=aH$.

At cosmological horizon exit, $d\log k/dN=1-ϵ\simeq 1$. Differentiating the slow-roll curvature power spectrum and using the Hubble slow-roll parameters givesThe supplied relations implyConsequently the scalar spectral index is

For $V=\lambda {\varphi }^4$,Thus $n_s-1=-6{ϵ}_V+2{\eta }_V=-3{ϵ}_V$. Combining this with the tensor-to-scalar ratio $r=16{ϵ}_V$ gives the quartic-inflation tilt-tensor relationThe measured tilt $n_s-1=-0.04$ would require $r\simeq 0.213$, whereas $r=0.001$ predicts $n_s-1\simeq -1.88\times 10^{-4}$. Therefore quartic slow-roll inflation is inconsistent with the stated measurements.