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Along cosmological horizon exit $k\propto aH$, differentiation gives $d\log k/d\log a=1-{ϵ}_H$, where ${ϵ}_H=-\dot{H}/H^2$. The chain rule gives the first expression. In reduced Planck units, the slow-roll equations $3H\dot{\varphi }\simeq -V^{\prime }$ and $3H^2\simeq V$ give the second expression to leading order. Applied to a quantity already of slow-roll order, the omitted factor changes the answer only at second order.

Denote the comoving box volume by $\mathcal{V}$, keeping it distinct from the inflaton potential $V(\varphi )$. For the Canonically rescaled de Sitter scalar mode $u_k=a\delta {\varphi }_k$, the equation in part (b) becomesDirect differentiation verifies the positive-frequency solutionFor $|k\tau |\gg 1$, $u_k$ tends to $e^{-ik\tau }/\sqrt{2k\mathcal{V}}$, the positive-frequency flat-space mode. It therefore selects the incoming Bunch-Davies vacuum. The Bunch-Davies mode normalization in a finite comoving volume is fixed by the Wronskian conditionAfter cosmological horizon exit, $|k\tau |\ll 1$, the mode tends to the time-independent value $iH/\sqrt{2k^3\mathcal{V}}$. The complex mode and its Hermitian conjugate together make the real quantum field.

A perturbation of the inflaton shifts its local clock: $\delta \varphi \simeq \dot{\varphi }\delta t$. Regions with different clock shifts reach the end of inflation at different times and hence undergo different amounts of expansion. The inflaton clock-shift origin of curvature perturbations gives a local expansion difference of magnitude $H\delta t=H\delta \varphi /\dot{\varphi }$. The sign depends on whether $\delta t$ denotes a clock advance or a delay to a fixed end value.

The comoving curvature perturbation has amplitude $\mathcal{R}\simeq H\delta \varphi /\dot{\varphi }$. For adiabatic evolution, superhorizon conservation of single-field comoving curvature carries this amplitude into the later growing density contrast, with the numerical conversion depending on the epoch and the density slice. Thus the approximate density-amplitude statement in the question follows; it does not identify the density contrast and $\mathcal{R}$ in every gauge.

The frozen scalar power per $d\log k$ is $(H/2\pi {)}^2$, so the slow-roll curvature power spectrum isConsequently $⟨{\mathcal{R}}^2⟩=\int (dk/k){\Delta }_{\mathcal{R}}^2(k)$, giving the requested approximate growing-density spectrum. Both $H$ and $\dot{\varphi }$ are evaluated separately when each Fourier mode undergoes cosmological horizon exit.

At cosmological horizon exit, $k$ is proportional to $aH$; the convention-dependent factor $2\pi$ has zero logarithmic derivative. Define the first Hubble slow-roll parameter ${ϵ}_H=-\dot{H}/H^2$. ThenThe chain rule gives $d/d\log a=(\dot{\varphi }/H)d/d\varphi$. In reduced Planck units, the slow-roll equations are $3H\dot{\varphi }\simeq -V_{,\varphi }$ and $3H^2\simeq V$. Therefore the horizon-exit logarithmic derivative in slow roll is

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

The creation and annihilation operators obeyThe two-point correlation function in their vacuum has Fourier amplitude $|f_k{|}^2/a^2$. After cosmological horizon exit, $|k\tau |\ll 1$, and the stated Bunch-Davies vacuum mode and $a=-1/(H\tau )$ giveThe dimensionless power spectrum is consequently

For each mode, the right side ofis evaluated at cosmological horizon exit, $k=aH$. Sincefirst order in the Hubble slow-roll parameters gives

The right-hand side of the slow-roll curvature power spectrum is evaluated at cosmological horizon exit, $k=aH$. Usingrewrites it asAt horizon exit, $d\log k=d\log (aH)=(1-ϵ)dN$, so $d/d\log k=d/dN$ to first slow-roll order. With the first Hubble slow-roll parameter and second Hubble slow-roll parameter,The supplied relations $ϵ={ϵ}_V$ and ${\eta }_V=2ϵ-\eta /2$ imply $\eta =4{ϵ}_V-2{\eta }_V$. Therefore

The right-hand side of the slow-roll curvature power spectrum must be evaluated separately for each mode at cosmological horizon exit, $k=aH$. Taking logarithms givesFor $N=\log a$,HenceThis is the scalar spectral index in Hubble slow-roll parameters.

Canonical quantization imposesThe two-point correlation function has mode power $P_{\delta \varphi }=|f_k{|}^2/a^2$. After cosmological horizon exit, $|k\tau |\ll 1$, soThereforeThis is the scale-invariant inflationary power spectrum of a light canonical scalar.

For a quadratic inflaton potential ending at the leading estimate ${ϵ}_V=1$, $\Phi /M_{\mathrm{Pl}}=\sqrt{4N+2}$. Cosmological horizon exit gives $d\log k/dN=-1+{ϵ}_H$. A narrow logarithmic interval therefore probes a field span $2\Delta \log k/(\sqrt{4N+2}(1-{ϵ}_H))$, to the accuracy of the slow-roll approximation. At sixty e-folds, five logarithmic wavenumber units probe about $0.64M_{\mathrm{Pl}}$ around a background field near $15.6M_{\mathrm{Pl}}$.