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Use primes for conformal-time derivatives in this question, reserving $V_{,\varphi \varphi }$ for a potential derivative. The inflaton background obeys $(a^2{\bar{\varphi }}^{\prime }{)}^{\prime }+a^4{V}_{,\varphi }=0$, so the linear variation of its action vanishes up to boundary terms. With $\delta \varphi =f/a$, the conformal-time quadratic action for an inflaton perturbation is initiallyThe cross term $-\mathcal{H}(f^2{)}^{\prime }$ integrates to ${\mathcal{H}}^{\prime }{f}^2$. Since ${\mathcal{H}}^{\prime }+{\mathcal{H}}^2=a^{\prime \prime }/a$, this becomesVarying $f$ gives $f^{\prime \prime }-{\nabla }^{2}f+(a^2{V}_{,\varphi \varphi }-a^{\prime \prime }/a)f=0$. The symmetric Fourier transform convention used in the question turns $-{\nabla }^2$ into $k^2$. Dropping the small mass term under the stated slow-roll inflation assumption yieldsFor de Sitter spacetime, $a=-1/(H\tau )$ with $\tau <0$, so $a^{\prime \prime }/a=2/{\tau }^2$. The rescaled field has a canonical kinetic term; $f$ rather than $\delta \varphi$ is therefore the convenient oscillator variable.

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.

The slow-roll approximation neglects $\ddot{\varphi }$ relative to $3H\dot{\varphi }$ and kinetic energy relative to the potential. It is self-consistent on the attractor when the potential slow-roll parameter $ϵ\ll 1$ and the second potential slow-roll parameter $|\eta |\ll 1$. ThusHere, as in the inflaton equations, $M_{\mathrm{pl}}$ denotes the reduced Planck mass, not the unreduced mass used in the thermal calculation. The slow-roll curvature power spectrum becomesso

To differentiate with respect to horizon-exit scale, write $\mathcal{N}=\ln a$, increasing with physical time. Along the slow-roll trajectory,The difference between differentiating with respect to $\ln k$ and $\mathcal{N}$ contributes only at second slow-roll order to the tilt. Since $\ln {\Delta }_{\mathcal{R}}^2=3\ln V-2\ln |V^{\prime }|+\mathrm{constant}$,Therefore the scalar spectral index in potential slow-roll parameters isAll background quantities in these expressions are evaluated when the particular mode exits the Hubble radius.

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.

Divide the primordial tensor power spectrum by the slow-roll curvature power spectrum:Differentiating the scalar-field Friedmann equation and using the inflaton equation gives $\dot{H}=-{\dot{\varphi }}^2/(2M_{\mathrm{pl}}^2)$. Hence $r=16{ϵ}_H\simeq 16ϵ$. For the quadratic model at 50 remaining e-folds,This is above the supplied upper bound, so the model fails the stated tensor constraint, despite its satisfactory scalar tilt and adjustable scalar amplitude. Calibrating $m$ does not reduce $r$, since the coupling cancels from the ratio.

The first Hubble slow-roll parameter measures the fractional change of $H$ per number of e-folds, and the second Hubble slow-roll parameter measures the fractional change of the first parameter. Since $d\log a/dt=H$,For a canonical inflaton, $ϵ={\dot{\varphi }}^2/(2M_{\mathrm{Pl}}^2{H}^2)\ge 0$. Expanding cosmic inflation means $\ddot{a}>0$, equivalently $\boxed{ϵ<1}$ because $\ddot{a}/a=H^2(1-ϵ)$. Small $\eta$ is not required for accelerated expansion itself, but helps maintain the slow-roll approximation over many e-folds. Its logarithmic definition assumes $ϵ>0$; in exact de Sitter expansion it is undefined unless interpreted as a limit.

For conventional slow-roll inflation, assume ${\dot{\varphi }}^2\ll V$, $|\ddot{\varphi }|\ll 3H|\dot{\varphi }|$, a smooth positive scalar potential, and slowly varying small parameters. The Friedmann equation and inflaton equation of motion then reduce toSubstitute these in the exact expression for $ϵ$:To obtain the potential and Hubble slow-roll parameter relation, differentiate $\log {ϵ}_V$ along the slow-roll trajectory $\dot{\varphi }/H\simeq -M_{\mathrm{Pl}}^2{V}_{,\varphi }/V$:HenceThe second quantity on the right is the second potential slow-roll parameter, not the logarithmic Hubble-flow $\eta$ itself. Consistency usually requires ${ϵ}_V\ll 1$ and $|{\eta }_V|\ll 1$; identities involving division by $V_{,\varphi }$ are used on rolling intervals and interpreted by limiting values at stationary points.

For the inflaton,The continuity equation therefore becomesAway from an isolated turning point division by $\dot{\varphi }$ gives, and continuity extends through such points,

The homogeneous inflaton obeysUnder the slow-roll approximation, the acceleration and kinetic-energy corrections may be neglected, so the field equation and Friedmann equation becomeThus a positive field rolls away from the hilltop while $H$ is approximately constant.

In an effective field theory, higher-power terms in an inflaton scalar potential can be important when $\varphi /M_{\mathrm{Pl}}$ is large, even if $V/M_{\mathrm{Pl}}^4$ is small. An assumed simple monomial inflation potential over such a field range needs a symmetry or ultraviolet completion controlling these operators and radiative corrections. A large field value alone does not establish Planckian energy density or invalidate every possible completion.

For a canonical inflaton under the slow-roll approximation, $\dot{\varphi }/H\simeq -M_{\mathrm{Pl}}^2{V}^{\prime }/V$ and $ϵ\simeq {ϵ}_V=(M_{\mathrm{Pl}}^2/2)(V^{\prime }/V{)}^2$. Differentiating $\log {ϵ}_V$ along this trajectory givesHere $\eta =d\log ϵ/d\log a$, not the alternate convention $-\ddot{\varphi }/(H\dot{\varphi })$. The logarithmic definition requires $ϵ>0$; exact de Sitter expansion is treated as a limit.

For a quadratic potential with the end set by ${ϵ}_V=1$, a mode leaving $N$ e-folds earlier has ${\varphi }_{\ast }^2=(4N+2)M_{\mathrm{pl}}^2$. Its curvature amplitude is $(m/M_{\mathrm{pl}}{)}^2(2N+1{)}^2/(24{\pi }^2)$. Thus the scalar amplitude fixes the inflaton mass while the tilt and tensor ratio depend only on the exit e-fold count at this order.