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At a fixed horizon-exit e-fold count and with the later cosmology fixed, rescaling a canonical single-field inflationary potential by a positive constant leaves its potential slow-roll parameters and field solution unchanged. The slow-roll curvature power spectrum is proportional to $V/{ϵ}_V$, so the matter power spectrum scales by the same factor while its leading tilt and transfer-function shape remain unchanged. A changed specified reheating history can also shift the horizon-exit count assigned to a fixed physical wavenumber.

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.

The slow-roll equations turn the slow-roll curvature power spectrum intoThe spectral tilt is the logarithmic derivative of this dimensionless spectrum. Apply the horizon-exit logarithmic derivative in slow roll:With the potential slow-roll parameter $ϵ$ above and the second potential slow-roll parameter $\eta =V_{,\varphi \varphi }/V$,This is the first-order departure of the scalar spectral index from unity, in reduced Planck units.

The slow-roll equations give $H^2\simeq V/(3M_{\mathrm{Pl}}^2)$ and $\dot{\Phi }\simeq -V^{\prime }/(3H)$. Substituting in the stated slow-roll curvature power spectrum yieldsWith the potential slow-roll parameter ${ϵ}_V=M_{\mathrm{Pl}}^2{V}^{\prime 2}/(2V^2)$,All quantities are evaluated at horizon exit. The two forms have the same normalization because $(\sqrt{8\pi }{)}^4=64{\pi }^2$.

At horizon exit $k=aH$, $d\log k/dt=H(1-{ϵ}_H)$, where ${ϵ}_H=-\dot{H}/H^2$ is the Hubble slow-roll parameter. To leading slow-roll order,The slow-roll curvature power spectrum is proportional to $V^3/V^{\prime 2}$. Therefore the scalar spectral index in potential slow-roll parameters isUsing the potential slow-roll parameter and the second potential slow-roll parameter givesThe neglected horizon-exit derivative correction contributes at higher slow-roll order.

The supplied slow-roll curvature power spectrum simplifies toAt a fixed number of e-folds before the end, multiplying the potential by a constant does not change $V^{\prime }/V$, $V^{\prime \prime }/V$ or the slow-roll field value. It multiplies ${\mathcal{P}}_{\mathcal{R}}$ by that constant. Thus, for an unchanged subsequent cosmology,The scalar spectral index, transfer-function shape, turnover and acoustic features are unchanged at this order. Root-mean-square density fluctuations increase by $\sqrt{2}$. The comparison uses the same exit e-fold count; a different specified reheating history could shift the detailed mapping between $k$ and that count.

The quantum harmonic oscillator has $\hat{H}=({\dot{\hat{q}}}^2+{\omega }^2{\hat{q}}^2)/2$, with $\omega >0$. Using $\hat{a}|0⟩=0$ and the canonical commutation relation givesFor $q=r{e}^{is}$, the Wronskian normalization isConsequentlyEquality requires $\dot{r}=0$ and $r^2=1/(2\omega )$. The minimum-energy normalized oscillator mode is thereforeThe constant phase $\alpha$ is irrelevant. This mode also solves the oscillator equation of motion; an arbitrary squeezed mode would have larger vacuum energy.

For inflation, introduce the Mukhanov-Sasaki variable $v=z\mathcal{R}$. Since $z$ depends only on conformal time,Integrating the cross term by parts gives the canonical bulk actionup to the boundary term $-\frac{1}{2}\int d^{3}x[(z^{\prime }/z)v^2{]}_{\mathrm{boundary}}$. The resulting Euler-Lagrange field equation and its Fourier transform areAt leading order in the slow-roll approximation, retain a small positive, nearly constant $ϵ$ in $z=a\sqrt{2ϵ}$, while approximating $a=-1/(H\tau )$ and $H$ as constant. Then $z^{\prime \prime }/z\simeq 2/{\tau }^2$, soStrict exact de Sitter spacetime would have $ϵ=0$ and would not supply the nonzero curvature kinetic coefficient assumed here; the calculation is the leading quasi-de Sitter limit, not a substitution of zero into $z$.

For $-k\tau \gg 1$, the expansion correction is negligible and each canonical mode is a quantum harmonic oscillator of conformal frequency $k$. The Bunch-Davies vacuum selects its positive-frequency, minimum-energy mode $e^{-ik\tau }/\sqrt{2k}$ in that early subhorizon regime. It does not minimize an instantaneous Hamiltonian after the effective squared frequency has become negative outside the horizon.

A basis of exact solutions to the leading mode equation is $e^{-ik\tau }(1-i/(k\tau ))$ and its complex conjugate. Write the normalized Bogoliubov transformation combination asThe Bunch-Davies vacuum boundary condition sets ${\beta }_k=0$ and ${\alpha }_k=1$ up to phase. HenceSubstitution verifies the equation, and $v_k{v}_k^{\ast \prime }-v_k^{\prime }{v}_k^{\ast }=i$ verifies the Wronskian normalization.

Dividing by $z$ gives the comoving curvature perturbation varianceThis is the dimensional slow-roll curvature power spectrum in the printed normalization, with the reduced Planck mass set to one. The corresponding dimensionless cosmological power spectrum is $k^3{P}_{\mathcal{R}}/(2{\pi }^2)=H^2/(8{\pi }^2ϵ)$, which is independent of $k$ at this order. Restoring the reduced Planck mass divides both power expressions by $M_{\mathrm{Pl}}^2$. Slowly varying background quantities are evaluated near each mode's horizon exit; their variation generates the small departure from exact scale invariance.

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.

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.

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 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.