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Apply the stated passive coordinate change to the background metric. Its spatial variation isCompare this with the spatial scalar decomposition $\delta g_{ij}=-2a^2\Psi {\delta }_{ij}+2a^2{\partial }_i{\partial }_j\chi$. It gives $\tilde{\Psi }=\Psi +\mathcal{H}T$ and $\tilde{\chi }=\chi -L$, where $\mathcal{H}=a^{\prime }/a$. The scalar's background depends only on time, soThe spatial displacement $L$ changes the shear potential but not these two scalar amplitudes in this decomposition. Direct cancellation of $T$ gives either of the useful invariant combinations$Q$ is the field fluctuation on flat gauge in cosmology slices: choosing $T=-\Psi /\mathcal{H}$ sets the transformed curvature potential to zero and leaves $\widetilde{\delta \varphi }=Q$. The comoving curvature perturbation $\mathcal{R}$ is the spatial curvature potential on uniform-field slices: choose $T=\delta \varphi /{\bar{\varphi }}^{\prime }$ and use (iii) to see that these slices also have zero scalar momentum density. For this spatial sign convention the intrinsic curvature perturbation is $\delta R^{(3)}=4a^{-2}{\nabla }^2\mathcal{R}$ in that gauge; some conventions define the curvature perturbation with the opposite sign. The canonically useful Mukhanov-Sasaki variable is $v=aQ$.

The interpretation of $\mathcal{R}$ requires ${\bar{\varphi }}^{\prime }\ne 0$, and the flat-slice expression requires $\mathcal{H}\ne 0$. The polynomial combination ${\bar{\varphi }}^{\prime }\Psi +\mathcal{H}\delta \varphi$ remains invariant without division; at a stationary field background $\delta \varphi$ itself is invariant under the time shift, but that field cannot serve as a comoving clock.

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.