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For a quantum harmonic oscillator with $\omega >0$, a complex mode satisfying the Wronskian normalization $q{\dot{q}}^{\ast }-\dot{q}{q}^{\ast }=i$ can be written $q=r{e}^{is}$ with $\dot{s}=-1/(2r^2)$. Its vacuum energy is ${\dot{r}}^2/2+1/(8r^2)+{\omega }^2{r}^2/2$. This is at least $\omega /2$, with equality exactly when $\dot{r}=0$ and $r^2=1/(2\omega )$. The same argument fixes the subhorizon positive-frequency normalization in the Bunch-Davies vacuum of a canonical cosmological mode.

For single-field adiabatic inflation, the Mukhanov-Sasaki variable rescales the comoving curvature perturbation by $z=a\sqrt{2ϵ}{M}_{\mathrm{Pl}}$, where $M_{\mathrm{Pl}}$ is the reduced Planck mass. Its quadratic bulk action is $\frac{1}{2}\int [(v^{\prime }{)}^2-(\nabla v{)}^2+(z^{\prime \prime }/z)v^2]$, after a boundary term is removed. A Fourier transform mode obeys $v_k^{\prime \prime }+(k^2-z^{\prime \prime }/z)v_k=0$ and has canonical Wronskian normalization. The mass scale and sound-speed factors depend on the underlying scalar theory; this definition assumes the canonical single-field action and positive slow-roll kinetic coefficient.

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.

Evaluate the field variance in the zero-occupation Fock vacuum, ${\hat{a}}_{\mathbf{k}}|0⟩=0$, with $[{\hat{a}}_{\mathbf{k}},{\hat{a}}_{{\mathbf{k}}^{\prime }}^{†}]={\delta }^3(\mathbf{k}-{\mathbf{k}}^{\prime })$. The canonical commutation relation alone does not specify a quantum state; this vacuum assumption is needed for the requested expression without occupation factors. In the Heisenberg picture, use the position-space mode expansion of a free fieldThis is the position-space form of the shorthand oscillator expansion in the question; equivalently its Fourier operator uses the creation operator with opposite wavevector. The annihilation operators give $⟨{\hat{a}}_{\mathbf{k}}{\hat{a}}_{{\mathbf{k}}^{\prime }}^{†}⟩={\delta }^3(\mathbf{k}-{\mathbf{k}}^{\prime })$, while all other vacuum pairings vanish. The mean field fluctuation is zero, so the two-point function is its coincident-point vacuum field variance. Assuming spatial isotropy, angular integration gives the variance per logarithmic wavenumber interval:Here the dimensionless power spectrum is denoted ${\Delta }_v$, as in the question, rather than the alternative notation ${\Delta }_v^2$. A coincident vacuum field product needs smearing or ultraviolet regulation: the unregulated full vacuum variance is generally divergent. With a specified finite wavenumber band, the same expression holds with those integration limits. This does not affect the mode-by-mode dimensionless power spectrum.

For the leading nonzero slower branch, put $x=-k\tau =k/(aH)>0$ and $v_k=c{x}^{1/2-\nu }/\sqrt{2k}$. Since the physical fluctuation is $\delta \chi =v/a$, its dimensionless power spectrum isFor a $k$-independent coefficient $c$, this is a blue spectral tilt, because $3-2\nu >0$ in the specified mass range. At fixed $k$ it decays as $a^{-(3-2\nu )}$, consistently with the field-amplitude decay found above.

To include the other branch explicitly, the general superhorizon expression isIf $c_{-}\ne 0$, it approaches the displayed target spectrum with $c=c_{-}$. A pure faster classical branch would instead have power proportional to $x^{3+2\nu }$. For a normalized quantum mode, however, the Wronskian normalization $v_k{v}_k^{\ast \prime }-v_k^{\ast }{v}_k^{\prime }=i$ requires both independent real solutions; its slower coefficient cannot vanish. Thus the target is the leading superhorizon quantum spectrum, while the faster classical solution remains part of the complete mode analysis.

Regulate the zero mode by a soft momentum $\mathbf{q}$ and write the free modes asThe Wronskian normalization $f_q{g}_q^{\ast }-f_q^{\ast }{g}_q=i$ follows from the canonical commutator, while the power spectrum is $P(q)=|f_q{|}^2$. Inserting the creation part of $Q$ on the ket and the annihilation part on the bra, their difference is precisely the Wronskian. For $\mathcal{O}=\varphi (\mathbf{k})\varphi ({\mathbf{k}}^{\prime })$ this gives, after taking $q\to 0$ and setting the irrelevant normalization $c=1$,Equivalently, before removing the regulator, the right-hand side is the soft three-point function divided by $P(q)$.

The positive-frequency solution for $u_k=a{f}_k$ is proportional to $e^{-ick\tau }$. Since $a=-1/(H\tau )$, its scalar-field mode function has the stated form $f_k=\mathcal{N}\tau e^{-ick\tau }$. The canonical momentum in conformal time is $\pi =a^2{\phi }^{\prime }$, so the canonical commutation relation requires the Wronskian normalizationSubstitution gives $2ick{a}^2|\mathcal{N}{|}^2{\tau }^2=i$. Hence, up to an irrelevant constant phase,The choice $e^{-ick\tau }$ is the Bunch-Davies vacuum condition at early conformal time.

Let $O=\phi ({\mathbf{k}}_1)\phi ({\mathbf{k}}_2)$. Since $Q_S=-2{\omega }_{ij}{\Pi }_{ij}(0)$,For a soft mode ${\gamma }^s=f_q{a}_s+f_q^{\ast }{a}_s^{†}$ and ${\Pi }^s=g_q{a}_s+g_q^{\ast }{a}_s^{†}$, the vacuum obeysThe mode-function Wronskian normalizationimplieswhere $P_{\gamma }(q)$ is the graviton power spectrum per polarization. The equal-time bispectrum is real in this parity-even configuration, and thereforeThis turns the charge insertion into the soft-graviton insertion used in the Soft graviton theorem.