= Solution
The <quantum harmonic oscillator> has $\hat H=(\dot{\hat q}^{\,2}+\omega^2\hat q^{\,2})/2$, with $\omega>0$. Using $\hat a|0\rangle=0$ and the <canonical commutation relation> gives
$$
E_0\equiv\langle0|\hat H|0\rangle=\frac12\left(|\dot q|^2+\omega^2|q|^2\right).
$$
For $q=re^{is}$, the <Wronskian normalization> is
$$
q\dot q^*-\dot q q^*=-2ir^2\dot s=i,
\qquad \dot s=-\frac1{2r^2}.
$$
Consequently
$$
E_0=\frac{\dot r^2}{2}+\frac1{8r^2}+\frac{\omega^2r^2}{2}
\geq\frac\omega2.
$$
Equality requires $\dot r=0$ and $r^2=1/(2\omega)$. The <minimum-energy normalized oscillator mode> is therefore
$$
\boxed{q(t)=\frac{e^{-i\omega t+i\alpha}}{\sqrt{2\omega}},\quad
E_0=\frac\omega2,\quad
\langle0|\hat q^\dagger\hat q|0\rangle=|q|^2=\frac1{2\omega}}.
$$
The 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>,
$$
z^2(\mathcal R')^2=\left(v'-\frac{z'}zv\right)^2,
\qquad z^2(\partial_i\mathcal R)^2=(\partial_iv)^2.
$$
Integrating the cross term by parts gives the canonical bulk <action>
$$
\boxed{S=\frac12\int d\tau\,d^3x\left[(v')^2-(\nabla v)^2+\frac{z''}zv^2\right]}
$$
up to the boundary term $-\tfrac12\int d^3x\,[(z'/z)v^2]_{\rm boundary}$. The resulting <Euler-Lagrange field equation> and its <Fourier transform> are
$$
v''-\nabla^2v-\frac{z''}zv=0,\qquad
\boxed{v_k''+\left(k^2-\frac{z''}z\right)v_k=0}.
$$
At leading order in the <slow-roll approximation>, retain a small positive, nearly constant $\epsilon$ in $z=a\sqrt{2\epsilon}$, while approximating $a=-1/(H\tau)$ and $H$ as constant. Then $z''/z\simeq2/\tau^2$, so
$$
v_k''+\left(k^2-\frac2{\tau^2}\right)v_k=0.
$$
Strict exact <de Sitter spacetime> would have $\epsilon=0$ 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 $-k\tau\gg1$, 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 as
$$
v_k=\frac1{\sqrt{2k}}\left[
\alpha_k e^{-ik\tau}\left(1-\frac{i}{k\tau}\right)
+\beta_k e^{ik\tau}\left(1+\frac{i}{k\tau}\right)\right],
\qquad |\alpha_k|^2-|\beta_k|^2=1.
$$
The <Bunch-Davies vacuum> boundary condition sets $\beta_k=0$ and $\alpha_k=1$ up to phase. Hence
$$
\boxed{v_k(\tau)=\frac{e^{-ik\tau}}{\sqrt{2k}}\left(1-\frac{i}{k\tau}\right)}.
$$
Substitution verifies the equation, and $v_kv_k^{*\prime}-v_k'v_k^*=i$ verifies the <Wronskian normalization>.
Dividing by $z$ gives the <comoving curvature perturbation> variance
$$
|\mathcal R_k(\tau)|^2
=\frac{H^2}{4\epsilon k^3}(1+k^2\tau^2),
\qquad
\boxed{P_{\mathcal R}(k)=\frac{H^2}{4\epsilon k^3}}.
$$
This 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 $k^3P_{\mathcal R}/(2\pi^2)=H^2/(8\pi^2\epsilon)$, which is independent of $k$ at this order. Restoring the <reduced Planck mass> divides both power expressions by $M_{\rm Pl}^2$. Slowly varying background quantities are evaluated near each mode's horizon exit; their variation generates the small departure from exact scale invariance.
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