= Solution
The annihilation operators kill the vacuum, and their commutator gives
$$
\langle0|\delta\hat\phi_{\mathbf k}^\dagger\delta\hat\phi_{\mathbf k'}|0\rangle
=\frac{|u_k|^2}{a^2}\delta^{(3)}(\mathbf k-\mathbf k')
=\frac{H^2}{2k^3}(1+k^2\tau^2)\delta^{(3)}(\mathbf k-\mathbf k').
$$
Therefore the <scale-invariant inflationary power spectrum> has the superhorizon limit
$$
\boxed{\langle\delta\hat\phi_{\mathbf k}^\dagger\delta\hat\phi_{\mathbf k'}\rangle
\longrightarrow\frac{H^2}{2k^3}\delta^{(3)}(\mathbf k-\mathbf k'),\qquad
\Delta_{\delta\phi}^2=\frac{k^3}{2\pi^2}\frac{H^2}{2k^3}=\left(\frac H{2\pi}\right)^2}.
$$
The 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>, $\mathcal R\simeq-H\delta\phi/\dot{\bar\phi}$ up to sign convention. Restoring the <reduced Planck mass> and using $\epsilon=\dot{\bar\phi}^{\,2}/(2M_{\rm pl}^2H^2)$ gives
$$
\boxed{\Delta_{\mathcal R}^2\simeq\frac{H^2}{8\pi^2\epsilon M_{\rm pl}^2}}.
$$
This 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 $\Delta_t^2=2H^2/(\pi^2M_{\rm pl}^2)$, so the single-field leading ratio is $r=16\epsilon$. 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.
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