= Solution
Evaluate the field variance in the zero-occupation <Fock vacuum>, $\hat a_{\mathbf k}|0\rangle=0$, with $[\hat a_{\mathbf k},\hat a_{\mathbf k'}^\dagger]=\delta^3(\mathbf k-\mathbf k')$. 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 field>
$$
\hat v(\tau,\mathbf x)=\int\frac{d^3k}{(2\pi)^{3/2}}
\left[v_k(\tau)e^{i\mathbf k\cdot\mathbf x}\hat a_{\mathbf k}
+v_k^*(\tau)e^{-i\mathbf k\cdot\mathbf x}\hat a_{\mathbf k}^\dagger\right].
$$
This 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 $\langle\hat a_{\mathbf k}\hat a_{\mathbf k'}^\dagger\rangle=\delta^3(\mathbf k-\mathbf k')$, 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 \b[variance per logarithmic wavenumber interval]:
$$
\boxed{\langle\hat v(\tau,0)\hat v^\dagger(\tau,0)\rangle
=\int\frac{d^3k}{(2\pi)^3}|v_k|^2
=\int_0^\infty\frac{dk}{k}\,\Delta_v(k,\tau),
\qquad \Delta_v=\frac{k^3}{2\pi^2}|v_k|^2.}
$$
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> is
$$
\begin{aligned}
\Delta_{\delta\chi}
&=\frac{k^3}{2\pi^2a^2}|v_k|^2
=|c|^2\frac{k^2}{4\pi^2a^2}x^{1-2\nu}\\
&=\boxed{|c|^2\left(\frac H{2\pi}\right)^2
\left(\frac{k}{aH}\right)^{3-2\nu}.}
\end{aligned}
$$
For 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 is
$$
\Delta_{\delta\chi}=\left(\frac H{2\pi}\right)^2x^3
\left|c_-(k)x^{-\nu}+c_+(k)x^\nu\right|^2.
$$
If $c_-\ne0$, 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> $v_kv_k^{*\prime}-v_k^*v_k^\prime=i$ 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.
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