The <creation and annihilation operators> obey the canonical bosonic commutation relations
$$
\boxed{[\hat a_{\mathbf k},\hat a^\dagger_{\mathbf k'}]
=(2\pi)^3\delta^{(3)}(\mathbf k-\mathbf k')},
$$
$$
\boxed{[\hat a_{\mathbf k},\hat a_{\mathbf k'}]
=[\hat a^\dagger_{\mathbf k},\hat a^\dagger_{\mathbf k'}]=0},
\qquad \hat a_{\mathbf k}|0\rangle=0.
$$
Since $\delta\phi=\hat f/a$, the vacuum <two-point correlation function> is
$$
\langle0|\delta\phi(\tau,\mathbf x)
\delta\phi(\tau,\mathbf x+\mathbf r)|0\rangle
=\frac1{a^2}\int\frac{d^3k}{(2\pi)^3}|f_k(\tau)|^2e^{-i\mathbf k\cdot\mathbf r}.
$$
For
$$
f_k=\frac{e^{-ik\tau}}{\sqrt{2k}}
\left(1-\frac{i}{k\tau}\right),
$$
one has $|f_k|^2=(1+1/(k^2\tau^2))/(2k)$. Comparison with the definition of the dimensionless <power spectrum> gives
$$
\Delta_{\delta\phi}^2
=\frac{k^3}{2\pi^2a^2}|f_k|^2
=\frac{H^2}{4\pi^2}(1+k^2\tau^2),
$$
where $a=-1/(H\tau)$. On <superhorizon scale>[superhorizon scales], $k\ll aH$ or $|k\tau|\ll1$, and therefore
$$
\boxed{\Delta_{\delta\phi}^2=\left(\frac{H}{2\pi}\right)^2}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 310 / 4 / a / Solution

A cosmological mode is on a superhorizon scale when its physical wavenumber is smaller than the Hubble expansion rate, $k/a\ll H$, equivalently $k\ll aH$.


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 Incoming links: Superhorizon scale