The right-hand side of the <slow-roll curvature power spectrum> must be evaluated separately for each mode at <cosmological horizon exit>, $k=aH$. Taking logarithms gives
$$
\log\Delta_{\mathcal R}^2=2\log H-\log\epsilon+\text{constant}.
$$
For $N=\log a$,
$$
\frac{d\log H}{dN}=-\epsilon,
\qquad
\frac{d\log\epsilon}{dN}=\eta,
\qquad
\frac{d\log k}{dN}=1-\epsilon.
$$
Hence
$$
n_s-1
=\frac{-2\epsilon-\eta}{1-\epsilon}
=\boxed{-2\epsilon-\eta+O(\epsilon^2,\epsilon\eta)}.
$$
This is the <scalar spectral index in Hubble slow-roll parameters>.

Solution

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

Canonical quantization imposes
$$
[a_{\mathbf k},a_{\mathbf k'}^\dagger]=(2\pi)^3\delta^{(3)}(\mathbf k-\mathbf k'),
\qquad [a_{\mathbf k},a_{\mathbf k'}]=[a_{\mathbf k}^\dagger,a_{\mathbf k'}^\dagger]=0.
$$
The <correlation function>[two-point correlation function] has mode power $P_{\delta\phi}=|f_k|^2/a^2$. After <cosmological horizon exit>, $|k\tau|\ll1$, so
$$
|f_k|^2\simeq\frac1{2k^3\tau^2},
\qquad a^2=\frac1{H^2\tau^2}.
$$
Therefore
$$
\Delta_{\delta\phi}^2=\frac{k^3}{2\pi^2}P_{\delta\phi}
=\boxed{\left(\frac H{2\pi}\right)^2}.
$$
This is the <scale-invariant inflationary power spectrum> of a light canonical scalar.

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 310 / 4 / a / Solution

A Fourier mode exits the Hubble horizon during inflation when $k=aH$. On superhorizon scales $k\ll aH$, the growing mode of a sufficiently light scalar perturbation approaches a constant.


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 Incoming links: Cosmological horizon exit