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.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-310/4/a/solution