Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 310 / 4

The creation and annihilation operators obeyThe two-point correlation function in their vacuum has Fourier amplitude $|f_k{|}^2/a^2$. After cosmological horizon exit, $|k\tau |\ll 1$, and the stated Bunch-Davies vacuum mode and $a=-1/(H\tau )$ giveThe dimensionless power spectrum is consequently

The exact mode functions satisfy the Wronskian normalizationUsing the Dirac delta function generated by the ladder-operator commutator therefore gives the exact canonical commutation relationThis remains true on superhorizon scales, so the exact operators do not literally commute. A claim of classical behavior must instead compare this fixed commutator with the growing statistical fluctuations, or appeal to quantum decoherence.

At leading order in $|k\tau |$, the inflationary Fourier mode and its derivative areThus the leading parts of $\hat{f}$ and $\hat{\pi }$ contain the same quadrature of the creation and annihilation operators and satisfyTheir commutator consequently vanishes in this leading approximation. The exact nonzero result in part b is carried by the subleading decaying mode. Relative to the rapidly growing anticommutator, its effect is suppressed by powers of $|k\tau |$, which is the squeezing captured by the classicality parameter of a cosmological perturbation. The perturbation can therefore be treated as an effectively classical stochastic field on superhorizon scales even though its exact quantum commutator is unchanged.

The supplied superhorizon evolution equation isFor an adiabatic cosmological perturbation, $\delta P=({\bar{P}}^{\prime }/{\bar{\rho }}^{\prime })\delta \rho$, so $\delta P_{\mathrm{nad}}=0$ and the comoving curvature perturbation $\mathcal{R}$ is conserved on superhorizon scales.

For photons and cold dark matter, put $A={\bar{\rho }}_{\gamma }$, $C={\bar{\rho }}_c$, and use $\bar{P}=A/3$, $A^{\prime }=-4\mathcal{H}A$, and $C^{\prime }=-3\mathcal{H}C$. ThenThe cosmological entropy perturbation $S={\delta }_c-3{\delta }_{\gamma }/4$ implies ${\delta }_c=S+3{\delta }_{\gamma }/4$, so the adiabatic part cancels andSince $\bar{\rho }+\bar{P}=C+4A/3$, the curvature evolves asIn the sign convention requested in the question, $d\mathcal{R}/d\ln a=-fS$, this means