Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 310 / 4

In conformal time, $\sqrt{-g}=a^4$ and $g^{\mu \nu }=a^{-2}\mathrm{diag}(-1,1,1,1)$. The inflaton action in an expanding universe becomesVary $\varphi$ with compact support, integrate the derivative terms by parts, and discard the boundary term. The Euler-Lagrange field equation isDividing by $a^2$ provesThe factor two is the conformal-time form of Hubble friction. Using $d\tau =dt/a$ converts it to the usual inflaton equation of motion $\ddot{\varphi }+3H\dot{\varphi }-a^{-2}{\nabla }^2\varphi +V_{,\varphi }=0$, providing a sign and normalization check.

Subtract the homogeneous background equation and linearize the inflaton equation of motion. The field perturbation satisfiesWith $f=a\delta \varphi$, direct differentiation cancels the first-derivative term and givesThe conformal-time quadratic action for an inflaton perturbation, after a boundary term is discarded, isFor the specified de Sitter spacetime background, $a^{\prime \prime }/a=2/{\tau }^2=2a^2{H}^2$. The light-field approximation $|V_{,\varphi \varphi }|\ll 2H^2$ therefore permits dropping the mass term. Taking a spatial Fourier transform yields the Canonically rescaled de Sitter scalar mode equationIts kinetic term is canonical, with momentum ${\pi }_f=f^{\prime }$. Each independent real Fourier component consequently has the canonical coordinate/momentum algebra of a quantum harmonic oscillator, with time-dependent squared frequency ${\omega }_k^2=k^2-a^{\prime \prime }/a$. This is why canonical quantization of a real scalar field proceeds by oscillator creation and annihilation operators. On superhorizon scales the squared frequency is negative: the system is then an inverted time-dependent oscillator, not a stationary oscillator with a positive frequency. The canonical commutation relations remain valid.

Use the Fourier normalization of the PDF. The creation and annihilation operators obeyThere is no $(2\pi {)}^3$ in this commutator because the expansion uses $(2\pi {)}^{-3/2}$. The Bunch-Davies vacuum satisfies ${\hat{a}}_{\mathbf{k}}|0⟩=0$. Write $u_k=f_k^{\ast }$ for the coefficient of the annihilation operator. Only the contraction $⟨0|{\hat{a}}_{\mathbf{k}}{\hat{a}}_{{\mathbf{k}}^{\prime }}^{†}|0⟩={\delta }^3(\mathbf{k}-{\mathbf{k}}^{\prime })$ contributes, soThe given mode hasSince $a^{-2}=H^2{\tau }^2$, the equal-time two-point function of a de Sitter scalar therefore has dimensional power spectrumComparing with the dimensionless power convention in the hint givesEquivalently, angular integration expresses the correlation as ${\int }_0^{\infty }(dk/k){\Delta }_{\delta \varphi }^2\sin (kr)/(kr)$, where $r=|\mathbf{r}|$.

On superhorizon scales, $k\ll aH$ is $|k\tau |\ll 1$, andThis scale-invariant inflationary power spectrum gives equal variance per logarithmic wavenumber interval. It is exactly scale invariant in the massless constant-$H$ approximation here; slow evolution of $H$ and small field mass produce a small spectral tilt. Such quantum primordial perturbations supply the initial fluctuations that subsequently seed cosmological structure after conversion to curvature and density perturbations. The curvature conversion itself requires the metric perturbations excluded from this calculation.

The continuum correlation is understood with the usual infrared regulator, such as finite volume or finite inflationary duration. The strictly massless infinite-volume model has a logarithmic infrared divergence. Coincident products also require ultraviolet regularization or smearing; the finite mode spectrum is unaffected by these qualifications.

Under the stated linear field expansion, the Bunch-Davies vacuum is a centered Gaussian state. By Wick theorem, vacuum correlators reduce to pair contractions; an odd product has no complete pairing. In particular,This holds for arbitrary spatial arguments as a regulated correlator or distribution. One can also use number parity: $\mathcal{P}=(-1{)}^{\hat{N}}$ leaves the vacuum unchanged and sends every creation or annihilation operator, hence the linear field $\hat{f}$, to its negative. An odd product consequently has expectation equal to its own negative. Thus vanishing odd correlators of a free Gaussian field givesFor a coincident product, impose a momentum regulator or smear the field before taking the expectation; the zero result holds for every such symmetry-preserving regulator. It is not a claim that the unregulated local power is an ordinary finite random variable. Interactions, a non-Gaussian state, or inclusion of a nonzero background mean can invalidate the odd-moment conclusion. Within this free perturbation calculation, the primordial bispectrum vanishes.