Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 48 / 4

Use primes for conformal-time derivatives in this question, reserving $V_{,\varphi \varphi }$ for a potential derivative. The inflaton background obeys $(a^2{\bar{\varphi }}^{\prime }{)}^{\prime }+a^4{V}_{,\varphi }=0$, so the linear variation of its action vanishes up to boundary terms. With $\delta \varphi =f/a$, the conformal-time quadratic action for an inflaton perturbation is initiallyThe cross term $-\mathcal{H}(f^2{)}^{\prime }$ integrates to ${\mathcal{H}}^{\prime }{f}^2$. Since ${\mathcal{H}}^{\prime }+{\mathcal{H}}^2=a^{\prime \prime }/a$, this becomesVarying $f$ gives $f^{\prime \prime }-{\nabla }^{2}f+(a^2{V}_{,\varphi \varphi }-a^{\prime \prime }/a)f=0$. The symmetric Fourier transform convention used in the question turns $-{\nabla }^2$ into $k^2$. Dropping the small mass term under the stated slow-roll inflation assumption yieldsFor de Sitter spacetime, $a=-1/(H\tau )$ with $\tau <0$, so $a^{\prime \prime }/a=2/{\tau }^2$. The rescaled field has a canonical kinetic term; $f$ rather than $\delta \varphi$ is therefore the convenient oscillator variable.

Promote the canonical field and momentum $\pi =f^{\prime }$ to operators satisfying the canonical commutation relation, $[\hat{f}(\tau ,\mathbf{x}),\hat{\pi }(\tau ,\mathbf{y})]=i{\delta }^{(3)}(\mathbf{x}-\mathbf{y})$. A real field has Fourier reality condition ${\hat{f}}_{\mathbf{k}}^{†}={\hat{f}}_{-\mathbf{k}}$. Its oscillator expansion isThe creation and annihilation operators add or remove excitations in the selected mode basis, withCanonical normalization requires the Wronskian $u_k{u}_k^{\ast \prime }-u_k^{\ast }{u}_k^{\prime }=i$.

For the Bunch-Davies vacuum, choose the positive-frequency short-wavelength behavior $u_k\sim e^{-ik\tau }/\sqrt{2k}$ as $k|\tau |\to \infty$. For the proposed solution, direct differentiation givesSubstitution cancels every term in $u_k^{\prime \prime }+(k^2-2/{\tau }^2)u_k$. Its Wronskian is $i$, and its large-$|k\tau |$ limit is the required flat-spacetime positive-frequency mode. ThusA term with the conjugate frequency is another allowed classical solution, but would describe a different quantum state, through a Bogoliubov transformation. Its coefficient is set to zero by the early-time vacuum condition, not by absence of a second solution to the differential equation.

The annihilation operators kill the vacuum, and their commutator givesTherefore the scale-invariant inflationary power spectrum has the superhorizon limitThe canonical mode grows as $1/|\tau |$ outside the Hubble radius, while the physical field perturbation $f/a$ freezes. A slowly varying $H$ produces an almost scale-invariant spectrum rather than exact scale invariance.

For a slowly rolling single field, fluctuations in the inflaton clock become the comoving curvature perturbation, $\mathcal{R}\simeq -H\delta \varphi /\dot{\bar{\varphi }}$ up to sign convention. Restoring the reduced Planck mass and using $ϵ={\dot{\bar{\varphi }}}^2/(2M_{\mathrm{pl}}^2{H}^2)$ givesThis conversion presumes nonzero background roll; a strictly constant test scalar in exact de Sitter spacetime does not by itself define a finite comoving curvature perturbation through this formula. The curvature perturbations seed the later density fluctuations.

When metric fluctuations and their Einstein-gravity normalization are restored, the two graviton polarizations have the same massless mode equation. The usual total primordial tensor power spectrum is ${\Delta }_t^2=2H^2/({\pi }^2{M}_{\mathrm{pl}}^2)$, so the single-field leading ratio is $r=16ϵ$. Tensor amplitudes therefore probe the inflationary energy scale, whereas scalar amplitudes also depend on the roll rate. These tensor statements explain the relevance of the mode result; they are not derived from a scalar action with metric perturbations omitted.