Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 310 / 4

For either scalar field $f$ with mass $m_f$, the supplied covariant equation in the FLRW metric is the Klein-Gordon equationSince $\dot{f}=f^{\prime }/a$, $\ddot{f}=(f^{\prime \prime }-\mathcal{H}{f}^{\prime })/a^2$ and $H=\mathcal{H}/a$, its conformal-time form isThe homogeneous scalar field backgrounds satisfy the same equations without spatial derivatives. After subtracting them, let $\delta \varphi =u/a$ and $\delta \chi =v/a$. The key derivatives areBecause $a^{\prime \prime }/a={\mathcal{H}}^{\prime }+{\mathcal{H}}^2$, the expansion-friction term cancels after this canonical rescaling of a scalar cosmological perturbation. The fluctuation equations areThe quadratic potential has no mixed derivative, and the stipulated neglect of metric perturbations removes gravitational mixing. Thus these two rescaled fluctuations evolve independently in this approximation.

The given symmetric Fourier transform convention sends $-{\nabla }^2$ to $k^2$, where $k=|\mathbf{k}|$. Hence the Fourier-mode equations areThey are harmonic-oscillator equations with time-dependent squared frequencies; a negative effective squared frequency for a rescaled field does not itself imply growth of the physical field $v/a$.

In the constant-$H$ de Sitter approximation, $a=-1/(H\tau )$ with $\tau <0$, so $a^{\prime \prime }/a=2/{\tau }^2$. The massive rescaled mode obeysFor a fixed mass in the specified open interval, at sufficiently late superhorizon scales with $|k\tau |\ll 1$, the $k^2$ term is negligible compared with the inverse-square term in the limiting equation. Try $v\propto (-\tau {)}^p$. The indicial equation is $p(p-1)+m_{\chi }^2/H^2-2=0$, givingThus both independent branches of the massive de Sitter superhorizon scalar mode are present inThe coefficients are constant in time and fixed by initial conditions. Writing $-k\tau >0$ avoids ambiguous fractional powers of the negative number $k\tau$; a chosen complex phase in the printed notation can be absorbed into the coefficients.

For the specified mass range, $1/2<\nu <\sqrt{5}/2<3/2$. The physical fluctuation is $\delta {\chi }_{\mathbf{k}}=v_{\mathbf{k}}/a=-H\tau v_{\mathbf{k}}$, so its two branches behave as $(-\tau {)}^{3/2-\nu }$ and $(-\tau {)}^{3/2+\nu }$. Both exponents are strictly positive. Therefore both physical-field branches decay as $\tau \to 0^{-}$:The slower branch of $v$ itself grows because $1/2-\nu <0$, but the expanding scale factor more than cancels that growth. A spatially homogeneous background field of this mass satisfies the corresponding $k=0$ equation and also has these decaying powers.

Evaluate the field variance in the zero-occupation Fock vacuum, ${\hat{a}}_{\mathbf{k}}|0⟩=0$, with $[{\hat{a}}_{\mathbf{k}},{\hat{a}}_{{\mathbf{k}}^{\prime }}^{†}]={\delta }^3(\mathbf{k}-{\mathbf{k}}^{\prime })$. The canonical commutation relation alone does not specify a quantum state; this vacuum assumption is needed for the requested expression without occupation factors. In the Heisenberg picture, use the position-space mode expansion of a free fieldThis is the position-space form of the shorthand oscillator expansion in the question; equivalently its Fourier operator uses the creation operator with opposite wavevector. The annihilation operators give $⟨{\hat{a}}_{\mathbf{k}}{\hat{a}}_{{\mathbf{k}}^{\prime }}^{†}⟩={\delta }^3(\mathbf{k}-{\mathbf{k}}^{\prime })$, while all other vacuum pairings vanish. The mean field fluctuation is zero, so the two-point function is its coincident-point vacuum field variance. Assuming spatial isotropy, angular integration gives the variance per logarithmic wavenumber interval:Here the dimensionless power spectrum is denoted ${\Delta }_v$, as in the question, rather than the alternative notation ${\Delta }_v^2$. A coincident vacuum field product needs smearing or ultraviolet regulation: the unregulated full vacuum variance is generally divergent. With a specified finite wavenumber band, the same expression holds with those integration limits. This does not affect the mode-by-mode dimensionless power spectrum.

For the leading nonzero slower branch, put $x=-k\tau =k/(aH)>0$ and $v_k=c{x}^{1/2-\nu }/\sqrt{2k}$. Since the physical fluctuation is $\delta \chi =v/a$, its dimensionless power spectrum isFor a $k$-independent coefficient $c$, this is a blue spectral tilt, because $3-2\nu >0$ in the specified mass range. At fixed $k$ it decays as $a^{-(3-2\nu )}$, consistently with the field-amplitude decay found above.

To include the other branch explicitly, the general superhorizon expression isIf $c_{-}\ne 0$, it approaches the displayed target spectrum with $c=c_{-}$. A pure faster classical branch would instead have power proportional to $x^{3+2\nu }$. For a normalized quantum mode, however, the Wronskian normalization $v_k{v}_k^{\ast \prime }-v_k^{\ast }{v}_k^{\prime }=i$ requires both independent real solutions; its slower coefficient cannot vanish. Thus the target is the leading superhorizon quantum spectrum, while the faster classical solution remains part of the complete mode analysis.