Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 62 / 3 / d

Table of contents
- Solution d

Solution

In the incoming Bunch-Davies vacuum, only

⟨ a_{k} a_{q}^{†} ⟩ = δ_{kq}

contributes to the coincident vacuum expectation value. Thus

⟨ δ \hat{ϕ}^{2} ⟩ = \sum_{k} ∣ δ ϕ_{k}^{(+)} ∣^{2} = \sum_{k} \frac{1}{2 k V} (\frac{1}{a ^{2}} + \frac{H ^{2}}{k ^{2}}) .

Replacing the box sum by

V \int d^{3} k / (2 π)^{3}

gives

⟨ δ \hat{ϕ}^{2} ⟩ = \int \frac{d ^{3} k}{( 2 π ) ^{3}} \frac{1}{2 k} (\frac{1}{a ^{2}} + \frac{H ^{2}}{k ^{2}}) = \frac{1}{4 π ^{2}} \int_{0}^{\infty} d k (\frac{k}{a ^{2}} + \frac{H ^{2}}{k}) .

The subhorizon vacuum contribution is the flat-space zero-point term; at fixed comoving

k

it redshifts as

a^{- 2}

. The other term gives frozen scalar power on superhorizon scales. Specifically, the dimensionless cosmological power spectrum per logarithmic interval is

Δ_{δ ϕ}^{2} (k, τ) = \frac{k ^{2} / a ^{2} + H ^{2}}{4 π ^{2}} ⟶ (\frac{H}{2 π})^{2} .

Its late-time value is independent of

k

, producing a scale-invariant inflationary power spectrum. Each fixed Fourier mode freezes separately as

τ \to 0

.

Figure 1.
Vacuum redshifting and frozen power of massless scalar modes during de Sitter expansion
. Details.

The coincident integral is formal: the vacuum part diverges at large

k

, and an unlimited scale-invariant part diverges at small

k

. The conclusion concerns the power in each finite logarithmic band. A finite inflationary history supplies a largest generated wavelength, as described by the infrared qualification of a scale-invariant covariance; taking the late-time limit inside an unregulated integral would not be justified.

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version