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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-53/4/ii/solution

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Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 53 / 4 / ii / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

E_{ij}^{T}

and the given polarization relation imply

ψ_{(p)} (- k) = ψ_{(p)} (k)^{*}

. Insert the Fourier expansion twice into each quadratic term. The spatial integral gives

\int d^{3} x e^{i (k + k^{'}) \cdot x} = (2 π)^{3} δ^{3} (k + k^{'}) .

This cancels the two factors of

(2 π)^{- 3/2}

k^{'} = - k

and use

M_{ij}^{(q)} (- k) = M_{ij}^{(q)} (k)^{*}

. The unit-normalization condition eliminates cross-polarization terms, giving

\int d^{3} x E_{ij}^{T'} E^{T ij'} = \sum_{p} \int d^{3} k ∣ ψ_{(p)}^{'} (k) ∣^{2} .

For the gradient term,

(i k_{l}) (i k_{l}^{'})

becomes

k^{2}

after the delta-function substitution. Thus

S^{(2)} = \frac{M _{Pl}^{2}}{2} p = \pm 2 \sum \int d η d^{3} k a^{2} [∣ ψ_{(p)}^{'} ∣^{2} - k^{2} ∣ ψ_{(p)} ∣^{2}] .

The full momentum-space integral and its reality relation are those of real fields; there is no additional independent copy of the negative-wavevector modes.

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