 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-310/4/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

With the symmetric Fourier normalization displayed in the question, the creation and annihilation operators obey the canonical commutation relation

[\overset{a}{^}_{k}, \overset{a}{^}_{k^{'}}^{†}] = δ^{(3)} (k - k^{'}), [\overset{a}{^}_{k}, \overset{a}{^}_{k^{'}}] = [\overset{a}{^}_{k}^{†}, \overset{a}{^}_{k^{'}}^{†}] = 0.

Since

δ \hat{ϕ} = \hat{f} / a

⟨ 0∣ δ \hat{ϕ} (τ, x) δ \hat{ϕ} (τ, x + r) ∣0 ⟩ = \int \frac{d ^{3} k}{( 2 π ) ^{3}} \frac{∣ f _{k} ( τ ) ∣ ^{2}}{a ^{2}} e^{- i k \cdot r} .

Comparison with the definition in the question gives

Δ_{δ ϕ}^{2} (k, τ) = \frac{k ^{3}}{2 π ^{2}} \frac{∣ f _{k} ∣ ^{2}}{a ^{2}} .

Using

∣ f_{k} ∣^{2} = (1 + 1/ (k^{2} τ^{2})) / (2 k)

and

a = - 1/ (H τ)

Δ_{δ ϕ}^{2} (k, τ) = (\frac{H}{2 π})^{2} (1 + k^{2} τ^{2}) .

k ≪ a H

∣ k τ ∣ ≪ 1

Δ_{δ ϕ}^{2} = (\frac{H}{2 π})^{2} .

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