 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-353/1/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

For

X = (x, p)^{T}

A = (0 λ - 1/ m γ / m), b b^{T} = (00 0 2 k_{B} T γ) .

Solving

A σ + σ A^{T} = b b^{T}

gives

σ = (k_{B} T / λ 0 0 m k_{B} T) .

e^{- H / (k_{B} T)}

for

H = p^{2} / (2 m) + λ x^{2} /2

factorizes into independent centered Gaussians. Its equipartition variances are exactly

⟨ x^{2} ⟩ = k_{B} T / λ

and

⟨ p^{2} ⟩ = m k_{B} T

, while the absence of an

x p

term gives

σ_{12} = 0

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