For $X=(x,p)^T$,
$$
\boxed{A=\begin{pmatrix}0&-1/m\\ \lambda&\gamma/m\end{pmatrix}},
\qquad
\boxed{bb^T=\begin{pmatrix}0&0\\0&2k_BT\gamma\end{pmatrix}}.
$$
Solving $A\sigma+\sigma A^T=bb^T$ gives
$$
\boxed{\sigma=
\begin{pmatrix}
k_BT/\lambda&0\\0&mk_BT
\end{pmatrix}}.
$$
The <canonical ensemble> density proportional to $e^{-H/(k_BT)}$ for $H=p^2/(2m)+\lambda x^2/2$ factorizes into independent centered Gaussians. Its equipartition variances are exactly $\langle x^2\rangle=k_BT/\lambda$ and $\langle p^2\rangle=mk_BT$, while the absence of an $xp$ term gives $\sigma_{12}=0$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 353 / 1 / b / i / Solution

 Solution

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