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

Stationarity makes the covariance depend only on $\tau =t-s$. For $\tau >0$,At the reversed lag, the $s>t$ formula givesand thereforeThe same identity follows directly by exchanging the two random variables in $G_{ij}(t,s)=⟨x_i(t)x_j(s)⟩$.

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Split the Fourier integral at zero and use the two stationary covariance branches:Equivalently, Fourier transforming the Multivariate Ornstein-Uhlenbeck process equation givesUnit white-noise covariance then yields the Ornstein-Uhlenbeck power spectrumso

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For $X=(x,p{)}^T$,Solving $A\sigma +\sigma A^T=b{b}^T$ givesThe canonical ensemble density proportional to $e^{-H/(k_{B}T)}$ for $H=p^2/(2m)+\lambda x^2/2$ factorizes into independent centered Gaussians. Its equipartition variances are exactly $⟨x^2⟩=k_{B}T/\lambda$ and $⟨p^2⟩=m{k}_{B}T$, while the absence of an $xp$ term gives ${\sigma }_{12}=0$.

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LetThe second column of $(A+i\omega I{)}^{-1}$ is ${\Delta }^{-1}(1/m,i\omega {)}^T$. HenceIn particular,This is the thermally broadened resonance of the damped harmonic oscillator.

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The stationary solution ofisIt follows from the Itô isometry thatThus $q$ is zero-mean colored noise with correlation time ${\nu }^{-1}$. The equation $\dot{x}=\mu (-\lambda x+q)$ is an overdamped harmonic particle of mobility $\mu$ driven by that correlated random force.

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For $X=(x,q{)}^T$,The Lyapunov equation gives ${\sigma }_{qq}=\alpha /2$, ${\sigma }_{xq}=\lambda {\sigma }_{xx}$, and $(\mu \lambda +\nu ){\sigma }_{xq}=\mu {\sigma }_{qq}$. Therefore

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Take $\nu \to \infty$ withSince $(\nu /2)e^{-\nu |\tau |}\to \delta (\tau )$,Thus $\mu q$ tends to thermal white forcing of covariance $2\mu k_{B}T\delta (\tau )$, as required by the fluctuation--dissipation relation. The equal-time position variance becomesthe equilibrium harmonic variance.

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