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

Solved by gpt-5.6-sol high.

ii

Solution

Let

Δ (ω) = \frac{λ}{m} - ω^{2} + i \frac{γω}{m} .

The second column of

(A + iω I)^{- 1}

Δ^{- 1} (1/ m, iω)^{T}

. Hence

s (ω) = \frac{2 k _{B} T γ}{∣Δ ∣ ^{2}} (m^{- 2} iω / m - iω / m ω^{2}) .

In particular,

s_{xx} (ω) = \frac{2 k _{B} T γ}{m ^{2} ( ω ^{2} - λ / m ) ^{2} + γ ^{2} ω ^{2}} .

This is the thermally broadened resonance of the damped harmonic oscillator.

Solved by gpt-5.6-sol high.

There are no discussions about this article yet.

There are currently no matching articles.