 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-1/33e/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Since

L^{T} = L

and

B^{T} = - B

[B, L]^{T} = (B L - L B)^{T} = L (- B) - (- B) L = [B, L],

so the commutator is symmetric. If

Lv = λ v

with

∥ v ∥ = 1

, symmetry permits a differentiable orthonormal eigenbasis locally, and

\dot{λ} = v^{T} \dot{L} v = v^{T} (B L - L B) v = λ v^{T} B v - λ v^{T} B v = 0.

H (q, p) = \frac{1}{2} (p^{2} + ω^{2} q^{2}), \overset{q}{˙} = p, \overset{p}{˙} = - ω^{2} q .

With the stated

L

, take

B = (0 ω /2 - ω /2 0) .

Direct calculation gives

[B, L] = (- ω^{2} q ω p ω p ω^{2} q) = \dot{L} .

L

are

λ_{\pm} = \pm p^{2} + ω^{2} q^{2} .

Their constancy proves conservation of the oscillator energy

H = λ_{+}^{2} /2

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