 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-1/33b/c/solution

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

Rotational invariance gives

{m_{y}, H} = {m_{z}, H} = 0

. Direct calculation gives

m_{x} = {m_{y}, m_{z}} = y p_{z} - z p_{y} .

Take

G = m_{x}^{2} + m_{y}^{2} + m_{z}^{2}

{G, m_{x}} = 0

, and central symmetry gives

{G, H} = 0

. Hence

(H, G, m_{x})

x = (1, 0, 0)

m_{y} = - p_{z}

m_{z} = p_{y}

. The differentials of

G

and

m_{x}

have independent components in the

p_{y}, p_{z}

and transverse position directions when

p_{y}, p_{z} \neq = 0

;

d H

supplies an independent radial/momentum component. Thus they are locally independent and the system is Liouville integrable.

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