 Solution

ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2015/paper-1/3/a/ii/solution

\frac{d O}{d t} = \frac{\partial O}{\partial t} + \sum_{i} (\frac{\partial O}{\partial q _{i}} \overset{q}{˙}_{i} + \frac{\partial O}{\partial p _{i}} \overset{p}{˙}_{i}) .

Substituting

\overset{q}{˙}_{i} = \partial H / \partial p_{i}

and

\overset{p}{˙}_{i} = - \partial H / \partial q_{i}

gives

\frac{d O}{d t} = \frac{\partial O}{\partial t} + {O, H} .

For the printed

O (q, p)

there is no explicit time term. In particular, an autonomous observable is conserved when its Poisson bracket with

H

vanishes. Taking

O = q_{i}

O = p_{i}

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