 Solution

ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2013/paper-1/3/b/solution

A locally invertible time-independent canonical transformation is canonical when

{Q_{i}, Q_{j}} = 0, {P_{i}, P_{j}} = 0, {Q_{i}, P_{j}} = δ_{ij} .

The proposed coordinates are locally independent only for

q_{1} \neq = 0

, since their configuration-space Jacobian has determinant

2 q_{1}

. Work on one sign branch of

q_{1}

The mixed brackets first require

P_{1} = \frac{p _{1} - p _{2}}{2 q _{1}} + A (q_{1}, q_{2}), P_{2} = p_{2} + B (q_{1}, q_{2}) .

The last bracket condition is

{P_{1}, P_{2}} = \partial_{q_{2}} A - \frac{\partial _{q_{1}} B - \partial _{q_{2}} B}{2 q _{1}} = 0.

Expressing

A, B

in the

Q

coordinates turns this into

\partial_{Q_{2}} A = \partial_{Q_{1}} B

. Thus their shift one-form is closed and locally is the differential of an arbitrary smooth function

S (Q_{1}, Q_{2})

. The canonical momenta for a squared coordinate therefore have the general local form

P_{1} = \frac{p _{1} - p _{2}}{2 q _{1}} + \frac{\partial S}{\partial Q _{1}}, P_{2} = p_{2} + \frac{\partial S}{\partial Q _{2}} .

Conversely, any such

S

satisfies all bracket conditions. This does not give one nonsingular transformation through

q_{1} = 0

, or identify the two sign branches globally.

Choose

S = Q_{2}^{3} /3

. Then

P_{1} = \frac{p _{1} - p _{2}}{2 q _{1}}, P_{2} = p_{2} + (q_{1} + q_{2})^{2}, H = P_{1}^{2} + P_{2} .

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