Discussion
The in order to have following transformation to be canonical
\[ \begin{aligned} Q &= q^{1/\alpha}\cos(\beta p) \\ P &= q^{1/\alpha}\sin(\beta p) \end{aligned} \]
what should be the value of \(\alpha\) and \(\beta\) ?
Answer: D
The transformation is canonical for \(\alpha = 2, \beta = 2\).
A transformation is canonical when the Poisson bracket \(\{Q, P\}_{q,p} = 1\). With \(Q = q^{1/\alpha}\cos\beta p\) and \(P = q^{1/\alpha}\sin\beta p\):
\[ \begin{aligned} \{Q,P\} &= \frac{\partial Q}{\partial q}\frac{\partial P}{\partial p} - \frac{\partial Q}{\partial p}\frac{\partial P}{\partial q} \\ &= \frac{\beta}{\alpha}q^{\frac{2}{\alpha} - 1}(\cos^{2}\beta p \\ &\qquad + \sin^{2}\beta p) \\ &= \frac{\beta}{\alpha}q^{\frac{2}{\alpha} - 1} \end{aligned} \]
For this to equal 1 for every \(q\), the power of \(q\) must vanish, so \(2/\alpha - 1 = 0\) and \(\alpha = 2\); then \(\beta/\alpha = 1\) gives \(\beta = 2\).
\(\alpha = \beta = 1\) leaves a factor \(q\); \(\alpha = 1/2\) leaves \(q^{3}\); and \(\alpha = 2, \beta = 1/2\) gives the bracket 1/4.
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