What Values of α and β Represent an Extended Canonical Transformation?

AI Thread Summary
The discussion focuses on determining the values of α and β for the transformation equations Q=q^α cos(βp) and P=q^α sin(βp) to represent an extended canonical transformation. The condition for this transformation is suggested to be PQ' = λpq', but the exact form of λ is uncertain. The user attempts to derive the condition using a small angle approximation for β, leading to the expression PQ' ≈ (α/2)βpq'. Ultimately, the user concludes that α should be 1/2, while the treatment of β remains ambiguous. The conversation seeks clarity on the correctness of this approach.
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Homework Statement



The transformation equations are:

<br /> Q=q^\alpha cos(\beta p)<br />

<br /> P=q^\alpha sin(\beta p)<br />

For what values of \alpha and \beta do these equations represent an extended canonical transformation?

The Attempt at a Solution


Well, just for a start, what is the condition for a canonical transformation to be an extended canonical transformation?
 
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I got a solution but it doesn't seems very satisfactory

I believe that the condition for a canonical transformation to be an extended canonical transformation is that

<br /> PQ^\prime = \lambda p q^\prime<br />

But I am not 100% sure.

Then I do

<br /> PQ^\prime = q^\alpha sen(\beta p)(\alpha q^{\alpha-1}q^\prime cos(\beta p)-q^\alpha \beta sen(\beta p) p^\prime)<br />

Now I do the small angle approximation saying that \beta is small. Is this point that I am not sure because the problem statement don't gives any information about this approximation.
However, doing this I get:

<br /> PQ^\prime \simeq \alpha q^{2\alpha -1} \beta p q^\prime - \beta^3 q^{2\alpha} p^2 p^\prime<br />

Using

<br /> \beta^3 \simeq 0<br />

and

<br /> \alpha=\frac{1}{2}<br />

I get

<br /> PQ^\prime \simeq \frac{\beta}{2} p q^\prime<br />

At least I get the condition of an extended canonical transformation

Am I thinking right?
Thanks for any suggestion.
 
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