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How can I determine the values of α and β for a canonical transformation?
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[QUOTE="Lambda96, post: 6846491, member: 696125"] Thank you Steve4Physics, vanhees71, vela and TSny for your help👍 👍 👍👍 I have now started to rewrite ##a_{12}## as follows: $$2p^{\alpha} \cos^2(q) \beta p^{\beta -1}+2 \alpha p^{\alpha -1} \sin^2(q) p^{\beta}$$ $$2 \cos^2(q) \beta p^{\beta +\alpha -1}+ 2 \alpha p^{\alpha +\beta -1} \sin^2(q)$$ After all, I want p to vanish or rather become 1, which is the case when ##\alpha + \beta -1=0## simultaneously, I also want the 2 to vanish and become a 1. I would achieve all this if ##\alpha= \frac{1}{2}## and ##\beta=\frac{1}{2}##. Regarding task b, I still have a question, if I have now understood correctly, the system has two degrees of freedom, must the Poisson bracket then be calculated as follows? $$\Bigl\{ q'_j, p'_k \Bigr\}=\sum\limits_{i=1}^{2}\frac{\partial q'_j}{\partial q_i}\frac{\partial p'_k}{\partial p_i}- \frac{\partial q'_j}{\partial p_i}\frac{\partial p'_k}{\partial q_i}=\frac{\partial q'_j}{\partial q_1}\frac{\partial p'_k}{\partial p_1}- \frac{\partial q'_j}{\partial p_1}\frac{\partial p'_k}{\partial q_1}+\frac{\partial q'_j}{\partial q_2}\frac{\partial p'_k}{\partial p_2}-\frac{\partial q'_j}{\partial p_2}\frac{\partial p'_k}{\partial q_2}$$ I have tried to determine ##\alpha, \beta, \delta, \gamma## by calculating the following $$\Bigl\{ q'_1,p'_1 \Bigr\}=1$$ $$\Bigl\{ q'_2,p'_2 \Bigr\}=1$$ $$\Bigl\{ q'_1,q'_2 \Bigr\}=0$$ $$\Bigl\{ p'_1,p'_2 \Bigr\}=0$$ With this I found out that ##\gamma=1##, ##\alpha=0##, ##\beta=0## and ##\delta=0## but since I got so many zeros I don't know if I defined the Poisson brackets correctly? [/QUOTE]
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How can I determine the values of α and β for a canonical transformation?
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