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Why is $\{Q, P\} = 1$ for a canonical transformation? Given $P(p,q)$ and $Q(p,q)$.

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    $\begingroup$ This is definition of a canonical transformation. It is a transformation which preserves the form $dp\wedge dq$ of symplectic 2-form. $\endgroup$ – user10001 Jun 27 '13 at 11:16
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    $\begingroup$ Check out the relations between P, Q and p,q on wikipedia "canonical transformation"(en.wikipedia.org/wiki/Canonical_transformation), e.g. $\left(\frac{\partial Q}{\partial p}\right)_{q,p}=-\left(\frac{\partial q}{\partial P}\right)_{Q,P}$ etc. Then apply chain rule of partial differentiation you'll see $\{P,Q\}_{p,q}=1$ $\endgroup$ – Jia Yiyang Jun 27 '13 at 11:23