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If $(q,p)$ to $(Q,P)$ is a canonical transformation, then does this imply $(Q,P)$ to $(q,p)$ is also, assuming Hamilton's equations hold for the coordinates $(q,p)$?

This seems like it should be true from the derivation for a canonical transformation using Poisson brackets, but I was wondering if anyone knew any better (or had some specific counter examples).

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  • $\begingroup$ Comment to the question (v2): Note that there exist various definitions of a canonical transformation (CT) in the literature, cf. e.g. this Phys.SE post and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Jan 3, 2015 at 13:53

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A canonical transformation leaves the symplectic form on the cotangent bundle invariant. Hence its determinant is 1 globally, which means that your coordinate transformation can be inverted everywhere. The inverse transformation then preserves the symplectic form too, hence is a canonical transformation.

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