Is the definition of a canonical transformation symmetric in the specification of old and new coordinates?

Question

Consider the following transformation:

$$q=P^\alpha \cos(\beta Q)$$

$$p=P^\alpha \sin(\beta Q)$$

for $\alpha=1/2$ and $\beta=2$.

Now, by convention, one takes $(q,p)$ to be the old coordinates and $(Q, P)$ to be the new coordinates such that if for a given transformation $$\{Q,P\}_{q,p}=\frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial Q}{\partial p}\frac{\partial P}{\partial q}=1$$

then the given transformation is canonical transformation.

If now we need to check whether the above-given transformation is canonical one might try to make $Q$ and $P$ subjects in the above equations such that $Q=Q(q,p)$ and $P=P(q,p)$ and then just check the Poisson bracket.

As one can see making $Q$ and $P$ subjects in the above equations and then taking the partial derivatives is an ugly job.

My friend suggested to me that we can instead simply check $$\{q,p\}_{Q, P}=1$$ and if it is true then the transformation is canonical since the system doesn't care which way we make the transformation.

Though this suggestion sounds plausible to me, I am not entirely convinced of it since I do not see how the "system doesn't care which way we make the transformation" argument justifies this since as far as I can remember in some cases an inverse transformation is not possible. (zero Jacobian determinant)

Is my friend's argument correct?

If yes, is there a better way (so that my dumb brain understands) of seeing it?

Answer 1

Canonical transformations preserve the volume and orientation in phase space, so the Jacobian is automatically unity. So your friend is correct.

Answer 2

1. Yes, the inverse of a symplectomorphism is also a symplectomorphism.

2. Here is perhaps a more down-to-Earth explanation using 2 local coordinate systems: If $$Z^I~=~f^I(z),\qquad I~\in~\{1,\ldots,2n\},$$ is a time-independent symplectomorphism, then by definition $$ \{Z^I,Z^J\}_{z}~=~\{z^I,z^J\}_{z}, \qquad I,J~\in~\{1,\ldots,2n\},$$ or equivalently, the Jacobian matrix $M^I{}_J=\frac{\partial Z^I}{\partial z^J}$ is a symplectic matrix. The inverse coordinate transformation $f^{-1}$ exists in a local neighborhood due to the inverse function theorem. One may show that $f^{-1}$ is in fact a symplectomorphism since the inverse Jacobian matrix is also a symplectic matrix. $\Box$

References:

1. H. Goldstein, Classical Mechanics, 2nd edition; Section 9.3.

2. H. Goldstein, Classical Mechanics, 3rd edition; Section 9.4.