 Solution

In one canonical degree of freedom it is enough to check $\{Q,P\}=1$ for a locally invertible transformation. Hereso $f{f}^{\prime }=1$ and $f(p{)}^2=2p+C$. To obtain exactly the requested Hamiltonian, choose $C=0$ and the positive branch for $p>0$:The identities ${\cosh }^{2}q-{\sinh }^{2}q=1$ and ${\cosh }^{2}q+{\sinh }^{2}q=\cosh 2q$ giveThis is a local canonical chart, with inverseIt is not a global identification of the real $(q,p>0)$ half-plane with the entire oscillator plane. The negative branch for $f$ gives the analogous lower wedge. This domain matters if an oscillator trajectory is continued beyond the boundary of the chart.