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I have to find the value $\alpha\in\mathbb{R}$ such that the following change of coordinates is symplectic:

$ \varphi(p,q)\rightarrow (P,Q)$

where

$Q = q^2 + \alpha\sqrt{q^2+p}$

$P = q + \sqrt{q^2+p}$

My first approach to solve it was to directly verify for which $\alpha$ it holds $dp\wedge dq = dP \wedge dQ$ and if comes out $\alpha=2$, which should be right.

But on some notes, I found that to prove that this holds it could be even imposed that the determinant of the Jacobian of the transformation $\varphi$ is equal 1.

Is this related to Liouville volume preserving theorem? I really don't get why we should have that this last verification implies that $\varphi$ is a symplectomorphism.

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I one sentence: You are in 2D, the symplectic form is the volume form.

In more words: Jacobian determinant measures (local) volume stretching, so is being 1 at every point means volume is preserved, but in 2D this is the same as being a symplectomorphism. You can see it very explicitly: one one hand, Jacobian determinant is $\frac{\partial{P}}{\partial{p}} \frac{\partial{Q}}{\partial{q}} - \frac{\partial{Q}}{\partial{p}} \frac{\partial{P}}{\partial{q}}$; on the other hand $dP= \frac{\partial{P}}{\partial{p}} dp+ \frac{\partial{P}}{\partial{q}} dq$ and $dQ= \frac{\partial{Q}}{\partial{p}} dp+ \frac{\partial{Q}}{\partial{q}} dq$ so $dP\wedge dQ= (\frac{\partial{P}}{\partial{p}} \frac{\partial{Q}}{\partial{q}} - \frac{\partial{Q}}{\partial{p}} \frac{\partial{P}}{\partial{q}}) dp \wedge dq$. Hence $dP\wedge dQ=dp\wedge dq$ is the same as Jacobian 1.

In higher dimensions, the volume form is $\omega^n$. Preserving $\omega$ (symplectomorphism) implies preserving volume $\omega^n$. This is Liouville theorem. However, when $n\geq 2$ (i.e. in 4D and up) the converse is not true.

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To begin just a definition: A symplectic linear structure on $\mathbb{R}^{2n}$ is a non-degenerate bilinear skew symmetric 2-form. This form is called the skew-scalar product such that $[a,b]=-[b,a]$.

Now a transformation $S:\mathbb{R}^{2n} \rightarrow \mathbb{R}^{2n}$ of the symplectic space is symplectic iff it preserves the differential 2-Form. In your case where $n=1$ this means that $dp \wedge dq$ is preserved and maps to $dP \wedge dQ$. Which is equivelent to $[Sa,Sb]=[a,b]$

Moreover a transformation is symplectic iff the Matrix Representation of S satisfies $S^T\Omega S=\Omega$ where $\Omega$ is the block matrix

\begin{bmatrix} 0 & -I \\ I & 0 \end{bmatrix}

and I is the identity matrix; becasue: $[\Omega Sa,Sb]=[\Omega a,b]=[S^T \Omega Sa,Sb]$.

Now then you could find the matrix representation of the transformation $\phi$ and then compute $\phi ^T \Omega \phi$, let it equal to $\Omega$ and solve for $\alpha$.

Furthermore I recommend the chapter on Symplectic Manifolds from V.I Arnold's book on Classical Mechanics.

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