Amplitude-phase decomposition as a canonical transformation I am studying a classical dynamical system defined on $\mathbb{CP}^2$: the phase space is parametrized in terms of three complex coordinates $\psi_i$ ($i=1,2,3$) and Hamilton's equations of motion take the form,
$$\imath \frac{d\psi_i}{dt} = \frac{\partial H}{\partial \psi_i^*},\quad \imath \frac{d\psi_i^*}{dt} = -\frac{\partial H}{\partial \psi_i}.$$
I would like to make an amplitude-phase decomposition, replacing the three complex coordinates and their conjugates, $\{\psi_i, \psi_i^*\}$, with six real ones $\{n_i, \phi_i\}$, with
$$\psi_i = \sqrt{n_i} \exp(\imath \phi_i)$$
But this transformation appears not to be canonical: instead of the usual,
$$\nabla_\xi \Theta \cdot \Omega \cdot \nabla_\xi \Theta = \Omega,$$
where $\nabla_\xi \Theta$ is the Jacobian of the transformation and $\Omega$ is the symplectic block matrix, I get,
$$\nabla_\xi \Theta \cdot \Omega \cdot \nabla_\xi \Theta = \frac{1}{\imath}\Omega.$$
Is the amplitude-phase decomposition not a canonical transformation?  Or did I make a mistake?
I'm sure this is a standard problem, but I am very new to the idea of classical dynamics on complex manifolds and haven't gotten my bearings yet.  Any reference suggestions would be welcome!
 A: For simplicity consider the 1-d case, with $\psi =\sqrt{n} e^{2i\phi}$, then
$$i \psi_t =\frac{i}{2} \frac{\dot{n}}{\sqrt{n}} e^{2i\phi} -\sqrt{n} e^{2i\phi} 2\dot{\phi}.$$
Similarly 
$$ \frac{\partial H}{\partial \psi^*} = \frac{\partial H}{\partial n}\frac{\partial n}{\partial \psi^*} + \frac{\partial H}{\partial \phi}\frac{\partial \phi}{\partial \psi^*} = 2\sqrt{n} e^{2i\phi} \frac{\partial H}{\partial n} + \frac{i}{2\sqrt{n}} e^{2i\phi} \frac{\partial H}{\partial \phi}.$$
Equating the real and imaginary parts (with H real), we have 
$$\frac{d n}{dt} = \frac{\partial H}{\partial \phi}; \quad \quad \frac{d \phi}{dt} = -\frac{\partial H}{\partial n}.$$
The other governing equation for $d \psi^*/dt$ gives the same information. Hence, $(n,\phi)$ are canonical variables. 
$\textbf{EDIT}$:
As Ted Pudlik correctly pointed out, the above reasoning is incorrect. Why? Well, it's because I was being sloppy and got bit. Let's try this again. 
As usual, we need to work at the order of the action in order to get coherent results. 
Consider 
$$ S = \int i\dot{\psi}\psi - H dt.$$
Hamilton's principle states the dynamics of the system are given when $S$ is stationary, and indeed this yields the set of Hamilton's equations you originally stated. 
Next, we consider a different action, $S'$ defined as
$$S' = \int -2n\dot{\phi}- 2H' dt$$
for some undetermined $H'$. Hamilton's principle yields (1) with $H\to H'$. 
For $S$ and $S'$ to give the same dynamics, they must differ by a constant, ie
$$S-S' =\int \frac{d f}{dt} \ dt$$ 
for some function $f$. Now, when we substitute in our two actions, we find 
$$ \int -2n\dot{\phi} +\dot{n} -H +2n\dot{\phi} + 2H' \ dt $$
The perfect derivative integrates to 0 (we assume the wave is compact in time) and we are left with the requirement that for the transformation to be canonical, $2H' \equiv H$, as you pointed out in your comment. 
A: After thinking about Nick P's answer and re-reading the relevant chapter of Sussman's Structure and Interpretation of Classical Mechanics, I came up with the following elaboration of Nick's argument.  It's not water-tight, but it convinced me, and perhaps it will help someone else.  I will use Sussman's unorthodox but precise notation.
The first step (and this is the part that I can't rigorously justify) is to expand the definition of the phase-space derivative operator.  The definition given by Sussman in Eq. (5.15) is,
$$D_s H(t,q,p) = (1,\partial_2 H(t,q,p),-\partial_1 H(t,q,p)).$$
The extension we'll make is to define $D_s$ for Hamiltonians that are functions of complex-conjugate coordinates and momenta as,
$$D_s H(t,\psi,\psi^*) = (1,-\imath \partial_2 H(t,\psi,\psi^*),\imath \partial_1 H(t,\psi,\psi^*)).$$
With this extension, Hamilton's equations can be written in the same form for both the usual real and the complex coordinates:
$$D \sigma = D_s H \circ \sigma,$$
where $\sigma(t) = (t, q(t), p(t))$ or $\sigma(t) = (t, \psi(t), \psi^*(t))$, a mapping from time to phase space position, represents a path.
Now, let $C$ be a phase space coordinate transformation: $\sigma = C \circ \sigma'$.  The transformation is canonical if there exists a new Hamiltonian $H'$ such that the equations of motion derived from it describe the same motion of the system.  A sufficient condition for this is Eq. (5.19),
$$D_s H \circ C = DC \cdot D_s H'.$$
I'll show that the transformation,
$$(t, \psi(t), \psi(t)^*) = C(t, n(t), \phi(t)) = (t, \sqrt{n} e^{\imath \phi}, \sqrt{n} e^{-\imath \phi})$$
satisfies this condition for any $H$, and that furthermore $H' = H \circ C$, i.e. the new Hamiltonian can be obtained from the old one simply by substituting $\sqrt{n} e^{\imath \phi}$ for $\psi$.  (This doesn't generally have to be the case: for example, it's not the case for the transformation $\psi = \sqrt{n} e^{2\imath \phi}$ discussed by Nick.)
The left-hand-side of the sufficient condition is,
$$D_s H(t, \psi, \psi^*) = (1, -\imath \partial_2 H(t,\psi,\psi^*), \imath \partial_1 H(t,\psi,\psi^*))$$
$$(D_s H \circ C)(t, n, \phi) = (1, -\imath (\partial_2 H) \circ C, \imath (\partial_1 H) \circ C)$$
Here, $\partial_1$ is the partial derivative with respect to the first argument (i.e., $\psi$: following Sussman, I'm using zero-based indexing, where time is the zeroth argument).
On the right-hand-side, the Jacobian of the transformation is,
$$DC = [\partial_0 C, \partial_1 C, \partial_2 C] = \begin{bmatrix}
\begin{pmatrix}
\partial_{0,0} C \\
\partial_{0,1} C \\
\partial_{0,2} C 
\end{pmatrix} &
\begin{pmatrix}
\partial_{1,0} C \\
\partial_{1,1} C \\
\partial_{1,2} C 
\end{pmatrix} &
\begin{pmatrix}
\partial_{2,0} C \\
\partial_{2,1} C \\
\partial_{2,2} C 
\end{pmatrix}
\end{bmatrix}$$
and so the right-hand-side reads,
$$DC \cdot D_s H' = \begin{pmatrix}
\partial_{0,0} C + \partial_{1,0} C \partial_2 H' - \partial_{2,0} C \partial_1 H' \\
\partial_{0,1} C + \partial_{1,1} C \partial_2 H' - \partial_{2,1} C \partial_1 H' \\
\partial_{0,2} C + \partial_{1,2} C \partial_2 H' - \partial_{2,2} C \partial_1 H'
\end{pmatrix}.$$
The only nonzero elements of the Jacobian are,
$$\partial_{1,2} C = \frac{1}{2\sqrt{n}} e^{-\imath \phi},$$
$$\partial_{2,2} C= -\imath \sqrt{n} e^{-\imath \phi},$$
$$\partial_{1,1} C = \frac{1}{2\sqrt{n}} e^{\imath \phi},$$
$$\partial_{2,1} C = \imath \sqrt{n} e^{\imath \phi},$$
$$\partial_{0,0} C = 1.$$
The canonicity condition reduces to the system of equations,
$$-\imath (\partial_2 H) \circ C = \partial_{1,1} C \partial_2 H' - \partial_{2,1} C \partial_1 H'$$
$$\imath (\partial_1 H) \circ C = \partial_{1,2} C \partial_2 H' - \partial_{2,2} C \partial_1 H'$$
Solving for $\partial_1 H'$ gives,
$$\imath (\partial_{2,2} C \partial_{1,1} C - \partial_{2,1} C \partial_{1,2} C) \partial_1 H' = \partial_{1,2} C (\partial_2 H) \circ C + \partial_{1,1} C (\partial_1 H) \circ C.$$
The quantity in parentheses on the left is exactly $-\imath$, so using the chain rule,
$$\partial_1 H' = ((\partial_2 H) \circ C ) \partial_{1,2} C + ((\partial_1 H) \circ C) \partial_{1,1} C = \partial_1 (H\circ C).$$
A similar relation holds for $\partial_2 H'$.  Therefore, the transformation can be made canonical using the "natural" choice of $H' = H \circ C$.
