classical mechanics - 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!

Answer

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.