The problem is the same as it was with the Lagrangian
<br />
L=\dot{x}y - x\dot{y} - x^2 - y^2<br />
Your "Lagrangian" is of the form
L = P_{x}\dot{x} + P_{y}\dot{y} - H
So, if you identify (x, P_{x}=y) and (y, P_{y}=-x) as two canonical conjugate pairs, you end up with degenerate (4-dimensional) phase-space. This is because; the fundamental (Poisson) brackets;
\{x , P_{x} \}_{P} \equiv \{x , y \}_{P} = 1
\{y , P_{y}\}_{P} \equiv \{y , -x\}_{P} = 1
are identical. This means that the "physical" phase-space is 2-dimensional. Phase-space reduction can be achieved by choosing an equivalent ( gauge-fixed) Lagrangian. In english, we use the fact that Lagrangians which differ by a total time derivative are equivalent, i.e., they lead to the same E-L equations and the same conserved quantities (the Hamiltonian is one of these quantities).
Now, your Lagrangian can be written as
L = \bar{L} - \frac{d}{dt}(xy)
where
\bar{L} = 2 x \dot{y} - V(x,y)
is now of the form
\bar{L} = P_{y}\dot{y} - H
and immediately identifies (y,2x) as a canonical pair. In this 2-dimensional phase space, the fundamental Poisson bracket describes non-commutative coordinates
\{y , x \}_{P} = \frac{1}{2}
Notice that Dirac bracket is not needed in this "physical" phase space, i.e., you do not need to worry about constraints.
Now, let us FORMALLY quantize the thing by the usual rule
<br />
\{x , y \}_{P} \rightarrow - i \left[ \hat{x},\hat{y} \right]_{-} = - \frac{1}{2}<br />
This means that we can not diagonalize \hat{x} and \hat{y} simultaneously, in better words, these operators posses no common eigenstates, or even better, it means that the wavefunction can not depend on both variables x and y. So, in the coordinate y-representation:
\hat{y}\rightarrow y
\hat{x} \rightarrow -\frac{i}{2} \partial_{y}
Schrodinger equation becomes
<br />
i\partial_{t}\Psi (y,t) = H( y ,\hat{x}) \Psi (y,t) = V( y , -\frac{i}{2}\partial_{y}) \Psi (y,t)<br />
.Do read my posts to learn more about the use of Dirac brackets in constraint systems.
