I was struck by the following line in Hall's Quantum Theory for Mathematicians (Ch. 23, p. 484):
In the case $N = T^*M$, for example, with the natural “vertical” polarization, geometric quantization does not allow us to quantize the kinetic energy function, at least not by the “standard procedure” of geometric quantization.
This sounds bad, because for a "free particle" system, the kinetic energy is the Hamiltonian, and without that we don't have a quantum system at all. I'd like to know if the procedure at least produces a "good" quantization for the following "simple" systems.
Let $M$ be a smooth manifold of dimension $1$ or $2$ embedded in $\mathbb R^3$. We will think of $M$ as the configuration space for a particle, say a bead on a wire (in the 1D case) or a particle on a curved surface (in the 2D case). $M$ inherits Riemannian structure from its embedding into $\mathbb R^3$, and there is a way to construct a classical Hamiltonian so that the particle moves along geodesics of $M$. Let $\mathscr S_0$ denote the classical system of a "free" particle constrained to move along $M$, meaning there is no potential function on $M$ that affects the particle's motion. Let $\mathscr S_V$ denote the same system but with a sufficiently nice potential function $V$ on $M$.
Does geometric quantization produce a "good" quantum system for either of these classical systems, meaning that for some (hopefully systematic) choice of polarization, the classical observables of position, momentum, and energy can be quantized? If yes, is the quantization "very good" in the sense that $H=p^2/2m+V$?
