Geometric quantization is a marvelous tool for understanding the relation between classical physics and quantum physics. However, it's a bit like a power tool - you have to be an expert to operate it without running the risk of seriously injuring your brain. Here's a brief sketch of how it goes. This is pretty terse; for the details you'll have to read the series of articles on geometric quantization on the sci.physics.research archive.
Warning: we can only do this step if ω satisfies the BOHR-SOMMERFELD CONDITION, which says that ω/2π defines an integral cohomology class. If this condition holds, L and D are determined to isomorphism, but not canonically.
Warning: for P to be a polarization, there are some crucial technical conditions we impose on the subspaces Px. First, they must be ISOTROPIC: the complexified symplectic form ω must vanish on them. Second, they must be LAGRANGIAN: they must be maximal isotropic subspaces. Third, they must vary smoothly with x. And fourth, they must be INTEGRABLE.
Modulo some technical trickery, we get this example when we run the above machinery and use a certain god-given real polarization on X = T*M, namely the one given by the vertical vectors.
Here are some definitions of important terms. Unfortunately they are defined using other terms that you might not understand. If you are really mystified, you need to read some books on differential geometry and the math of classical mechanics before proceeding.
ω(.,v(f)) = df
In other words, for any vector field u on X we have
ω(u,v(f)) = df(u) = u f
The vector field v(f) is guaranteed to exist by the fact that ω is nondegenerate.
This trick works as follows: given any smooth function f we can take its differential df, which is a 1-form. Then there is a unique vector field v(f), the Hamiltonian vector field associated to f, such that
ω(-,v(f)) = df
Using this we define
{f,g} = ω(v(f),v(g))
It's easy to check that we also have {f,g} = dg(v(f)) = v(f) g
so {f,g} says how much g changes as we differentiate it in the direction of the Hamiltonian vector field generated by f.
In the familiar case where M is R2n with momentum and position coordinates pi, qi, the Poisson brackets of f and g work out to be
{f,g} = sumi df/dpi dg/dqi - df/dqi dg/dpi
© 2000 John Baez
baez@math.removethis.ucr.andthis.edu