From: baez@galaxy.ucr.edu (John Baez)
Subject: Re: Geometric Quantization
Date: 18 Aug 2000 01:20:25 GMT
Newsgroups: sci.physics.research
Summary: [missing]

In article <399B305C.869B051C@wall.org>, Aron Wall  <aron@wall.org> wrote:

>John Baez has been threatening to tell us how the observables work out
>in geometric quantization.  I suggest he make good this threat.

I've been putting this off for a long time.  Okay, here goes!

One step at a time.   For now I'll only say how we *prequantize*
classical observables.  In other words:

Suppose we have a classical phase space: a manifold M with symplectic 
structure w.  To "prequantize" this classical phase space means to 
find a line bundle L with a U(1) connection D whose curvature is iw:

[D_u, D_v] s = i w(u,v) s

for any vector fields u and v, and any section s of L.  We've discussed
ad nauseum the conditions under which we can do this.  Suppose we can.
The "prequantum Hilbert space" is then the space of square-integrable 
sections of L.

Now, suppose we have a classical observable, that is, a smooth function
f: M -> R.  How do we turn it into a prequantum observable, that is, an 
operator Q(f) on the prequantum Hilbert space?  Ideally, we would like 
this process to turn Poisson brackets into commutators:

[Q(f),Q(g)] = i Q({f,g})                                 [1]

for any pair of smooth functions f,g : M -> R.   After all, that's
what they always told us quantization is supposed to do.  Quantization
is not really that simple, but *prequantization* is!

Here's how we do it: we define

Q(f) s = - i D_{v(f)} + f s                              [2]

where v(f) is the vector field generated by f.  Remember, in 
classical mechanics, observables give vector fields on phase
space.  Here's the formula for *that*:

v(f) g = {f, g}

Now, before I say anything more, somebody had better post the
proof that definition [2] implies equation [1]!  And, just to
make sure this task isn't too easy, I've carefully randomized
equation [2] so that the signs and factors of i have a 37.5% 
chance of being wrong.  If they're wrong, you'll have to change
them until equation [1] actually follows.   

Equation [2] seems a bit mysterious.  I think there is
a nice conceptual way to explain where it COMES FROM, but I'll
have to think about it a bit more before talking about that.

At least one thing should be clear: it's nice that equation 
[2] involves both "multiplication by f" and "differentiation
by v(f)", since we all know that ultimately observables are 
supposed to become operators that involve multiplication and
differentiation.