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I was wondering how one obtains geometric quantization from a path integral. It's often assumed that something like this is possible, for example, when working with Chern-Simons theory, but rarely explained in detail. One problem I run into is that, when trying to repeat the usual derivation of the path integral, we want to insert a complete set of states, but typically here the Hilbert space is finite dimensional, so I don't see how to interpret this as the integral over some manifold, as in the usual case. The simplest case I can think of is $S^2$ with $j$ times its usual symplectic form, which gives the spin $j$ representation of $SU(2)$. Is there a way to recover this from an integral over paths on $S^2$?

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I like your minimalistic use of tags. What level of rigour are we speaking of here? Are you looking for a purely mathematical answer? –  NikolajK Oct 26 '12 at 7:11
    
I'd appreciate any kind of answer that could be provided. I don't need an especially high level of rigor, but of course I wouldn't mind such an answer. –  user6013 Oct 26 '12 at 13:08
    
I only have an undergrads understanding of it, but as far as I understand, as both are quantization procedures, why do you even want to obtain "geometric quantization from a path integral"? Both viewed as going from classical to quantum (not completed theories if you're talking field theory), can't you just deduce the classical (path) information (say the action) and go up again to the geometric quantization? I think for example with deformation quantization, there is not unique/right way of doing it. Are you asking "which path integral measure corresponds to which field operator algebra"? –  NikolajK Oct 26 '12 at 13:24
    
My point was that one often uses arguments motivated by a path integral construction in theories like Chern-Simons theory, and it is mysterious to me how these can be connected to the finite Hilbert space, yet they seem to agree. At the rigorous level, it is difficult to make sense of the path integral, while the Hilbert space is perfectly well-defined, but I'm even just looking for a non-rigorous argument how they are connected. –  user6013 Oct 26 '12 at 22:18

2 Answers 2

In the special case of geometric quantization with respect to a Kähler polarization, (which covers the path integral for spin over $S^2$ mentioned in the question), there exits a rigorous way to define a path integral, i.e., with respect to a well defined measure on the space of paths. Please see for example the following article by Laurent Charles. This type of path integral was proposed by F.A. Berezin in his famous article on covariant and contravariant operator symbols. (There is an on-line version in Russian in which a discretized version is given in the last page (before the references)).

Actually, Witten used this path integral in his seminal work Quantum field theory and the Jones polynomial, but did not show it explicitly. The form used by Witten , which will be described here, is a path integral representation of a Wilson loop. Here I'll write this special case of the path integral in a more comprehensible form and try to explain the physical intuition behind it.

$\left < tr_{\mathcal{H}}T\{exp(i\int_0^T B^{a}(t) \sigma_a)\}\right > = \mathrm{lim}_{m\rightarrow \infty}\int exp\big (i\int _0^T \alpha^{\mathcal{H}}_i\dot{z}^i - \bar{\alpha}^{\mathcal{H}}_i \dot{\bar{z}}^i + \frac{m}{2}g_{i\bar{j}}\dot{z}^i \dot{\bar{z}}^j+B^{a} (t)\Sigma^{\mathcal{H}}_a(z, \bar{z})\big) \mathcal{D}z\mathcal{D}\bar{z}$

The left hand side is the trace of a time ordered product of a spin (or more generally an element of a Lie algebra $\sigma_a)$) coupled to an external magnetic field $B^{a}$ in the representation $\mathcal{H}$. As explained by Witten according to the Borel Weil Bott theorem, to every representation there corresponds a coadjoint orbit with a given Kähler form $ \omega^{\mathcal{H}}$ depending on the representation $\mathcal{H}$. The path integral is performed on this manifold. It is a Riemannian path integral whose action contains 3 types of terms: The first terms depending on the symplectic potentials $ \alpha^{\mathcal{H}}$ satisfying: $ \omega^{\mathcal{H}} =d\alpha^{\mathcal{H}}$ . This term is of the form of an interaction with a magnetic monopole.

The second term is the Riemannian kinetic energy term. The third term is proportional to the Hamiltonian functions $\Sigma^{\mathcal{H}}_a(z, \bar{z})$ whose Poisson brackets satisfy the Lie algebra on the coadjoint orbit. This term is of the type of interaction with an external time varying magnetic field.

Thus in summary the path integral is a nonrelativitic path integral over a Riemannian manifold in the presence of magnetic fields. In other words the corresponding Hamiltonian is a magnetic Schrödinger operator. As very well known the solutions to these types of problems are Landau levels a and the lowest landau level is degenerate in general.

The crucial observation is that when the particle mass $m$ tends to infinity, the excited Landau level energies become very high and decouple, thus we remain with the lowest Landau level which happens to be exactly the representation $\mathcal{H}$ we started from.

It is worthwhile to mention that this form of the path integral has other applications and was used in works on Heisenberg ferromagnets and quark confinement.

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I have to admit, that I'm not fully understand your question - but I try to answer it :)

After performing the geometric quantization procedure (prequantization and polarization) you get a well-defined hilbert space (which in general is infinite-dimensional). Furthermore, under suitable conditions you can quantize the classical observables to obtain quantum operators. Assume you get a Hamilton-operator $H$. Then you are able to reformulate this "quantum" theory in the path-integral formalism. A discussion of path integral in the framework of geometric quantization can be found in the standard book: Woodhouse: Geometric quantization!

Side remark: Path integrals are defined for finite dimensional Hilbert spaces, too. They are even simpler as the "integral over all paths" reduces to a "sum over all possible intermediate states". Google should help you with path integrals in two/many state systems (eg http://theory.tifr.res.in/~sgupta/courses/qm2008/lec22.pdf).

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