Geometric Quantization

Of course, I know the procedure: Start with a classical phase space $T^{*}X$, $X$ is the configuration space, then do prequantization by creating a prequantum (complex) line bundle (of course, the symplectic structure must satisfy the Bohr-Sommerfeld condition). The space of square integrable sections of this prequantum line bundle is the prequantum Hilbert space, so choose a polarization $P$. The space of all square-integrable sections of the prequantum line bundle that gives zero when we their covariant derivative at any point $x\in T^{*}X$ in the direction of any vector in a Lagrangian subspace modeled by the polarization is the quantum Hilbert space.

My question is as follows:

Why does geometric quantization work?

Edit: More specifically, let $\mathcal{E}$ be the configuration space of the classical theory on $X$. I.e., if $T^{*}\mathcal{E}$ denotes the phase space, why is the space of square-integrable sections of the complex line bundle over $T^{*}\mathcal{E}$ that gives zero when we compute their covariant derivative at any point $x∈T^{*}\mathcal{E}$ in the direction of any vector in a Lagrangian subspace modeled by the polarization the quantum Hilbert space?

Cross-posted from: "http://math.stackexchange.com/questions/698297/geometric-quantization".

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Dear Sanath. It is not clear to me what kind of answer you want. Could you be more specific, and mention some properties of the geometric quantization construction that you are interested in, and about which you'd like to ask "why"? –  André Henriques Mar 25 at 23:18
@SanathDevalapurkar, the answer to your literal question is "by definition". Within geometric quantization, the "quantum Hilbert space" is a primitive notion, defined as you have outlined it. A deeper answer is that this terminology happens to agree with what physicists have done in a number of key examples (like those discussed in basic QM books). This agreement is best seen by explicit calculation, comparing GQ with some traditional approach. Whether GQ is "right" or "works" in all possible situations, is still somewhat up for debate. –  Igor Khavkine Mar 26 at 0:29
@SanathDevalapurkar are you able to see why if you start with $X = \mathbb R^n$ you get the usual quantization of first year quantum courses? –  Eric O. Korman Mar 26 at 4:49
@SanathDevalapurkar this is no general method of quantization. I would check out these mathoverflow posts mathoverflow.net/questions/6200/what-is-quantization mathoverflow.net/questions/8606/… –  Eric O. Korman Mar 26 at 15:05
@SanathDevalapurkar Sorry, what I meant is that for a general configuration space there is no known canonical way to quantize it. The point is that geometric quantization gives one method to quantize a space that satisfies certain conditions. That this is a "correct" approach comes down to it satisfying certain axioms a quantization should have and agreeing in the simple cases with what physicists expect (e.g. on $\mathbb R^n$ or $S^2$). –  Eric O. Korman Mar 26 at 20:27

Perhaps we can approach something like an answer, by following the lines set out in Ritter's exposition of geometric quantization (2002). Geometric quantization works because the Heisenberg equations of motion in quantum mechanics have the same structure as Hamilton's equations in classical mechanics, and to exploit that similarity the quantization procedure should satisfy the three conditions of (1) linearity, (2) a constant classical observable corresponds to a real number times the quantum mechanical identity operator, and (3) commutators of quantum mechanical operators correspond to Poisson brackets of classical observables. One more postulate (irreducibility) requires the restriction to polarized operators, and that completes the description of geometric quantization.

As Ritter emphasises, it's helpful to keep in mind that we are dealing with the mathematical structure of a physical theory: Based ultimately on physical experiment, Dirac formulated these prescriptions for the mathematical structure of quantization around 1925, long before mathematicians knew that the procedure was possible.

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In the pre-quantum Hilbert space, Poisson bracket algebra of pre-quantum operators is reducible because pre-quantum wave-functions depend on all phase space variables. To get an irreducible representation of the algebra one introduces a polarization by using a covariant derivative where the connection is the symplectic potential. This covariant derivative acting on the pre-quantum wave-function equals zero is called the polarization condition. This procedure forces the wave-function to depend on half of the phase space variables that commute with each other. So the quantum wave-function cannot be a simultaneous eigenstate of non-commuting variables. Generally there are three options: choosing the coordinate Hilbert space where the quantum wave-function is $\psi(x_i)$, choosing the momentum Hilbert space where $\psi(p_i)$ or choosing the Segal-Bargmann space $\psi(z_i)$(holomorphic polarization), where $z=x+iy$.

So basically the covariant derivative is zero in the polarized section because that is the condition that defines the polarized section.

I hope I understood your question right.

Refs:

[1] V.P. Nair. Quantum Field Theory: A modern perspective. Springer, 2005.

[2] B. C. Hall. Quantum Theory for Mathematicians. Springer, 2013.

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