I'm curious about 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:

Whydoes geometric quantization work?

As I've said, I'm curious about this, so any help would be appreciated.

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?Edit:

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

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