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My question is related to Geometric Quantization. I don't undrestand the philosophy of following assertion

If $(V,\omega)$ be a symplectic vector space then the quantizations of $V$ corresponds to choices of Lagrangian subspaces of $V$,

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  1. The first attempt to "quantize" a dynamical variable $u$ on a symplectic manifold $(M,\omega)$, that is, to associate a linear operator $\hat u$ on the space of square summable smooth function $\psi : M \to \bf{C}$, is to define $$ \hat u(\psi) = {\partial \psi \over \partial x} {\rm grad}_\omega(u), $$ where ${\rm grad}_\omega$ denotes the symplectic gradient. Then, the Poisson bracket of dynamic variables is mapped to the bracket of operators: $$ \widehat{\{u,v\}} = [\hat u,\hat v ]. $$ But this doesn't satisfy the Dirac program because Dirac wants the constant function $1 : x \mapsto 1$ be quantized by the identity $\hat 1 = [ {\bf 1}:\psi \mapsto \psi]$, and the procedure above gives $\hat 1 = {\bf 0}$.

  2. To resorb this failure one introduces a supplementary dimension to the space $M$, that is, a circle bundle $Y$ over $M$, equipped with a connexion form $\lambda$ with curvature $\omega$(*). Then, an infinitesimal symplectomorphism ${\rm grad}_\omega(u)$ can be lifted on $Y$ by an infinitesimal quantomorphism (means preserving the connexion form), $$\xi_u = {\rm grad}_\omega(u) \oplus u \xi, $$ where I denote here by ${\rm grad}_\omega(u)$ the horizontal lift on $Y$ of the symplectic gradient ${\rm grad}_\omega(u)$ on $M$, and by $\xi$ the infinitesimal generator of the action of $S^1$ on $Y$. Then, we consider some subspace of the space of $S^1$-equivariant map $\psi : Y \to {\bf C}$ (which is a candidate for our Hilbert space) and define $$ \hat u(\psi) = {\partial \psi \over \partial y} (\xi_u). $$ This correspondence still satisfies the Dirac first condition $\widehat{\{u,v\}} = [\hat u,\hat v ]$ but also, as you can easily check (because of the $S^1$-equivariance), $\hat 1(\psi) = \psi$.

  3. Does that mean that we are done yet? Unfortunately no, because the wave function $\psi$ is assumed to depend only on the configuration variables that is half the dimension of $M$. And here begins the mess. How do we overcome this difficulty? We don't... We have no algorithmic answer, we have just now a few different attempts.

  4. The first idea that came in mind was to assume that we have a polarization of the symplectic manifold $M$, that is a Lagrangian foliation $\pi : M \to Q$ and to consider only polarized functions on $M$, that is, functions that depend only on $Q$. Why a polarization? Because a polarization splits by 2 the number of variables an that is what we want, and secondly in the classical case of a free particle the configuration space is exactly a space of leaves of a polarization, that is, $\pi = (q,p) \mapsto q$.

  5. In the case of a symplectic vector space $(V,\omega)$ the prequantized bundle is trivial: $Y = V \times S^1$, and the connection form on $Y$ is $\lambda = \alpha + {dz \over iz}$, where $\alpha$ is a primitive of $\omega$, that is, $\omega = d\alpha$. The simplest polarizations are linear: they are given by Lagrangian affine subspaces parallel to a given Lagrangian vector subspace $L$. That is one of a few reasons why you can say: "Quantizations of $V$ corresponds to choices of Lagrangian subspaces of $V$".

  6. But this is not the end of the story. Unfortunately, when you want to represent the 1-parameter group of symplectomorphisms by the process described above you need the polarization to be invariant, but it is almost never the case. In particular, the 1-parameter group associated to the harmonic oscillator hamiltonian makes the polarization rotate. This is a big, big difficulty of the geometric quantization process. We can solve the problem for the harmonic oscillator by using the Pairing Method (Blattner-Kostant-Sternberg). It works in that case and in some other cases but it's not a universal method, unfortunately. It's why Geometric Quantization is still not achieved.

Hope it helps.

As References the Book of Souriau "Structure des systèmes dynamiques" and also his paper "Construction explicite de l'indice de Maslov, Applications". Also the book "Symplectic techniques in physics" by Sternberg and Guillemin. And you can also Google "Blattner-Kostant-Sternberg".

P.S. Actually the Hilbert space is made of half-densities, but that is technical and do not affect essentially the global picture.

(*) This is what implies that $\omega$ is integral, see here and there.

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    $\begingroup$ I had trouble understanding your first displayed formula, which defines the operator $\hat{u}$, but I think it's because I'm not familiar with your notation and terminology. It appears to me that what you call the symplectic gradient is also called the Hamiltonian vector field and often denoted $H_u$. And by $\partial\psi/\partial x$ you mean the $1$-form $d\phi$. So the operator $\hat{u}$ is the natural contraction of the $1$-form with the vector field. In the notation I'm familiar with, it would be: $$ \hat{u}(\psi) = \langle d\psi, H_u\rangle. $$ $\endgroup$ – Deane Yang Nov 16 '13 at 15:53
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    $\begingroup$ @DeaneYang - Yes, these are the notations I am used to. Indeed, $\hat u(\psi)$ can be written also $d\psi(H_u)$. The choice of $\xi_u$ is because the infinitesimal generator of the circle is often denoted by $\xi$ and corresponds to $\xi_1$. The use of ${\rm grad}_\omega$ for the symplectic gradient is almost standard in France. $\endgroup$ – Patrick I-Z Nov 16 '13 at 16:00

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