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  Is there a relationship between quantization of coherent maps and the Kostant Souriau prequantum operator?

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Suppose $(M,\omega)$ is a symplectic manifold with symplectic form $\omega$. Suppose at first that $\omega$ is exact, meaning that there is a globally defined ''symplectic potential'' $\theta$ with $d\theta=\omega$. We can consider the "prequantum Hilbert space" of square-integrable functions on $M$ (with respect to the Liouville volume measure). For each smooth function $f$ on $M$, we can define the Kostant–Souriau prequantum operator

\tilde{\Gamma}(f):=i\hbar\left( X_f -\frac{i}{\hbar}\theta(X_f)\right) +f

where $X_f$ is the Hamiltonian vector field associated to $f$.

When comparing geometric to coherent quantization, what is the exact relationship between the Souriau prequantum operator and the coherent quantization operator as defined in Theorem 3.12 of this paper? The theorem says that for $Z$ be a coherent space, $Q(Z)$ a  quantum space
of a coherent space $Z$, and  $A$ be a coherent map on $Z$.

(i) There is a unique linear map $\Gamma(A)\in Lin \,Q(Z)$ such that
\Gamma(A)|z>=|Az> \forall z\in Z.
(ii) For any adjoint map $A^*$ of $A$,
<z|\Gamma(A)=<A^*z| \, \forall z\in Z,
(iii) $\Gamma(A)$ can be extended to a linear map
$\Gamma(A):=\Gamma(A^*)^*\in Lin \, Q^x(Z)$, and this extension maps
$\overline{Q}(Z)$ into itself.

As I naively understand it, the coherent space $Z$ should correspond to the symplectic manifold $M$, the coherent states to the square integrable functions and the smooth functions on $M$ to the coherent maps. But I have some difficulties relating the Souriau prequantum operator $\tilde{\Gamma}(f)$ to the quantization operator $\Gamma(A)$. For example what would take the role of the Hamiltonian vector field and the potential $\theta$ in the coherent quantization picture?

asked Oct 1, 2018 in Theoretical Physics by Dilaton (6,240 points) [ revision history ]

a nice ncatlab page on the question; perhaps the answer is in 4.Examples , equation (4)

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