Let $(X,\omega)$ be a symplectic manifolds. 



> 
> For $f∈C^∞(M,ℂ)$ a function on phase space, the corresponding quantum
>  operator(prequantized observable function) is the linear map
>  $$\mathcal{O}\_f:Γ\_X(L)→Γ\_X(L)$$ given by $$ψ↦−i\hbar∇ \_{X\_f}ψ+f⋅ψ$$
>  Where
> 
> 
> 1. $X\_f$ is the Hamiltonian vector field corresponding to $f$;
> 
> 
> 2.$∇\_{X\_f}:Γ\_X(L)→Γ\_X(L)$ is the covariant derivative of sections along $X\_f$ for the given choice of pre-quantum connection;
> 
> 
> 3.$f⋅(−):Γ\_X(L)→Γ\_X(L)$ is the operation of degreewise multiplication pf sections.
> 
> 
> 


Recall that a global flow is one whose flow domain is all of $\mathbf{R} \times M$ and a vector field is complete if it generates a global flow.


Then I can not see why 
when the Hamiltonian vector field $X\_f$ is complete the pre-quantized observable functions can be written as $$\mathcal{O}\_fs=-\frac{i\hbar}{2\pi}\frac{d}{dt}(\exp(tX\_f)s)\vert\_{t=0}$$. 


which $s$ is a section of Line bundle $L$ and $exp$ means "exponential" . Is there any referrence . What about when Hamiltonian vector field is not complete. Can we write the quantum operators as flow?
see[here](http://tqft.net/other-papers/Geometric%20Quantization%20-%20Kirillov.pdf) for more details



Pre-quantized observable functions can be written as flow
Conceptually, it's much easier to see what is going on if you focus not on the line bundle $L$, but on the principal $\mathrm{U}(1)$-bundle $\pi:P\rightarrow M$ to which it is associated (so that $L=P\times\_{\mathrm{U}(1)}\mathbb{C}$; $P$ can be thought of as the subset $L$ of unit norm with respect to the Hermitian form). In his [1970 lectures](http://books.google.co.uk/books/about/Lectures_in_Modern_Analysis_and_Applicat.html?id=ZgUvAAAAIAAJ), Kostant showed how to lift the Hamiltonian vector fields $X\_f$ on $M$ to $U(1)$-invariant vectors fields $A\_f$ on $P$ while preserving the commutation relations (so $[A\_f,A\_g] = -A\_{\lbrace f,g\rbrace}$ on $P$ covers $[X\_f,X\_g] = -X\_{\lbrace f,g\rbrace}$ on $M$). Explicitly,
$$
A\_f = X\_f^\mathrm{h} - \frac{i}{\hbar}(f\circ\pi)
$$
where the first term is the horizontal lift of $X\_f$ (with respect to a connection $\alpha$ on $P$ of curvature $\frac{i}{\hbar}\omega$), while the second term is a vector lying along the $\mathrm{U}(1)$-direction in $P$. BTW, the classical observables $f, g$ should be real-valued, not complex-valued as you wrote. It is not too difficult to see that $A\_f$ is complete iff $X\_f$ is complete, and by construction the flow $\phi\_t^{A\_f}:P\rightarrow P$ covers $\phi\_t^{X\_f}:M\rightarrow M$.


Now, a section $s\in\Gamma(L)$ can be thought of as a $\mathrm{U}(1)$-equivariant function $\tilde{s}:P\rightarrow \mathbb{C}$. The flow $\phi\_t^{A\_f}$ induces a flow on this space of sections, given by
$$
\tilde{s}\mapsto \tilde{s}\circ\phi\_{-t}^{A\_f}.
$$
Differentiating this expression wrt $t$ (and multiplying by $i\hbar$ to produce a Hermitian operator) yields the corresponding quantum observable $\mathcal{O}\_f$
$$
\mathcal{O}\_f\tilde{s} = i\hbar\frac{\mathrm{d}}{\mathrm{d}t}\Bigg\vert\_{t=0}\left(\tilde{s}\circ\phi\_{-t}^{A\_f}\right) = -i\hbar\, A\_f\tilde{s} = \left(-i\hbar\, X\_f^\mathrm{h} + f\circ\pi\right)\widetilde{s}.
$$
Thinking of sections $s\in\Gamma(L)$ instead of $\mathrm{U}(1)$-equivariant functions $\tilde{s}$, this says that
$$
\mathcal{O}\_f s = (-i\hbar \nabla\_{X\_f} + f)s.
$$


I believe this also answers your [other question](https://mathoverflow.net/questions/152932/alternative-definition-of-quantum-operators-by-using-reeb-vector-fields) - the Reeb vector field in this case is the vector field along the $\mathrm{U}(1)$ direction of $P$ (the contact form is the connection form $\alpha$).


Kostant's lecture notes are still the best place to learn all this, IMO.


Edit: of course, the whole point of requiring $[A\_f,A\_g] = A\_{-\lbrace f,g\rbrace}$ is so that
$$
[\mathcal{O}\_f,\mathcal{O}\_g] = [-i\hbar\,A\_f, -i\hbar\,A\_g] = (i\hbar)^2 A\_{-\lbrace f,g\rbrace} = i\hbar\,\mathcal{O}\_{\lbrace f,g\rbrace},
$$
the usual Heisenberg ``Poisson brackets go to commutators'' relationship.


