# Pre-quantized observable functions can be written as flow

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? seehere for more details

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Could you please edit the question to explain the set-up and notation? Is $\omega$ the curvature of a given connection in a hermitian line bundle $L\to X$? If so, what exactly does $exp(tX_f)s$ mean when $s$ is a section of $L$? Also, please include a reference for this formula. –  Tim Perutz Dec 28 '13 at 14:42
(I know that a reference is what you are asking for, but presumably the formula came from somewhere.) –  Tim Perutz Dec 28 '13 at 15:04
yes, all of your notations are correct. But the main part is the second part which Hamiltonian vector field is not complete? –  Hassan Jolany Dec 28 '13 at 15:07
Is $\exp(tX_f)$ the flow of $X_f$ through some point? Also, as @TimPerutz asked, could you explain what is meant by $\exp(tX_f)s$? –  Peter Crooks Dec 28 '13 at 18:30
It has been edited now –  Hassan Jolany Dec 28 '13 at 18:59

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, 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 - 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 $\alpha$).
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.