Let $(M,\omega)$ be a symplectic manifold, and let $X_f$ denote the Hamiltonian vector field for $f\in C^{\infty}(M)$. I know that the integral curve $$(X_H)_{\gamma (t)}=\dot{\gamma}(t)$$ means the Hamiltonian equation. If the symplectic form $\omega$ is exact, that is, $\omega=d\theta$ for some $1$-form $\theta$ on $M$, then we can find the prequantization $Q$ defined by $$Q(f)=i\hbar \left(X_f-\frac{i}{\hbar}\theta(X_f) \right)+f$$ for each $f\in C^{\infty}(M)$. And it satisfies the relation $$\frac{i}{\hbar}[Q(f),Q(g)]=Q(\{f,g\})$$ (see the page469 of Brain.C.Hall), where $\{f,g\}=\omega(X_f,X_g)$ is the Poisson bracket.
On the other hand, in the quantum mechanics, we consider the Heisenberg equation $$\frac{i}{\hbar}[\hat{H},A(t)]=\frac{d}{dt} A(t).$$ Since $X_H(f)=\{H,f\}$ holds, it seems likely that the Heisenberg picture would be quantized by applying the prequantization $Q$ to the Hamiltonian equation, but it is not rigorous. Are there any relation between them?
My attempt. For a curve $\gamma(t)$, we consider the Dirac delta function $\delta_{\gamma(t)}$. If the operator $X_H$ can act distributions and if the Poisson bracket $\{\cdot,\cdot\}$ can be also defined for delta functions, then it might be true that $$(X_H)_{\gamma(t)}=X_H(\delta_{\gamma(t)})=\{H,\delta_{\gamma(t)} \}.$$ If it holds, then the Hamiltonian equation might turn into $$\{H,\delta_{\gamma(t)}\}=\frac{d}{dt}\delta_{\gamma(t)}.$$ Letting $A(t)=Q(\delta_{\gamma(t)})$ and applying $Q$ to the previous equation, then we could have the Heisenberg picture $$\frac{i}{\hbar}[Q(H),A(t)]=\frac{d}{dt}A(t).$$ Will it be mathematically justified? In addition, the $Q(H)$ is just self-adjoint? Furthermore, would the Heisenberg picture induce the classical mechanics $(X_H)_{\gamma(t)}=\dot{\gamma}(t)$?
The bracket equation gives us a straightforward way to quantize into the Heisenberg picture. If $Q$ has $\{f,g\}=-\frac{i}{\hbar}[Q(f),Q(g)]$ then for Heisenberg picture if $H,f\in C^\infty(M)$ and $H$ is the Hamiltonian, $$\frac{d}{dt}Q(f)=\frac{i}{\hbar}[Q(H),Q(f)]=Q(\{f,H\})$$ Recall that in Hamiltonian mechanics, $\frac{d}{dt}f=\{f,H\}$ meaning that in this case we have $$\frac{d}{dt}Q(f)=Q\left(\frac{d}{dt}f\right)$$ This tells us that the time evolution commutes with the quantization, and that in any system which satisfies the quantization condition the time evolution of quantized observables corresponds to the quantization of their time evolution.
In your discussion, I believe you can make most of the equations and notions rigorous through weak derivatives and functional analysis, but it is not clear that what you have written gives any indication to the correspondence between the classical observable picture and the quantized Heisenberg picture.