Why does the classical Noether charge become the quantum symmetry generator? It is often said that the classical charge $Q$ becomes the quantum generator $X$ after quantization. Indeed this is certainly the case for simple examples of energy and momentum. But why should this be the case mathematically?
For clarity let's assume we are doing canonical quantization, so that Poisson brackets become commutators. I assume that the reason has something to do with the relationship between classical Hamiltonian mechanics and Schrodinger's equation. Perhaps there's a simple formulation of Noether's theorem in the classical Hamiltonian setting which makes the quantum analogy precisely clear?
Any hints or references would be much appreciated!
Mathematical Background
In classical mechanics a continuous transformation of the Lagrangian which leaves the action invariant is called a symmetry. It yields a conserved charge $Q$ according to Noether's theorem. $Q$ remains unchanged throughout the motion of the system.
In quantum mechanics a continuous transformation is effected through a representation of a Lie group $G$ on a Hilbert space of states. We insist that this representation is unitary or antiunitary so that probabilities are conserved. 
A continuous transformation which preserves solutions of Schrodinger's equation is called a symmetry. It is easy to prove that this is equivalent to $[U,H] = 0$ for all $U$ representing the transformation, where $H$ is the Hamiltonian operator.
We can equivalently view a continuous transformation as the conjugation action of a unitary operator on the space of Hermitian observables of the theory
$$A \mapsto UAU^\dagger = g.A$$
where $g \in G$. This immediately yields a representation of the Lie algebra on the space of observables
$$A \mapsto [X,A] = \delta A$$
$$\textrm{where}\ \    X \in \mathfrak{g}\ \ \textrm{and} \ \  e^{iX} = U \ \  \textrm{and} \ \  e^{i\delta A} = g.A$$ 
$X$ is typically called a generator. Clearly if $U$ describes a symmetry then $X$ will be a conserved quantity in the time evolution of the quantum system.
Edit
I've had a thought that maybe it's related to the 'Hamiltonian vector fields' for functions on a symplectic manifold. Presumably after quantization these can be associated to the Lie algebra generators, acting on wavefunctions on the manifold. Does this sound right to anyone? 
 A: Consider a quantum field formalism, where fields are operators.
For instance, consider, for simplicity, a charged scalar  field , with action $S = \int d^4x \partial_\mu \phi \partial^\mu \phi^*$, with a field : 
$$\phi(x) = \int ~ \frac{d^3k}{2E_k} ~(a(p)e^{-ip.x} + b^+(p)e^{ip.x} )\tag{1}$$
$$\phi^*(x) = \int ~ \frac{d^3k}{2E_k} ~(b(p)e^{-ip.x} + a^+(p)e^{ip.x} )\tag{2}$$
where $a^+(p), a(p)$ are creation/anihilation operators for the particle, and $b^+(p), b(p)$ creation/anihilation operator for the anti-particle.
If we have a infinitesimal  transformation,  $\delta \phi(x)$ , leaving unchanged the action $S$, the conserved current is $j^\mu(x) = [\frac{\partial L}{\partial(\partial_\mu \phi)}~\delta \phi(x) + \frac{\partial L}{\partial(\partial_\mu \phi^*)}~\delta \phi^*(x)]$ (skipping the infinitesimal parameters), and the conserved generalized charge is $Q =~:\int d^3x~ j^0(x): ~= ~:\int d^3x ~[\Pi(x)~ \delta\phi(x) +\Pi^*(x)~ \delta\phi^*(x)]:$ - where the $\Pi(x),\Pi^*(x)$ are the conjugated momenta of $\phi(x),\phi^*(x)$, and the sign $:$ is for normal ordered product (putting anihilation operators at right).
We see, of course, that $Q$ is an operator.
An example : the standard (electric) charge is the conserved quantity which corresponds to the (global) transformation $\phi(x) \rightarrow e^{-i\Lambda}\phi(x)$, here the infinitesimal transformation  is $\delta \phi(x) = -i\epsilon ~\phi(x)~$, so we have : 
$$Q = -i:\int ~ d^3x ~(\Pi(x) ~\phi(x) - \Pi^*(x) ~\phi^*(x) ): \tag{3}$$
Or, equivalently : 
$$Q = \int ~ \frac{d^3k}{2E_k} ~(a^+(p)a(p) - b^+(p)b(p) )\tag{4}$$
And we have : $$[Q,\phi(x)]=-\phi(x) \tag{5}$$
(This can be checked from the fundamental commutation relations between operators, like $[a(k),a^+(k')]=\delta^3(k-k')$) 
A: The canonical quantization after Dirac should fulfill the following axioms:


*

*Q1: The map $f \to \hat f$ that assigns a operator to every function on the phase space is linear and the constant 1-functions get mapped to the 1-operator

*Q2: The Poisson bracket maps to the commutator decorated with $\hbar$

*Q3: A complete system of functions in involution maps to a complete system of commutative operators.
It is the last condition which ensures that $G$ is a symmetry on the quantum side (the assignment  $f \to \hat f$ needs to be a irreducible representation of the symmetry generators).
But the No-Go theorems of Groenwald und Van Hove shows that a quantization for all observables with Q1-Q3 is not possible. The two mayor solutions are: Weaken Q2 and only require that it holds only up to first order of $\hbar$ - this leads to deformation quantization. On the other hand, geometric quantization modifes Q3 in the sense that it should hold only for some reasonable subalgebra of functions (eg which contains momentum ect.). 
