This question has many sub-questions. I'll try to give you brief explanations and good references.
How to perform symplectic reduction in "Quantum mechanics"
Symplectic reduction is basically equivalent to Dirac's first class constraints, i.e., reduction of gauge symmetry. The prototype model used in QFT obtained by symplectic reduction is the $CP^{N}$ model, which is a symplectic reduction of $C^{N+1}$ by $U(1)$:
We start from $N+1$ scalar fields $\phi_i$ with its standard Lagrangian. (Everything can be performed in $0+1$ space-time dimensions i.e., quantum mechanics, however, this is not obligatory, we can work in any dimension). This model has $U(N)$ global symmetry. If we want to reduce the $U(1)$ symmetry $\phi_i \rightarrow e^{i\theta} \phi_i$, we perform two operations:
first, we minimally couple the model to a non-dynamical gauge field $A_{\mu}$.
Secondly, we observe that the Noether charge of this symmetry is $\phi_i \phi_i$ (Einstein's convention - this is the analog of the Gauss' law in electrodynamics). This charge must be a constant. Thus we add a constraint term to the Lagrangian: $\lambda(\phi_i \phi_i-\mathrm{const.}) $.
Now, we solve the classical equations of motion for the gauge field and the Noether charge constraint and substitute the solutions and voilà, we get the $CP^N$ sigma model.
Please see BKMU, (section 5.1.) for the explicit construction. The solution of the gauge field removes one real parameter and the constraint removes an additional real parameter, thus the resulting model target space is one complex dimension less.
How to perform symplectic reduction and quantize the 2+1 D Chern Simons theory
In the Abelian Chern-Simons theory, (in the Hamiltonian form with $A_0 = 0$). The number of degrees of freedom before the reduction is $2 $ , ($A_1$ and $A_2$) and as above the reduction removes two real degrees of freedom, thus we are left with zero degrees of freedom and this means that even though we started with a field theoretical model, the residual model can be at most quantum mechanical (in $0+1$ dimensions). Said differently, the moduli space is finite dimensional. Thus the above method does not directly work. The same reasoning works also for the non-Abelian case.
What we can do, is to substitute the most general classical solution with the residual quantum mechanical degrees of freedom (i.e., the moduli space), compatible with the boundary conditions. All the other degrees of freedom, but the moduli space, will disappear from the action due to the gauge symmetry (and the Gauss law) and we will reach a quantum mechanical model.
This is what was exactly performed by EMSS, who obtained a lot of exact solutions to the Chern Simons theory with and without sources, in particular they identified the phase space and the resulting Hilbert space. Please see other equivalent methods Bos and Nair and Dunne.
On the origin of the quantum group symmetry
The finite dimensional moduli spaces which emerge from the symplectic reduction of the Chern-Simons theory are symplectic.( In finite number of dimensions, It was proved by Marsden and Weinstein that a symplectic reduction of a symplectic manifold results a symplectic space. In our case where the initial space is infinite dimensional (Chern-Simons), there is no such a general result, but it can be proved case by case). The symplectic form on the reduced theory can be considered as a magnetic field (Closed but non-exact $2-$ form).
It is a general result that the reduced quantum mechanical theory of the Chern-Simons theory always describe the restricted to the lowest Landau level dynamics of a nonrelativistic particle on the moduli space in the presence of a magnetic field equal to $k$ (the level) times a basic symplectic structure (generator of $H^2(M, \mathbb{Z})$) of the moduli space.
The quantum group symmetry is the algebra satisfied by the magnetic finite translation operators, having the form:
$$e^{i \alpha^i (p_i-A_i)}$$
Please see the following article by Sato with a concise explanation of this fact.
The solution of the theory on the moduli space, thus for the full Chern-Simons model will be given by a representation of the quantum group. In particular, it fixes the dimension of the Hilber sapce of the theory.
This post imported from StackExchange Physics at 2018-03-22 21:50 (UTC), posted by SE-user David Bar Moshe