The question is about:

(1) whether giving a Lagrangian is sufficient enough to (uniquely) well-define a Hamiltonianian quantum theory with a Hilbert space?

The answer should be Yes, or No.

If yes, then I suppose one can write any of its state function $|\Psi \rangle$ in the Hilbert space, with a Hamiltonian $H$, satisfies:

$ i \partial_t |\Psi \rangle=H |\Psi \rangle $

Then, whether the situation changes if one replacing Lagrangian $L$ to an action $S$ or a path integral or partition function $Z$, i.e.

(2) whether this $L$, $S$ or $Z$ to which level (uniquely) well-defines a Hamiltonianian quantum theory with a Hilbert space? Or, they cannot well-define it yet?

(3) The even further question is whether a Lagrangian is enough to well-define Hamiltonianian quantum theory with a Hilbert space on the discrete lattice? From this arXiv paper 1305.1045, it is obvious that even if we have a Lagrangian for the standard model, it is NOT enough to well-define a Hamiltonianian quantum theory with a Hilbert space on the lattice NON-PERTURBATIVELY.

[comments]

It will be important to explain why this is so. I may be wrong, but I had impression reading some post comment about questions by an expert at Physics Stackexchange(Dr. Luboš Motl?) stating that knowing Lagrangian/action is everything about knowing the physical system. Sorry if I made a mistake here.

But my interactions with other experts in the field of condensed matter, often stated that Lagrangian is not yet enough. The quantization, Hamiltonian and Hilbert space etc is needed.

It will be nice to touch the issues of gauge theory, whether giving a gauge theory Lagrangian can define a Hamiltonianian quantum theory with a Hilbert space? As I know the 2+1D $Z_N$ gauge theory written as 2+1D U(1)xU(1) Chern-Simons theory is discussed in this post. In that sense givging a U(1)xU(1) Chern-Simons theory can mean $Z_N$ gauge theory (with discrete $Z_N$ symmetry) or something else(two copies of U(1) symmetry).

Thanks.