# Quantization from Poisson brackets

It is known that given a symplectic manifold, one can define different Poisson brackets. I am trying to see whether in classical sense ($\hbar=0$) given two different Poisson brackets (i.e. a symplectic manifold can be equipped with non equivalent different Poisson bracket structures), the quantized structures are totally different in both observables and other algebraic structures. And say one can perform quantization by tuning $\hbar=1$. Would the result different due to different Poisson brackets employed? I guess this answer would also apply to field theory content as well. In addition to that question, is the quantization unique up to some equivalence?

• Oct 23, 2016 at 22:01
• Could read up on Todorov . Note $f(0)\neq g(0)$ could not lead to $f(\hbar)=g(\hbar)$ unless the limits were singular. Conversely, $f(0)$ could not fully specify $f(\hbar)$ unless you stacked extra conditions on f, which nature finds ways to bypass, quite often. Oct 23, 2016 at 22:31
• Note that, on a symplectic manifold, there is a canonical symplectic form, which is the one used in mechanics Oct 23, 2016 at 22:40