In deformation quantization there is a full classification available: let us first focus on the symplectic case which is easier. If $(M, \omega)$ is a symplectic manifold (like the $\mathbb{R}^2$ in your example) then the equivalence classes of star products are classified by formal series in the second deRham cohomology of $M$, a purely topological quantity.

Here equivalence means that $\star$ and $\star'$ are called equivalent if there is a formal series $T = \mathrm{id} + \sum_{n = 1}^\infty \hbar^r T_r$ of differential operators $T_r$ starting with the identity in zeroth order such that
\begin{equation}
T(f \star g) = Tf \star'Tg
\end{equation}
This implies clearly that the deformations yield isomorphic algebras. But it is sligtly stronger, as one requires $T$ to start with the identity in zeroth order. This means that $T$ is invisible in the classical limit. If you are interested in a physical interpretation, this is very important as the interpretation of the elements of $C^\infty(M)$ should not be changed under quantization: the same function should correspond to the same physical observable, both in the classical and the quantum world.

There is a good reason that the above notion of equivalence can be interpreted as the freedom of having possible different ordering descriptions as they show up in other quantization schemes.

So in your example, there is only one deformation up to equivalence since the second deRham cohomology of $\mathbb{R}^2$ is trivial.

If $M$ admits a non-trivial deRham cohomology then the higher order terms in the classifying "characteristic class" $c(\star)$ of the star product have the physical interpretation of a magnetic field. Their non-triviality then gives a magnetic monopol. So from that point of view (which can be justified physically more striclty) quantization requires to be told what magnetic charges are around and then it is unique up to equivalence, which means physically, up to the choice of an ordering description.

Names involved in this classification results are Bertelson-Cahen-Gutt, Nest-Tsygan, Fedosov, Weinstein-Xu, and many others.

Let me add here the precise definition of the characteristic class. On a symplectic manifold, given a star product $\star$ there is an intrinsic way to construct a formal series
\begin{equation}
c(\star) \in \frac{[\omega]}{i\hbar} + \mathrm{H}^2_{dR}(M, \mathbb{C})[[\hbar]]
\end{equation}
of second de Rham cohomology classes in such a way that $\star$ is equivalent to $\star'$ iff their classes coincide. Moreover, for every such formal series there is also a star product $\star$ having this series as characteristic class. The choice of the origin in this affine space is by convention, and given by the class of the symplectic form itself.

There are several ways to construct $c(\star)$, all leading to the same result. The easiest is probably the construction in Gutt-Rawnsley link text. As a result: there is always the way to choose the class $c(\star) = \frac{[\omega]}{i\hbar}$ without higher order contributions. This is the somehow canonical choice on a symplectic manifold. However, having this choice does of course not mean to have a unique quantization (up to equivalence): depending on the non-triviality of the deRham cohomology, there are hiogher orders possible and then one get's non-trivial equivalence classes of quantizations.

Now for the Poisson case: here the situation is similar in so far as in the symplectic case you can interprete the higher order terms of the class as perturbation of the symplectic form. Kontsevich proved using his formality theorem that the possible star products are classified by formal deformations of the Poisson tensor $\pi = \hbar \pi_1 + \hbar^2 \pi_2 + \cdots$ up to the action of the formal diffeomorphism group. This is given by exponentiating the formal vector fields $X = \hbar X_1 + \hbar^2 X_2 + \cdots$.

In the symplectic case, this of course matches with the "characteristic class" classification. One can even show that, under correct identification, the classes are equal (this was done by Bursztyn, Dolgushev, and myself).

In particular, not all deformations will be isomorphic in general, this depends now really very much on the underlying manifold and the Poisson tensor.

Surprisingly, and this point I don't really understand well, in the Poisson case, the classification, i.e. the identification of the class of $\star$ and the class of $\pi$ depends to some extend on the choice of the formality map (and on those, GT acts). This is funny, because in the symplectic case, the classification can be done quite intrinsically, not depending on any sort of choices. Here the construction of Deligne, explained later also in Gutt-Rawnsley, is very nice.

Note added: I fear that for the refined question I have no answer. The situation is the following: one has two types of data needed to construct a quantization (in the sense of star products) of a Poisson manifold together with a meaningful classification:

First is to choose a quantization machinery. In the generic Poisson case, the only one we know is a formality map, which can be obtained by e.g. Kontsevich's or Tamarkin's methods. These constructions require certain choices, like the ones mentioned by Alexander Chervov_ a propagator in Kontsevich's construction or an associator for Tamarkin.

Second, having a formality we can choose a classical deformation $\pi = \hbar \pi_1 + \hbar^2 \pi_2 + \cdots$ of the given Poisson structure $\pi_1$ and plug it into the formality. This will give a star product and any two obtained this way will be equivalent iff the classical deformations $\pi$ and $\pi'$ are equivalent in the sense of the action of the formal diffeos.

So the question, how the notion of equivalence depends on the choice of the formality seems to be much more subtle. I have no idea about that. ONe knows that there are really different formalities (in a sense of being non-homotopic) but whether and how they yield a really different classification scheme seems to be not known (please correct me). I had some discussions about that with Vasiliy Dolgushev some time ago, but we didn't come to a real conclusion...

The somehow surprising thing is that in the symplectic case, the classification is independent of any construction, and done by this characteristic class. Since the classifying space is the deRham cohomology $H_{dR}^2(M, \mathbb{C})[[\hbar]]$, we have a canonical choice for the trivial class. So in this sense, we really have a canonical quantization up to equivalence. Note however, that there are good reasons for considering also nontrivial classes in the symplectic case.

From that point if view, the original question in the Poisson case can also be formulates as folows. given $\pi = \hbar \pi_1$ and a formality yielding $\star$, is there another formality such that $\star$ has a corresponding class $\pi' = \hbar \pi_1 + \hbar^2 \pi_2 + \cdots$ which is not equivalent to $\pi$ by means of the formal diffeos?

OK, not much of a help, but just a refomulation of the problem.

This post imported from StackExchange MathOverflow at 2017-09-17 21:00 (UTC), posted by SE-user Stefan Waldmann