Physical interpretation of $q$-deformation

I am currently reading the paper Quantum Group Particles and Non-Archimedean Geometry by Volovich and Aref'eva. Here they discuss the difference between $q$-deformation and $\hslash$-deformation. In a classical system, one has \begin{align} x_ix_j&=x_jx_i\\ p_ip_j&=p_jp_i\\ p_ix_j&=x_jp_i, \end{align} where $x_i$ and $p_i$ are position and momentum, respectively. Then an $\hslash$-deformation gives \begin{align} [x_i,x_j]&=0\\ [p_i,p_j]&=0\\ [p_i,x_j]&=i\hslash \delta_{ij}, \end{align} where $\delta_{ij}$ is the Kronecker delta. This is a quantum system, and I am familiar with this. Then they say that a $q$-deformation gives \begin{align} x_ix_j&=qx_jx_i, \quad i<j,\\ p_ip_j&=qp_jp_i\\ p_ix_j&=qx_jp_i + ..., \end{align} which is different from an $\hslash$-deformation. This seems to be in connection with what is called a quantum line or quantum plane.

My questions are:

1) What is the physical interpretation of $q$-deformation? It doesn't look like it is the same as going from classical mechanics to quantum mechanics. I have seen $q$-deformation in the case of quantum groups, though nothing was said about the physical interpretation.

2) What is the physical interpretation of a quantum line or quantum plane?

• This kind of algebra (but not applied directly to $x$ or $p$) appears in the quantization of Chern-Simons theories, see a baby example ($U(1)$ Chern-Simons theory) in this question/answer – Trimok Oct 10 '13 at 17:42

1) What is the physical interpretation of q-deformation? It doesn't look like it is the same as going from classical mechanics to quantum mechanics. I have seen q-deformation in the case of quantum groups, though nothing was said about the physical interpretation.

A relation between the two notions is

If $[x,y]=h$ (a scalar or central element) then $e^x e^y = q e^y e^x$ with $q = e^h$ (assume the exponential series converge)

This is consistent with commutative algebras being the case $\hbar = 0$ and $q=1$ respectively.

2) What is the physical interpretation of a quantum line or quantum plane?

There is no standard physical interpretation at this time. In some physical models, turning on a magnetic field is a form of noncommutative deformation, and there is braiding in some experimentally realizable 2-d systems, and probably other examples, but there is no physical system you can learn about and then use as an intuitive model for quantum groups.

• Thanks, but I have a few questions: So I see there is a difference since in a quantum system we have $[x_i, x_j] = 0$, while we have $x_ix_j=qx_jx_i$ for $q$-deformation. But why are there no exponentials here (like in your comparison)? Shouldn't there be a physical interpretation of $x_ix_j=qx_jx_i$? If it depends on the situation, I would like to see some references to that. Also, could you please give me references about your answer to my first question? – N.U. Sep 17 '13 at 20:07