Relationship between $\star$-products in phase-space QM and NC geometry

Question

What exactly is the relationship between $\star$-products in phase-space quantum mechanics, i.e.

$$ (f \star g) (x,p) = f(x,p) e^{\frac{i \hbar}{2} ( \overleftarrow{\partial_x} \cdot \overrightarrow{\partial_p} - \overleftarrow{\partial_p} \cdot \overrightarrow{\partial_x})} g(x,p) ~,$$ and $\star$-products in $2n$-dimensional non-commutative spaces, i.e.,

$$ (f \star g) (X) = \left . e^{\frac{i\hbar}{2} \theta^{\mu\nu} \frac{\partial}{\partial X^\mu}\frac{\partial}{\partial Y^\nu}} f(X) g(Y)\right \vert_{Y \to X} ~~,$$ where $\theta^{\mu\nu}$ is a constant anti-symmetric tensor?

To be a bit more specific, is there a way to reinterpret/reformulate the phase-space non-commutativity as spatial non-commutative geometry? Or something of that sort.

Answer 1

If the phase space is $\mathbb{R}^{2r+k}$, and $\theta^{\mu\nu}$ has rank $2r$ then there exists a bijective linear coordinate transformation that brings the anti-symmetric real matrix $\theta^{\mu\nu}$ on standard form $$ \theta^{\prime \mu\nu}~=~\begin{pmatrix} \mathbb{0}_{r\times r} & \mathbb{1}_{r\times r}& \cr -\mathbb{1}_{r\times r} & \mathbb{0}_{r\times r}& \cr && \mathbb{0}_{k\times k} \end{pmatrix}. $$ If furthermore $\theta^{\mu\nu}$ is non-degenerated (that is: $k=0$) then its $\star$-product is equivalent to the standard Groenewold-Moyal $\star$-product in Darboux/canonical coordinates.

Finally, we should stress that there are important conceptional issues to be sort out if time $x^0$ is non-commutative. In contrast, this does not happen in QM, where time is a parameter (as opposed to an operator).

Answer 2

I cannot pretend to understand the question (I would answer what I understand as a clumsier version of @Qmechanic 's answer), but, as a courtesy of the casual reader, I'd rewrite the first equation into the equivalent format of the second one, so as to make it easier to compare and contrast apples with apples.

The classic Groenewold star product $$ (f \star g) (x,p) = f(x,p)~ e^{\frac{i \hbar}{2} ( \overleftarrow{\partial_x} \cdot \overrightarrow{\partial_p} - \overleftarrow{\partial_p} \cdot \overrightarrow{\partial_x})} g(x,p)=\left . e^{\frac{i \hbar}{2} ( \partial_x \cdot \partial_{p'} - \partial_p \cdot \partial_{x'})} f(x,p)g(x',p') \right \vert_{x'=x,~ p'=p} $$ is routinely recast as $$ (f \star g) (X) =\left . e^{\frac{i\hbar}{2} \epsilon^{\mu\nu} \frac{\partial}{\partial X^\mu} \frac{\partial} {\partial Y^\nu}} f(X) g(Y) \right \vert_{Y \to X}\\ X^\mu\equiv (x,p) ~. $$

It should be apparent how to generalize this to multidimensional phase spaces, with r space coordinates, wrinkle their geometry, and append degenerate dimensions, as the proper answer details. You may be suggesting this with the dots dotting r x components to r p components, as with the symbol used, in which the expression is already generalized to higher r. That is, you may give $X^\mu$ another index, j, ranging from 1 to r, in a direct product tensor space, so $X^\mu \to X^\mu_j$, in which case the exponent becomes $\frac{i\hbar}{2} \epsilon^{\mu\nu} \frac{\partial}{\partial X^\mu_j } \frac{\partial} {\partial Y^\nu_j}$; and then, finally, omit the j indices, understanding a dot product in their space. In your conventions, if you really mean the dot in your starting expression, just write $\frac{i\hbar}{2} \epsilon^{\mu\nu} \frac{\partial}{\partial X^\mu }\cdot \frac{\partial} {\partial Y^\nu}$. The fine real answer I am deferring to blends the j indices into an extension of $\mu,\nu,...$ to those of a 2r-dimensional symplectic vector.

In the mathematics literature, the * product has been extended to recondite spaces that more than cover any NCG ever contemplated.