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I have the following definition of star product, \begin{equation} \star=\exp\left[\frac{i\hbar}{2}\left(\frac{\overleftarrow{\partial}}{\partial Q^{I}}\frac{\overrightarrow{\partial}}{\partial P_{I}}-\frac{\overleftarrow{\partial}}{\partial P_{I}}\frac{\overrightarrow{\partial}}{\partial Q^{I}}\right)\right];\,\,\,I=1,\ldots,M \end{equation} So if $A(P,Q)$ and $B(P,Q)$ are matrix observables, whose poisson brackets are written as, \begin{equation} \left\{A,B\right\}=\frac{\partial A}{\partial Q^{I}}\frac{\partial B}{\partial P_{I}}-\frac{\partial A}{\partial P_{I}}\frac{\partial B}{\partial Q^{I}}, \end{equation} How can I write an expression for $\{A,B\}_{\star}$?

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  • $\begingroup$ What exactly are you after? There are several such expressions. $\endgroup$
    – Qmechanic
    Commented Mar 2, 2018 at 19:38
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    $\begingroup$ So you want $ A\star B - B\star A$ ? $\endgroup$
    – DanielC
    Commented Mar 2, 2018 at 21:00
  • $\begingroup$ I wanting an expression for $\{A,B\}_{\star}$ in terms of $\star$. $\endgroup$
    – D.Silva
    Commented Mar 2, 2018 at 21:00
  • $\begingroup$ I want to know if there is an analogue to the poisson brackets considering the star product. $\endgroup$
    – D.Silva
    Commented Mar 2, 2018 at 21:04
  • $\begingroup$ Related: physics.stackexchange.com/q/19770/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Mar 2, 2018 at 21:43

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The quantum extension (deformation) of the PB is the scaled commutator expressed in phase space, conventionally dubbed the Moyal bracket, $$ \frac{1}{i \hbar} \left(A \star B - B \star A \right) \equiv \{\{A,B\}\} = \frac{2}{\hbar} A ~~ \sin \left ( {{\frac{\hbar }{2}}(\overset{\leftarrow}{\partial_x} \overset{\rightarrow}{\partial_p}-\overset{\leftarrow}{\partial_p}\overset{\rightarrow}{\partial_x})} \right )~~ B = \{A,B\} + O(\hbar^2),$$ as expected from the correspondence principle, the limit ħ → 0.

Many of its properties related to associativity are more easily proved in Baker's integral representation, $$ \{ \{ A,B \} \}(x,p) = {2 \over \hbar^3 \pi^2 } \int\! dp' \, dp'' \, dx' \, dx'' A(x+x',p+p') B(x+x'',p+p'')\sin \left( \tfrac{2}{\hbar} (x'p''-x''p')\right)~. $$

The $O(\hbar^2)$ higher derivatives over and above the PB often probe nonlinearity in the potential of the relevant problem and deform classical Liouville flows into dramatically different characteristic quantum configurations in phase space. In sharp contrast to classical mechanics, they render the quantum probability fluid compressible.

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  • $\begingroup$ Is star produc distributive? Is correct says that,$A\star(B+C)=A\star B+A\star C$? $\endgroup$
    – D.Silva
    Commented Mar 6, 2018 at 11:42
  • $\begingroup$ Yes, of course: star products do what matrix products do. They are associative, noncommutative, linear operations. Expand them to leading order in hbar and observe their action. $\endgroup$ Commented Mar 6, 2018 at 14:09
  • $\begingroup$ Taking advantage of the opportunity, I would like someone to help me interpret how I would write, $\{M,\{N,L\}_{\star}\}_{\star}$ in the context of article [arxiv.org/pdf/0909.1448.pdf], where $\star$ is still given by the expression above. Thank you in advance. $\endgroup$
    – D.Silva
    Commented Mar 13, 2018 at 11:21
  • $\begingroup$ ? You pug into the first or the 2nd formula I am writing down in the answer, and compute away. Actually, associativity entails the Jacobi identity, so their two formulas following their (5) work to all orders in $\hbar$. That's why I gave you the integral representation, so you may plug in, shift variables (too many!) and check. $\endgroup$ Commented Mar 13, 2018 at 14:58

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