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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 $\{A,B\}_{\star}$?

Qmechanic
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D.Silva
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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.

Cosmas Zachos
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  • Is star produc distributive? Is correct says that,$A\star(B+C)=A\star B+A\star C$? – D.Silva Mar 06 '18 at 11:42
  • 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. – Cosmas Zachos Mar 06 '18 at 14:09
  • 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 [https://arxiv.org/pdf/0909.1448.pdf], where $\star$ is still given by the expression above. Thank you in advance. – D.Silva Mar 13 '18 at 11:21
  • ? 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. – Cosmas Zachos Mar 13 '18 at 14:58