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Can someone please explain me what is a Moyal product?

Also, how does putting $$X_a(\psi) ~=~ x_a\star\psi$$ realise $$[X_a,X_b]=i\theta_{ab}{\bf 1}?$$

Ref: Quantum mechanics on non-commutative plane

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1 Answer 1

up vote 2 down vote accepted

I) The associative non-commutative Moyal/Groenewold/star product $f\star g$ is explained on Wikipedia. The corresponding $\star$-commutator is defined as

$$\tag{1} [f\stackrel{\star}{,} g]~:=~f\star g-g\star f.$$

In particular, the Jacobi identity for the $\star$-commutator is a consequence of the associativity of the $\star$-product.

II) On one hand there is the algebra of function, say, the algebra $\mathbb{C}[[x]]$ of powerseries in indeterminates $x_a$. We equip it with a unit $1$ and the $\star$-product$^1$ so that

$$\tag{2} [x_a\stackrel{\star}{,}x_b]~=~i\theta_{ab}.$$

III) On one hand, there is the Heisenberg algebra $({\cal A}, +, \circ)$ generated by

$$\tag{3} [X_a\stackrel{\circ}{,}X_b]~=~i\theta_{ab}{\bf 1}.$$

Here the elements of the Heisenberg algebra are (linear) operators acting on functions; the algebra product $\circ$ is composition; the algebra unit ${\bf 1}$ is the identity operator; and

$$\tag{4} [A \stackrel{\circ}{,}B]~:=~A \circ B - B \circ A$$

is the usual composition commutator of two operators $A$ and $B$.

IV) There is a unique algebra isomorphism

$$\tag{5} (\mathbb{C}[[x_a]],+, \star) ~\stackrel{\Phi}{\longrightarrow}~({\cal A}, +, \circ) $$

generated by

$$\tag{6} \Phi(x_a)~:=~X_a.$$

It follows that the algebra isomorphism $\Phi$ maps the (2) into (3).

V) The Heisenberg algebra acts on the algebra $\mathbb{C}[[x]]$, i.e. an operator $A$ acts on a function $\psi$ and produce a new function $A(\psi)$. Concretely, for an element $A\in {\cal A}$ define

$$\tag{7} A(\psi)~:=~\Phi^{-1}(A) \star \psi. $$

Equivalently,

$$\tag{8} \Phi(f) (g)~:=~f \star g. $$

It is not hard to see that the definition (7) is consistent with that $\Phi$ is an algebra isomorphism.

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$^1$ There is also the standard commutative and associative pointwise multiplication $\cdot$ of function, which plays almost no role here.

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Could you explain this to someone who has just done a basic level course in quantum mechanics (till the level of 'Principles of Quantum Mechanics' by R Shankar) –  Kamal May 1 '13 at 11:02
    
I updated the answer. –  Qmechanic May 1 '13 at 21:05
    
After reading Qmechanic's nice answer +1, here and here are some cool physics applications of noncommutative geometry and the star product. –  Dilaton May 11 '13 at 22:22
    
Corrections to the answer (v3): The word function in the last sentence should be functions. –  Qmechanic May 11 '13 at 22:32

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