A Short Course on Quantum Mechanics and Methods of Quantization 63(4) Iff=qandg=p(or vice versa), one obtains:(q∗p)(q,p)=qp+ı
2;(p∗q)(q,p)=qp−ı
2. (1.333)
(5) Notice that Eq.(1.330) implies:̂Ω(q)·Ω(̂ g)=Ω(̂ qg)+ı
2Ω̂
(
∂g
∂p)
(1.334)
and similarly for the others.
Using the Moyal product, we can define theMoyal Bracket{·,·}∗as:
{·,·}∗:F(
R^2
)
×F
(
R^2
)
→F
(
R^2
)
(1.335)
{f,g}∗≡1
i(f∗g−g∗f)={f,g}+O(
^2
)
, (1.336)
We notice that the difference betweenthe Moyal and Poisson brackets isO(^2 ),
since the differencef∗g−g∗fcontains only odd powers of.
Being defined in terms of an associative product, the Moyal bracket fulfils
all the properties of a Poisson bracket (linearity, anti-symmetry and the Jacobi
identity), and defines a new Poisson structure on the (non-commutative) algebra of
functions endowed with the Moyal product. In particular, just as for the ordinary
Poisson brackets, the Jacobi identity implies that {f,·}∗is a derivation (with
respect to the∗-product) on the algebra of functions:
{f,g∗h}∗={f,g}∗∗h+g∗{f,h}∗, (1.337)Now{f,·}∗is not necessarily a vector field, contrary to what happens with the
standard Poisson bracket{f,·}. This can be seen by explicitly writing down the
second term in (1.336) as:{f,g}∗={f,g}+^2 {f,g} 2 +...,to obtain:
{f,g} 2 (q,p)=1
24
{
∂^3 f
∂q^3∂^3 g
∂p^3− 3
∂^3 f
∂p∂q^2∂^3 g
∂q∂p^2+3
∂^3 f
∂p^2 ∂q∂^3 g
∂q∂q^2−
∂^3 f
∂p^3∂^3 g
∂q^3}
,
showing that{f,g}∗contains also higher-order derivatives. The reason for that
is precisely that the Moyal bracket is non-local. It is only whenf is at most
a quadratic polynomial that{f,·}∗becomes a derivation on the usual pointwise
product. Indeed, if this is the case, the Moyal and Poisson brackets offwith other
functions coincide.
Finally, using the definitions of the Weyl and Wigner maps, we have in general:
{f,g}∗=iΩ−^1
(
Ω(̂ f)·Ω(̂g)−Ω(̂g)·̂Ω(f))/
, (1.338)
i.e.:
[
Ω(̂ f),Ω(̂ g)
]
=−iΩ(̂ {f,g}∗). (1.339)