TITLE: what's the motivation of Weyl calculus ?
QUESTION [11 upvotes]: In the pseudo-differential operator theory, we can define a  pseudo-differential operator by $$a(x,D)u=(2\pi)^{-n}\int{a(x,\xi)e^{i\langle x-y,\xi \rangle}u(y)dyd\xi}$$ with $a(x,\xi)$ belong to some particular function space (denoted by $S^m$).In the Weyl calculus one adopts the symmetric compromise  $$a^{w}(x,D)u=(2\pi)^{-n}\int{a((\frac{x+y}{2}),\xi)e^{i\langle x-y,\xi \rangle}u(y)dyd\xi}$$ again defined in the weak sense. From this one can see that the adjoint of $a^w$ is equal to $\bar a^{w}$. In particular, $a^w$ is its own adjoint when a is real valued. Is this convenience making Weyl calculus more applicable for physics? In mathematics, are there other reasons to the motivation of Weyl calculus? Furthermore, Can anyone show some problems which are solved by using this tool?

REPLY [6 votes]: In the Weyl calculus there is closer agreement between operator and symbol composition than in the standard Kohn-Nirenberg calculus. For example, $(a^w)^2\equiv (a^2)^w$ holds modulo order zero if $a$ has order one and is real-valued. Similarly, $(f\circ a)^w$ approximates $f(a^w)$, which is defined by the functional calculus, to higher order than the corresponding construction using the standard calculus. This is related to the symplectic invariance properties of the Weyl calculus, more specifically to the appearance of the Poisson bracket on the subprincipal level of the composition formula. Note that the Poisson bracket $\{f\circ a,a\}$ vanishes for real-valued symbols $a$.