# what's the motivation of Weyl calculus ?

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?

• In addition to answers below, G. Folland's "Harmonic Analysis on Phase Space" exhibits and explains the relation between the Weyl map from symbols to operators (as opposed to Kohn-Nirenberg) in terms of function algebras on Heisenberg groups. Feb 28, 2013 at 22:53

On the other hand, the symplectic invariance of the Weyl calculus was discovered much later by André Weil: for $\chi\in Sp(n)$ ($Sp(n)$ is the linear symplectic group), there exists $U\in Mp(n)$ ($Mp(n)$ is the metaplectic group) such that $$(a\circ \chi)^w=U^* a^w U.$$ There are many generalizations of that formula where $\chi$ is a canonical transformation not necessarily linear and the equality is replaced by some asymptotic equivalence. This result is as important as the change of variable formula in integrals.
• thank you very much for the historical references.By the way,what does Mp(n) mean ? In hormander's book,U is a unitary tansformation in $L^{2}(V)$,where $a \in \varphi'(W)$,and $W=V \oplus V'$. are these two expression essentially equivalent ? May 1, 2012 at 14:02
• The metaplectic group is a double cover of the symplectic group and it has a representation as a subgroup of the group of unitary transformations of $L^2(\mathbb R^n)$. May 2, 2012 at 13:35
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 $a$.