Associating a pseudo-differential operator to the symbol in the SG setting

Question

We all know that given a symbol $a(x,\xi) \in S^{\mu,\rho}(\mathbb{R}^n,\mathbb{R}^n)$, a pseudo-differential operator can be defined as
\begin{equation} Op(a)u(x)=(2\pi)^{-n}\int \int e^{i(x-x')\cdot \xi}a(x,\xi)u(x')dx'd\xi \end{equation} for $u$ in Schwartz class of test functions, $\mathcal{S}(\mathbb{R}^n)$. $Op(a)$ is Fourier inverse operator.

Now, for symbols in the class $S^{\mu,\mu',\rho, \rho'}(\mathbb{R}^{2n} \times \mathbb{R}^{2n})$, which are defined as fuctions $a(x,x',\xi, \xi')\in C^{\infty}(\mathbb{R}^{2n}_{x,x'}\times \mathbb{R}^{2n}_{\xi,\xi'})$ satisfying

\begin{equation} |D_x ^{\alpha}D_{x'} ^{\alpha'}D_\xi ^{\beta} D_{\xi'} ^{\beta'}a(x,x',\xi, \xi')|\leq c(1+|\xi|)^{\mu-|\beta|}(1+|\xi'|)^{\mu'-|\beta'|}(1+|x|)^{\rho-|\alpha|}(1+|x'|)^{\rho'-|\alpha'|} \end{equation} The associated pseudo-differential operator is defined as

\begin{equation} Op(a)u(x)=(2\pi)^{-2n}\int \int \int \int e^{-i(y\eta+y'\eta')}a(x,x+y,\eta,\eta')u(x+y+y')dydy'd\eta d\eta' \end{equation} My question is what is the intution or idea behind the above definition. In the first definition, it is clear that $Op(a)$ is Fourier inverse operator. But in the latter definition what meaning do we associate?

Reference : Boundary Value Problems and Singular Pseudo-Differential Operators by Bert-Wolfgang Schulze, page 143.

Answer 1

The point is that the term arises if you want to compose pseudo-differential operators (the theorem on p.144). The symbol space is just the space of products of SG-ψdo's. This kind of "quantization" comes from the representation $Op(a)Op(b) = Op(c)$, where $$c(x,\xi) = e^{i<D_\xi, D_y>} a(x,\xi) b(y,\eta)|_{y=x,\eta=\xi}.$$ (my $\xi$ and $y$ might be wrong and have to be replaced by $\eta$ and $x$). Two (in my opinion) easier references are Nicola-Rodino (they also have the SG-calculus) and Zworski (he has only semiclassical pseudos, but composition works formally the same).