when a pseudo-differential operators to be compact? In the theory of Pseudo-differential operators,when a symbol $a(x,\xi)\in S^{0}$,then the operator $a(x,D)$ defined by$$a(x,D)u=\int{e^{ix\xi}a(x,\xi)\widehat{u}}d \xi$$ is $L^2$ bounded.$ $
  My question is when add what condition (as least as possible) to the symbol $a(x,\xi)$ can assure the operator $a(x,D)$ to be compact on $L^2$ ?Or is there a equivalent condition of this ?I guess some decayed assumption on $a(x,\xi)$ (about $\xi$) is necessary.but I'm not sure.some references about this are also appreciated
Added:There is a equivalent condition of a pseudo-differential operators to be compact.Assume that $g \leq g^{\sigma}$,that g is $\sigma$ temperate.and that m is $\sigma$,g temperate.then the operators $a^{w}(x.D)$with $a\in S(m,g)$are compact (bounded) in $L^{2}$ if and only if $m \to 0 $ at$\infty$ (m is bound).When we let g to be the metric $|dx|^{2}+|d \xi|^{2}/(1+|\xi|^2)$,and $m=(1+|\xi|^{2})^{\frac{\mu}{2}}.$then the class $S(m,g)$ become the usual$S^{\mu}$,Is it implying that when $\mu<0$,then $a^{w}(x.D)$ is compact in $L^{2}$ without further assuming the kernel to be compact supported ?
More precisely, considering the symbol $a(x,\xi)=V(x)(1+|\xi|^2)^{-1}$(it appears in the scattering problem of the schr\ddot{o}dinger operators),now we have the differential condition (rather than the integral condition)$\partial^{\alpha}{V} \leq (1+|x|)^{-\beta-|\alpha|}$,with $|\beta|>\frac{1}{2}$. Then does the operator $a(x,D)$ compact in $L^{2}$ ? 
I try to prove it by decomposing both the $\xi-space$ and $x-space$ with a partition of unity as the almost orthogonality method. But it seems without the use of the decay of $x-space$,the compactness wouldn't be obtained(just think about the case $(1-\triangle)^{-1}$.
 A: First a simple remark: in the formulation of the question $\mu$ should be replaced by $\mu/2$ to get 
$S(m,g)=S^\mu_{1,0}$. 
Next the "if and only if" is correct but misleading since it is a condition on all the class of operators with symbols in $S(m,g)$. It is of course possible that a given operator is compact without that condition: take for instance the resolvent of the 2D operator
$$
-∆_{x,y}+x^2y^2.
$$
It is compact in $L^2(\mathbb R^2)$. That example is due to Barry Simon and was extended to a more general setting by Charles Fefferman (Bull. AMS, 1983) in his discussion of Schrödinger equation with polynomial potentials.
A: The following sufficient condition for operators on $\mathbb{R}^n$ is well-known: If the kernel of $a(x,D)$ has compact support and $\mathop{sup}_{x} |a(x,\xi)| \to 0$ as $\xi \to \infty$,
then A extends to a compact operator on $L^2$.
A possible reference is M. Shubin, Pseudodifferential operators and spectral theory, Second Edition, Springer. 
It follows easily that on a compact manifold a classical pseudodifferential operator is compact if and only if it has negative order. 
