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 $(1-\triangle)^{-1}$.