Given that $$\psi_{(x_0,\xi_0)}(x)=(\pi h)^\frac{-n}{4}e^{\frac{i}{h}(x-\frac{x_0}{2}).\xi_0}e^{-\frac{1}{2h}(x-x_0)^2}$$ Then the wavefront set $\text{WS}(\psi_{(x_0,\xi_0)})=\{(x_0,\xi_0)\}$
In the textbook of semi-classical analysis they give this example but I don't know why. Can somebodies help me to explain it? Thanks alot.
Let us call $\psi_{(x_0,\xi_0)}(x)=u_h(x).$
One can compute (see e.g. Zworski) that the semiclassical defect measure associated to $(u_h)$ is precisely $\delta_{(x_0,\xi_0)},$ and it is a fact that the support of a semiclassical defect measure assoociated to a sequence of normalized states is a subset of the semiclassical wavefront set, i.e. $$\{(x_0,\xi_0)\}=\operatorname{supp} \delta_{(x_0,\xi_0)}\subseteq \text{WF}_h(u_h).$$ To see that they're equal, suppose that $(x',\xi')\neq (x_0,\xi_0).$ To show that $(x',\xi')\notin \text{WF}_h(u_h),$ it suffices to show that there exist smooth cutoffs $\chi\equiv 1$ near $x'$ and $\psi\equiv 1$ near $\xi'$ for which $$\psi\mathcal{F}_h(\chi u_h)=\mathcal{O}(h^\infty).$$ Note that
$$\psi\mathcal{F}_h(\chi u_h)(\xi)=2^{-n/2}(\pi h)^{-3n/4}\psi(\xi)\int e^{\frac{i}{h}\left((x-x_0)\cdot \xi_0+\frac{i}{2}|x-x_0|^2-x\cdot\xi\right)}\chi(x)\, dx.$$ You can choose the cutoffs $\psi,\chi$ straightforwardly and apply the non-stationary phase lemma to conclude that the above is $\mathcal{O}(h^\infty).$
You can see the whole thing from the later work, but the fact on semiclassical defect measures tells you what you should be looking for (and it might be useful if you're familiar with them; if not, I'd recommend learning a bit about them).
EDIT: Here's the statement of non-stationary phase. Let $a\in C_c^\infty$, $\varphi\in C^\infty$ be real-valued. Define the oscillatory integral $$I_h=\int e^{i\varphi(x)/h} a(x)\, dx.$$ The non-stationary phase lemma states that if $\partial \varphi\neq 0$ on $\operatorname{supp} a,$ then $I_h=\mathcal{O}(h^\infty).$ Very important result!! It tells us that if our phase $\varphi$ has no critical points on the support of the amplitude $a$, then we have rapid decay.