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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.

Hoang Skadi
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1 Answers1

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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.

cmk
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  • Thank you for your idea. It helps a lot. I am not quite familiar with the non-stationary phase lemma, so can you suggest some references or clarify it a little bit for me? Thank you. – Hoang Skadi Jul 13 '21 at 01:33
  • Sure, if you have Zworski's book, then it's Lemma 3.14. I'll edit my post to add it in. – cmk Jul 13 '21 at 01:45
  • Thank you for your answer. I have just obtained the Zworski's book as you suggested. To verify my understanding of your answer, so here, we have to specify the form of $\psi$ and $\chi$ first right? Also, why it is sufficient to show "that there exist smooth cutoffs ... $O(h^\infty)$". Is this based on some lemma or it is just common sense? Thank you. – Hoang Skadi Jul 13 '21 at 06:58
  • Yeah, we need to choose the cutoffs first. What I was doing was showing what $\psi\mathcal{F}_h(\chi u_h)$ looks like in general. From there, it's straightforward to choose the cutoffs so that you can apply non-stationary phase. As for your question on why what I wrote was sufficient, this is often taken as the definition of the semiclassical wavefront set. Other definitions are sometimes given, as well, but you can prove that they're all equivalent. – cmk Jul 13 '21 at 14:42