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Let $\Omega \subseteq \mathbb{R}^n$ be a non-empty domain with piece-wise smooth boundary and consider a Dirichlet eigenfunction $\varphi$ on $\Omega$. That is, $\varphi : \Omega \to \mathbb{C}$ is a non-trivial solution to $\Delta \varphi = \lambda \varphi$ for some $\lambda \geq 0$ and $\varphi = 0$ on $\partial\Omega$.

Consider the propagator $U(t) = e^{-it\sqrt{-\Delta}}$. Is it true that $U(t)\varphi = \varphi$? This fact seems to be part of a proof I'm reading but I am not sure why this is true.

Note: It is known that $U(t)$ commutes with the Laplacian, hence $$ \Delta \left(U(t)\varphi\right) = U(t)\left(\Delta \varphi\right) = \lambda\left(U(t)\varphi\right). $$ That is, $U(t)\varphi$ is itself an eigenfunction.

I'm relatively new to this subject, so I might be misunderstanding what is needed in the proof. I appreciate any input!

cmk
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Quoka
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  • I'm still thinking about it. Are you sure that it's true? Could you give the place that you saw it? – cmk Apr 16 '20 at 00:36
  • It's not directly written in the paper, but I'm guessing that it's needed. On the other hand, there is a very similar statement in these notes. See the very first line in the proof of Lemma 2.2. I would be perfectly content if I could understand why it works there! – Quoka Apr 16 '20 at 00:55
  • Okay, I get it there. I can’t answer at the moment, but I will tomorrow. – cmk Apr 16 '20 at 01:39
  • That's awesome! I really appreciate you taking the time. – Quoka Apr 16 '20 at 02:00
  • You may be interested to look at sections 6.2 and 6.3 of http://canzani.web.unc.edu/files/2016/08/Laplacian.pdf. This deals with the heat propagator on compact manifolds in clear detail. – cmk Apr 16 '20 at 13:37

1 Answers1

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My answer will be related to the link. I'll provide the necessary context.

Here, $\Delta_g u_j=\lambda_j^2 u_j,$ $p=\frac{1}{2}\left(|\xi|_g-1\right)+\mathcal{O}(h),$ $P:=\text{Op}_h(p)=\frac{1}{2}\left(h^2\Delta_g^2-1\right),$ and $U(t;h)=\exp (-itP/h).$ Note that if we let $h_j=1/\lambda_j,$ then $\text{Op}_{h_j}(p)u_j=0,$ and we can compute directly that $$\partial_t (U(t;h_j)u_j-u_j)=0.$$ So, the difference is constant in time. Evaluating at $t=0$ yields that the constant is zero, so that $U(t;h_j)u_j=u_j.$ It was very important that $h_j=\lambda_j^{-1}.$ For example, if $P(h)=-\Delta_g+V(x)$ with eigenfunctions $$P(h)u_j(h)=E_j(h)u_j(h)$$ normalized in $L^2,$ then $$U(t;h)u_j(h)=\exp \left(\frac{-itE_j}{h}\right) u_j(h).$$

Also, note that this is using the Schrodinger propagator, not the wave propagator (like in the original).

cmk
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  • I can write out more of the computation of the derivative if you'd like. – cmk Apr 16 '20 at 02:21
  • Thank you! To get that $\partial_t\left(U(t; h_j)u_j\right) = 0$, I used the fact that $h_jD_t U(t; h_j) + U(t; h_j) P(h_j) = 0$ where $P(h_j) = \frac{1}{2}\left(h_j^2\Delta - 1\right)$. Is that right? – Quoka Apr 16 '20 at 02:46
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    Indeed! $ \ \ $ – cmk Apr 16 '20 at 13:21