Applying propagator to Laplace eigenfunctions 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!

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).
