Formulas for Schrödinger unitary groups of operators

Question

Let $\Omega$ an open set of $\mathbb{R}^n$. Consider the Hilbert space $X=L^{2}\left(\Omega\right)$ and the Schrödinger operator $A=i\Delta$ defined on the domain $D(A)=H^2(\Omega)$. Is there any explicit formula for the unitary group $U(t)$ generated by the operator $A$, in the case $\Omega$ bounded and in the case $\Omega=\mathbb{R}^n$? My intuition tells me that if $\Omega$ is bounded, then it must be something like $$U(t)=\sum_n e^{it\lambda_n}P_n,$$ with $P_n$ being projections. For $\Omega=\mathbb{R}^n$ it must be something like the Bochner integral $$U(t)=\int e^{it\lambda}P_{\lambda}d\lambda.$$ But I don't know if such formulas exist.

Answer 1

For a densely-defined selfadjoint operator $A$ on a Hilbert Space $X$, the spectral theorem gives you $$ Ax = \int_{-\infty}^{\infty} \lambda dE(\lambda)x, $$ where you can interpret the above through measure theory, or as a Riemann-Stieltjes integrable with respect to the non-decreasing selfadjoint projection-valued function $E(\lambda)$, where $E(-\infty)=0$ and $E(+\infty)=I$. This is John von Neumann's Spectral Theorem that he developed for Quantum Mechanics. In either case, $$ U(t)x=e^{itA}x = \int e^{it\lambda}dE(\lambda)x. $$ That's what you have here after you correctly define the operator domain to include boundary restrictions. The unconstrained $A=-\Delta$ on $\mathcal{D}(A)=H^{2}$ is not selfadjoint on a bounded region without restricting the domain to functions satisfying some type of homogeneous boundary condition. For example, $\mathcal{D}(A)=H^{2}\cap H^{1}_{0}$ is a common restriction. Von Neumann also described an abstract theory of defining conditions to do this.