Wave Operators: Reducibility

Question

Given Hilbert spaces $\mathcal{H}_0$ and $\mathcal{H}$.

Consider Hamiltonians: $$H_\#:\mathcal{D}(H_\#)\to\mathcal{H}_\#:\quad H_\#=H_\#^*$$

Denote their evolutions: $$U_\#(t)^*=U_\#(-t)=U_\#(t)^{-1}$$

Regard a bounded operator: $$J:\mathcal{H}_0\to\mathcal{H}:\quad\|J\|<\infty$$

Assume the limit: $$\Omega\varphi:=\lim_{t\to\infty}U(t)^*JU_0(t)\varphi\quad(\varphi\in\mathcal{H})$$

Then one has: $$1_{\mathcal{N}\Omega}H_0\subseteq H_01_{\mathcal{N}\Omega}\quad1_{\overline{\mathcal{R}\Omega}}H\subseteq H1_{\overline{\mathcal{R}\Omega}}$$

How can I prove this?

Answer 1

Meanwhile I got it...

Note that one has: $$\overline{\mathcal{R}\Omega}=\left(\mathcal{N}\Omega^*\right)^\perp\quad\left(\overline{\mathcal{R}\Omega}\right)^\perp=\mathcal{N}\Omega^*$$

By functional calculus: $$\varphi_0\in\mathcal{N}\Omega:\quad\Omega E_0(A)\varphi_0=E(A)\Omega\varphi_0=0\quad(A\in\mathcal{B}(\mathbb{R}))$$ $$\varphi\in\mathcal{N}\Omega^*:\quad\Omega^*E(A)\varphi=E_0(A)\Omega^*\varphi=0\quad(A\in\mathcal{B}(\mathbb{R}))$$

By selfadjointness: $$E_0(A)\mathcal{N}\Omega\subseteq\mathcal{N}\Omega\implies E_0(A)(\mathcal{N}\Omega)^\perp\subseteq(\mathcal{N}\Omega)^\perp$$ $$E(A)\mathcal{N}\Omega^*\subseteq\mathcal{N}\Omega^*\implies E(A)\left(\mathcal{N}\Omega^*\right)^\perp\subseteq\left(\mathcal{N}\Omega^*\right)^\perp$$

Concluding reducibility.*

*See the thread: Reducibility