# Wave Operators: Hamiltonian

Reference

This is taken out of M. Reed and B. Simon, Scattering Theory.

Problem

Given a Hilbert space $\mathcal{H}$.

Consider a free Hamiltonian $H_0$ and a perturbed one $H$.

Introduce the wave operators: $$\Omega^\pm(H,H_0):=\mathrm{s-lim}_{\sigma\to\pm\infty}e^{i\sigma H}e^{-i\sigma H_0}P_\text{ac}(H_0)$$ (The existence being implicitely assumed!)

First, note that they're partial isometries: $$\mathcal{H}_0:=\mathcal{N}\Omega^\pm(H,H_0)^\perp$$ $$\mathcal{R}^\pm:=\mathcal{R}\Omega^\pm(H,H_0)$$ (Especially, they're continuous!)

Now, a straightforward calculation shows: $$e^{-itH}\Omega^\pm(H,H_0)=\Omega^\pm(H,H_0)e^{-itH_0}$$ So by Stone's theorem one has: $$\varphi\in\mathcal{D}(H_0):\quad \Omega^\pm(H,H_0)H_0\varphi=\Omega^\pm(H,H_0)\lim_{\tau\to0}\frac{1}{\tau}\left(e^{-i\tau H_0}-\mathbb{1}\right)\varphi\\=\lim_{\tau\to0}\frac{1}{\tau}\left(e^{-i\tau H_0}-\mathbb{1}\right)\Omega^\pm(H,H_0)\varphi=H\Omega^\pm(H,H_0)\varphi$$ which gives the inclusion: $$H\Omega^\pm(H,H_0)\supseteq\Omega^\pm(H,H_0)H_0$$ (Nothing more is stated according to Reed and Simon.)

But then it is stated that: $$H\restriction_{\mathcal{R}^\pm}\cong H_0\restriction_{\mathcal{H}_0}$$ How can this be possible?

Edit

Ok, so I figured out that this must be abuse of notation, only.

Next, this shows that the Møller operators map into the absolutely continuous subspaces: $$\mathcal{R}^\pm\subseteq\mathcal{H}_\text{ac}(H)$$ How does this follow from the restricted unitary equivalence?
(That seems trivial on first sight but involves a lot of technical subtleties due to proper inclusion...)

• ... corrected ! – C-Star-W-Star Nov 10 '14 at 19:30
• An isometry $U$ always satisfies $U^{\star}U=I$. And, it has the property that $UU^{\star}=I$ on $\mathcal{R}(U)=\mathcal{N}(U)^{\perp}$. And the operators $\Omega^{\pm}$ are isometries. – DisintegratingByParts Nov 10 '14 at 19:42
• Correction : The previous comment should have read $\mathcal{R}(U)=\mathcal{N}(U^{\star})^{\perp}$ – DisintegratingByParts Nov 10 '14 at 20:01
• Yep, right, but still the above only reads $H\restriction_{\mathcal{R}_0}\cong H_0\restriction_{\mathcal{D}_0\cap\mathcal{H}_0}$ for $\mathcal{D}_0:=\mathcal{D}(H_0)$ and $\mathcal{H}_0:=\mathcal{H}_\text{ac}(H_0)$ as well as $\mathcal{R}_0:=\Omega(\mathcal{D}_0\cap\mathcal{H}_0)$ and with the unitary $U:\mathcal{D}_0\cap\mathcal{H}_0\to\mathcal{R}_0:U\varphi:=\Omega\varphi$. – C-Star-W-Star Nov 11 '14 at 15:44
• My point is that I do have only a restricted version of $H\cong H_0$. – C-Star-W-Star Nov 11 '14 at 15:45

What you have are partial isometries $\Omega^{\pm}(H,H_{0})$ which are isometries on $\mathcal{H}_{0}$, with $$e^{-itH}\Omega^{\pm}(H,H_{0})=\Omega^{\pm}(H,H_{0})e^{-itH_{0}}.$$ Because of this, $e^{-itH}y$ is strongly differentiable for $y \in \mathcal{R}^{\pm}$ iff $e^{-itH_{0}}x$ is strongly differentiable for $x=\Omega^{\pm}(H,H_{0})^{\star}y$. However, $e^{-itH}y$ is strongly differentiable iff $y\in\mathcal{D}(H)$, and $e^{-itH_{0}}x$ is differentiable iff $x\in\mathcal{D}(H_{0})$. Therefore, $$\mathcal{D}(H)\cap\mathcal{R}^{\pm}=\Omega^{\pm}(H,H_{0})\left[\mathcal{D}(H_{0})\cap \mathcal{H}_{0}\right],$$ and the restriction of $H$ to $\mathcal{D}(H)\cap\mathcal{R}^{\pm}$ is unitarily equivalent to the restriction of $H_{0}$ to $\mathcal{D}(H)\cap\mathcal{H}_{0}$. And, really, by definition of restriction, you would say that the restriction of $H$ to $\mathcal{R}^{\pm}$ is unitarily equivalent to the restriction of $H_{0}$ to $\mathcal{H}_{0}$.
For the last part of your question, Let $x \in \mathcal{H}_{0}$, and let $y$ be the corresponding element of $\mathcal{R}^{\pm}$. Then, in terms of the spectral measures $E_{H}$ and $E_{H_{0}}$, $$\|E_{H}(\cdot)y\|^{2}=\|E_{H_{0}}(\cdot)x\|^{2}.$$ One measure is absolutely continuous with respect to Lebesgue measure iff the other is.
Added: It takes a little while to get used to a partial isometry. The map $U$ is isometric on $\mathcal{H}_{0}$ with range $\mathcal{R}^{\pm}$ and $U^{\star}$ is isometric on $\mathcal{R}^{\pm}$ with range $\mathcal{H}_{0}$, and the two are inverses on these spaces. The intertwining relations make sure that things stay in the right spaces. For example, if $x\in \mathcal{H}_{0}$, $$e^{-itH}Ux = Ue^{-itH_{0}}x$$ then the right side is in $\mathcal{R}(U)=\mathcal{R}^{\pm}$ which gives the isometric inverse: $$U^{\star}e^{-itH}Ux=e^{-itH_{0}}x,\;\;\; x\in\mathcal{H}_{0}.$$ Because of the isometric correspondence, $e^{-itH}Ux$ has a strong derivative at $0$ iff $e^{-itH_{0}}x$ has a strong derivative at $0$ for $x\in\mathcal{H}_{0}$, and $$U^{\star}HUx = H_{0}x,\;\;\; x\in\mathcal{D}(H_{0})\cap \mathcal{H}_{0},\\ Hy = UH_{0}U^{\star}y,\;\;\; y\in\mathcal{D}(H)\cap \mathcal{R}^{\pm}.$$ For the spectral theorem $H=\int \lambda dE$ and $H_{0}=\int \lambda dE_{0}$, and $$(e^{-itH}x,x)=(Ue^{-itH_{0}}U^{\star}x,x) \\ \int e^{-its}d\|E_{H}(s)x\|^{2}=\int e^{-its}d\|E_{H_{0}}(s)U^{\star}x\|^{2}, \;\;\; x \in \mathcal{R}^{\pm}.$$ The above holds for all $t \ge 0$ as well as $t < 0$ by conjugating. It follows that the two measures are the same. Consequently, $x \in \mathcal{R}^{\pm}$ is in the absolutely continuous spectral space of $H$.
• Ok so the relation $e^{-itH}\Omega=\Omega e^{-itH_0}$ gives additionally $\Omega^*e^{-itH}=e^{-itH_0}\Omega^*$ and therefore $\varphi\in\mathcal{D}_0\cap \mathcal{H}_{ac}\implies\Omega\varphi\in\mathcal{R}^\pm\cap\mathcal{D}$ as well as $\psi\in\mathcal{D}\cap\mathcal{R}^\pm\implies\Omega^*\psi\in\mathcal{D}_0\cap H_{ac}\implies\psi=\Omega\Omega^*\in\Omega(\mathcal{D}_0\cap\mathcal{H}_{ac})$. So far right? Moreover, the restricted Møller operators become unitary. Next how does the relation between the spectral measures evolve from the unitary equivalence? (I'm sorry - I never saw this.) – C-Star-W-Star Nov 11 '14 at 18:32