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


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?


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...)

  • $\begingroup$ ... corrected ! $\endgroup$ – C-Star-W-Star Nov 10 '14 at 19:30
  • $\begingroup$ 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. $\endgroup$ – DisintegratingByParts Nov 10 '14 at 19:42
  • $\begingroup$ Correction : The previous comment should have read $\mathcal{R}(U)=\mathcal{N}(U^{\star})^{\perp}$ $\endgroup$ – DisintegratingByParts Nov 10 '14 at 20:01
  • $\begingroup$ 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 $U:\mathcal{D}_0\cap\mathcal{H}_0\to\mathcal{R}_0:U\varphi:=\Omega\varphi$. $\endgroup$ – C-Star-W-Star Nov 11 '14 at 15:44
  • $\begingroup$ My point is that I do have only a restricted version of $H\cong H_0$. $\endgroup$ – 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 $H$.

  • $\begingroup$ Ok let me think. $\endgroup$ – C-Star-W-Star Nov 11 '14 at 18:04
  • $\begingroup$ 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.) $\endgroup$ – C-Star-W-Star Nov 11 '14 at 18:32
  • $\begingroup$ I already gave you (+1) for your effort!! $\endgroup$ – C-Star-W-Star Nov 11 '14 at 18:36
  • $\begingroup$ @Freeze_S : I've added more detail for you. $\endgroup$ – DisintegratingByParts Nov 11 '14 at 21:04
  • $\begingroup$ So you say the only measure that gives trivial integral for all exponential functions is the trivial measure? Ok but then one didn't need the unitary equivalence of the Hamiltonians, don't you? (Simon and Reed took that, however, somehow as argument.) $\endgroup$ – C-Star-W-Star Nov 11 '14 at 23:25

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