Compact-support short-range scattering closure

Assume the Axiom of Choice and Countable Choice. Let $V\in L^{\mathrm{\infty }}({\mathbb{R}}^3;\mathbb{R})$ and suppose there is a compact set $K\subset {\mathbb{R}}^3$ such that $V=0$ almost everywhere outside $K$. Let $\mathcal{H}=L^2({\mathbb{R}}^3;\mathbb{C})$, and set $H_0=-\mathrm{\Delta }$ and $H=H_0+M_V$ on $D(H_0)=D(H)=H^2({\mathbb{R}}^3;\mathbb{C})$. Use

$\widehat{f}(p)=(2\pi {)}^{-3\mathrm{/}2}{\int }_{{\mathbb{R}}^3}{e}^{-ip\cdot x}f(x)dx,{\mathrm{\Omega }}_{\pm }={\mathrm{s-lim}}_{t\to \pm \mathrm{\infty }}{e}^{itH}{e}^{-it{H}_0},S={\mathrm{\Omega }}_{+}^{\ast }{\mathrm{\Omega }}_{-}\mathrm{.}$

Here $M_{V}f=Vf$, and the strong limits act on every $f\in \mathcal{H}$. Then:

(a) The wave operators exist, are isometries, and satisfy

$\mathrm{ran}{\mathrm{\Omega }}_{+}=\mathrm{ran}{\mathrm{\Omega }}_{-}={\mathcal{H}}_{\mathrm{a}\mathrm{c}}(H)={\mathcal{H}}_{\mathrm{c}}(H),{\mathcal{H}}_{\mathrm{s}\mathrm{c}}(H)=\{0\}\mathrm{.}$

For $u\in \mathcal{H}$, write ${\mu }_u(B)=\mathrm{\parallel }{E}_H(B)u{\mathrm{\parallel }}^2$. The subspace ${\mathcal{H}}_{\mathrm{a}\mathrm{c}}(H)$ consists of vectors with ${\mu }_u\ll dE$; ${\mathcal{H}}_{\mathrm{s}\mathrm{c}}(H)$ is the subspace with atomless scalar spectral measures singular with respect to Lebesgue measure; ${\mathcal{H}}_{\mathrm{p}\mathrm{p}}(H)$ is the closed span of eigenspaces, and ${\mathcal{H}}_{\mathrm{c}}(H)={\mathcal{H}}_{\mathrm{p}\mathrm{p}}(H{)}^{\perp }$. These are the canonical spectral-type subspaces of the PVM.

(b) The scattering operator $S$ is unitary on $\mathcal{H}$ and commutes with $e^{-it{H}_0}$ for every $t\in \mathbb{R}$.

Use time in the corresponding $\mathrm{\hslash }=1$ evolution convention and the standard scaled Schrödinger normalization in which the kinetic coefficient ${\mathrm{\hslash }}^2\mathrm{/}(2m)$ is set to $1$. When length is dimensional, $E$ and $V$ then have units $L^{-2}$ and $k=\sqrt{E}$ has units $L^{-1}$; the proof is an operator deduction in this normalization.

(c) Put $\mathfrak{h}=L^2(S^2,d\sigma )$, where $d\sigma (S^2)=4\pi$, and define the normalized free energy transform

$(F_{0}f)(E,\omega )=\frac{E^{1\mathrm{/}4}}{\sqrt{2}}\widehat{f}(\sqrt{E}\omega ),F_0:\mathcal{H}\to L^2((0,\mathrm{\infty }),dE;\mathfrak{h})\mathrm{.}$

Fix a bounded ball $B$ containing $K$. For $E>0$, $k=\sqrt{E}$, define bounded maps $Z_0(E):L^2(B)\to \mathfrak{h}$ and $Z_0(E{)}^{\ast }:\mathfrak{h}\to L^2(B)$ by

$(Z_0(E)f)(\omega )=\frac{\sqrt{k}}{4{\pi }^{3\mathrm{/}2}}{\int }_B{e}^{-ik\omega \cdot x}f(x)dx,(Z_0(E{)}^{\ast }a)(x)=\frac{\sqrt{k}}{4{\pi }^{3\mathrm{/}2}}{\int }_{S^2}{e}^{ik\omega \cdot x}a(\omega )d\sigma (\omega )\mathrm{.}$

Let $R_H(E+i0):L^2(B)\to L^2(B)$ be the operator-norm limit ${\lim }_{ϵ\downarrow 0}{\mathbf{1}}_B(H-(E+iϵ)I{)}^{-1}{\mathbf{1}}_B$; existence and continuity for $E>0$ are part of the stated supplier input. The products below are these bounded maps on $B$. Then $F_0$ is unitary and there is a measurable unitary field $S_{\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{t}}(E)\in \mathcal{B}(\mathfrak{h})$ such that

$F_{0}S{F}_0^{\ast }=M_{S_{\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{t}}(E)},S_{\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{t}}(E)=I_{\mathfrak{h}}-2\pi i{Z}_0(E)[V-V{R}_H(E+i0)V]Z_0(E{)}^{\ast },E>0.$

The formula uses only positive energies; it makes no claim at the threshold $E=0$, and no long-range Coulomb potential is included.

The Axiom of Choice

The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)

Wave, scattering and channel operators

Real multiplication operators and free plus bounded-potential Schrödinger operators

Compact-support asymptotic completeness

Stationary and dynamical scattering in the full energy-angular representation

Canonical decomposition into pure point, absolutely continuous and singular continuous parts