Scattering channels, stationary resolvents and time delay — Examples

Since $\delta (E)$ is real, $S(E{)}^{\ast }S(E)=e^{-2i\delta (E)}{e}^{2i\delta (E)}=1$; thus this is a one-channel unitary matrix on all of $I$. Differentiation gives $S^{\mathrm{\prime }}(E)=2i{\delta }^{\mathrm{\prime }}(E)e^{2i\delta (E)}$.

Substitution into [F1] gives $Q(E)=-i\mathrm{\hslash }{e}^{-2i\delta (E)}2i{\delta }^{\mathrm{\prime }}(E)e^{2i\delta (E)}=2\mathrm{\hslash }{\delta }^{\mathrm{\prime }}(E)$, which is real. Since $\delta$ is dimensionless and $d\delta \mathrm{/}dE$ has units ${\mathrm{J}}^{-1}$, $Q$ has units $\mathrm{s}$. No pointwise value of $Q$ is interpreted as a measured time.

By [F6], the Borel sigma-algebra on $(0,1)$ is countably generated, and by [F7] its measure is finite, hence sigma-finite. [F8] gives separability of ${\mathcal{H}}_0$; a direct sum of two separable spaces is separable because pairs from the product of countable dense subsets are countable and dense, so $\mathcal{H}$ is separable. Since $0<x<1$, $\mathrm{\parallel }{H}_{0}f{\mathrm{\parallel }}^2={\int }_0^1{x}^2\mathrm{\mid }f{\mathrm{\mid }}^{2}d\beta \le \mathrm{\parallel }f{\mathrm{\parallel }}^2$, and $⟨H_{0}f,h⟩=⟨f,H_{0}h⟩$ for every $f,h\in {\mathcal{H}}_0$. Thus every $h$ belongs to $D(H_0^{\ast })$, with $H_0^{\ast }h=H_{0}h$, proving self-adjointness by [F2]. The same componentwise pairing on the full domain $\mathcal{H}$ proves $H^{\ast }=H$. Its PVM is the multiplication family in [F3], so bounded Borel functional calculus gives $e^{it{H}_0}=M_{e^{itx}}=U_0(t)$ and $e^{itH}=U(t)$, consistently with [F4].

By [F3], every scalar spectral measure of $H_0$ is absolutely continuous, so $P_{\mathrm{a}\mathrm{c}}(H_0)=I$. Thus the projection in [F5] drops out. Using $Jf=(f,0)$, the two-space wave-operator approximants are exactly $U(t)J{U}_0(-t)f=(e^{itx}{e}^{-itx}f,0)=(f,0)$ for every $f\in {\mathcal{H}}_0$ and every real $t$. Since $HJ=J{H}_0$ and both generators are bounded, direct differentiation gives $\frac{d}{dt}(e^{itH}J{e}^{-it{H}_0})=i{e}^{itH}(HJ-J{H}_0)e^{-it{H}_0}=0$. Thus the approximants are constant and both strong limits exist with ${\mathrm{\Omega }}_{\pm }f=(f,0)$. Their norm equals $\mathrm{\parallel }f\mathrm{\parallel }$, so they are isometries.

By [F3], every vector $f\oplus g\in \mathcal{H}$ has scalar spectral measure $B⟼{\int }_{B\cap (0,1)}(\mathrm{\mid }f(x){\mathrm{\mid }}^2+\mathrm{\mid }g(x){\mathrm{\mid }}^2)dx$, which is absolutely continuous with respect to Lebesgue measure. Therefore ${\mathcal{H}}_{\mathrm{a}\mathrm{c}}(H)=\mathcal{H}$.

The wave-operator range is ${\mathcal{H}}_0\oplus \{0\}$, which is a proper subspace of $\mathcal{H}$ because the constant function $1\in L^2((0,1))$ has norm one, so $0\oplus 1$ is a nonzero vector outside the range. Hence the wave operators are isometric but not complete, refuting the statement.