Wave Operators: Calculus

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: $$\eta\in\mathcal{B}(\mathbb{C}):\quad\Omega\eta(H_0)\subseteq\eta(H)\Omega$$

How can I prove this?

I assume you are looking at the step to the last equation from the previous? If so, have you looked at Fourier transforms for any class of functions $f$ as a starting point? — Disintegrating By Parts, Apr 11 '15 at 09:54
@T.A.E.: Yes, that step. I thought about it however I doubted this works since the measure is in general not the Lebesgue measure, or? — C-star-W-star, Apr 11 '15 at 10:12
If you start with some nice $f$ which can be written as a Fourier transform that converges classically pointwise and uniformly on finite intervals, you should be able to extend to that $f$ ... I think. Then you could bootstrap from that class of functions. — Disintegrating By Parts, Apr 11 '15 at 10:21
@T.A.E.: Sounds like real work. However, the paper I read says "We easily conclude". Am I maybe missing something? — C-star-W-star, Apr 11 '15 at 10:25
If $\int e^{itx}d\mu(t)=\int e^{itx}d\nu(t)$ for all $x \in \mathbb{R}$ for finite complex Borel measures $\mu$ and $\nu$, then ... . — Disintegrating By Parts, Apr 11 '15 at 10:32

score 0 · Accepted Answer · edited Apr 13 '17 at 12:20

Thanks really alot to T.A.E.!!!

Inclusion

They are bounded: $$\|\Omega\varphi\|=\lim_{t\to\infty}\|U(t)^*JU_0(t)\varphi\|\leq\|J\|\cdot\|\varphi\|$$

By intertwining relations: $$\int_{-\infty}^{+\infty}e^{-it\lambda}\mathrm{d}\langle E(\lambda)\Omega\varphi,\chi\rangle=\langle U(t)\Omega\varphi,\chi\rangle=\langle\Omega U_0(t)\varphi,\chi\rangle=\langle U_0(t)\varphi,\Omega^*\chi\rangle\\=\int_{-\infty}^{+\infty}e^{-it\lambda}\mathrm{d}\langle E_0(\lambda)\varphi,\Omega^*\chi\rangle=\int_{-\infty}^{+\infty}e^{-it\lambda}\mathrm{d}\langle\Omega E_0(\lambda)\varphi,\chi\rangle$$

By Fourier uniqueness:* $$\Omega E_0(A)=E(A)\Omega\quad(A\in\mathcal{B}(\mathbb{R}))$$

By measurable calculus:** $$\Omega\eta(H_0)\subseteq\eta(H)\Omega\quad(\eta\in\mathcal{B}(\mathbb{R}))$$

Concluding inclusion.

Strictness

Consider a Hamiltonian: $$H_0:\mathcal{D}(H_0)\to\mathcal{H}_0:\quad\mathcal{D}(H_0)\subsetneq\mathcal{H}_0$$

Then for trivial operator: $$J=0:\quad\mathcal{D}(\Omega H_0)=\mathcal{D}(H_0)\subsetneq\mathcal{H}_0=\mathcal{D}(0)=\mathcal{D}(H\Omega)$$

Concluding strictness.

*See the thread: Fourier Uniqueness

**See proof of: Reducibility

Wave Operators: Calculus

1 Answers1

Linked