1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – DisintegratingByParts Apr 11 '15 at 9:54
  • $\begingroup$ @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? $\endgroup$ – C-Star-W-Star Apr 11 '15 at 10:12
  • $\begingroup$ 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. $\endgroup$ – DisintegratingByParts Apr 11 '15 at 10:21
  • $\begingroup$ @T.A.E.: Sounds like real work. However, the paper I read says "We easily conclude". Am I maybe missing something? $\endgroup$ – C-Star-W-Star Apr 11 '15 at 10:25
  • $\begingroup$ 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 ... . $\endgroup$ – DisintegratingByParts Apr 11 '15 at 10:32
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.