# Wave Operators: Isometry

This thread is only Q&A.

Given Hilbert spaces $\mathcal{H}_0$ and $\mathcal{H}$.

Consider Hamiltonians: $$H_\#:\mathcal{D}H_\#\to\mathcal{H}_\#:\quad H_\#=H_\#^*$$

Denote their evolutions: $$U_\#(t):=e^{itH_\#}\in\mathcal{B}(\mathcal{H}_\#)$$

Regard an operator: $$J\in\mathcal{B}(\mathcal{H}_0,\mathcal{H})$$

Assume the limit: $$\Omega\varphi:=\lim_{t\to\infty}U(t)JU_0(t)\varphi\quad(\varphi\in\mathcal{H}_0)$$

Suppose one has: $$\{J^*J-1\}U_0(t)\varphi\stackrel{t\to\pm\infty}{\to}0\quad(\varphi\in\mathcal{H})$$

Then one obtains: $$\|\Omega\varphi\|=\|\varphi\|\quad(\varphi\in\mathcal{H})$$

How can I check this?

## 1 Answer

By unitarity one has: $$\|\Omega\varphi\|=\lim_{t\to\infty}\|JU_0(t)\varphi\|$$

It can be written as: $$\|JU_0(t)\varphi\|^2=\langle\{J^*J-1\}U_0(t)\varphi,U_0(t)\varphi\rangle+\|\varphi\|^2$$

But it is bounded: $$\|U_0(t)\varphi\|_{t\in\mathbb{R}}\equiv\|\varphi\|_{t\in\mathbb{R}}<\infty$$

So one obtains: $$\langle\{J^*J-1\}U_0(t)\varphi,U_0(t)\varphi\rangle\stackrel{t\to\infty}{\to}0$$

Concluding the assertion.