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Consider the Møller operator

$$ \Omega_+ = \lim_{t \rightarrow -\infty } e^{i H t } e^{- i H_0 t } , $$

Now, suppose a state $\psi $ is located far away from the potential $V = H- H_0$. I feel that $\Omega_+ \psi $ is close to $\psi $ in norm, i.e.,

$$ || \Omega_+ \psi - \psi || \rightarrow 0 . $$

To make it more precise, let us define the translation operator

$$ T(\vec{a}) \psi(x)= \psi(x- \vec{a}) .$$

Then, it is conjectured that

$$ \lim_{|\vec{a} | \rightarrow \infty} || \Omega_+ T(\vec{a}) \psi - T(\vec{a}) \psi || = 0 $$

for arbitrary $\psi \in \mathcal{H}$.

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    $\begingroup$ What is your question about this? $\endgroup$ – ACuriousMind Feb 17 '15 at 21:12
  • $\begingroup$ Is it right or wrong? $\endgroup$ – Jiang-min Zhang Feb 17 '15 at 21:15
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The moller operator take the wave function and send to $t=-\infty$ with the influence of potential $V$ and send back to $t=0$ without the influence of $V$. $$ \Omega_+ = \lim_{t \rightarrow -\infty } e^{i H t } e^{- i H_0 t } $$ If we make this with a far away wave function $\psi$, this wave function don't feel the potential $V$ at time $t=0$. But we don't have any assumption that this wave function can't feeling the potential in some time $t=-t_0$ (propagated with $H$). $$ \psi(t)=e^{-i H t }\psi $$ We don't know if this wave function came of some scatering process or if this wave function still there, far away. Your intuition only make sense if we guarantee that the wave function still far away of the potential for any time.

Now, if you make a translation in an usual wave function towards in far of the potential region and state that this is equivalent of moller operator, then your state is completely wrong.

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