QS 5: Moller Operators

$\newcommand{\h}{{\mathcal H}}
\newcommand{\molp}{\Omega^{(+)}}
\newcommand{\molm}{\Omega^{(-)}}
\newcommand{\molpm}{\Omega^{(\pm)}}$

Moller operators are the basic tools of scattering theory. They have the following properties almost obvious from definition.

1. $\molpm$ are independent of time.\\ This is because the limit $t\to\pm\infty$ remains the same even if the origin of $t$ is changed by a finite constant.
2. They preserve norm, as noted above : \begin{eqnarray*} (\psi(t),\psi(t))=(\phi(t),\phi(t))\qquad\mbox{for all }t,\end{eqnarray*}therefore, taking the limit $t\to\infty$,\begin{eqnarray*} (\molp\phi,\molp\phi)=(\phi,{\molp}^\dagger\molp\phi)=(\phi,\phi)\qquad\mbox{for all }\phi\in\h.\end{eqnarray*}With a similar argument for $\molm$, it follows that\begin{eqnarray}{\molpm}^\dagger\molpm =1. \end{eqnarray}
3. ${\molpm}^\dagger$ annihilates bound states :\begin{eqnarray}{\molpm}^\dagger\psi_{\rm bd}=0 \end{eqnarray}This can be seen as follows. Let $\phi$ be any free state and $\psi=\molp\phi$ the corrsponding the scattering state. As all scattering states are orthogonal to the bound states: $(\psi,\psi_{\rm bd})=0$, it follows that $(\phi,{\molp}^\dagger\psi_{\rm bd})=0$ for any $\phi\in \h$. Similarly for $\molm$.
4. They are not unitary in general. Actually, \begin{eqnarray}\molpm{\molpm}^\dagger=P_{\rm scatt}=1-P_{\rm bd}\end{eqnarray} where the operators $P_{\rm scatt}$ and $P_{\rm bd}$ are projection operators on the subspaces of scattering and bound states.
5. $U(t)\molpm =\molpm U_0(t)$We begin with, for any fixed $t$,\begin{eqnarray}U(t)\molpm &=& \lim_{s\to\mp\infty}U(t)U(s)^{-1}U_0(s)\nonumber\\&=& \lim_{s\to\mp\infty}U(t-s)U_0(s-t)U_0(t)\nonumber\\&=& \lim_{u\to\mp\infty}U(u)^{-1}U_0(u)U_0(t)\nonumber\\&=&\molpm U_0(t) \label{mol6}\end{eqnarray}Differentiating with $t$ and putting $t=0$ gives another useful result, \begin{eqnarray}H\molpm=\molpm H_0. \label{mol5}\end{eqnarray}Physically, this equation means that the energy spectrum of free particle states is contained in the spectrum of the total hamiltonian : if $\phi_E$ is an eigenstate of $H_0$ with energy $E$, $\molpm\phi_E$ is an eigenstate of $H$ with the same eigenvalue.