QS 4: Scattering States

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

Let $U(t)$ and $U_0(t)$ be the evolution operators corresponding to the total and free hamiltonians $H$ and $H_0$ respectively :
\begin{eqnarray} U(t)=\exp(-itH/\hbar) \qquad U_0(t)=\exp(-itH_0/\hbar). \end{eqnarray}
A {\bf scattering state} $\psi$ at time $t=0$ is such that its "ancient history" $\psi(t)\equiv U(t)\psi$ for large negative times coincides with that of {\em some}  free state $\phi(t)\equiv U_0(t)\phi$ :
\begin{eqnarray} \lim_{t\to -\infty} \|\psi(t)-\phi(t)\| =0 \end{eqnarray}
For systems in which we are interested, such a scattering state {\em also looks like a free state in remote future}. That is, there is a vector $\chi$ such that
\begin{eqnarray} \lim_{t\to \infty} \|\psi(t)-\chi(t)\| =0 \end{eqnarray}
where $\chi(t)=U_0(t)\chi$. The existence of these limits is equivalent to the limits of {\bf Moller wave operators} $\molpm$ defined as follows
\begin{eqnarray} \molp \equiv \lim_{t\to -\infty} U(t)^{-1}U_0(t) \end{eqnarray}
\begin{eqnarray} \molm \equiv \lim_{t\to +\infty} U(t)^{-1}U_0(t) \end{eqnarray}
For any time $t$ the scattering state $\psi(t)$ is obtained from the free state $\phi(t)$ from which it has evolved, by $\psi(t)=\molp \phi(t)$. Similarly the free state $\chi(t)$ into which it will eventually settle down in remote future is related as $\psi(t)=\molm \phi(t)$. Thus the system starting out with a free state $\phi(t)$ in remote past becomes $\chi(t)$ in remote future.  The two free states,  final and initial, are related by
\begin{eqnarray} \chi(t)= S \phi(t) \equiv {\molm}^\dagger \molp \phi(t) \end{eqnarray}
The operator $S$ is called the {\bf Scattering Matrix} ot {\bf S-matrix} and contains all information about scattering.

It is important to remind that although $U(t)^{-1}U_0(t)$ are  unitary for all $t$ their limits $\molpm$ may not be. The reason is, that although the domain of definition is all $\h$, that is, $\molpm$ are defined for all vectors $\phi\in\h$, their range is not. For an operator to be unitary the inverse should exist. But we know that for every scattering state there is a free state, there are, for example, bound states which do not go over to free states in remote past or future. Therefore the mappings $\molpm$ are not ``one-to-one and onto" that they would be if they were unitary.

It is equally important to note that $\molpm$ are {\em norm preserving } operators :
\begin{eqnarray} \|\molp \phi\| &=& \|\phi \|, \qquad  \mbox{and} \\
 \|\molm \phi\| &=& \|\phi \| \qquad  \forall \phi\in\h \end{eqnarray}
 because they are the limits of norm preserving operators.

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