# About the definition of the Møller operator

The Møller operator is defined as $$\Omega_+ = \lim_{t\rightarrow -\infty} U^\dagger (t) U_0(t).$$ Does the operator $$\lim_{t\rightarrow -\infty} U_0^\dagger (t) U(t)$$ also make sense? Is it important?

• Anyway, this operator has a big defect. Its domain does not cover the full Hilbert space. The action on the bound state does not converge. – senator Nov 17 '14 at 11:42
• These operators are usually defined with the projection on the absolutely continuous spectrum, to avoid problems with bound states (as you can see in my answer). – yuggib Nov 17 '14 at 12:11
• The second operator is just the adjoint of the first. – Arnold Neumaier May 13 '15 at 15:57

The overall idea is the following. You have a state evolving with the full interacting theory $\Psi(t) = U_t\Psi$. If the theory admits an asymptotic description for $t\to -\infty$, there must be a state $\Psi_0$ evolving with the free theory $\Psi_{0}(t) = U_0(t)\Psi_0$ such that the two evolutions coincide'' at large time in the past: $$\lim_{t\to -\infty}||U_0(t)\Psi_0- U(t) \Psi|| =0\:.$$ Using unitarity of $U$ you can write, form the above equation, $$\lim_{t\to -\infty}||U(t)^\dagger U_0(t) \Psi_0 - \Psi||=0\:.\tag{1}$$ In other words $$\Psi = \Omega_- \Psi_0$$ (I prefer to define that operator as the Moller operator $\Omega_-$ instead of $\Omega_+$) where $$\Omega_- := \mbox{s -}\lim_{t\to -\infty} U(t)^\dagger U_0(t) \tag{2}$$ that s denoting the strong operator topology involved in (1), assuming that there is an interacting scattering state for every free state of the Hilbert space. Usually the image of $\Omega_-$ is not the full Hilbert space $\cal H$, but just a closed subspace ${\cal H}_-$ because, in particular, there may exist bound states which do not admit a scattering description.

However (1) is also equivalent to $$\lim_{t\to -\infty}||\Psi_0 - U_0(t)^\dagger U(t)\Psi||=0\:.\tag{3}$$ and this define the inverse of $\Omega_-$ associating interacting states $\Psi \in {\cal H}_-$ to free states $\Psi_0 \in \cal H$. So

$$(\Omega_-)^{-1} \Psi := \lim_{t\to -\infty} U_0(t)^\dagger U(t) \Psi\tag{4}\:.$$ with $\Psi \in {\cal H}_-$. (Here the inverse $(\Omega_-)^{-1}$ is the one of the isometric map $\Omega_- : {\cal H} \to {\cal H}_-$.) That is $$(\Omega_-)^{-1} := \mbox{s}-\lim_{t\to -\infty} U_0(t)^\dagger U(t) \tag{5}\:.$$ where $s$ again denotes the strong operator topology for a class of operators defined on the Hilbert space ${\cal H}_-$ and values in the Hilbert space ${\cal H}$. (${\cal H}_-$ is closed due to a well-known elementary result on partial isometries in a Hilbert space. Since it is closed in $\cal H$, ${\cal H}_-$ is a Hilbert space in its own right.)

REMARKS

(1) The fact that the limit (4) exists for $\Psi \in {\cal H}_-= Ran(\Omega_-)$ is evident from the fact that every sequence $\Psi_0 - U_0(t_n)^\dagger U(t_n) \Psi$ is Cauchy since the isometric sequence $U(t_n)^\dagger U_0(t_n) \Psi_0 - \Psi$ is Cauchy by hypotheses.

(2) The fact that the strong limit in (4) produces the inverse of $\Omega_-$ on ${\cal H}_-$ is equally obvious since $\Omega_-$ associates $\Psi_0 \in {\cal H}$ to a certain $\Psi \in {\cal H}_-$ and the limit operator $\mbox{s -}\lim_{t\to -\infty} U_0(t)^\dagger U(t)$ associated that $\Psi$ to the initial $\Psi_0$.

• Is it in general true that $(\Omega_{-})^{-1}=\Omega_{-}^{\dagger}$ or does the latter one have a different domain? the whole Hilbert space for instance. – L.R. Oct 16 '17 at 17:51
• Indeed the domains are in general different. $\Omega_-^\dagger$ is everywhere defined and $\Omega_-^{-1}$ is defined on $\cal H_-$ that may be smaller than $\cal H$. – Valter Moretti Oct 16 '17 at 18:08

Well, the general theory is that you define the generalized wave operators for any couple of self-adjoint operators $A$ and $B$ to be: $$\Omega^{\pm}=\mathrm{s-lim}_{t\to \mp\infty} e^{iAt}e^{-iBt}P_{ac}(B)\; ,$$ where $\mathrm{s-lim}$ stands for the limit in the strong operator topology, and $P_{ac}(B)$ is the projection on the absolutely continuous spectrum of $B$ (because discrete spectrum corresponds to bound states and these states are not expected to scatter, i.e. behave as free states in the limit).

The investigation on the existence of these operators and their range, often denoted by $\mathscr{H}_{in}=\mathrm{Ran} \Omega^+$ and $\mathscr{H}_{out}=\mathrm{Ran}\Omega^-$, are crucial for the understanding of quantum scattering. A very important notion (e.g. to prove unitarity of the S-matrix) is that of completeness:

Suppose $\Omega^{\pm}(A,B)$ exist. We say they are complete if and only if $$\mathrm{Ran}\Omega^+=\mathrm{Ran}\Omega^-=\mathrm{Ran}P_{ac}(A)\; .$$

The following proposition can be proved:

Suppose that $\Omega^\pm(A,B)$ exist. Then they are complete if and only if $\Omega^\pm(B,A)$ exist.

So you see that given $\Omega^{\pm}(A,B)$, the existence of $\Omega^\pm(B,A)$ is quite important, since it is sufficient to prove completeness.

For further information you can consult the reference book on scattering from a mathematical physics point of view, i.e. the third volume of Reed and Simon.

Yes it makes sense, and I guess you are asking for a clarification as to the physical properties of such limits. In ordinary 1D scattering the Moller operators $\widehat{\Omega}_{\pm}$ defined by the strong limits \begin{equation} \widehat{\Omega}_{\pm} = \lim_{t \rightarrow \mp \infty}e^{i \widehat{H}t/\hbar}e^{-i \widehat{H_{0}}t/\hbar} \end{equation} link the actual state $\psi$ to its asymptotic free motion reference states $\phi_{{in}}$ and $\phi_{out}$, such that \begin{eqnarray} \lim_{t \rightarrow -\infty}||\psi(t) - \phi_{in}(t)|| &=& 0 \\ \lim_{t \rightarrow +\infty}||\psi(t) - \phi_{out}(t)|| &=& 0 \end{eqnarray} The operator $\widehat{\Omega}_{+}$(respectively $\widehat{\Omega}_{-}$ ) provides the scattering state by acting on the incoming (resp. outgoing) asymptote $\phi_{in}$,(resp. $\phi_{out}$) \begin{equation} \widehat{\Omega}_{+/-} |\phi_{in/out}(t) > = |\psi(t)> \end{equation} for all $t$.

• No, this is not my question. My question is about another operator, something similar to the moller operator but not. – senator Nov 17 '14 at 10:48