I am trying to understand the Moller operator in quantum scattering theory. An important result concerning this operator is the following useful property that can be used to determine bound states.
The Møller operator $\Omega^{(+)}$ maps in-states that belong to the continuum spectrum of the free Hamiltonian $H_0$ into states that belong to the continuum spectrum of the exact Hamiltonian $H$. For potential scattering, $H_0$ has no bound states but $H$ may have them. If $\phi$ is a bound state of $H$ then $(\phi ,\Omega^{(+)} \psi^{(+)}) = 0$ for any in-state $\psi^{(+)}$.
I am trying to understand why this property is true. Here are my thoughts.
$$ \psi^{(\pm)}_\alpha = \Omega^{(\pm)}\phi_\alpha\;,\tag{1} $$ where $\Omega^{(\pm)}$ are Moller wave operators, as discussed further, for example, in Joachain's "Quantum Collision Theory" at chapter 14.
The function $$ \psi^{(\pm)}_\alpha $$ is an eigenfunction of the full Hamiltonian $H$, including the scattering potential $V$.
The function $$ \phi_\alpha $$ is an eigenfunction of the "unperturbed" free particle Hamiltonian $H_0 = H-V$.
The index $\alpha$ on each of the two above functions represents a collection of numbers defining the state of the system and we have $$ H_0 \phi_\alpha = E_\alpha \phi_\alpha\;. $$
We can also write Eq. (1) as: $$ \psi^{(\pm)}_\alpha = \lim_{\epsilon\to 0}\frac{\pm i\epsilon}{E_\alpha - H \pm i\epsilon}\phi_\alpha \tag{2}\;. $$ $$ =\phi_\alpha + \lim_{\epsilon\to 0}\frac{1}{E_\alpha - H \pm i\epsilon}V\phi_\alpha $$
The $S$ matrix is given by: $$ S_{\beta,\alpha} = \langle \phi_\beta|S|\phi_\alpha\rangle $$ $$ =\langle \phi_\beta|{\Omega^{(-)}}^{\dagger}\Omega^{(+)}|\phi_\alpha\rangle $$ $$ =\langle \psi^{-}_\beta|\psi^{+}_\alpha\rangle \;, $$ which provides a simple expression for the $S$ matrix in terms of the $\psi$ states.
After this I am stuck. How can I show the If $\phi$ is a bound state of $H$ then $(\phi ,\Omega^{(+)} \psi^{(+)}) = 0$ for any in-state $\psi^{(+)}$?
