QS 8: $\Omega^{(+)}$ and $S$ in energy basis

$\newcommand{\kk}[1]{|#1\rangle}
\newcommand{\bb}[1]{\langle #1}
\newcommand{\dd}[1]{\delta_{#1}}
\newcommand{\molp}{\Omega^{(+)}}
$

The Lippmann-Schwinger equation can be written in the energy eigenvalue basis as follows.

Let $\kk{E\alpha}$ be eigenvectors of $H_0$ with normalization
\begin{eqnarray*} H_0\kk{E\alpha}=E\kk{E\alpha}\qquad
\bb{E\alpha}\kk{E'\alpha'}=\delta(E-E')\dd{\alpha\alpha'}  \end{eqnarray*}
where $\alpha$ are observables other than energy needed to from a complete
set of commuting observables.

Sandwich the Lippmann-Schwinger equation between these states
\begin{eqnarray}\bb{E\alpha}|\molp\kk{E'\alpha'}=\delta(E-E')\dd{\alpha\alpha'}
+\frac{\bb{E\alpha}|V\molp\kk{E'\alpha'}}{E'-E+i\epsilon} \end{eqnarray}
where we have interpreted the singular integral
\begin{eqnarray*} \int_{-\infty}^0\exp[i(E-E')t/\hbar]dt &=&
\lim_{\epsilon\to 0}\int_{-\infty}^0\exp[i(E-E'-i\epsilon)t/\hbar]dt\\
&=& \frac{i\hbar}{E'-E+i\epsilon}.\end{eqnarray*}
Note that the sign of $i\epsilon$ is determined by the requirement of the convergence of the integral.  Similarly the equation for the S-matrix can be written
\begin{eqnarray}\bb{E\alpha}|S\kk{E'\alpha'} &=& \delta(E-E')\dd{\alpha\alpha'}\nonumber\\
&-&2\pi i\delta(E-E')\bb{E\alpha}|V\molp\kk{E'\alpha'} \end{eqnarray}