Lippmann-Schwinger equation and $T$ expansion

Question

Lippmann-Schwinger equation, in operator form, is: $$ T=V+V\frac{1} {E-H_0+i \hbar \varepsilon} T=:V+V\Theta_0T, $$ where $H_{tot}=H_0+{V}$ is the hamiltonian ($H_0$ is the free particle hamiltonian and $V$ a potential).
We can also obtain, by recursion, the following expansion: $$ T= V+V\Theta_0V+V\Theta_0V\Theta_0V+V\Theta_0V\Theta_0V\Theta_0V+ \cdots $$ Gathering $V$: $$ T=V(1+\Theta_0V+(\Theta_0V)^2+(\Theta_0V)^3+\cdots)=V\sum_{k=0}^\infty (\Theta_0V)^k $$ so, if $\| \Theta_0V\|<1$, and if what I'm doing makes sense..\begin{equation} T=V\frac{1} {1-\Theta_0V}. \end{equation} My questions: was all this legitimate? I've never really seen the last expression in action, in scattering theory so far. One usually uses the $T$ expansion and leaves just the needed terms. Maybe the last form isn't so convenient? Is it correct? Also, is the convergence request $\| \Theta_0V\|<1$ always fulfilled?

Answer 1

Your final expression is correct except that $V$ should be in front since it does not commute with $Θ_0$ in general.

In potential scattering ${Θ_0}V$ is often a compact operator and for large positive imaginary part of $z=E+iℏε$ its norm becomes arbitrarily small so the series converges. In certain cases one can do things a little differently by considering ${V^{1/2}}{Θ_0}{V^{1/2}}$ instead of ${Θ_0}V$ which then has a limit as $Imz$ tends to the real axis (so-called Rollnik potentials).See Simon "Hamiltonians as quadratic forms" or the series "Methods of Modern Mathematical Physics" by Reed cand Simon.

The full expression for $T$ can easily be evaluated further for rank one potentials, $V=|f><f|$ where $f$ is an element of the underlying Hilbert space.