Pole in reflection/transmission coefficient and bound states I was working on a scattering problem in a quantum mechanical system with Hamiltonian $$H_1=A^{\dagger}A=(-\partial_x+W(x))((\partial_x+W(x))).$$ One can show that a 'supersymmetric' partner to this Hamiltonian is given by $H_2=A A^{\dagger}$. The energy eigenfunctions of $H_1$ are mapped to eigenfunctions of $H_2$ (except for the groundstate): $\psi^{(2)}_n=A\psi^{(1)}_n$, leading to the aforementioned susy. 
Setting up the scattering problem, one can relate the reflection and transmission coefficients of the two systems in a very direct way. It turns out that: $$R_1(E)=\frac{ik_-+W_-}{-ik_-+W_-}R_2(E)$$ and $$T_1(E)=\frac{-ik_++W_+}{-ik_-+W-}T_2(E)$$ where $W_{\pm}=W(x\to\pm \infty)$ and $k_{\pm}=\sqrt{E-W^{2}_{\pm}}$. Hence, one concludes that for $W_-<0$ the reflection and transmission coefficients of system 1 have an additional pole for $k_-$ in the upper half plane ($k_-=-iW_-$). The reason we want $k_-$ to be in the upper half plane is because the wavefunction will be normalizable. Of course, imaginary $k_-$ signals the wavefunction of some confined particle/bound state.
However, it is not entirely clear to me why this bound state shows up as a pole in the reflection and transmission coefficients. Clearly, those coefficients are ill-defined concepts for  a bound state: we cannot send in a state at $-\infty$ with $E<W_-$ (or cannot detect a state at $+\infty$ with $E<W_+$), so one may expect something weird happening in $R(E),T(E)$ for those energies. Yet it is not obvious to me the bound state shows up as a pole in $R(E),T(E)$. 
My question then: is there a simple/intuitive proof of the fact that a bound state appears as a pole in the reflection/transmission coefficients?
 A: TL;DR: The reflection and transmission coefficients directly inherit poles of the Green's function, so they have all the information about bound states and resonances.


(1) The Free Retarded Green's function as Counter-Propagating Plane Waves
The retarded Green's function for a free particle in one dimension is given by
\begin{equation}
G_0(x, x'; E) = \frac{e^{ik|x-x'|}}{2ik}
\end{equation}
with $k=\sqrt{E}$, which satisfies
\begin{equation}
(\partial_x^2 + k^2)\,G_0(x,x';E) = \delta(x-x').
\end{equation}
Note that $G_0$ corresponds to counter-propagating plane waves generated by a unit source at $x'$. That is, for $x > x'$ ($x < x'$), $G_0$ is a right- (left-) moving plane wave $e^{ikx}$ ($e^{-ikx}$) multiplied by an amplitude $e^{-ikx'}/2ik$ ($e^{ikx'}/2ik$). The plane waves are depicted by the red and blue wavy arrows in Figure 1.


Figure 1. The full retarded Green's function when the unit source is located far to the left of the scattering region.

(2) The Retarded Green's Function in the Presence of a Scattering Potential
Next, suppose that a localized scattering potential is present around $x=0$ and that a unit source is located far to the left of the scattering region ($x'\to-\infty$). Then, the region around $x'$ is a free space, in which the full retarded Green's function $G(x,x';E)$ should satisfy the same equation as the free retarded Green's function $G_0$.
\begin{equation}
(\partial_x^2 + k^2)\,G(x,x';E) = \delta(x-x')\qquad(x,x' \to -\infty).
\end{equation}
Hence, we can see that just like $G_0$, $G$ is also in the form of two counter-propagating plane waves, except that now, the right-moving wave must be partially reflected by and partially transmitted through the scattering potential with some reflection and transmission coefficients $R(E)$ and $T(E)$. From this observation, we see that the following relations hold:
\begin{equation}
G(x, x';E) \  \to \ \frac{1}{2ik}\Big(e^{ik|x-x'|} + R(E)e^{-ik(x + x')}\Big)\qquad(x,x'\to-\infty),
\end{equation}
\begin{equation}
G(x, x';E) \  \to \ \frac{1}{2ik} T(E)e^{ik(x - x')}\qquad(x\to\infty,\,x'\to-\infty).
\end{equation}
(See Figure 1.)

(3) The Reflection and Transmission Coefficients
By reorganizing the above relations, we obtain the expressions for the reflection and transmission coefficients:
\begin{equation}
R(E) = \lim_{x,x' \to -\infty}2ik \,e^{ik(x+x')} \Big(G(x,x';E) - G_0(x,x';E)\Big),
\end{equation}
\begin{equation}
T(E) = \lim_{\substack{x\to\infty\\x'\to-\infty}} 2ik \, e^{-ik(x-x')}\,G(x, x';E).
\end{equation}
Thus, we have shown that all poles of $G(x, x';E)$, corresponding to bound states or resonances, are also present in $R(E)$ and $T(E)$.
A: I cannot speak to this specific case, but, in general, the poles of the reflection coefficient correspond to bound states. It is helpful to think about a similar classical problem. 
In electrodynamics, the bound states of a planar system (polaritons) are at the poles of the reflection coefficient. Consider the definition of reflectance.
$$R = |r|^2 = \left|\frac{E_r}{E_i}\right|^2 $$
The poles occur when $E_i \rightarrow 0 $, that is, there is no incident light. The bound state (polariton) is a self-sustained coupling of light with the material. It sustains itself without any incident field, thus it is bound to the material.
Similarly, replace "light" with "probability current" and "material" with "potential" and you can think of quantum mechanics.
