$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\supp}{\operatorname{supp}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\mes}{\operatorname{mes}}$
Here we discuss idea of scattering. Basically there are two variants of the Scattering Theory--non-stationary and stationary. We start from the former but then fall to the latter. We assume that there is unperturbed operator $L_0$ and perturbed operator $L=L_0+V$ where $V$ is a perturbation. It is always assumed that $L_0$ has only continuous spectrum (more precisely--absolutely continuous) and the same is true for $L$ (otherwise our space $\mathsf{H}$ is decomposed into sum $\mathsf{H}=\mathsf{H}_{\mathsf{ac}}\oplus \mathsf{H}_{\mathsf{pp}}$ where $L$ acts on each of them and on $\mathsf{H}_{\mathsf{ac}}$, $\mathsf{H}_{\mathsf{pp}}$ it has absolutely continuous and pure point spectra respectively. Scattering happens only on the former.
Now consider $u=e^{itL}u_0$ be a solution of the perturbed non-stationary equation. In the reasonable assumptions it behaves as $t\to \pm \infty$ as solutions of the perturbed non-stationary equation: \begin{equation} \|e^{itL}u_0- e^{itL_0}u_\pm\|\to 0\qquad \text{as }\ t\to \pm \infty \label{eq-13.5.1} \end{equation} or, in other words the following limits exist \begin{equation} u_\pm=\lim_{t\to \pm \infty} e^{-itL_0}e^{itL}u_0. \label{eq-13.5.2} \end{equation} Then operators $W_\pm : u_0\to u_pm$ are called wave operators and under some restrictions they are proven to be unitary operators from $\mathsf{H}$ onto $\mathsf{H}_{\mathsf{ac}}$. Finally $S=W_+W_-^{-1}$ is called a scattering operator.
Despite theoretical transparency this construction is not very convenient and instead are considered some test solutions which however do not belong to space $\mathsf{H}_{\mathsf{ac}}$.
Let us consider on $\mathsf{H}=L^2(\mathbb{R})$ operators $L_0u:= -u_{xx}$ and $L=L_0+V(x)$. Potential $V$ is supposed to be fast decaying as $|x|\to \infty$ (or even compactly supported). Then consider a solution of \begin{equation} u_t =iLu= -iu_{xx} + V(x)u \label{eq-13.5.3} \end{equation} of the form $e^{ik^2t}v(x,k)$; then $v(x,k)$ solves \begin{equation} v_{xx}- V(x)v + k^2v=0 \label{eq-13.5.4} \end{equation} and it behaves as $a_{\pm} e^{ikx}+ b_{\pm} e^{-ikx}$ as $x\to \pm \infty$.
Consider solution which behaves exactly as $e^{ikx}$ as $x\to -\infty$: \begin{equation} v(k,x)= e^{ikx}+o(1)\qquad \text{as } x\to -\infty; \label{eq-13.5.5} \end{equation} then \begin{equation} v(k,x)= A(k)e^{ikx}+ B(k)e^{-ikx}+o(1)\qquad \text{as } x\to +\infty. \label{eq-13.5.6} \end{equation} Complex conjugate solution then \begin{align} &\bar{v}(k,x)= e^{-ikx}+o(1)\qquad \text{as } x\to -\infty, \label{eq-13.5.7}\\ &\bar{v}(k,x)= \bar{A}(k)e^{-ikx}+ \bar{B}(k)e^{ikx}+o(1)\qquad \text{as } x\to +\infty. \label{eq-13.5.8} \end{align} Their Wronskian $W(v,\bar{v})$ must be constant (which follows from equation to Wronskian from ODE) and since $W(v,\bar{v})= W(e^{ikx},e^{-ikx})+o(1)=-2ik+o(1)$ as $x\to -\infty$ and \begin{align} &W(v,\bar{v})= W(A(k)e^{ikx}+ B(k)e^{-ikx},\label{eq-13.5.9}\\ &\bar{A}(k)e^{-ikx}+ \bar{B}(k)e^{ikx})+o(1)= -2ik \bigl( |b(k)|^2-|a(k)|^2\bigr)+o(1) \label{eq-13.5.10} \end{align} as $x\to +\infty$ we conclude that \begin{equation} |A(k)|^2-|B(k)|^2 =1. \label{eq-13.5.11} \end{equation} We interpret it as the wave $A(k)e^{ikx}$ at $+\infty$ meets a potential and part of it $e^{ikx}$ passes to $-\infty$ and another part $B(k)e^{-ikx}$ reflects back to $+\infty$.
We observe that (\ref{eq-13.5.11}) means that the energy of the passed (refracted) and reflected waves together are equal to the energy of the original wave. We can observe that \begin{equation} A(-k)=\bar{A}(k), \qquad B(-k)=\bar{B}(k). \label{eq-13.5.12} \end{equation} Functions $A(k)$ and $B(k)$ are scattering coefficients and together with eigenvalues $-k_j^2$ \begin{equation} \phi_{j,xx} -V_j(x)\phi_j -k_j^2\phi_j=0, \qquad \phi_j\ne 0 \label{eq-13.5.13} \end{equation} they completely define potential $V$.
Consider $-\Delta$ as unperturbed operator and $-\Delta + V(x)$ as perturbed where $V(x)$ is smooth fast decaying at infinity potential. We ignore possible point spectrum (which in this case will be finite and discrete). Let us consider perturbed wave equation \begin{equation} u_{tt}-\Delta u +V(x)u=0; \label{eq-13.5.14} \end{equation} it is simler than Schrödinger equation. Let us consider its solution which behaves as $t\to -\infty$ as a plane wave \begin{equation} u\sim u_{-\infty} = e^{ik (\boldsymbol{\omega}\cdot \mathbf{x}-t)}\qquad \text{as } t\to -\infty. \label{eq-13.5.15} \end{equation} with $\boldsymbol{\omega}\in \mathbb{S}^2$ (that means $\boldsymbol{\omega}\in \mathbb{R}^3$ and $|\boldsymbol{\omega}|=1$), $k\ge 0$.
Theorem 1. If (\ref{eq-13.5.15}) holds then \begin{equation} u\sim u_{+\infty} = e^{ik (\boldsymbol{\omega}\cdot \mathbf{x}-t)}+ v(x) e^{-ikt}\qquad \text{as } t\to +\infty. \label{eq-13.5.16} \end{equation} where the second term in the right-hand expression is an outgoing spherical wave i.e. $v(x)$ satisfies Helmholtz equation (9.1.19) and Sommerfeld radiation conditions (9.1.20)--(9.1.21) and moreover \begin{equation} v(x)\sim a(\boldsymbol{\theta}, \boldsymbol{\omega}; k )|x|^{-1} e^{ik|x|} \qquad\text{as } x= r\boldsymbol{\theta}, r\to \infty, \boldsymbol{\theta}\in \mathbb{S}^2. \label{eq-13.5.17} \end{equation}
Sketch of Proof Observe that $(u-u_{-\infty})_{tt}-\Delta (u-u_{-\infty})= f:=-V u$ and $(u-u_{-\infty})\sim 0$ as $t\to -\infty$ and then applying Kirchhoff formula (9.1.21) with $0$ initial data at $t=-\infty$ we arrive to \begin{equation} u-u_{-\infty}= \frac{1}{4\pi} \iiint |x-y|^{-1} f(y, t-|x-y|)\,dy \label{eq-13.5.18} \end{equation} and one can prove easily (\ref{eq-13.5.17}) from this.
Definition 1. $a(\boldsymbol{\theta}, \boldsymbol{\omega}; k)$ is Scattering amplitude and operator $S(k):L^2 (\mathbb{S}^2)\to L^2 (\mathbb{S}^2)$, \begin{equation} (S(k)w)(\boldsymbol{\theta})= w(\boldsymbol{\theta})+ \iint _{\mathbb{S}^2} a(\boldsymbol{\theta}, \boldsymbol{\omega}; k) w(\boldsymbol{\omega})\, d\sigma (\boldsymbol{\omega}) \label{eq-13.5.19} \end{equation} is a scattering matrix.
It is known that
Theorem 2. Scattering matrix is a unitary operator for each $k$: \begin{equation} S^*(k)S(k)=S(k)S^*(k)=I. \label{eq-13.5.20} \end{equation}
Remark 1.