$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\supp}{\operatorname{supp}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\mes}{\operatorname{mes}}$

1. Introduction
2. One dimensional scattering
3. Three dimensional scattering

13.5. Continuous spectrum and scattering

Introduction

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 $\mathsf{H}_{\mathsf{ac}}$.

One dimensional scattering

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 $V$.

Three dimensional scattering

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.

1. The similar results are proven when the scatterer is an obstacle rather than potential, or both.
2. Determine scatterer from scattering amplitude is an important Inverse scattering problem.
3. In fact fast decaying at infinity potential means decaying faster than Coulomb potential; for the latter theory needs to be heavily modified.

$\Leftarrow$  $\Uparrow$  $\Rightarrow$