Scattering Amplitude
Spinless Particle
we are dealing with quantum description of scattering.
- Elastic Scattering $ \rightarrow $ between two spinless, non-relativistic particles of masses m1 and m2. During the scattering process, the particles interact with one another. If the interaction is time independent, we can describe the two-particle system with stationary states.
- \begin{equation}\Psi\left(\vec{r}_{1}, \vec{r}_{2}, t\right)=\psi\left(\vec{r}_{1}, \vec{r}_{2}\right) e^{-i E_{T} t / n}\end{equation}
\begin{equation}\left[-\frac{\hbar^{2}}{2 m_{1}}
\vec{\nabla}_{1}^{2}-\frac{\hbar^{2}}{2 m_{2}}
\vec{\nabla}_{2}^{2}+\hat{V}\left(\vec{r}_{1}, \vec{r}_{2}\right)\right]
\psi\left(\vec{r}_{1}, \vec{r}_{2}\right)=E_{T} \psi\left(\vec{r}_{1},
\vec{r}_{2}\right) \end{equation}
defining $ r= \mid \vec{r_1}- \vec{r_2}\mid \implies V(r_1, r_2) =
V(r)$
and hence this makes the two body problem as to two decoupled eigenvalue
problems: one for the center of mass (CM), which moves like a free particle of
mass $M=m_{1}+m_{2}$ and which is of no concern to us here, and another for a
fictitious particle with a reduced mass $\mu=m_{1} m_{2}
/\left(m_{1}+m_{2}\right)$ which mores in the potential $\hat{V}(r)$
\begin{equation}\label{wave} -\frac{\hbar^{2}}{2 \mu} \vec{\nabla}^{2}
\psi(\vec{r})+\hat{V}(r) \psi(\vec{r})=E \psi(\vec{r})\end{equation}
The problem thus reduced to solve this equation. we need to connect
differential cross section in the CM frame can be obtained from an asymptotic
form of the solution of Differential equation. Its solutions can then be used
to calculate the probability per unit solid angle per unit time that the
particle $\mu$ is scattered into a solid angle $d \Omega$ in the direction
$(\theta, \varphi) ;$ this probability is given by the differential cross
section $d \sigma / d \Omega$. In quantum mechanics the incident particle is
described by means of a wave packet that interacts with the target. The
incident wave packet must be spatially large so that spreading during the
experiment is not appreciable. It must be large compared to the target's size
and yet small compared to the size of the Lab so that it does not overlap
simultaneously with the target and detector. After scattering, the wave
function consists of an unscattered part propagating in the forward direction
and a scattered part that propagates along some direction $(\theta, \varphi)$
We can view as representing the scattering of a particle of mass $\mu$
from a fixed scattering center that is described by $V(r),$ where $r$ is the
distance from the particle $\mu$ to the center of $V(r) .$ We assume that
$V(r)$ has a finite range $a$. Thus the interaction between the particle and
the potential occurs only in a limited region of space $r \leq a$, which is
called the range of $V(r),$ or the scattering region. Outside the range,
$r>a$, the potential vanishes, $V(r)=0 ;$ the eigenvalue problem then
becomes \[ \left(\nabla^{2}+k_{0}^{2}\right) \phi_{i n c}(\vec{r})=0 \] where
$k_{0}^{2}=2 \mu E / \hbar^{2} .$ In this case $\mu$ behaves as a free
particle before collision and hence can be described by a plane wave \[
\phi_{i n c}(\vec{r})=A e^{i \vec{k}_{0} \cdot \vec{r}} \] where $\vec{k}_{0}$
is the wave vector associated with the incident particle and $A$ is a
normalization factor. Thus, prior to the interaction with the target, the
particles of the incident beam are independent of each other; they move like
free particles, each with a momentum $\vec{p}=\hbar \vec{k}_{0}$
(a) Angle between the incident and scattered wave vectors $\vec{k}_{0}$ and $\vec{k}$.(b) Incident and scattered waves: the incident wave is a plane wave, $\phi_{i n c}(\vec{r})=A e^{i \vec{k}_0 \cdot \vec{r}},$ and the scattered wave, $\phi_{s c}(\vec{r})=A f(\theta, \varphi) \frac{e^{i \vec{k} \cdot \vec{r}}}{r},$ is an outgoing wave.
When the incident wave collides or interacts with the target, an outgoing wave
$\phi_{s c}(\vec{r})$ is scattered out. In the case of an isotropic
scattering, the scattered wave is spherically symmetric, having the form $e^{i
\vec{k} \cdot \vec{r}} / r .$ In general, however, the scattered wave is not
spherically symmetric; its amplitude depends on the direction $(\theta,
\varphi)$ along which it is detected and hence \begin{equation}\phi_{s
c}(\vec{r})=A f(\theta, \varphi) \frac{e^{i \vec{k} \cdot
\vec{r}}}{r}\end{equation} where $f(\theta, \varphi)$ is called the scattering
amplitude, $\vec{k}$ is the wave vector associated with the scattered
particle, and $\theta$ is the angle between $\vec{k}_{0}$ and $\vec{k}$ as
displayed in Figure 1. a. After the scattering has taken place (Figure $1
\mathrm{b}$ ), the total wave consists of a superposition of the incident
plane wave and the scattered wave: \begin{equation}\psi(\vec{r})=\phi_{i n
c}(\vec{r})+\phi_{s c}(\vec{r}) \simeq A\left[e^{i \vec{k}_{0} \cdot
\vec{r}}+f(\theta, \varphi) \frac{e^{i \vec{k} \cdot
\vec{r}}}{r}\right]\end{equation} We now need to determine $f(\theta,
\varphi)$ and $d \sigma / d \Omega .$
Scattering Amplitude and Differential Cross Section
In order to find the $ f(\theta , \varphi), $ we need to take into
account flux densities. \begin{equation}\begin{aligned} \vec{J}_{\text {inc}}
&=\frac{i \hbar}{2 \mu}\left(\phi_{\text {inc}} \vec{\nabla} \phi_{\text
{inc}}^{*}-\phi_{\text {inc}}^{*} \vec{\nabla} \phi_{\text {inc}}\right) \\
\overrightarrow{J_{s c}} &=\frac{i \hbar}{2 \mu}\left(\phi_{s c}
\vec{\nabla} \phi_{s c}^{*}-\phi_{s c}^{*} \vec{\nabla} \phi_{s c}\right)
\end{aligned}\end{equation}
Putting $ \phi_{i n c} \; \& \; \phi_{sc} $ and we will get ,
\begin{equation}J_{i n c}=|A|^{2} \frac{\hbar k_{0}}{\mu}, \quad J_{s
c}=|A|^{2} \frac{\hbar k}{\mu r^{2}}|f(\theta, \varphi)|^{2}\end{equation}
Now, we may recall that the number $d N(\theta, \varphi)$ of particles
scattered into an element of solid angle $d \Omega$ in the direction
$(\theta, \varphi)$ and passing through a surface element $d A=r^{2} d
\Omega$ per unit time
\begin{equation}d N(\theta, \varphi)=J_{S C} r^{2} d
\Omega\end{equation}
\begin{equation}\frac{d N}{d \Omega}=J_{s c} r^{2}=|A|^{2} \frac{\hbar
k}{\mu}|f(\theta, \varphi)|^{2}\end{equation}
Note: The number of particles scattered into an element of solid angle $d \Omega(d \Omega=\sin \theta d \theta d \varphi)$ : the differential cross section which is $d \sigma(\theta, \varphi) / d \Omega,$ is defined as the number of particles scattered into an element of solid angle $d \Omega$ in the direction $(\theta, \varphi)$ per unit time and incident flux: \[ \frac{d \sigma(\theta, \varphi)}{d \Omega}=\frac{1}{J_{i n c}} \frac{d N(\theta, \varphi)}{d \Omega} \] where $J_{i n c}$ is equal to the number of incident particles per area per unit time.
Now, Therefore we have \begin{equation}\frac{d \sigma}{d \Omega}=\frac{1}{J_{i n c}} \frac{d N}{d \Omega}=\frac{k}{k_{0}}|f(\theta, \varphi)|^{2}\end{equation} For elastic scattering $k_{0} = k $ hence reduces to \begin{equation} \boxed{\frac{d \sigma}{d \Omega}=|f(\theta, \varphi)|^{2}} \end{equation} Our job is to find the $ f(\theta, \varphi) $ and this will solve our problem. To do that we need to use the Green's Function.
Its really an amazing article
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