$

\newcommand{\intp}{\int \frac{{\rm d}^3p}{2p^0}}

\newcommand{\intpp}{\int \frac{{\rm d}^3p'}{2{p'}^0}}

\newcommand{\intx}{\int{\rm d}^3{\rm r}}

\newcommand{\tp}{\otimes}

\newcommand{\tpp}{\tp\cdots\tp}

\newcommand{\kk}[1]{|#1\rangle}

\newcommand{\bb}[1]{\langle #1}

\newcommand{\dd}[1]{\delta_{#1}}

\newcommand{\ddd}[1]{\delta^3(#1)}

\newcommand{\vv}[1]{{\bf #1}}

\newcommand{\molp}{\Omega^{(+)}}

$

A beam of particles gets scattered by a fixed immovable target sitting at the origin and represented by a potential $V$. In this case total energy is conserved but momentum is not.

We consider the beam of N spin-less particles with initial state representing particles with sharp momentum around $\vv{k}$ and such that these particles pass near the origin where the scatterer is located.

We want to find out the number of scattered particles moving with momenta in a narrow range $\Delta$ around $\vv{k'}$.

This number is given by the formula given in the previous section, with the projection operator $\kk{\xi(t)}\bb{\xi(t)}|$ ocurring in $B(t)={\molp}^\dagger\kk{\xi(t)}\bb{\xi(t)}|V\molp$ replaced by

\begin{eqnarray*} \kk{\xi(t)}\bb{\xi(t)}|=\int_{\Delta}{\rm d}^3\vv{k'}

\kk{\vv{k'}}\bb{\vv{k'}}| \end{eqnarray*}

Therefore, we must calculate $\bb{\vv{k}}|B\kk{\vv{k}}$.

\begin{eqnarray*} \bb{\vv{k}}|B\kk{\vv{k}}=\int_{\Delta}{\rm d}^3\vv{k'}

\bb{\vv{k}}|{\molp}^\dagger\kk{\vv{k'}}\bb{\vv{k'}}|V\molp\kk{\vv{k}} \end{eqnarray*}

From the Lippmann-Schwinger equation in energy eigenstates basis, we get

\begin{eqnarray*} \bb{\vv{k'}}|\molp\kk{\vv{k}}=\frac{\bb{\vv{k'}}|V\molp\kk{\vv{k}}}

{E_{\vv{k}}-E_{\vv{k'}}+i\epsilon} \end{eqnarray*}

the term corresponding to identity being zero as $\bb{\vv{k'}}\kk{\vv{k}}=0$.

Substituting the complex conjugate of the matrix element of $\molp$ in the previous formula we get,

\begin{eqnarray*} \bb{\vv{k}}|B\kk{\vv{k}}=\int_{\Delta}{\rm d}^3\vv{k'}

\frac{|T(\vv{k'},\vv{k})|^2}{E_{\vv{k}}-E_{\vv{k'}}+i\epsilon} \end{eqnarray*}

where the transition amplitude $T$ is

\begin{eqnarray*} T(\vv{k'},\vv{k})\equiv\bb{\vv{k'}}|V\molp\kk{\vv{k}}\end{eqnarray*}

Using the well known formula

\begin{eqnarray}\frac{1}{x-i\epsilon}=P\left(\frac{1}{x}\right) +i\pi\delta(x) \end{eqnarray}

we get

\begin{eqnarray*} {\rm Im}\bb{\vv{k}}|B\kk{\vv{k}}=\pi\int_{\Delta}{\rm d}^3\vv{k'}

|T(\vv{k'},\vv{k})|^2\delta(E_{\vv{k}}-E_{\vv{k'}}) \end{eqnarray*}

We can further simplify the formula by writing ${\rm d}^3\vv{k'}=m|\vv{k'}|dE_{\vv{k'}}d\Omega_{\vv{k'}}$ and integerating over the Dirac delta function. Therefore

\begin{eqnarray*} n_\xi=(2\pi)^4\hbar^2\rho m|\vv{k}|\int_\Delta d\Omega_{\vv{k'}}

|T(\vv{k'},\vv{k})|^2 \end{eqnarray*}

where it is to be noted that in the transition amplitude now $|\vv{k}|=|\vv{k'}|$.

The number of particles scattering per unit time into the required final states is proportional to the {\bf flux} given by the product of density $\rho$ and velocity of particles $|\vv{k}|/m$. The ratio of $n_\xi$ to flux is a quantity of the dimensions of area called {\bf cross-section}.

The cross section is \begin{eqnarray}\sigma=(2\pi)^4\hbar^2 m^2\int_\Delta d\Omega_{\vv{k'}}

|T(\vv{k'},\vv{k})|^2 \end{eqnarray}

If the range $\Delta$ is very narrow then we have the formula for the {\bf differential cross-section}

\begin{eqnarray}d\sigma=(2\pi)^4\hbar^2 m^2|T(\vv{k'},\vv{k})|^2d\Omega_{\vv{k'}}

\end{eqnarray}

### Exclude node summary :

### Exclude node links:

Pankaj Sharan

Physics Department, Jamia Millia Islamia

New Delhi

{1981-2013}

- QS: Lecture 1
- QS: Lecture 2
- QS: Lecture 3
- QS: Lecture 4
- QS: Lecture 5
- QS: Lecture 6