# Differential Cross Section and the Green's Function Formulation of Scattering

Much of what we know about forces and interactions in atoms and nuclei has been learned from scattering experiments, in which say atoms in the target are bombarded with beams of particles. These particles are scattered by the target atoms and then detected as a function of a scattering angle and energy. From a theoretical point of view, we are now concerned with the continuous part of the energy spectrum. We are free to choose the value of the incident particle energy and by a proper choice of the zero of energy, this corresponds to $E>0\!$ and to eigenfunctions of the unbound states. Before, when we were studying bound states, our focus was on the discrete energy eigenvalues, which allows a direct comparison of theory and experiments. In the continuous part of the spectrum, which comes into play in scattering, the energy is given by the incident beam, and intensities are the object of measurement and prediction. The intensity measures of the likelihood of finding a particle traveling in a given direction, and is related to the eigenfunctions, rather than eigenvalues. Relating observed intensities to calculated wave functions is the first problem in scattering theory.

Figure 1: Collimated homogeneous beam of monoenergetic particles, long wavepacket which is approximately a planewave, but strictly does not extend to infinity in all directions, is incident on a target and subsequently scattered into the detector subtending a solid angle $d\Omega\!$. The detector is assumed to be far away from the scattering center.

If $I_0\!$ is the number of particles incident from the left per unit area per unit time, $I(\theta,\phi)\,d\Omega\!$ the number of those scattered into the cone with solid angle $d\Omega\!$ per unit time, and if the density of particles in the incident beam is so small that we can neglect the interaction of the particles with each other and consider their collisions to be independent events then these two quantities are proportional to each other. With these considerations the differential cross section is defined as:

$\frac{d\sigma}{d\Omega}=\frac{I(\theta,\phi)}{I_0}$

There exist two different types of scattering; elastic scattering, in which the incident energy is equal to the detected energy and inelastic scattering that arises from, say, lattice vibrations within the sample. For inelastic scattering, one would need to tune the detector to detect particles of energy $E-dE,\!$ where $dE\!$ is the energy loss due to lattice vibrations. For simplicity, we will only discuss elastic scattering.

$\left(-\frac{\hbar^2}{2m}\nabla^2+V(\mathbf r)\right)\psi(\mathbf r)=E\psi(\mathbf r),$

or

$(\nabla^2+k^2)\psi(\mathbf r)=\frac{2mV(\mathbf r)}{\hbar^2}\psi(\mathbf r),$

where $E=\frac{\hbar^2k^2}{2m}$ and $V(\mathbf r)\!$ will be assumed to be finite in a limited region of space $r. This is called the range of the force; e.g., for nuclear forces, $d\sim 10^{-15}\,\text{m}\!$ and, for atomic forces, $d\sim10^{-10}\,\text{m}.\!$ Outside this range of forces, the particles move essentially freely. Our problem consists in finding those solutions of the above differential equation that can be written as a superposition of an incoming wave and an outgoing, scattered, wave. We found such solutions by first rewriting the Schrödinger equation as an integral equation:

$\psi_k(\mathbf r)=\psi_k^{(0)}( \mathbf r ) +\int d^3\mathbf{r}'\,G_k(\mathbf{r},\mathbf{r}')\frac{2m}{\hbar^2}V(\mathbf r')\psi_k(\mathbf r')$

where the Green's function $G_k(\mathbf{r},\mathbf{r}')$ satisfies

$(\nabla^2+k^2)G_k(\mathbf{r},\mathbf{r}')=\delta(\mathbf{r}-\mathbf{r}')$

and

$(\nabla^2+k^2)\psi^{(0)}_k(\mathbf{r})=0.$

The solution $\psi^0_k(\mathbf{r})$ is chosen such that the second term in the above wave function corresponds to an outgoing wave. The Green's function can then be written as

$G_k(\mathbf{r},\mathbf{r}')=-\frac{1}{4\pi}\frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|}.$

The full wave function solution then becomes

$\psi_{k}(\mathbf{r})=\psi_{k}^{(0)}(\mathbf{r})-\frac{m}{2\pi \hbar ^{2}}\int d^3\mathbf{r}'\,\frac{e^{ik\left|\mathbf{r}-\mathbf{r}' \right |}}{\left |{\mathbf{r-r'}} \right |}V(\mathbf{r'})\psi _{k}(\mathbf{r'}),$

where the first term represents the incident plane wave and the second term represents the scattered wave.

The detector is located far away from the scattering potential and we need to discuss the asymptotic behaviour in the limit of $r\to\infty.\!$. In this limit,

$k \left| \mathbf{r}-\mathbf{r}'\right| \sim kr - k\mathbf{r}' \cdot \mathbf{\hat{r}} = kr - \mathbf{k}' \cdot \mathbf{r}',$

where the wave vector, $\mathbf{k}' = k \mathbf{\hat{r}}$, is seen far from the scattering potential. We can therefore write

\begin{align} \lim_{r \to \infty} \psi_k(\mathbf{r}) &= \psi_k^{(0)}(\mathbf{r}) -\frac{m}{2\pi\hbar^2 }\int d^3\mathbf{r}'\,e^ {-i\mathbf{k}'\cdot\mathbf{r'}}V(\mathbf{r'})\psi_k(\mathbf{r'})\frac{e^{ikr}}{r} \\ &= \psi_k^{(0)}(\mathbf{r})+f_k(\theta,\phi)\frac{e^{ikr}}{r}, \end{align}

where the scattering amplitude $f_k(\theta,\phi)\!$ is given by

$f_k(\theta,\phi)=-\frac{m}{2\pi\hbar^2 }\int d^3\mathbf{r}'\,e^{-i\mathbf{k}'\cdot\mathbf{r'}}V(\mathbf{r'})\psi_k(\mathbf{r'})$

and the angles $\theta\!$ and $\phi\!$ are the angles between $\mathbf{\hat{r}},\!$ the vector defining the direction of the outgoing particle, and $\mathbf{k},\!$ the wave vector of the incoming wave.

Now the differential cross section is written through the ratio of the outgoing radial current density $j_r\!$ and the incident current density $j_{inc}\!$ as

$\frac{d\sigma}{d\Omega}=\frac{j_r r^2}{j_{inc}}$

The radial current density is given by

$j_r=\frac{\hbar}{2mi}\left(\psi_{sc}^\ast\frac{\partial \psi_{sc}}{\partial r}-\psi_{sc}\frac{\partial \psi_{sc}^\ast}{\partial r}\right)=\frac{\hbar k}{mr^2}|f_k(\theta,\phi)|^2,$

so that

$\frac{d\sigma}{d\Omega}=|f_k(\theta,\phi)|^2.$

## The Born Approximation

For a small $V(\mathbf{r}),$ the full wave function $\psi_k(\mathbf{r})$ can be obtained by substituting $\psi_k(\mathbf{r})$ into the right-hand side of the integral equation iteratively, obtaining

\begin{align} \psi_k(\mathbf{r}) =\psi_k^{(0)}(\mathbf{r}) + \left(\frac{2m}{\hbar^2}\right) \int d^3\mathbf{r}'\,G_k(\mathbf{r},\mathbf{r}')V(\mathbf{r'})\psi_k^{(0)}(\mathbf{r'}) \\ + \left(\frac{2m}{\hbar^2}\right)^{2} \int d^3\mathbf{r}'\,\int d^3\mathbf{r}''\,G_k(\mathbf{r},\mathbf{r}') & V(\mathbf{r'}) G_k(\mathbf{r}',\mathbf{r}'')V(\mathbf{r''}) \psi_k^{(0)}(\mathbf{r''})+\ldots \end{align}

In what is known as the first Born approximation, we keep only the first-order term in $V(\mathbf{r}).$ For large $r\!$ we have

$\psi_k(\mathbf{r})\approx\psi_k^{(0)}(\mathbf{r}) -\frac{m}{2\pi\hbar^2 }\int d^3\mathbf{r}'\,e^{-i\mathbf{k}'\cdot\mathbf{r'}}V(\mathbf{r'})\psi_k^{(0)}(\mathbf{r'})\frac{e^{ikr}}{r}$

and

$f_k(\theta,\phi)=-\frac{m}{2\pi\hbar^2 }\int d^3\mathbf{r}'\,e^{-i\mathbf{k}'\cdot\mathbf{r'}}V(\mathbf{r'})e^{i\mathbf{k}\cdot \mathbf r'}=-\left(\frac{m}{2\pi\hbar^2}\right)\langle\mathbf k_{sc}|V|\mathbf k_{inc}\rangle$

For a central potential $V(\mathbf{r}) = V(r),$ the Born scattering amplitude reduces to

$f_k(\theta) = -\frac{m}{2 \pi \hbar^2}\int d^3\mathbf{r}'\,V(r') e^{-i\mathbf{q}\cdot\mathbf{r}'},$

where $\mathbf{q} = \mathbf{k}' - \mathbf{k}$

is known as the momentum transfer (in units of $\hbar$). Keep in mind that $\mathbf{k}'$ is the wave vector pointing in the incident direction and $\mathbf{k}$ is in the scattered direction.

Note that all the angular dependence is carried by q. The integral over the solid angle is thus easily carried out and yields the following result:

$f_\text{Born}(\theta) = -\frac{2m}{\hbar^2} \int_0^\infin dr'\,V(r') \frac{\sin(qr')}{qr'} {r'}^2$

Here we have denoted the scattering angle between $\mathbf{k}$ and $\mathbf{k'}$ by θ, and note that k' = k for elastic scattering, so that

$q = 2k \sin\left(\frac{\theta}{2}\right).$

As an example, consider the screened Coulomb, or Yukawa, potential,

$V(r) = V_0 \frac{e^{-\alpha r}}{\alpha r}.$

In the Born approximation, we find that

\begin{align} f_\text{Born}(\theta) &= -\frac{2m}{\hbar^2} V_0 \int_0^\infin dr' {r'}^2 \frac{e^{-\alpha r}}{\alpha r'} \frac{\sin(q r')}{qr'} \\ &= -\frac{2m}{\alpha \hbar^2} V_0 \frac{1}{q^2 + \alpha^2} \\ &= -\frac{2mV_0}{\hbar^2 \alpha} \frac{1}{4k^2 \sin^2(\theta/2) + \alpha^2} \end{align}

The differential scattering cross section is obtained simply by taking the square of this amplitude:

$\frac{d \sigma}{d \Omega} = \left (\frac{2mV_0}{\hbar^2 \alpha}\right )^2\frac{1}{[4k^2 \sin^2(\theta/2) + \alpha^2]^2}$

The Coulomb potential between two charges $q_1\!$ and $q_2\!$ is a limiting case of the potential that we just considered for $\alpha \rightarrow 0 \!$ and $V_0 \rightarrow 0 \!$ in such a way that $\frac{V_0}{\alpha} = q_1 q_2.\!$ Thus, in the Born approximation,

\begin{align} \frac{d \sigma}{d \Omega} &= \frac{m^2 {q_1}^2 {q_2}^2 }{4\hbar^4 k^4 \sin^4(\theta/2)} \\ &= \frac{{q_1}^2 {q_2}^2}{16E^2 \sin^4(\theta/2)}. \end{align}

We will see later that the exact Coulomb scattering amplitude differs from the Born amplitude by a phase factor, and thus the differential cross section that we just obtained is in fact exact.

As a second, simpler, example, consider the scattering amplitude from a Gaussian potential,

$V(r)=Ae^{-\alpha r^2}.$

The scattering amplitude in the Born approximation is

\begin{align} f_{\text{Born}}(\theta) &= -\frac{2mA}{\hbar^2q}\int_0^\infty r'e^{-\alpha r'^2}\sin(q r')dr' \\ &= -\frac{2mA}{\hbar^2q}\int_0^\infty \frac{\partial}{\partial r'}\left(-\frac{1}{2\alpha}e^{-\alpha r'^2}\right)\sin(q r')dr' \\ &= \frac{mA}{\alpha\hbar^2 q}\left(0 + q\int_0^\infty e^{-\alpha r'^2}\cos(q r')dr'\right) \\ &= \frac{mA}{\alpha\hbar^2}\frac{\sqrt{\pi}}{2\sqrt{\alpha}}e^{-\frac{q^2}{4\alpha}} \\ &= \frac{mA\sqrt{\pi}}{2\hbar^2\alpha^{\frac{3}{2}}}e^{-\frac{q^2}{4\alpha}} \end{align} .

## Problems

(1) Using the Born approximation, find the differential cross section for the exponential potential,

$V(r)=-V_0 e^{-r/a}.\!$

(2) Use the Born approximation to find the differential and total cross sections for a particle of mass $m\!$ and a delta function scattering potential, $V(\mathbf{r})=g\delta^3(\mathbf{r}).$