5. Scattering in classical and in quantum mechanics

Scattering experiments are perhaps the most important tool for obtaining detailed information on the structure of matter, in particular the interaction between particles. Examples of scattering techniques include neutron and X-ray scattering for liquids, atoms scattering from crystal surfaces, elementary particle collisions in accelerators. In most of these scattering experiments, a beam of incident particles hits a target which also consists of many particles. The distribution of scattering particles over the different directions is then measured, for different energies of the incident particles. This distribution is the result of many individual scattering events. Quantum mechanics enables us, in principle, to evaluate for an individual event the probabilities for the incident particles to be scattered off in different directions; and this probability is identified with the measured distribution.

Suppose we have an idea of what the potential between the particles involved in the scattering process might look like, for example from quantum mechanical energy calculations (programs for this purpose will be discussed in the next few chapters). We can then parametrise the interaction potential, i.e. we write it as an analytic expression involving a set of constants: the parameters. If we evaluate the scattering probability as a function of the scattering angles for different values of these parameters, and compare the results with experimental scattering data, we can find those parameter values for which the agreement between theory and experiment is optimal. Of course, it would be nice if we could evaluate the scattering potential directly from the scattering data (this is called the inverse problem), but this is unfortunately very difficult (if not impossible) as many different interaction potentials can have similar scattering properties as we shall see below.

Many different motivations for obtaining accurate interaction potentials can be given. One is that we might use the interaction potential to make predictions about the behaviour of a system consisting of many interacting particles, such as a dense gas or a liquid.

Scattering might be elastic or inelastic. In the former case the energy is conserved, in the latter energy disappears. This means that energy transfer takes place from the scattered particles to degrees of freedom which are not included explicitly in the system (inclusion of these degrees of freedom would cause the energy to be conserved). In this chapter we shall consider elastic scattering.

5.1. Classical analysis of scattering

A well known problem in classical mechanics is that of the motion of two bodies attracting each other by a gravitational force whose value decays with increasing separation $r$ as $1/r^2$. This analysis is also correct for opposite charges which feel an attractive force of the same form (Coulomb's law). When the force is repulsive, the solution remains the same -- we only have to change the sign of the parameter $A$ which defines the interaction potential according to $V(r) = A/r$. One of the key experiments in physics which led to the notion that atoms consist of small but heavy kernels, surrounded by a cloud of light electrons, is Rutherford scattering. In this experiment, a thin gold sheet was bombarded with $\alpha$-particles (i.e. helium-4 nuclei) and the scattering of the latter was analysed using detectors behind the gold film. In this section, we shall first formulate some new quantities for describing scattering processes and then calculate those quantities for the case of Rutherford scattering.

Rutherford scattering is chosen as an example here -- scattering problems can be studied more generally; see Griffiths, chapter 11, section 11.1.1 for a nice description of classical scattering.

We consider scattering of particles incident on a so-called 'scattering centre', which may be another particle. The scattering centre is supposed to be at rest. This might not always justified in a real experiment, but in a standard approach in classical mechanics, the full two-body problem is reduced to a one-body problem with with a reduced mass, which is the present case (in problem 5.1 we will perform the same procedure for a quantum two-body system). The incident particles interact with the scattering centre located at ${\bf r}= \pmb 0$ by the usual scalar two-point potential $V(r)$ which satisfies the requirements of Newton's third law. Suppose we have a beam of incident particles parallel to the $z$-axis. The beam has a homogeneous density close to that axis, and we can define a flux, which is the number of particles passing a unit area perpendicular to the beam, per unit time. Usually, particles close to the $z$-axis will be scattered more strongly than particles far from the $z$-axis, as the interaction potential between the incident particles and scattering centre falls off with their separation $r$. An experimentalist cannot analyse the detailed orbits of the individual particles -- instead a detector is placed at a large distance from the scattering centre and this detector counts the number of particles arriving at each position. You may think of this detector as a photographic plate which changes colour to an extent related to the number of particles hitting it. The theorist wants to predict what the experimentalist measures, starting from the interaction potential $V(r)$ which governs the interaction process.

FIGURE 5.1: Geometry of the scattering process. b is the impact parameter and $\varphi$ and $\vartheta$ are the angles of the orbit of the outcoming particle

In figure 5.1, the geometry of the process is shown. In addition a small cone, spanned by the spherical polar angles $d\vartheta$ and $d\varphi$, is displayed. It is assumed here that the scattering takes place in a small neighbourhood of the scattering centre, and for the detector the orbits of the scattered particles all seem to be directed radially outward from the scattering centre. The surface $dA$ of the intersection of the cone with a sphere of radius $R$ around the scattering centre is given by $dA = R^2 \sin\vartheta d\vartheta d\varphi$. The quantity $\sin\vartheta d\vartheta d\varphi$ is called spatial angle and is usually denoted by $d\Omega$. This $d\Omega$ defines a cone like the one shown in figure 5.1. Now consider the number of particles which will hit the detector within this small area per unit time. This number, divided by the total incident flux (see above) is called the differential scattering cross section, $d\sigma/ d\Omega$: $$\frac{d\sigma(\Omega)}{d\Omega} = \frac{{\text{Number of particles leaving the scattering centre through the cone } d\Omega \text{ per unit time}}}{{\text{Flux of incident beam } \times d\Omega}}.$$ The differential cross section has the dimension of area (length$^2$).

First we realise ourselves that the problem is symmetric with respect to rotations around the $z$-axis, so the differential scattering cross section only depends on $\vartheta$. The only two relevant parameters of the incoming particle then are its velocity and its distance $b$ from the $z$-axis. This distance is called the impact parameter -- it is also shown in figure 5.1.

We first calculate the scattering angle $\vartheta$ as a function of the impact parameter $b$. We perform this calculation for the example of Rutherford scattering, for which we have the standard Kepler solution which is now a hyperbola (see your classical mechanics lecture course). The potential for the Kepler problem is $V(r) = -A/r$. The orbits are given by specifying $r(t)$, $\vartheta(t)$. However, for the Kepler problem it is more convenient to focus on the shape of the orbitals, which is given as $$r = \lambda \frac{1+\epsilon}{\epsilon \cos(\vartheta-C) - 1}$$ with $$\tag{5.3}\label{Eq:KepHelp} \epsilon = \sqrt{ 1 + \frac{2E \ell^2}{\mu A^2}};$$ this parameter is called eccentricity -- for a hyperbola, we have $|\epsilon|>1$. Here, $\ell$ is the angular momentum and $\mu$ the reduced mass. The integration constant $C$ reappears in the cosine because we have not chosen $\vartheta=0$ at the perihelion -- the closest approach occurs when the particle crosses the dashed line in figure 5.1 which bisects the in- and outgoing particle direction.

We know that for the incoming particles, for which $\vartheta=\pi$, $r\rightarrow\infty$, we have $$\cos(\pi - C) = 1/\epsilon.$$ Because of the fact that cosine is even $$\cos x = \cos (-x)$$ we can infer that the other value of $\vartheta$ for which $r$ goes to infinity, and which corresponds to the outgoing direction occurs when the argument of the cosine is $C-\pi$, so that we find $$\vartheta_\infty - C = C - \pi,$$ or $\vartheta_\infty = 2C-\pi$. The subscript $\infty$ indicates that this value corresponds to $t \rightarrow \infty$. From the last two equations we find the following relation between the scattering angle $\vartheta_\infty$ and $\epsilon$: $$\tag{5.6}\label{Eq:ThetaEps} \sin(\vartheta_\infty/2) = \cos(\pi/2-\vartheta/2) = \cos(\pi - C) = 1/\epsilon.$$

We want to know $\vartheta_\infty$ as a function of $b$ rather than $\epsilon$ however. To this end we note that the angular momentum is given as $$\ell = \mu v_{\text{inc}} b,$$ where 'inc' stands for 'incident', and the total energy as $$E = \frac{\mu}{2} v^2_{\text{inc}},$$ so that the impact parameter can be found as $$b = \frac{\ell}{\sqrt{2\mu E}}.$$ Using Eq. \eqref{Eq:KepHelp} and the fact that $\cot(x) = \sqrt{1-\sin^2(x)}/\sin(x)$, we can finally write \eqref{Eq:ThetaEps} in the form: $$\tag{5.10}\label{Eq:BTheta} \cot(\vartheta_\infty/2) = \sqrt{\epsilon^2 - 1} = \frac{2 E b}{|A|}.$$

From the relation between $b$ and $\vartheta_\infty$ we can find the differential scattering cross section. The particles scattered with angle between $\vartheta$ and $\vartheta+d\vartheta$, must have approached the scattering centre with impact parameters between particular boundaries $b$ and $b+db$. The number of particles flowing per unit area through the ring segment with radius $b$ and width $db$ is given as $j 2\pi b db$, where $j$ is the incident flux. We consider a segment $d\varphi$ of this ring. Hence: $$d\sigma(\Omega) = b(\vartheta) db d\varphi.$$ Relation \eqref{Eq:BTheta} can be used to express the right hand side in terms of $\vartheta_\infty$: $$d\sigma(\Omega) = \left( \frac{A}{2E} \right)^2 \cot(\vartheta/2) \; d\cot(\vartheta/2)\; d\varphi = \left( \frac{A}{2E} \right)^2 \cot(\vartheta/2) \frac{d\cot(\vartheta/2)}{d\vartheta}\frac{d\vartheta}{d\cos\vartheta} d \cos\vartheta d\varphi.$$ This can be worked out straightforwardly to yield: $$\frac{d\sigma(\Omega)}{d\Omega} = \left( \frac{A}{4E} \right)^2 \frac{1}{\sin^4\vartheta/2}.$$ This is known as the famous Rutherford formula.