Chapter 18

Theory of Scattering

True laws of nature cannot be linear.

—Albert Einstein

18.1  INTRODUCTION

It may be emphasized that the most direct information about the nature of interactions in atomic and sub-atomic particles has been obtained from the study of collisions between particles. For example, it was concluded from the famous Rutherford scattering experiment of α–particles by atoms, that the positive charge and almost all mass of an atom is confined in a very small dimension (the nucleus). It was also concluded from this experiment that the interaction between α–particle and the positive charge of the atom (the nucleus) is very strong and of very short range. In fact, the systematic study of scattering process is the main source of information about the strong, electromagnetic and weak interactions. The modern picture of matter as being ultimately consisted of quarks and leptons is based on the evidences provided by the scattering experiments.

In a simple scattering experiment, a beam of particles is directed at a target containing the scattering material. The energy and angular distribution of the scattered particles are measured by suitable experimental technique. In this Chapter, we shall study the quantum theory of scattering. We shall designate and evaluate (quantum mechanical) quantities which can be compared with the experimental results. But firstly, we shall have a brief description of scattering of classical particles.

18.2  SCATTERING EXPERIMENTS AND SCATTERING CROSS-SECTION

Let us consider a classical particle of energy E, incident with impact parameter b on a scattering centre fixed at the origin 0, as shown in Figure 18.1. The impact parameter is defined as the minimum distance to which the incident particle would approach the scatterer if there were no interaction between the particle and the scatterer. Let the interaction potential between the incident particle and the scatterer be isotropic and given by V(r), where r = |r| is the distance of the particle from the scatterer. The particle will be scattered at some angle θ, called as the scattering angle (the scattering is assumed to take place in x–z plane). In actual scattering experiments, an incident beam of monoenergetic particles from a source S is collimated by slits C and is scattered by a fixed scatterer at the origin 0 (as schematically shown in Figure 18.2). The incident beam may be specified by the incident flux I0 defined as the number of particles incident per unit time crossing unit area put normal to the direction of incidence. It may be explicitly mentioned here that we are assuming

Figure 18.1

Figure 18.1 (a) The classical scattering problem. Showing classical trajectory, impact parameter b and the scattering angle θ (taking the plane of paper as xz) (b) The scattering angle θ decreases with increasing impact parameter b

Figure 18.2

Figure 18.2 A monoenergetic beam of particles scattered by a target at the origin 0. Particles scattered into solid angle dΩ about (θ, ϕ) direction, are detected at D

  1. the incident beam to be uniform, and
  2. the density of particles to be sufficiently small (i.e. the flux I0 be small) so that the interaction between particles is negligible.

We are at the moment considering the case of classical particles, where the motion of each particle in the beam is along a well-defined path and each particle is scattered in some direction specified by (θ, ϕ). In a scattering experiment, the number of scattered particles in a particular direction is counted by a detector D placed at a relatively large distance from the scatterer along that direction. If the detector subtends a solid angle dΩ (at the scatterer) in the direction (θ, ϕ), the number of particles I(θ, ϕ)dΩ entering the detector per unit time can be recorded. Here I(θ, ϕ) denotes the number of particles scattered per unit time per unit solid angle in direction (θ, ϕ). In fact, this number is proportional to the incident flux I0. The scattering can be characterized by the differential scattering cross-section dσ/dΩ defined as

equation

which is the ratio of the number of particles I(θ, ϕ) scattered per unit time per unit solid angle in direction (θ, ϕ) and the incident flux I0. Now I(θ, ϕ) is the number of particles per unit time, per unit solid angle whereas I0 is the number of particles per unit time per unit area, so dimensions of [I(θ,ϕ)/I0] and hence that of dσ/dΩ is an area. That is why dσ/dΩ is termed as scattering cross-section. The total scattering cross-section σ is defined as the integral of the differential scattering cross-section dσ/dΩ over all solid angles.

equation

The total scattering cross-section measures only the occurrence of the scattering process, and not any special feature of scattered particles.

In this chapter, we shall be mainly concerned with elastic scattering. In the process of elastic scattering the internal energies of the incident and target particles do not change. In other words, total kinetic energy of the incident and target particles before and after the scattering process remains same. In case of fixed target particle (which we have considered above) kinetic energy of the incident particle remains same after the scattering has taken place.

18.3  CLASSICAL THEORY OF SCATTERING: RUTHERFORD SCATTERING

We firstly discuss, in brief, the case of classical particles scattered by a fixed scatterer. As shown schematically in Figure 18.3 a beam of classical particles with homogeneous flux I0, is incident in z-direction on a fixed scatterer at the origin 0. The interaction energy between the incident and target particles is considered to be spherically symmetric V(r). In the geometry shown in Figure 18.3, the pattern has cylindrical symmetry about z-axis. Neither the incident flux nor the scattered flux depend on azimuthal angle ϕ.

Still we include variable ϕ in our discussion as we have to specify scattered direction by (θ, ϕ). The particles incident at impact parameter b and azimuthal angle ϕ are scattered in direction (θ, ϕ). There is a definite relation between impact parameter b and scattering angle θ. The angle θ decreases with increasing impact parameter b, as shown schematically in Figure 18.1. For the interaction energy, for example between incident α-particle and the nucleus of charge Ze, V(r) = 2Ze2/4π0 r (the Rutherford scattering) the relation between θ and b is [see Goldstein, ‘Classical Mechanics’]:

Figure 18.3

Figure 18.3 A beam of classical particles is incident on a fixed scatterer at 0. The particles incident at impact parameter b and azimuthal angle θ are scattered in direction (θ, ϕ)

equation

where E is the kinetic energy of incident α-particles. It is now straightforward to find the expression of scattering cross-section dσ/dΩ using Eq. (18.1).

All the particles passing through an area dA = bdbdϕ of incident beam (Figure 18.3) are scattered through solid angle dΩ in (θ, ϕ) direction. So the number of particles scattered through solid angle dΩ about(θ, ϕ) is

 

I(θ, φ) dΩ = I0dA (I0 = incident flux)
= I0 b db dϕ        (18.4)

 

Putting values of b and db from Eq. (18.3), it gives

equation
equation

Now dΩ = sinθdθdϕ

So

equation

This is the expression for the Rutherford scattering cross section.

18.4  QUANTUM THEORY OF SCATTERING

Till now we have discussed the scattering of classical particles which have well-defined trajectories during the whole scattering process. We now proceed to discuss the scattering process of quantum particles. We know the quantum particles are described through wave functions. In fact, when we consider a monoenergetic incident beam of quantum particles moving towards the scatterer, we shall have to represent this beam by free particle wave functions. The wave functions of free particles moving in (say) z-direction with well-defined (kinetic) energy E (and therefore well-defined linear momentum equation wave-length equation wave vector equation are generally plane waves, written as (t = 0)

 

ϕk(r) = eik,r        (18.6)

 

Now we know that when classical plane waves reach a scatterer, these are scattered or diffracted in all possible directions in the form of spherical waves with the scatterer as their centre. In a similar way, we expect the quantum (matter) plane waves reaching the scatterer to be scattered in all possible directions in the form of spherical waves with possible different amplitudes in different directions, [see Figure 18.4]. As we are considering only elastic scattering here, the energy, and therefore, the magnitude of the wave vector, of the spherical wave is the same as that of the incident plane wave. The spherical wave of amplitude (say) f(θ, ϕ) and of wave vector (magnitude) k may be written as [f(θ, ϕ)/r] eikr. Now whereas the wavefunction of the incident beam is represented by plane wave of Eq. (18.6), the wave function of the scattered beam should be a mixture of the outgoing spherical waves and the plane waves. So after scattering the wave function should be of the form

equation
Figure 18.4

Figure 18.4 Scattering of an extended plane wave. The plane wave is scattered in the form of spherical waves

Let us now consider the process of detecting the scattered particles by the detector D put along (θ, ϕ) direction of the scatterer. We know that quantum particles propagate in the form of waves described by the wave functions but whenever detected by a detector, these are detected as particles. So the incident beam of quantum particles, which is described by the wave function Eq. (18.6) must be having well-defined flux of particles I0 [= number of particles crossing unit (perpendicular) area per unit time]. The detector D detects the number of scattered particles reaching the detector covering solid angle dΩ in direction (θ, ϕ) per unit time. This number may be represented by I(θ, ϕ) dΩ, where I(θ, ϕ) represents the number of particles scattered per unit time per unit solid angle about the direction (θ, ϕ). And, therefore, the definition of the differential scattering cross-section is the same as in classical case

equation

Let us elaborate what we have discussed above by finding flux in following two cases:

Case I: Let us consider a (homogeneous) incident beam of particles moving in z-direction described by the wave function

equation

We may find out the value of the probability current density J corresponding to the wave function ϕk0(r) using relation (4.33). We get

equation

The normalization factor equation in wave function (18.8) means that (every time) one particle is there in volume V. Particles are moving in z-direction with velocity v(= ħk0/m). So the flux is I0 = (v/V) (see Figure 18.5).

Case II: Consider an outgoing spherical wave of amplitude f(θ, ϕ) in (θ, ϕ) direction, given by

equation
Figure 18.5

Figure 18.5 Flux of a beam represented by a plane wave

The probability current density corresponding to this wavefunction is

equation

where terms of order (1/r3) have been neglected, keeping in mind that we are finding J(r, t) for large r. Now |J(r, t)|, the magnitude of current density, is nothing but the number of particles moving radially outwards per unit time through a unit surface area (on the spherical surface of radius r) in direction (θ, ϕ). Therefore, the number of particles moving radially outwards per unit time through a unit solid angle in direction (θ, ϕ), I(θ, ϕ) is

 

I(θ, ϕ) = |J(r, t)|r2
equation

(see Figure 18.6).

Therefore, differential scattering cross-section is

equation

So the experimentally measured quantity, dσ/dΩ is related with mode square of the quantity f(θ, ϕ), generally known as scattering amplitude. After this preliminary qualitative discussion, we shall now proceed to solve Schrodinger equation for particles scattered by the spherically symmetric potential V(r) to find the scattered wave function, that is, to find the scattering amplitude f(θ, ϕ).

Figure 18.6

Figure 18.6 Flux in (θ, ϕ) direction of an outgoing beam represented by a spherical wave

18.5  SOLUTION OF SCHRODINGER EQUATION FOR SCATTERING PROBLEM: GREEN’S FUNCTION

In this section, we shall develop the quantum theory of scattering of a beam of particles by a target particle fixed at the origin 0, with spherically symmetric interaction potential V(r). The system is described by the time-dependent Schrodinger equation

equation

Here m is the mass of the particle. It may be mentioned here that in case of incident particle of mass m being scattered by the target particle of mass M, Eq. (18.14) remains valid provided m is replaced by the reduced mass µ. When the incident beam of particles is switched on for a long time (in comparison to the time, a particle would take to cross the interaction region), the steady state condition will reach. Then we need to solve time-independent Schrodinger equation.

equation

where E is the energy of the particle

equation

p = mv = ħk is the linear momentum of the incident particle. Equation (18.15) may be re-written in the form

 

(∇2 + k2)ψ(r) = U(r) ψ(r)        (18.17)

 

where k2 = (2mE/ħ2) and U(r) is given by

equation

U(r) is generally called as the reduced potential. Equation (18.17) is an inhomogeneous differential equation and to solve it we firstly replace it by an integral equation.

Let us define a function G(k, r, r′) by

 

(∇2 + k2)G(k, r, r′) = –4πδ(rr′)        (18.19)

 

where δ(rr′) is the three-dimensional Dirac delta function. The function G(k, r, r′) so defined is called the Green’s function of operator (∇2 + k2). It can be easily seen that solution ψ(r) of Eq. (18.17) may be written in the form

equation

To see this, let us start with L.H.S. of Eq. (18.17)

 

L.H.S. = (∇2 + k2) ψ(r)
equation
= U(r)ψ(r) = R.H.S.

 

To this particular solution [Eq. (18.20)], we can also add an arbitrary solution of the homogeneous equation

 

(∇2 + k2) ϕk(r) = 0        (18.21)

 

which is Schrodinger equation of a free particle and has the solution

 

ϕk(r) = eik · r        (18.22)

 

Therefore, Eq. (18.20) should be replaced by

equation

So the differential equation (18.17) has been replaced by the integral equation (18.23) along with Eq. (18.19). Therefore, our purpose now is to solve Eq. (18.23) for ψk(r). There are two-fold problems with Eq. (18.23). Firstly, the form of the Green’s function is still unknown. Secondly, even when G (k, r, r′) is known, the wave function ψk(r) (to be found out) appears on the R.H.S. also, that too in the integrand. We shall have to resort to some approximations to solve Eq. (18.23) for ψk(r). Let us find out form of G(k, r, r′).

Let us seek for the Green’s functions of the form

 

G(k, r, r′) = G(k, r r′)        (18.24)

 

which may be obtained from

 

(∇2 + k2) G(k, r) = 4πδ(r)        (18.25)

 

the simplified form of Eq. (18.19). Firstly, we find a Fourier transformation of Eq. (18.25) by introducing the three-dimensional Fourier integral of G(k, r) and δ(r) as

equation

and

equation

Substituting Eqs (18.26) and (18.27) in Eq. (18.25) we have

equation

or

equation

or

equation

Substituting Eq. (18.28) in Eq. (18.26), we get

equation

To evaluate the integral (18.29), let us choose the direction of polar axis (i.e., the k′z axis) along r,

equation

Now so long as k′ is real, the integral (18.30) shows singularities and may not exist. However, we may give meaning to integral by regarding this as contour integral in the complex k′-plane. Thus we replace (18.30) by the integral

equation

with η a small positive number. Now after evaluating Eq. (18.31) we shall let η → 0. The result G+(r) shall be the solution of Eq. (18.25). The poles of the integrand in Eq. (18.31) are at

equation

As shown in Figure 18.7, we take anticlockwise contour of integration in the upper half k′-plane that leads the path of integration on the real axis from –∞ to +∞. The contour encloses the pole k= k + ( / 2k) in the right half plane. In the limit η → 0, and evaluating the residue at k= k, the integral gives

equation

However, if one replaces η in Eq. (18.31) by –η(η > 0), one obtains a second value of the Green’s function

equation

It must be emphasized here that, in fact, we may get still different Green’s function if we treat the singularities in Eq. (18.30) differently, corresponding to different physical situations.

It may be easily seen here that if one includes time dependence in Eqs (18.33), the two Green’s functions become

equation
Figure 18.7

Figure 18.7 Contour of integration in the complex k′-plane

and

equation

and G+ and G represent the outgoing and incoming spherical waves, respectively. For the present case (of plane waves incident on the scatterer) the scattered waves are the outgoing spherical waves. Therefore, we shall use G+(k, r) in Eq. (18.23), which gives

equation

Let us remember here that the original Schrodinger equation (18.14) of the scattering problem was firstly re-written in the integral form (18.23) and then inserting value of Green’s function in (18.23), finally written in the form (18.35). So Eq. (18.35) is exact replica of Schrodinger equation (18.14), with the unknown equation appearing on both sides. We shall discuss below the approximations under which Eq. (18.35) shall be solved. Let us emphasize that we are interested only in the asymptotic forms of equation, (i.e., the form of equation for r → ∞). If we consider that U(r′) ≠ 0 only for a finite region, say for values of r′ < a, the exponent in the integrand of Eq. (18.35) may be expanded in powers of (r′/r) to give

equation

So, we get

equation

or

equation

where

 

k′ = ker        (18.37)

 

is a vector along the direction of scattering, that is along(θ, ϕ) direction. This integral equation is known as Lippmann–Schwinger equation of potential scattering. This is equivalent to the Schrodinger equation (18.17) plus the boundary condition taken care through Green’s function G+(r) [Eq. (18.33a)]. The asymptotic expression (18.36) may be written as

equation

where

equation

is called the scattering amplitude. Here we have omitted + sign on ψk(r) but (we mean) this represents the case of outgoing spherical waves.

18.6  THE BORN APPROXIMATION

In Section 18.5, the Schrodinger equation for the scattering problem [Eq. (18.17)] has been converted into the integral equation (18.36) [re-written in the form of Eq. (18.38) alongwith Eq. (18.39)]. In finding the solution ψk(r) of Eq. (18.36), the difficulty is that ψk(r) appears also on R.H.S. of the equation (in the integrand). So it seems impossible to solve this equation exactly. Therefore, some approximate methods shall have to be used. We shall use the process of iteration and shall find the solution of ψk(r) in the form of a series in powers of interaction potential V(r). The process of iteration should be valid only if the V(r) is relatively small.

Replacing r and r′ in Eq. (18.36), we have

equation

Substituting ψk(r′) from (18.40) in (18.36), we get

equation

If we put value of ψk(r″) from Eq. (18.40) in the last term of Eq. (18.41) (and in successive equations) we shall get a series of infinite terms. Now if the interaction energy V(r) is small (in comparison to energy E of incident particles), we can neglect U2 term in Eq. (18.41)and ψk(r) can be approximated as

equation

The next approximation can be obtained by putting value of ψk(r″) in the integrand of third term of Eq. (18.41), which is

equation

The solutions equation and equation are known as first Born approximation and second Born approximation, respectively. In frequent use is the first Born approximation which is referred to as the Born approximation. equation may also be written in terms of the scattering amplitude equation as

equation

where

equation

equation is known as (first) Born approximation to the scattering amplitude. Putting the value of U(r)in terms of V(r), we get

equation

or

equation

where

equation

Here Vq is the q-th Fourier component of potential V(r). ħq (k′ k) is the momentum transfer in scattering (in fact, in the elastic scattering we have |k′| = |k|). To perform angular integration in Eq. (18.46a) we take q as the polar axis and denote by (α, β), the polar angles of r with respect to q. We have

equation

As shown in Figure 18.8

equation

where θ is the scattering angle. Integration in Eq. (18.47) may be performed

equation
Figure 18.8

Figure 18.8 Momentum transfer in an elastic scattering

It can be seen that this quantity is real and depends on k (i.e., on energy E) and on the scattering angle θ [as q = 2ksin(θ/2)]. The corresponding differential scattering cross-section (in the first Born approximation) is given by

equation
18.7  METHOD OF PARTIAL WAVES AND PHASE SHIFTS

In the previous sections, we solved Schrodinger equation (18.17) by first converting it into an integral equation (18.36) and then using Born approximation. In this section, we shall solve Schrodinger equation by using an alternative method: the method of partial waves. We know that in case of scattering by a spherically symmetric potential V(r), the system is completely symmetrical about the direction of incidence (the z-axis). Therefore, the wave functions (and hence the scattering amplitudes) depends on the polar angle θ and not on the azimuthal angle ϕ. Formally also, we see from Eq. (18.36) that if V is a function of r only, the wave function ψk(r) depends on k and r and the angle θ (between k and r) only. Therefore, both ψk(r, θ) and fk(θ) may be expanded in series of Legendre polynomials P1(cos θ) which form a complete set. For spherically symmetric potential, the angular momentum of the scattered particle is a constant of motion. For this reason also, it is advantageous to express solutions in terms of angular momentum eigenstates. Therefore, we start with the expansion of ψk(r, θ) and fk(θ) as

equation

and

equation

The expansion coefficients Rl(k, r), which are really radial functions, are known as partial waves and fl(k) are known as partial wave amplitudes. It can be easily checked that substituting the expansion of (18.51a) in Schrodinger equation (18.17) gives the radial part of the equation as

equation

Introducing the new radial function, ul(k, r) = rRl(k, r) the radial equation for ul(k, r) is

equation

Now for spherically symmetric potential the radial function Rl(k, r) is finite at r = 0, which dictates that

 

ul(k, 0) = 0        (18.54)

 

Let us consider the potential V(r) of relatively short range such that V(r) = 0 for r > a. So for sufficiently large r, the potential U(r) may be neglected and the radial equation for Rl(k, r) becomes

[the same as Eq. (10.54)]:

equation

As discussed in Section 10.6, Eq. (18.55) has linearly independent solutions, for each l in the form of spherical Bessel functions jl(kr) and spherical Neumann functions nl(kr). Therefore, in the region r > a, Rl(kr) is given as a linear combination of the functions jl(kr) and nl(kr):

 

Rl(k, r) = Bl(k) jl(kr) + Cl(k)nl(kr)        (18.56)

 

where coefficients Bl(k) and Cl(k) are independent of position r. Using the asymptotic expressions of jl(kr) and ηl(kr) from Eqs (10.65) and (10.66), we have

equation

This equation may be re-cast in the form

equation

where

equation

and

equation

The quantities δl(k) are known as phase shifts. The phase shifts characterise the strength of the scattering in the lth partial wave by the potential V(r) at energy E = (ħ2k2/2m). In fact, it can be seen that for V(r) = 0, the solution of Schrodinger equation (18.52) is in terms of jl(kr) [Eq. (10.76)], whereas for V(r) 0, the solution is in terms of Rl (k, r) [Eq. (18.51a)]. And comparing expression of R1 (k, r) [Eq. (18.58)] with that of equation, it is clear that δl(k) = 0 for V(r) = 0, as it should be.

Now let us start with the asymptotic form of the wave function ψk(r, θ)

equation

and expand it in terms of Legendre polynomials. In Section 10.6.2, we had already found the expansion of eik·r (where k is pointing in z-direction) in terms of Legendre polynomials as

equation

For large r, jl(kr) [from Eq. (10.65)] behaves like (1/r) sin [kr – (lπ / 2)]. Then Eq. (10.76) becomes

equation

Substituting (18.61) and (18.51b) in (18.38), we get

equation

Comparing Eq. (18.62) with Eq. (18.51a), we get

equation

By equating Eq. (18.63) with Eq. (18.58), we find that

 

Al(k) = il(2l + 1)eiδ1(k)        (18.64)

 

and

equation

Putting value of fl(k) in Eq. (18.51b), the scattering amplitude fk(θ) becomes

equation

or

equation

The total scattering cross-section comes out to be

equation
EXERCISES

Exercise 18.1

Show that if the integral in Eq. (18.30) is replaced by its Cauchy principal value, the Green’s function obtained is

equation

Exercise 18.2

Find out differential scattering cross-section, under Born approximation, in case a particle is scattered by the potential V(r) given as

  1.  
    V(r) = –V0   for   r < r0
    = 0   for   r > r0
  2.  
     V(r) = –V0 e–r2/a2
  3.  
    equation

Exercise 18.3

Neutrons and protons are spin equation fermions. Let their spin state be represented by equation and equation or s denotes ↑ and ↓. Let the scattering amplitude f(θ) for neutron–proton scattering be given by

equation

where equation and equation, the initial and final spin states of the neutron–proton system (i, j = 1,2,3,4), are given as

equation

Calculate the scattering amplitudes fii(θ).

Exercise 18.4

With the given information in above exercise, calculate the scattering amplitude fii(θ) (ij).

Exercise 18.5

With the results obtained in above two exercises, calculate the scattering amplitude for the triplet state → triplet state, and singlet state → singlet state scattering.

Show that cross-section for the triplet state → singlet state scattering is zero.

SOLUTIONS

Solution 18.2

  1. From Eq. (18.50)
    equation
  2. From Eq. (18.50)
    equation
  3.  
    equation

    Now

    equation

    so

    equation

Solution 18.3

Let us work in a representation in which

equation

We may write

equation

and

equation

Here

equation

We know that

equation

So one can easily get

 

f11 = (a + b)
f22 = (a – b)
f33 = (a – b)
f44 = (a + b)

Solution 18.4

equation

Solution 18.5

Scattering amplitude for

 

triplet (↑↑) → triplet (↑↑) = (a + b)
triplet (↓↓) → triplet (↓↓) = (a + b)
singlet (↑↓) → singlet (↑↓) = (ab)
singlet (↑↓) → singlet (↓↑) = 2b
triplet (↑↑) → singlet (↑↓) = 0
REFERENCES
  1. Goldstein, H. 1980. Classical Mechanics, Mass: Addison-Wesley.
  2. Bohm, D. 1951. Quantum Theory. New York: Prentice Hall.
  3. Merzbacher, E. 1999. Quantum Mechanics. 3rd edn., New York: John Wiley.
  4. Gasiorowicz, S. 1996. Quantum Physics. 2nd edn., New York: John Wiley.
  5. Griffiths, D.J. 1995. Introduction to Quantum Mechanics. New Jersey: Prentice Hall.
  6. Powell, J.L. and Crasemann, B. 1961. Quantum Mechanics. Mass: Addison-Wesley.
  7. Bransden, B.H. and Joachain, C.J. 2000. Quantum Mechanics, Singapore: Pearson Education Ltd.
  8. Baym, G. 1969. Lectures on Quantum Mechanics. New York: W.A. Benjamin.
  9. Sakurai, J.J. 1985. Modern Quantum Mechanics. Mass: The Benjamin/Cummings.
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