Elastic scattering

The differential scattering cross section

We will study the elastic scattering of two isolated spinless particles.  The only interaction we take into account is the interaction between the two particles.  We ignore the internal structure of the particles.  If they have internal structure we assume that it does not influence the elastic scattering process.  If we ignore the center of mass motion, we can study the scattering of particles m₁ and m₂ with interaction energy V(r₁ - r₂) by studying the scattering of a fictitious particle of reduced mass μ = m₁m₂/(m₁ + m₂) by a potential (energy) V(r).

In a typical scattering experiment a target is struck by a beam of mono-energetic particles.
Let F_i be the incident flux, i.e. the number of particles per unit area per unit time. 
F_i = n_pv, where n_p is the number of particles per unit volume. 
Typically n_p is very small and we can neglect any interaction between different incident particles.
We measure the number ΔN_p of particles scattered per unit time into a solid angle ΔΩ about the direction defined by the spherical coordinates θ and φ. 
We expect ΔN_p ∝ F_i, and ΔN_p ∝ ΔΩ .
We define ΔN_p = σ_t(θ,φ)F_iΔΩ.  Here σ_t(θ,φ) is the differential scattering cross section of the target.  It has the units of an area.  Commonly used units are cm² and barn = 10^-24 cm².

Often the target is made up of a large number N_t of scattering centers.  If it is sufficiently thin, so that we can neglect multiple scattering, and if the distances between the scattering centers are large compared to the wavelength of the incident particles, so that we may neglect coherence between waves scattered from different centers, we have ΔN_p ∝ N_t.  Here we assume that we have a big beam and a small target, so that the beam covers the target completely.  We then write ΔN_p = σ(θ,φ)N_tF_iΔΩ, where σ(θ,φ) is the differential scattering cross section of a target particle.
The total scattering cross section is given by σ = ∫σ(θ,φ)dΩ.

The scattering amplitude

In quantum mechanics we describe the scattering of a particle by a potential V(r) by describing the time evolution of the wave packet ψ(r,t) representing the state of the particle.  Assume that the particle approaches the target from the left, along the z-axis.  Assume that at t --> -∞ the characteristics of the wave packet are known.  The subsequent evolution can be calculated using the Schroedinger equation

(-ħ/(2m))∇²Ψ(r,t) + V(r,t) Ψ(r,t) = (iħ∂/∂t)Ψ(r,t),
or
HΨ(r,t) = (iħ∂/∂t)Ψ(r,t).

The eigenstates of the Hamiltonian form a basis for the state space E, {|E_k>}, the basis of stationary states.  An arbitrary state vector may be expanded in terms of these basis vectors.
|Ψ> = ∑_ka_k|E_k>.
The time evolution of the eigenstates of H can be easily found.
HΨ_k(r,t) = (iħ∂/∂t)Ψ_k(r,t) = E_kΨ_k(r,t).  Ψ_k(r,t) = Φ_k(r)exp(-iE_kt/ħ).

If we solve the eigenvalue equation HΦ_k(r) = E_kΦ_k(r), we can obtain the evolution of any wave packet by expressing it as a superposition of stationary states.  We will therefore first attempt to find the stationary state solutions of the scattering problem.  We are trying to solve the equation
(-ħ/(2m))∇²Φ_k(r) + V(r,t)Φ_k(r) = E_kΦ_k(r)  for E_k > 0.
Let E_k = ħ²k²/(2m) and V(r) = (ħ²/(2m))U(r).
Then we may write [∇² + k²]Φ_k(r) = U(r)Φ_k(r),

For each value of k an infinite number of solutions exist.  We must choose from them the one solution that corresponds to the physical problem to be studied.  To do this, we look at the asymptotic behavior of the solutions.  We assume that we have a finite range potential, i.e. that U(r) --> 0 as r --> ∞ faster than r^-1.  Note that this excludes the Coulomb potential.  Then

[∇² + k²]Φ_k(r)  = 0 as r --> ∞.

Plane waves of the form exp(ik∙r) moving into an arbitrary direction are solutions to this equation.  In addition outgoing and incoming spherical waves of the form exp(±ikr)/r are solutions as r --> ∞.  In our example, at large negative times t --> -∞, the particle is a free particle described by a wave packet build from plane waves moving towards the origin along the z-axis from z = -∞.  We therefore look for a plane wave term e^ikz in the asymptotic stationary state solution.  For large positive values of t --> ∞ the particle is described by transmitted wave packet build from plane waves and a scattered wave packet.  We expect the scattered wave packet to be a superposition of outgoing stationary states.  We therefore look for a term e^ikr/r in the asymptotic stationary state solution.  Since, in general, scattering is not isotropic, the amplitude of the scattered wave can be a function of θ and φ.  The asymptotic form of the eigenfunctions of H is therefore given by

Φ_k(r) = A[e^ikz + f_k(θ,φ) e^ikr/r] as r --> ∞.

Here f_k(θ,φ) is defined as the scattering amplitude.

Wave packets

We will now check that a wave packet with the proper asymptotic behavior can indeed be build out of energy functions of this form.

Let Ψ(r,t) = ∫₀^∞g(k) Φ_k(r) exp(-iω_kt) dk  with E_k = ħ²k²/(2m) = ħω_k.
Let g(k) = |g(k)|e^iα(k).  This is the most general form.  For simplicity let us assume that α(k) is zero and g(k) is real.

Note: By labeling our eigenfunctions with a single index k, we have neglected the degeneracy of the energy eigenvalues.  Plane waves with the same energy, traveling in different directions should really be labeled with an index k.  Limiting ourselves to an index k in building the wave packet means taking care of the energy dispersion, (the spread of the wave packet along the z-axis), but not of the angular dispersion (the spread of the wave packet perpendicular to the z-axis).

The asymptotic wave packet is
Ψ(r,t) = ∫₀^∞g(k) e^ikz exp(-iω_kt) dk + ∫₀^∞g(k) f_k(θ,φ) (e^ikr/r) exp(-iω_kt) dk.
Assume g(k) has a pronounced peak at k = k₀ and is non zero only in a small interval Δk about k₀.  We want to find the maximum of each of the two packets making up the asymptotic wave packet. 

Let β(k) = -ω_kt.
Use a Taylor series expansion,
β(k) = β(k₀) + dβ(k)/dk|_k0 (k - k₀) + ... .
i.e. assume β varies little in the small interval Δk.

For the first wave packet we then have
∫₀^∞g(k) e^ikz exp(-iω_kt)) dk = exp(i(k₀z + β(k₀)))∫₀^∞g(k) exp(i(k - k₀)(z - z₀)) dk,
where z₀ = -dβ(k)/dk|_k0.
For |z - z₀| >> Δk^-1, the period of exp(i(k - k₀)(z - z₀)) = exp(i2π(k - k₀)/τ) with τ = 2π/(z - z₀) is much smaller than Δk, and the function exp(i(k - k₀)(z - z₀)) oscillates very rapidly in the integration interval and the integral yields approximately zero.  
If |z - z₀| ≈ 0, then the period of exp(i(k - k₀)(z - z₀)) = exp(i2π(k - k₀)/τ) approaches infinity, and the function hardly oscillates at all in the integration interval.  The integral then yields its maximum value.  The first wave packet has its maximum value at z_M = z₀ = -dβ(k)/dk|_k0.
-dβ(k)/dk|_k0 = -dω(k)/dk|_k0t = ħk₀t/m = v_gt.
v_g = ħk₀/m, v_g is the group velocity of the wave packet.

Consider the second wave packet.
∫₀^∞g(k) f_k(θ,φ) (e^ikr/r) exp(-iω_kt)) dk
= |f_k(θ,φ)/r| exp(i(k₀r + β'(k₀)))∫₀^∞g(k) exp(i(k - k₀)(r - r₀)) dk,
where β'(k) = α_k(θ,φ) + β(k),  r₀ = -dβ'(k)/dk|_k0, and α_k(θ,φ) is the phase of the scattering amplitude.
For a given θ and φ the second wave packet has its maximum value at
r_M(r,θ,φ) = -dβ'(k)/dk|_k0 = -dα_k(θ,φ)/dk|_k0 + v_gt.

For large negative values of t, r_M is negative, there is no scattered wave packet.  All we find is the incident wave packet.  For large positive values of t, both packets are present.  One packet moves along the z-axis and the other diverges in all directions.  We can therefore build a proper asymptotic wave packet out of our asymptotic energy eigenfunctions and conclude that they are well suited for the description of the scattering process.

The scattering cross section

The energy eigenfunctions {Φ_k(r)} form a basis for the space of square integrable functions, L², but for E_k > 0 the functions Φ_k(r) are not square integrable, i.e. Φ_k(r) ∉ L².  The basis can only be "orthonormalized in the Dirac sense".  To find the scattering cross section we therefore do not look at probabilities, but probability currents.  We compare the incident and the scattered probability current.  The probability current density is given by

j(r,t) = (ħ/(2mi))[Ψ*(r,t)∇Ψ(r,t) - Ψ(r,t)∇Ψ*(r,t)], or
j(r,t) =  (1/m) Re[Ψ*(r,t) (ħ/i) ∇Ψ(r,t)].
In the asymptotic region the plane wave term represents a wave with incident current density

j_i =  (z/z) (1/m) Re[e^-ikz (ħ/i) (∂/∂z)e^ikz] = (z/z) ħk/m.

To find the current density represented by the term f_k(θ,φ) (e^ikr/r) we evaluate
∇(f_k(θ,φ) (e^ikr/r)) = [(r/r)∂/∂r + (θ/θ)(1/r)∂/∂θ + (φ/φ)(1/(r sinθ))∂/∂φ](f_k(θ,φ) (e^ikr/r)).

Retaining only the lowest order terms in 1/r, we find that (f_k(θ,φ) (e^ikr/r)) represents an outgoing wave of current density
j_s =  (r/r) (ħk/m)|f_k(θ,φ)|²(1/r²).

In practice, interference terms are negligible except near the forward direction, since in an actual experiment the detector is generally placed in such a way that it can only be reached by scattered particles.  We then have  dN_p = Cj_s∙dA_s = Cj_sr² dΩ = C (ħk/m)|f_k(θ,φ)|²dΩ.
F_i = C (ħk/m).
dN_p = F_i|f_k(θ,φ)|²dΩ.
But dN_p= σ_k(θ,φ)F_idΩ.  The scattering cross section is equal to the square of the scattering amplitude.
σ_k(θ,φ) =|f_k(θ,φ)|².

To calculate the cross section we need to know the asymptotic form of the stationary state wave function Φ_k(r).  To this effect we have to solve the eigenvalue equation  [∇² + k² - U(r)]Φ_k(r)  = 0,  for the given potential U(r).

Scattering by a central potential V(r)

A basis {Φ_klm(r,θ,φ)} of eigenfunctions of H, L², and L_z exists.  The eigenvalues are ħ²k²/(2m), l(l + 1)ħ², and mħ, respectively.  We call these wave functions partial waves.

For large r the partial waves will resemble the common eigenfunctions of H₀, L², and L_z, where H₀ is the Hamiltonian of a free particle.  These eigenfunctions have a well defined energy and magnitude and z-component of angular momentum.  We denote these eigenfunctions by Φ⁰_klm(r,θ,φ) and call them free spherical waves.  We assume that the asymptotic behaviors of Φ_klm(r,θ,φ) and Φ⁰_klm((r,θ,φ) are identical except for a difference in phase, since the solutions satisfy the same differential equation as r goes to infinity.  We will show that for a fixed k the value of this phase shift for all l is all we need to calculate the scattering cross section.