### The Born approximation

#### Problem:

An electron of incident momentum ki is scattered elastically by the electric field of an atom of atomic number Z.  The potential due to the nucleus is of the form U0(r) = -Ze2/r.  This potential is screened by the atomic electron cloud.  As a result, the total potential energy of the incident electron is U(r) = (-Ze2/r)exp(-r/a), where a is the radius of the atom.  Let kf be the final momentum of the electron and q = kf - ki be the momentum transfer.
(a)  Calculate the differential cross section, dσ/dΩ, in the Born approximation for the scattering of the electron by the atom (using the screened potential U(r)).
(b)  Now calculate the differential cross section, dσ/dΩ, in the Born approximation for the scattering of the electron by the nucleus only (using the unscreened potential U0(r)).
(c)  Plot the ratio of the two cross sections as a function of x = a|q| and briefly discuss the limits x --> 0 and x --> ∞.

Solution:

• Concepts:
The Born approximation
• Reasoning:
We are asked to evaluate the scattering cross section in the Born approximation.
• Details of the calculation:
(a)  In the Born approximation we have
σkB(θ,φ) = σkB(k,k') = [μ2/(4п2ħ4)]|∫d3r' exp(-iq∙r')U(r')|2,
where q = k' - k, k = μv0/h, k' = μv0/ħ (k'/k'), and μ is the reduced mass.
Here μ = me.
k is the incident wave vector, and (k'/k') is a unit vector pointing in the direction (θ,φ).
With U(r) = U(r) = (-Ze2/r)exp(-r/a) we have
∫d3r' exp(-iq∙r')U(r') = -Ze2∫∫∫r'dr'sinθ'dθ'dφ'exp(-r'/a)exp(-iqr'cosθ')
= -Ze22π∫0r'dr'exp(-r'/a)∫-11exp(-iqr'cosθ')dcosθ'
= -(Ze22π/q)∫0dr'exp(-r'/a)∫-qr'qr'exp(-ix)dx
= -(Ze22π/q)∫0dr'exp(-r'/a)[-i(eiqr' - e-iqr')]
= -(Ze24π/q)∫0dr'exp(-r'/a)sin(qr') = -(Ze24π/q2)q2a2/(1 + q2a2).
Therefore σkB(θ,φ) = [4μ2Z2e44][1/((1/a2) + q2)]2
With  q = 2ksin(θ/2) we have σkB(θ,φ) = [4μ2Z2e44][1/((1/a2) + 4k2sin2(θ/2))]2.
(b)  Let a --> ∞, then  U(r) = (-Ze2/r)exp(-r/a) -->  U0(r) = -Ze2/r.
Then  σkB(θ,φ) = [4μ2Z2e44]/(16k4sin4(θ/2)) = [Z2e4]/(16E2sin4(θ/2)).
This is the Rutherford cross section.
(c)  Ratio: σkB(θ,φ)part akB(θ,φ)part b = (q2a2/(1 + q2a2))2 = (x2/(1 + x2))2 with x = |qa|.
As x --> 0, the ratio approaches 0.
As a --> 0, x --> 0, the nuclear charge is completely screened, we have a neutral object.
As x --> ∞, the ratio approaches 1.
As a --> ∞, x --> ∞, the screening vanishes, we have a bare nucleus.

#### Problem:

Consider an electron of energy E0 and velocity v0 in the z-direction incident on an ionized Helium atom He+, with just one electron in its ground state.  Compute the differential cross section dσ/dΩ for the incident electron to scatter into the solid angle dΩ about the spherical angles (θ,φ).  Explain your assumptions and approximations.

Solution:

• Concepts:
The Born approximation
• Reasoning:
For fast electrons we can use the Born approximation to calculate the scattering cross section.
• Details of the calculation:
In the Born approximation we have
σkB(θ,φ) = σkB(k,k') = [μ2/(4п2ħ4)]|∫d3r' exp(-iq∙r')U(r')|2,
where q = k' - k, k = μv0/h, k' = μv0/ħ (k'/k'), and μ is the reduced mass.
Here μ = me, k is the incident wave vector, and (k'/k') is a unit vector pointing in the direction (θ,φ).
U(r) = -2e2/r + e2∫d3r' ρ(r')/|r - r'| = -e2∫d3r' ρt(r')/|r - r'|,
where ρt(r') = 2 δ(r') - ρ(r').

∫d3r exp(-iq∙r)U(r = -e2∫d3r exp(-iq∙r)∫d3r' ρt(r')/|r - r'|
= -e2∫d3r' ρt(r')exp(-iq∙r')∫d3r exp(-iq∙(r - r')/|r - r'|
= -(4πe2/q2)∫d3r' ρt(r')exp(-iq∙r') = -(4πe2/q2)[2 - F(q)],
where F(q) = ∫d3r' ρ(r')exp(-iq∙r') is the atomic scattering form factor.

[To evaluate ∫d3r exp(-iq∙(r - r')/|r - r'| we have used
∫d3r e-λr exp(ik∙r)/r = 2π∫0dr r e-λr-11d(cosθ)eikrcosθ
= 2π∫0dr e-λr(eikr - e-ikr)/(ik) = (4π/k)Im(∫0∞dre-(λ-ik)r)
= (4π/k)Im(1/(λ - ik)) = 4π/(λ2 + k2).
Take the limit as λ --> 0.]

σkB(θ,φ)  = [4me2e4/(q4ħ4)][2 - F(q)]2.
q2 = 4k2sin2(θ/2).
σkB(θ,φ)  = [4m2e4/(16ħ4k4sin4(θ/2))][2 - F(q)]2
= (e4/(16E2sin4(θ/2))[2 - F(q)]2.
If we assume that F(q) = 1, i.e.  that the incoming electron sees a screened nucleus of charge 1, we obtain the Rutherford scattering cross section.

#### Problem:

Evaluate the differential scattering cross section for a particle of mass m in a repulsive potential U(r) = A/r2 in the Born approximation.

Solution:

• Concepts:
The Born approximation
• Reasoning:
We are asked to evaluate the scattering cross section in the Born approximation.
• Details of the calculation:
In the Born approximation we have
σkB(θ,φ) = σkB(k,k') = [μ2/(4π2ħ4)]|∫d3r' exp(-iq∙r')U(r')|2,
where q = k' - k, k = μv0/ħ, k' = μv0/ħ (k'/k'), and μ is the reduced mass.
Here μ = m,  k is the incident wave vector, and (k'/k') is a unit vector pointing in the direction (θ,φ).
With U(r) = A/r2 we have
∫d3r' exp(-iq∙r')U(r') = A∫∫∫dr'sinθ'dθ'dφ'exp(-iqr'cosθ')
= 2πA∫0dr'∫-11exp(-iqr'cosθ')dcosθ'
= (4πA/q)∫0dr'sin(qr')/r' = 2π2A/q.
σkB(θ,φ) = [μ2/(4π2ħ4)][4π4A2/q2] = μ2π2A2/(ħ4q2).
With  q = 2ksin(θ/2) we have σkB(θ,φ) = μ2π2A2/(4ħ4k2sin2(θ/2)).

#### Problem:

Evaluate, in the Born Approximation, the differential cross section for the scattering of a particle of mass m by a delta-function potential U(r) = Bδ(r).  Comment on the angular and velocity dependence.  Find the total cross section.

Solution:

• Concepts:
The Born Approximation
• Reasoning:
We are asked to evaluate the scattering cross section in the Born approximation.
• Details of the calculation:
In the Born approximation
σkB(θ,φ) = σkB(k,k') = [μ2/(4п2ħ4)]|∫d3r' exp(-iq∙r')U(r')|2.

σkB(θ,φ) = [Bμ2/(4π2ħ4)]|∫d3r' exp(-iq∙r')δ(r')|2 = Bμ2/(4π2ħ4).
The differential scattering cross section is independent of scattering angle and velocity.
σkB = Bμ2/(πħ4) is the total scattering cross section.

#### Problem:

Consider two distinguishable spin-½ particles interacting via the interaction
Hint = gS1·S2δ3(r),
where g is a constant, Si are the spin operators of the two particles, and r is the separation of the two particles.
(a)  Using the Born approximation, calculate the differential cross section in the center of mass frame for particles in the spin-singlet and spin-triplet states, assuming the particles are non-identical.
(b)  What is the total cross section if the particles are unpolarized?

Solution:

• Concepts:
Born approximation, two spin ½ particles
• Reasoning:
In the CM frame we are only interested in the relative motion of the particles.  They are treated as a fictitious particle with wave function Φ(r)⊗Χ(S,Ms), where r = r1-r2 is the relative coordinate and Χ(S,Ms) characterizes the spin of the system.
For this fictitious particle In the Born approximation, neglecting spin, we have
kB/dΩ = σkB(θ,φ) = σkB(k,k') = [μ2/(4π2ħ4)]|∫d3r' exp(-iq∙r')U(r')|2,
where q = k' - k, k = μv0/ħ, k' = μv0/ħ (k'/k'), and μ is the reduced mass.
Here ħq is the momentum transfer and q = 2ksin(θ/2).

The differential scattering cross section is proportional to the Fourier transform of the potential.

Note:
We can derive the above expression for dσkB/dΩ using time-dependent perturbation theory and Fermi's golden rule.  If H = H0 + Wexp(±iωt), then the transition probability per unit time to a group of states nearly equal in energy  E = Ei + ħω is given by
w(i,βE) = (2π/ħ)ρ(β,E)|WEi|2δE-Ei,ħω, where WEi = <ΦE|W|Φi>.
Here we consider an initial state with momentum ħk and a final state with momentum ħk'.  The box normalized wave functions are given as
Φi(r) = L-3/2exp(ik∙r),   Φf(r) = L-3/2exp(ik'∙r),
where L is the dimension of the box.
We let W = U(r), ω = 0, and β = dΩ.  Then
WEi = <ΦE|U|Φi> = L-3∫d3r' exp(iq∙r')U(r').
The energy eigenstates of a particle confined to a cubical box with periodic boundary conditions are
Φnx,ny,nz(x,y,z) = L-3exp(i2π(nxx + nyy + nzz)/L),
with nx, ny, nz = 0, ±1, ±2, ...  .
We have kx = 2πnx/L,  ky = 2πny/L,  kz = 2πnz/L.
If k is large, then the number of states with wave vectors whose magnitudes lie between k and k + dk within an element of solid angle dΩ is
dN = k2dΩdk/(2π/L)3  = L3k2dΩdk/(2π)3.
dN/dk = L3k2dΩ/(2π)3.
The density of states is ρ(E,dΩ) = dN/dE = (dN/dk)(dk/dE).
With E = ħ2k2/(2μ) we have dN/dE = μħkdΩL3/(2πħ)3.
We now can insert WEi and ρ(E,dΩ) into the expression for the transition probability per unit time.
w(i,dΩE) = [μħkdΩ/(4π2ħ4L3)]|∫d3r' exp(-iq∙r')U(r')|2.
Here w(i,dΩE) is the probability per unit time that a particle scatters into the solid angle dΩ.  The cross section is the transition rate per incident flux.  The incident flux is the incident probability density times the velocity, |Φi(r)|2v = L-3v = L-3ħk/μ.  Therefore we have
σkB(k,k')dΩ =  [μ2/(4π2ħ4)]|∫d3r' exp(-iq∙r')U(r')|2dΩ.
This derivation makes it clear how to incorporate a spin-dependent potential into the Born approximation.

• Details of the calculation:
(a)  Assume  U = gS1·S2 d3(r) and the two particles are in an eigenstate of S2 and Sz
Then σkB(k,k') =  [g2m2/(4π2ħ4)]|∫d3r' exp(-iq∙r')δ3(r')<S,Ms|S1·S2|S,Ms>|2.
S1S2= ½(S2 - S12 - S22).
S1S2|S,Ms> = (ħ2/2)(S(S+1) - ½(3/2) - ½(3/2))|S,Ms> = (ħ2/2)(S(S+1) - (3/2))|S,Ms>.
For the singlet state we have <0,0|S1·S2|0,0> = -3ħ2/4, and for the triplet state we have <1,Ms|S1·S2|1,Ms> = ħ2/4.
Therefore σkB(k,k') = <S,Ms|S1·S2|S,Ms>2[g2m2/(4π2ħ4)]|∫d3r' exp(-iq∙r')δ3(r')|2.
[g2m2/(4π2ħ4)]|∫d3r' exp(-iq∙r')δ3(r')|2 = g2m2/(4π2ħ4).
For particles in the singlet state we have σkB(k,k')singlet = 9g2μ2/(64π2).
For particles in the triplet state we have σkB(k,k')triplet = g2μ2/(64π2).
(b)  For unpolarized particles we have
σkB(k,k')random = (σkB(k,k')singlet + 3σkB(k,k')triplet)/4 = 3g2μ2/(64π2).

#### Problem:

Consider a periodic scattering potential with translational invariance U(r + R) = U(r), where R is a constant vector.  Show that in the Born approximation scattering occurs only in the directions defined by (ki - kf)·R =  2πn where ki is the initial wave vector, kf is the final wave vector and n is an integer.

Solution:

• Concepts:
The Born approximation
• Reasoning:
We are asked to use the Born approximation
• Details of the calculation:
The elastic scattering cross section in the Born Approximation is
σkB(θ,φ) = σkB(ki,kf) = [μ2/(4π2ħ4)]|∫d3r' exp(-iq∙r')U(r')|2,
where q = kf - kfi, ki = μv0/ħ, kf = μv0/ħ (kf /kf), and μ is the reduced mass.
U is a periodic potential, we can write it in terms of a Fourier series.

U(r) = ∑-∞+∞C(km)exp(ikm∙r).

Here kn = n2π/R(R/R), Δk = kn+1 - kn = 2π/R, n = integer.
(Note: R/R denotes the unit vector in the direction of R.)

∫d3r' exp(-iq∙r')U(r') = ∑-∞+∞C(km)∫d3r' exp(-i(q - km)∙r') = 2π ∑-∞+∞C(km)δ( km - q)
The integral is zero unless ( km - q) = 0, (kf - ki) = km, (kf - ki)·R =  m2π.
Scattering occurs only in the directions defined by (kf - ki)·R =  2πm.