Integral scattering equation for stationary states

Integral expression for the scattering amplitude

The eigenvalue equation HΦ(r) = EΦ(r),  H = p²/(2m) + (ħ²/(2m))U(r),  E > 0,
can be put into the form [∇² + k² - U(r)]Φ(r) = 0,  with  E = ħ²k²/(2m).

The solutions to this equation may be written in the form
Φ(r) = Φ₀(r) + ∫d³r'G(r - r')U(r')Φ(r'),
where Φ₀(r) is a solution of the homogeneous equation
[∇² + k²]Φ₀(r) = 0, and G(r) is a solution of  [∇² + k²]G(r) = δ(r).

Verify:
[∇² + k²]Φ(r) = [∇² + k²]Φ₀(r) + [∇² + k²]∫d³r'G(r - r')U(r')Φ(r')
                                0              acts only on r

= ∫d³r' [∇² + k²]G(r - r')U(r')Φ(r') = U(r)Φ(r).
                 δ(r - r')

The solutions of [∇² + k²]G(r) = δ(r) are G^±(r) = -(4π)^-1e^±ikr/r.  These are two linearly independent solutions of a second order differential equation.

To verify that these are the solutions use ∇(ab) =  a∇b + b∇a.
∇²(ab) =  ∇∙∇(ab) = ∇a∙∇b + b∇²a + ∇b∙∇a + a∇²b,
and  ∇²(1/r) = -4πδ(r),
with a = e^±ikr/r and b = 1/r.

For a stationary scattering state one can choose Φ₀(r) = e^ikz and G(r) = G⁺(r).
Φ_k(r) = e^ikz + ∫d³r'G⁺(r - r')U(r')Φ_k(r').

If |r'| << |r| then |r - r'| ≈  r - (r/r)∙r', and G⁺(r - r') ≈ -(4π)^-1e^ikr/r exp(-ik(r/r)∙r').
Therefore
Φ_k(r) = e^ikz - (e^ikr/(4πr))∫d³r' exp(-ik(r/r)∙r')U(r')Φ_k(r')
= e^ikz + f_k(θ,φ) e^ikr/r.
This yields an integral expression for the scattering amplitude f_k(θ,φ).

The Born approximation

Φ_k(r) = e^ikz - (e^ikr/(4πr))∫d³r' G⁺(r - r')U(r' e^ikz' + ∫d³r'∫d³r''G⁺(r - r')U(r')G⁺(r - r'')U(r'')Φ_k(r'').

This procedure can be repeated and yields the Born expansion.
The Born approximation is the first term in the Born expansion.
The scattering amplitude in the Born approximation is given by

f_k^B(θ,φ) = -(4π)^-1∫d³r' exp(-ik(r/r)∙r') U(r') e^ikz'
= -(4π)^-1∫d³r' exp(-ik_s∙r') U(r') exp(ik_i∙r')
= -(4π)^-1∫d³r' exp(-i(k_s - k_i)∙r') U(r')
= -(4π)^-1∫d³r' exp(-iK∙r') U(r').
Here k_s = k(r/r) and K = k_s - k_i = momentum transfer/ħ.

The scattering amplitude is proportional to the Fourier transform of the potential energy.  The differential scattering cross section in the Born approximation is

σ_k^B(θ,φ) = σ_k^B(k_s,k_i) = [m²/(4п²ħ⁴)] |∫d³r' exp(-iK∙r')V(r')|².

The differential scattering cross section is proportional to the square of the Fourier transform of the potential energy.
We often want to know the scattering cross section as a function of the scattering angle and not as a function of the momentum transfer.  Let θ be the angle between k_s and  k_i .  Then K = 2ksin(θ/2).

The Yukawa potential

Consider a spherically symmetric potential energy function,  V(r) = V₀ e^-ar/r,  a > 0.
V(r) --> V₀/r as a --> 0.  Then the scattering amplitude in the Born approximation is given by
f_k^B(θ,φ) = -(4π)^-1(2mV₀/ħ²)∫∫∫r'dr'sinθ'dθ'dφ' exp(-ar') exp(-iK∙r').
Choose the coordinate system such that K = K(z/z).  Then

∫∫∫r'dr'sinθ'dθ'dφ' exp(-ar') exp(-iK∙r')
= 2π∫₀^∞r'dr'exp(-ar')∫_-1¹exp(-iKr'cosθ')dcosθ'
= -(2π/K)∫₀^∞dr'exp(-ar')∫_-Kr'^Kr'exp(-ix)dx
= -(2π/K)∫₀^∞dr'exp(-ar')[-i(e^iKr' - e^-iKr')]
= -(4π/K)∫₀^∞dr'exp(-ar')sin(Kr') = -4π/(a² + K²).

f_k^B(θ,φ) = f_k^B(θ) = -(2mV₀/ħ²)/(a² + K²).
σ_k^B(θ,φ) = σ_k^B(θ) = [4m²V₀²/ħ⁴]/(a² + K²)².

In terms of the scattering angle we have
σ_k^B(θ) = [4m²V₀²/ħ⁴]/(a² + 4k²sin²(θ/2))².

The total scattering cross section is
σ_k^B = ∫∫σ_k^B(θ)sinθdθdφ =  [4m²V₀²/ħ⁴] 4π/(a²(a² + 4k²)).

The Coulomb potential

Let a --> 0, V₀ = Z₁Z₂e². Then

σ_k^B(θ,φ) = [4m²Z₁²Z₂²e⁴/ħ⁴]/(16k⁴sin⁴(θ/2)) = Z₁²Z₂²e⁴/(16E²sin⁴(θ/2)).

This is Rutherford’s formula.  The Born approximation for the Yukawa potential gives Rutherford’s formula as a a --> 0.  The total cross section σ_k^B --> ∞.  The total cross section is infinite, because the Coulomb potential has infinite range.

When is the Born approximation a good approximation?

In the integral ∫d³r' exp(-ik(r/r)∙r') U(r') Φ_k(r')we are replacing the exact solution Φ_k(r') by the asymptotic incident wave e^ikz'.  Therefore Φ_k(r') and e^ikz' should not be too different inside the range of the potential, i.e. in the region where U(r) contributes appreciably to the integral.

We therefore need that |(4π)^-1∫d³r'( exp(-ik|r - r'|)/|r - r'|)U(r')Φ_k(r')| ≈ |(4π)^-1∫d³r'( exp(-ik|r - r'|)/|r - r'|)U(r')e^ikz'| << 1.
In the high k limit this inequality is easily satisfied because the integrand oscillates rapidly.  We can also satisfy the condition if the scattering potential is weak.

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
  σ_k^B(θ,φ) = σ_k^B(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 = m_e, 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' ρ_t(r')/|r - r'|,
  where ρ_t(r') = 2 δ(r') - ρ(r').

  ∫d³r exp(-iq∙r)U(r) = -e²∫d³r exp(-iq∙r)∫d³r' ρ_t(r')/|r - r'|
  = -e²∫d³r' ρ_t(r')exp(-iq∙r')∫d³r exp(-iq∙(r - r')/|r - r'|
  = -(4πe²/q²)∫d³r' ρ_t(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(iq∙r)/r = 2π∫₀^∞dr r e^-λr∫_-1¹d(cosθ)e^iqrcosθ
  = 2π∫₀^∞dr e^-λr(e^iqr - e^-iqr)/(iq) = (4π/q)Im(∫₀∞dre^-(λ-iq)r)
  = (4π/q)Im(1/(λ - iq)) = 4π/(λ² + q²).
  Take the limit as λ --> 0.]

  σ_k^B(θ,φ)  = [4m_e²e⁴/(q⁴ħ⁴)][2 - F(q)]².
  q² = 4k²sin²(θ/2).
  σ_k^B(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.