The partial wave method

Problem:

Find σ_k(θ) and σ_k for the scattering of a particle from a perfectly rigid sphere (an infinitely repulsive potential) of radius a.  Choose the energy of the particle such that ka << 1.

Solution:

• Concepts:
  Elastic scattering, the method of partial waves, s-wave scattering.
• Reasoning:
  For very slow particles or very short-range potentials the method of partial waves is the preferred method of calculating the scattering cross section, because only s-waves need to be considered. 
• Details of the calculation:
  V(r) = 0, for r > a, and V(r) = ∞ for r < a.  Since ka << 1, ka << (l(l + 1))^½ for all l except l = 0.  We can neglect all phase shifts except that of the s-wave. 
  Then σ_k(θ) = |f_k(θ)|² = (1/k²)sin²δ₀ and σ_k = (4π/k²)sin²δ₀. 
  To calculate δ₀ we must solve the radial equation for l = 0. 
  (∂²/∂r² + k²)u_k0(r) = 0,  u_k0(a) = 0
  yields u_k0(r) = Csin(kr - ka), r > a and u_k0(r) = 0, r < a.
  As r --> ∞ we expect u_k0(r) to be of the form Csin(kr + δ₀).  We therefore have δ₀ = -ka.  The phase shift is negative, the wave function is "pushed out".
  Since ka << 1 we now have (1/k²)k²a² = a² and σ_k = 4πa².  The total cross section is independent of energy (as long as ka << 1) and equal to four times the geometrical cross section of the hard sphere.  Classical mechanics would have yielded πa².  Low energy scattering means very large wavelength scattering, and we do not necessarily expect a classically reasonable result.
  [But for high energy scattering off a hard sphere we might expect the classical result.  However we obtain σ_k = 2πa² twice the geometrical cross section.  This is called shadow scattering.  If the wavelength of the incident particle is very small compared to a, then the sphere will cast a shadow.  Directly behind the sphere we will find no scattered particles, but the shadow will extend only up to a finite distance.  Very far away from the sphere we will not see the shadow at all, so it must get filled in by scattering of some of the waves at the edges of the sphere.  The scattered flux must have the same magnitude as the flux that is taken out of the incident beam by the geometrical cross section of the sphere in order to fill in the shadow.  The total scattering cross section therefore must have twice the magnitude of the geometrical cross section.]

Problem:

A slow particle is scattered by a spherical potential well of the form V(r) = -V₀, for r < a, V(r) = 0, for r > a. 
(a)  Write down the radial wave equation for this potential and boundary conditions that apply at r = 0, r = a, and r = ∞.
(b)  Assume that the de Broglie wavelength exceeds the dimension of the well, so that s-wave scattering dominates and write solutions both inside and outside the r = a sphere.  Using the continuity conditions at r = a, calculate the phase shift that occurs at this boundary.

Solution:

• Concepts:
  Elastic scattering, the partial wave method
• Reasoning:
  For low energy scattering from a finite-range, central potential, the partial wave method is the preferred method for calculating the scattering cross section.
  We find the scattering cross section by calculating the phase shifts of the partial waves.
• Details of the calculation:
  (a)  (∂²/∂r² + k² - U(r) - l(l + 1)/r²)u_kl(r) = 0 is the radial equation.
  Here E = ħ²k²/(2m),  V(r) = ħ²U(r)/(2m).
  We need u_kl(0) = 0, u_kl(r) and ∂u_kl(r)/∂r to be continuous at r = a, and u_kl(r) to stay finite at infinity.
  (b)  To calculate δ₀ we must solve the radial equation for l = 0.
  (∂²/∂r² + k² + U₀)u_k0(r) = 0,  u_k0(0) = 0, for r < a.
  Therefore  u_k0(r) = C₁sin((k² + U₀)^½r).
  (∂²/∂r² + k²)u_k0(r) = 0, for r > a.
  u_k0(r) = C₂sin(kr + δ₀).
  We need u_k0(r) and ∂u_k0(r)/∂r to be continuous at r = a.  This yields
  C₁sin((k² + U₀)^½a) = C₂sin(ka + δ₀)
  and
  (k² + U₀)^½C₁cos((k² + U₀)^½a) = kC₂cos(ka + δ₀)
  or
  tan(ka + δ₀) = (k/(k² + U₀)^½)tan((k² + U₀)^½a) = C.
  We have
  δ₀ = nπ + tan^-1(C) - ka.

Problem:

A slow particle is scattered by a spherical potential well of the form V(r) = -V₀, for r < a, V(r) = 0, for r > a. 
What must V₀a² be for a 3-dimensional square well potential in order that the scattering cross section be zero at zero bombarding energy (Ramsauer-Townsend effect)?

Solution:

• Concepts:
  Elastic scattering, the partial wave method
• Reasoning:
  As E --> 0, k --> 0, ka --> 0.  If ka << 1 then the scattering is dominated by s-wave scattering and we use the partial wave method to calculate the scattering cross section.
  Then σ_k(θ) = (1/k²)sin²δ₀ and σ_k = (4π/k²)sin²δ₀.
• Details of the calculation:
  When δ₀ = π, then σ_k(θ) = σ_k = 0.
  To find what V₀a² has to be for δ₀ = π, we need to find the expression for a 3-dimensional square-well potential.
  To calculate δ₀ we must solve the radial equation for l = 0.
  Let E = ħ²k²/(2m),  V₀ = ħ²U₀/(2m).  Then
  (∂²/∂r² + k² + U₀)u_k0(r) = 0,  u_k0(0) = 0, for r < a.
  Therefore  u_k0(r) = C₁sin((k² + U₀)^½r).
  (∂²/∂r² + k²)u_k0(r) = 0, for r > a.
  u_k0(r) = C₂sin(kr + δ₀).
  We need u_k0(r) and ∂u_k0(r)/∂r to be continuous at r = a.  This yields
  C₁sin((k² + U₀)^½a) = C₂sin(ka + δ₀)
  and
  (k² + U₀)^½C₁cos((k² + U₀)^½a) = kC₂cos(ka + δ₀)
  or
  tan(ka + δ₀) = (k/(k² + U₀)^½)tan((k² + U₀)^½a).
  We want δ₀ = π.  Then
  tan(ka) = (k/(k² + U₀)^½)tan((k² + U₀)^½a).
  As ka --> 0 this becomes
  ka = (k/U₀^½)tan(U₀^½a)  or  U₀^½a = tan(U₀^½a),
  U₀^½a = nπ + b.
  We are not interested in the solution n = 0, b = 0 for the free particle.
  n = 1, b = 1.352,  U₀^½a = 4.494,  tan(U₀^½a) = 4.497.
  V₀a² = ħ²U₀a²/(2m) = 20.2 ħ²/(2m).
  The Ramsauer-Townsend effect is analogous to the perfect transmission found at particular energies in a one dimensional problem.  It is the explanation for the extremely low minimum observed in the scattering cross section of electrons by rare gas atoms at about 0.7 eV bombarding energy.  The Ramsauer-Townsend effect cannot occur with a repulsive potential, since ka would have to be at least π to make |δ₀| = π.  The perfectly rigid sphere produces the largest |δ₀| for any repulsive potential with range a, |δ₀| = ka.   A potential of this large range produces higher l phase shifts.

Problem:

Determine the differential scattering cross section σ(θ) in units of cm²/sr for a particle of mass m = 9.1*10^-31kg incident on a spherically symmetric potential
V(r) = 0,  0 < r < a,  V(r) = V₀,  a < r < b,  V(r) = 0, r > b,
with a = 0.05 nm and b = 0.1 nm.  Let E = 1 eV and V₀ = 0.8 eV.

Solution:

• Concepts:
  Elastic scattering, the method of partial waves, s-wave scattering.
• Reasoning:
  E = ħ²k²/(2m), k² = 2mE/ħ², k = 5.1*10⁹/m, kb = 0.51 < 1.
  Only s-wave scattering is important.  We have σ_k(θ) = (1/k²)sin²δ₀.
• Details of the calculation:
  To calculate δ₀ we must solve the radial equation for l = 0.
  (∂²/∂r² + k² - U(r))u_k0(r) = 0,  V(r) = ħ²U(r)/(2m).
  r < a:  U(r) = 0, u_k0(r) = C₁sin(kr).
  a < r < b: U(r) = U₀,  V₀ = ħ²U₀/(2m),  E > V₀.
  Therefore  u_k0(r) = C₂sin((k² - U₀)^½r + δ') = C₂sin((k₁r + δ').
  r > a: U(r) = 0, u_k0(r) = C₃sin(kr + δ₀).
  We need u_k0(r) and ∂u_k0(r)/∂r to be continuous at r = a and r = b. 
  This yields at r = a
  C₁sin(ka) = C₂sin(k₁a + δ'),
  kC₁cos(ka) = k₁C₂cos(k₁a + δ'),
  or
  δ' = tan^-1((k₁/k)tan(ka)) - k₁a.
  k₁/k = ((E-V₀)/E)^½ = 0.447,  ka = 0.256,  δ' = 0.002.
  At r = b we have
  C₂sin((k₁b + δ') = C₃sin(kb + δ₀),
  k₁C₂cos((k₁b + δ') = kC₃cos(kb + δ₀),
  or
  δ₀ = tan^-1((k/k₁)tan(k₁b + δ')) - kb.
  δ₀ = -2.8*10^-2.
  σ_k(θ) = (1/k²)sin²δ₀ = 3*10^-19 cm²/sr.

Problem:

Calculate the differential and total elastic scattering cross section for an electron of energy 0.25 eV incident on a spherical positive potential of form V(r) = 2 eV (r < 4a₀) and V(r) = 0 (r > 4a₀).  Justify your method.  Give numerical answers.

Solution:

• Concepts:
  Elastic scattering, the partial wave method
• Reasoning:
  E = ħ²k²/(2m), k² = 2mE/ħ², k = 2.56*10⁹/m, k4a₀ = 0.54 < 1.
  Only s-wave scattering is important.  We have σ_k(θ) = (1/k²)sin²δ₀.

• Details of the calculation:
  To calculate δ₀ we must solve the radial equation for l = 0.
  (∂²/∂r² + k² - U(r))u_k0(r) = 0,  V(r) = ħ²U(r)/(2m).
  Let a = 4a₀.
  r < a:  U(r) = U₀,  V₀= ħ²U₀/(2m),  E = ħ²k²/(2m),  E < V₀.
  Therefore  u_k0(r) = C₁sinh((U₀ - k²)^½r).
  r > a:  U(r) = 0,  u_k0(r) = C₂sin(kr + δ₀).
  We need u_k0(r) and ∂u_k0(r)/∂r to be continuous at r = a. 
  This yields at r = a
  C₁sinh((U₀ - k²)^½a) = C₂sin(ka + δ₀),
  (U₀ - k²)^½C₁cosh((U₀ - k²)^½a) = kC₂cos(ka + δ₀),
  or
  tan(ka + δ₀) = (k/(U₀-k²)^½)tanh((U₀ - k²)^½a) = C.
  δ₀ = tan^-1(C) - ka.
  U₀ = 2mV₀/ħ²  = 5.24*10¹⁹ m^-2.
  k² = 6.55*10¹⁸ m^-2.
  ka = 0.54.
  (U₀ - k²)^½a = 1.43,  k/(U₀ - k²)^½ = 0.376.
  C = 0.335,  tan^-1(C) = 0.323,  δ₀ = -0.21.
  σ_k(θ) = (1/k²)sin²δ₀ = 7*10¹⁷ cm²/sr.
  σ_k = 8.85*10^-16 cm².

Problem:

Determine the total cross section for the scattering of slow particles (ka < 1) by a potential V(r) = Cδ(r-a).

Solution:

• Concepts:
  Elastic scattering, the method of partial waves, s-wave scattering.
• Reasoning:
  For very slow particles or very short-range potentials the method of partial waves is the preferred method of calculating the scattering cross section, because only s-waves need to be considered.
• Details of the calculation:
  σ_k = (4π/k²)sin²δ₀.
  We need to find the phase shift δ₀ for V(r) = Cδ(r-a).
  
  To calculate δ₀ we must solve the radial equation for l = 0.
  (∂²/∂r² + k² - 2mV(r)/ħ²)u_k0(r) = 0.
  Region 1:  V(r) = 0, u¹_k0(r) = C₁sin(kr).
  Region 2:  V(r) = 0, u²_k0(r) = C₂sin(kr + δ₀).
  Boundary conditions:
  u¹_k0(a) = u²_k0(a), C₁sin(ka) = C₂sin(ka + δ₀).
  ∂u²_k0(r)/∂r|_a+ε - ∂u¹_k0(r)/∂r|_a-ε = (2m/ħ²)Cu(a).
  kC₁cos(ka) + (2mC/ħ²)C₁sin(ka) = kC₂cos(ka + δ₀).
  kcot(ka) + 2mC/ħ² = kcot(ka + δ₀).
  δ₀ = cot^-1(cot(ka) + 2mC/(kħ²)) - ka.