Partial waves

Now let us study the situation when V(r) is not zero.  In any central potential energy function V(r), the common eigenfunctions of H, L² and L_z are of the form
Φ_klm(r) = R_kl(r)Y_lm(θ,φ) = [u_kl(r)/r]Y_lm(θ,φ).
[-(ħ²/(2m))(∂²/∂r²) + ħ²l(l + 1)/(2mr²) + V(r)]u_kl(r) = [ħ²k²/(2m)]u_kl(r).

With V(r) = (ħ²/(2m))U(r) we have
[(∂²/∂r²) - l(l + 1)/(r²) - U(r) + k²]u_kl(r) = 0.
with the boundary condition u_kl(0) = 0.
If U(r) --> 0 as r --> ∞ the equation reduces to [(∂²/∂r²) +  k²]u_kl(r) = 0|_r-->∞,
with the general solution u_kl(r) =Ae^ikr + Be^-ikr.

[We expect the solution of the eigenvalue equation to behave as u_kl(r) = F(r)e^±ikr for large r, with F(r) a slowly varying function of r, as long as U(r) --> 0 as r --> ∞.  If rU(r) --> 0 for large r, it can be shown that F(r) converges and u_kl(r) = Ce^±ikr for large r.
For the Coulomb potential U(r) = 2a/r,  F(r) ∝ e^{∓i(a/k)ln(r)} and u_kl(r) = Ce^{±i(kr - (a/k)ln(r))} as r --> ∞.
The radial solution never approaches the sinusoidal free particle solution, since there is always a logarithmic contribution to the phase at great distances, that cannot be neglected.]

For a potential that falls off faster than the Coulomb potential as r approaches infinity, the asymptotic behavior of the stationary state partial waves is u_kl(r) = Ae^ikr + Be^-ikr.  u_kl(r)/r is a superposition of an incoming and outgoing spherical wave.  Since we have a stationary state, we need |A|=|B|.  We therefore have
u_kl(r) = |A|(e^ikre^iφA + e^-ikre^iφB) = Csin(kr - β_l), with β_l = (π + φ_B - φ_A)/2 and C = -2|A|exp(i(φ_A + φ_B)/2).
The real phase β_l is found by matching the asymptotic solution to the solution in the region where U(r) is not zero.
If U(r) is the null potential, (a free particle), then u_kl⁰(r) = Csin(kr - lπ/2) as r --> ∞.
We take this value as a reference point and write

u_kl(r) = Csin(kr - lπ/2 + δ_l) as r --> ∞.
δ_l is called the phase shift of the l^th partial wave.

The asymptotic form of a free spherical wave is
Φ_klm⁰(r) = (2k²/π)^½(kr)^-1sin(kr - lπ/2))Y_lm(θ,φ).
At infinity it is a superposition of a incoming and an outgoing wave with equal amplitudes an a phase difference of lπ.
The asymptotic form of a partial wave is
Φ_klm(r) = C(kr)^-1sin(kr - lπ/2 + δ_l))Y_lm(θ,φ).
At infinity it is a superposition of an incoming and an outgoing wave, but the outgoing wave has accumulated a phase shift 2δ_l relative to the free outgoing wave in a null potential V(r) = 0.

If V(r) = 0 for r > r₀ and b_l(k) = (1/k)(l(l + 1))^1/2 > r₀, then the potential has virtually no effect on the partial wave Φ_klm(r).  The incoming wave turns back before reaching the zone of influence of V(r).  For partial waves Φ_klm(r) such that (l(l + 1))^1/2 > kr₀,  the phase shift δ_l is therefore approximately zero.  The shorter the range of the potential or the lower the incident energy, the fewer partial waves have non zero phase shift.

The scattering cross section in terms of phase shifts

We now want to find a linear superposition of partial waves whose asymptotic behavior is of the form
Φ_k(r,θ,φ) = e^ikz + f_k(θ,φ) e^ikr/r.
The plane wave e^ikz = <r|p(z/z)> may be expanded in terms of free spherical waves.  It does not depend of φ.
|p(z/z)> = Σ_l=0^∞c_kl|k,l,0>, since e^{ikz =} e^ikrcosθ is an eigenfunction of L_z with eigenvalue mħ = 0.  We therefore find that
e^ikz =  Σ_l=0^∞c_kl<r|k,l,0> = Σ_l=0^∞c_kl(2k²/π)^½j_l(kr)Y_lm(θ,φ) = Σ_l=0^∞c'_kl j_l(kr) P_l(cosθ)

Multiplying by P_l’(cosθ) and integrating over the solid angle yields
∫₀^πsinθ dθ e^ikrcosθ P_l’(cosθ) = Σ_l=0^∞c'_kl j_l(kr)∫₀^πsinθ dθ P_l’(cosθ)P_l(cosθ).
∫₀^πsinθ dθ P_l’(cosθ)P_l(cosθ) = δ_ll' 2/(2l + 1).
Therefore
∫₀^πsinθ dθ e^ikrcosθ P_l’(cosθ) = c'_kl' j_l'(kr)2/(2l' + 1).
∫₀^πsinθ dθ e^ikrcosθ P_l’(cosθ) is a well-known integral representation of the spherical Bessel function j_l’(kr).
∫₀^πsinθ dθ e^ikrcosθ P_l’(cosθ) = 2i^l' j_l’(kr).

We therefore have c'_kl = (2l + 1)i^l and e^ikz = Σ_l=0^∞(2l + 1)i^l j_l(kr) P_l(cosθ).
As r --> ∞ e^ikz = Σ_l=0^∞(2l + 1)i^l (kr)^-1sin(kr - lπ/2) P_l(cosθ).

Symmetry demands that in a spherically symmetric potential Φ_k(r,θ,φ) and f_k(θ,φ) are also independent of φ.  We write
Φ_k(r,θ) = Σ_l=0^∞b_l (kr)^-1sin(kr - lπ/2 + δ_l) P_l(cosθ),
i.e. Φ_k(r,θ) is an expanded in terms of Legendre polynomials.  We want
Σ_l=0^∞b_l (kr)^-1sin(kr - lπ/2 + δ_l) P_l(cosθ = Σ_l=0^∞(2l + 1)i^l (kr)^-1sin(kr - lπ/2) P_l(cosθ) + f_k(θ) e^ikr/r.
Using sinx = (e^ix - e^-ix)/(2i) this equation becomes
e^ikr[2ik f_k(θ) + Σ_l=0^∞(2l + 1)i^l e^-ilπ/2 P_l(cosθ)] - e^-ikrΣ_l=0^∞(2l + 1)i^l e^ilπ/2 P_l(cosθ)
= e^ikrΣ_l=0^∞b_l e^-ilπ/2 exp(iδ_l) P_l(cosθ) - e^-ikrΣ_l=0^∞b_l e^ilπ/2 exp(-iδ_l) P_l(cosθ).

The constants b_l are found by equating the coefficients of e^-ikr P_l(cosθ) on the two sides of this equation.
b_l = (2l + 1)i^l exp(iδ_l).
We can then find f_k(θ) by equating the coefficients of e^ikr on the two sides of the equation.
f_k(θ) = (2ik)^-1∑_l=0^∞(2l+1)(exp(2iδ_l) - 1)P_l(cosθ) = (1/k)∑_l=0^∞(2l+1)exp(iδ_l)sinδ_lP_l(cosθ).

Therefore
σ_k(θ) = dσ_k/dΩ = |f_k(θ)|² = (1/k²)|∑_l=0^∞(2l + 1)exp(iδ_l)sinδ_lP_l(cosθ)|²,
and
σ_k = ∫∫σ_k(θ)sinθdθdφ = (2п/k²)∑_l=0^∞∫[(2l + 1)sinδ_lP_l(cosθ)]²,
σ_k = (4п/k²)∑_l=0^∞(2l+1)sin²δ_l.

Since  ∫₀^πsinθ dθ P_l’(cosθ)P_l(cosθ) = δ_ll' 2/(2l+1), the cross products in the integral over the summation vanish.

σ_k  can also be written as σ_k = (4п/k²) Im(f_k(0)).
The relationship between the total cross section and the forward scattering amplitude is called the optical theorem.  It also remains true for inelastic scattering processes.  The total cross section represents the removal of flux from the incident beam.  This removal results from destructive interference between the incident and the scattered beam in the forward direction and is proportional to the imaginary part of f(0).

Note: Finding the cross section by the method of partial waves necessitates finding the phase shift δ_l for all l.  The method is only practical if there are only a small number of non zero phase shifts.
The scattering cross section σ(θ) depends on the energy of the incident particle and on the scattering angle θ.  Rapid variations of the total cross section in the neighborhood of certain energies are called scattering resonances.

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.  T
  hen 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:

In three dimensions, find the s-wave phase shift of an electron with kinetic energy E = 1 eV scattering from the repulsive square well potential V(r) = V₀, for r < a, V(r) = 0, for r > a, where V₀ > E.  Find both the analytical and numerical result (in radians).  Let a = 2a_B = 1.06 Å, m_e = 9.1*10^-31 kg, V₀ = 5 eV.

Solution:

• Concepts:
  Elastic scattering, the partial wave method
• Reasoning:
  E = ħ²k²/(2m), k² = 2mE/ħ², k = 5.19*10⁹/m, ka = 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).
  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  = 1.31*10²⁰ m^-2.
  k² = 2.62*10¹⁹ m^-2.
  (U₀ - k²)^½a = 1.09,  k/(U₀ - k²)^½ = 0.5.
  C = 0.398,  tan^-1(C) = 0.378,  δ₀ = -0.165.
  The phase shift is negative, the wave function is "pushed out".

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.

Low energy scattering

Let us look in more detail at low energy scattering.  We want to investigate the dependence of δ₀ on the depth of the potential well U₀.  If ka << 1 and δ₀ < 1, then tan(ka + δ₀) ≈ tanδ₀ + ka.  We have
tanδ₀ = [ka/((k² + U₀)^½a)]tan((k² + U₀)^½a) - ka = ka[tan((k² + U₀)^½a)/((k² + U₀)^½a) - 1].

If U₀ is very shallow then tanδ₀ ≈ δ₀ is positive and very small, δ₀ lies in the first quadrant.  If we increase the depth of the well the phase shift increases.  The phase shift is positive, the wave function is "pulled in".  At some point (k² + U₀)^½a will go through π/2.  The well becomes just deep enough to support one bound state when U₀^½a = π/2.  When (k² + U₀)^½a = π/2, we have δ₀ = π/2.  The phase shift will go through π/2.  The scattering cross section then is proportional to k^-2 and goes to infinity as the energy goes to zero.  This is called a resonance.  It occurs when there is a bound state with E approximately zero.

If we further increase the depth of the well, the curvature of the wave function inside the well will increase and the wave function will be" pulled in" further.  δ₀ will increase and at some point will become π.  The cross section is zero when the phase shift is π, s-wave scattering does not contribute to the scattering cross section.  This is the explanation for the Ramsauer-Townsend effect, the extremely low minimum observed in the scattering cross section of electrons by rare gas atoms at about 0.7 eV bombarding energy.  This effect is analogous to the perfect transmission found at particular energies in a one dimensional problem.  The Ramsauer-Townsend effect cannot occur with a repulsive potential, since ka would have to be at least π to make δ₀ = π.  A potential of this large range produces higher l phase shifts.

As the depth of the potential increases further, we again have
tanδ₀ = ka[tan((k² + U₀)^½a)/((k² + U₀)^½a) - 1].
The phase shift δ₀, however now lies in the third quadrant.  We write
tan(δ₀ - π) =  ka[tan((k² + U₀)^½a)/((k² + U₀)^½a) - 1].
δ₀ = π +  tan^-1[ka tan((k² + U₀)^½a)/((k² + U₀)^½a)] - ka.
If we make the potential deeper still, the well will be able to support a second bound state when U₀^½a = 3π/2.  Again we have a resonance when (k² + U₀)^½a = 3π/2.  In general we write
δ₀ = nπ +  tan^-1[ka tan((k² + U₀)^½a)/((k² + U₀)^½a)] - ka.
where n is the number of bound states supported by the potential well.