Now let us study the situation when V(r) is not zero. In any central
potential V(r), the common eigenfunctions of H, L2 and
Lz are of the form 
Here
ukl(r) is the solution to the radial equation
With 
we have
with the boundary condition ukl(0)=0. If U(r)®0 as r®¥
the equation reduces to
with the general solution ukl(r)=Aeikr+Be-ikr.
[We expect the solution of the eigenvalue equation to behave as ukl(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 ukl(r)=Ce±ikr for large r.
For the Coulomb potential
Here C is a constant. 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 ukl(r)=Aeikr+Be-ikr.
is a superposition of an incoming and
outgoing spherical wave. Since we have a stationary state, we need |A|=|B|.
We
therefore have
The real phase bl 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 
We take this value as a reference point and write
d
l is called the phase shift of the lth partial wave.The asymptotic form of a free spherical wave is
At infinity it is a superposition of a incoming and an outgoing wave with equal amplitudes an a phase difference of lp. The asymptotic form of a partial wave is
with 
At infinity it is a
superposition of an incoming and an outgoing wave, but the outgoing wave has accumulated a
phase shift 2dl relative to the free outgoing
wave in a null potential V(r)=0.
If V(r)=0 for r>r0 and bl(k)=(1/k)(l(l+1))1/2>r0,
then the potential has virtually no effect on the partial wave fklm(r).
The incoming wave turns back before reaching the zone of influence of V(r).
For
partial waves fklm(r) such that 
the phase shift dl
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.
We now want to find a linear superposition of partial waves whose asymptotic behavior is of the form
The plane wave 
may be expanded in
terms of free spherical waves. It does not depend of f.
is an eigenfunction of Lz
with eigenvalue 
We therefore find
that
Multiplying by Pl’(cosq) and integrating over the solid angle yields
Since 
the right hand side yields 
is a well-known integral
representation of the spherical Bessel function jl’(kr).
We therefore have 
and 
As 
Symmetry demands that in a spherically symmetric potential fk(r,q,f) and fk(q,f) are also independent of f. We write
i.e. fk(r,q) is an expanded in terms of Legendre polynomials. We want
Using 
this equation becomes
The constants bl are found by equating the coefficients of e-ikrPl(cosq) on the two sides of this equation. 
We can then find fk(q) by equating the coefficients of eikrPl(cosq) on the two sides of the equation.
Therefore
and
Since 
the cross products in
the summation vanish and we have
This can also be written as
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 dl for all l. The method is only practical if there are only a small number of non zero phase shifts.
The scattering cross section s(q) depends on the energy of the incident particle and on the scattering angle q. Rapid variations of the total cross section in the neighborhood of certain energies are called scattering resonances.
 | Solution: V(r)=0, for r>a, and V(r)= ¥ for r<a. Since ka <<1,  for all l
except l=0. We can neglect all phase shifts except that of the s-wave.
Then  and 
To calculate d0
we must solve the radial equation for l=0.
yields uk0(0)=Csin(kr-ka), r>a and uk0(0)=0, r<a.
As r ®¥ we expect ukl(r) to be of
the form We now have [But for high energy scattering off a hard sphere we might expect the classical result. However we obtain sk=2pa2 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.] |
| Solution: To calculate d0 we must solve the radial equation for l=0.
Here Therefore
We need
and
or
We have
ka=0.54, ka<1, s-wave scattering dominates.
The phase shift is negative, the wavefunction is "pushed out". |
| Solution:
|
Let us look in more detail at low energy scattering. We want to investigate the dependence of d0 on the depth of the potential well. If ka<<1 and d0<1, then tan(ka+d0) »tand0+ka. We have
If U0 is very shallow then tand0 »d0 is positive and very small, d0 lies in the first quadrant.
If we increase the
depth of the well the phase shift increases. The phase shift is positive, the wavefunction
is "pulled in". At some point 
will go through p/2.
This is the
condition for the appearance of a bound state. The well is now just deep enough to support
one bound state. When 
we have d0=p/2.
The phase
shift will go through p/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 wavefunction inside the well will increase and the wavefunction will be" pulled in" further; d0 will increase and at some point will become p. The cross section is zero when the phase shift is p, 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 p to make d0=p. A potential of this large range produces higher l phase shifts.
As the depth of the potential increases further, we again have
The phase shift d0, however now lies in the third quadrant. We write
If we make the potential deeper still, a second bound state will appear when 
Again we have a resonance. In general we
write
where n is the number of bound states supported by the potential well.
We therefore have d0 =-ka. The phase shift is negative, the
wavefunction is "pushed out".
and sk=4pa2.
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 pa2. Low energy scattering means
very large wavelength scattering, and we do not necessarily expect a classically
reasonable result.
to be continuous at r=a.
This yields 
is the radial equation. 
to be continuous at r=a, and ukl(r)
to stay finite at infinity.
to be continuous at r=a.
This yields 
