Physics 472 Solutions Problem Set 12 Spring 2007
48. Squaring the expression for f(θ) and integrating over dΩ gives
ZdΩ|f(θ)|2= 2π1
k2
∞
X
`, ` 0=0
(2`+ 1)(2`0+ 1)ei(δ`−δ`0)sin δ`sin δ`0Zπ
0dθ sin θP`(cos θ)P`0(cos θ)
= 2π1
k2
∞
X
`, ` 0=0
(2`+ 1)(2`0+ 1)ei(δ`−δ`0)sin δ`sin δ`0
2
(2`+ 1)δ`` 0
σT=4π
k2
∞
X
`=0
(2`+ 1) sin2δ`.
The imaginary part of the scattering amplitude for θ= 0 is
Im[f(θ= 0)] = 1
k
∞
X
`=0
(2`+ 1)Im[eiδ`] sin δ`P`(cos θ= 0)
=1
k
∞
X
`=0
(2`+ 1) sin2δ`P`(1)
Im[f(θ= 0)] = k
4πσT
since P`(1) = 1 for all `.
49. (a) The equation for the radial wave function u0(r) in this case is
d2u0
dr2+k2u0−α
aδ(r−a)u0= 0 .
Since the δ-function only affects the boundary condition at r=a, the solution for
u0(r) is
u0(r) = (Asin(kr) + Bcos(kr) for 0 ≤r < a
Csin(kr) + Dcos(kr) for r > a
The vanishing of u0(r) at r= 0 gives B= 0 and the boundary conditions at r=a
are
u0(a+) = u0(a−)
u0
0(a+)−u0
0(a−) = α
au0(a),
which give the equations
Asin(ka) = Csin(ka) + Dcos(ka)
Aµcos(ka) + α
ka sin(ka)¶=Ccos(ka)−Dsin(ka).
