Chapter 8 Scattering Theory - Particle Physics Group

CHAPTER 8. SCATTERING THEORY 146
we can write the radial equation inside the well as
[ d
2
dr + 2 ]
d l(l + 1)
− + K 2 R 2 r dr r 2 l (k,r) = 0 for r < a (8.78)
with K 2 = k 2 + U 0 . Inside the well the regular solution is thus
R in
l (K,r) = N l j l (Kr), r < a (8.79)
where N l is related to the exact solution in the exterior region
R ext
l (k,r) = B l (k)[j l (kr) − tanδ l (k)n l (kr)], r > a (8.80)
by the matching condition at r = a. Continuity of R and R ′ at r = a hence implies
N l j l (Ka) = B l (j l (ka) − tanδ l n l (ka)), (8.81)
KN l j ′ l(Ka) = kB l (j ′ l(ka) − tanδ l n ′ l(ka)). (8.82)
The ratio of these two equations yields an equation for tanδ l (k) whose solution is
with K = √ k 2 + U 0 .
tanδ l (k) = kj′ l (ka)j l(Ka) − Kj l (ka)j ′ l (Ka)
kn ′ l (ka)j l(Ka) − Kn l (ka)j ′ l (Ka) (8.83)
In the low energy limit k = √ 2mE/ → 0 we can insert the leading behavior j l (ρ) ∝ ρ l
and n l (ρ) ∝ ρ −l−1 , and thus for the derivatives j ′ l (ρ) ∝ ρl−1 and n ′ l (ρ) ∝ ρ−l−2 , to conclude that
tanδ l (k) goes to zero like a constant times k l /k −l−1 = k 2l+1 . In this limit the cross section,
σ l ∝ k 4l , (8.84)
is dominated by l = 0 so that the scattering probability approximately goes to a θ-independent
constant. With
j 0 (ρ) = sin ρ
ρ ,
n 0(ρ) = − cos ρ
ρ ,
j′ 0(ρ) =
ρcos ρ−sin ρ
ρ 2 , n ′ 0(ρ) =
ρsin ρ+cos ρ
ρ 2 (8.85)
and the abbreviations x = ka, X = Ka we find
tan δ 0 =
((xcos x−sin x) sin X−sin x(X cos X−sin X))/(xX)
(xsin x+cos x) sin X+cos x(X cos X−sin X))/(xX)
Dividing numerator and denominator by cosxcos X we obtain the result
tanδ 0 (k) =
=
xcos xsin X−X sin x cos X
xsin x sin X+X cos x cos X . (8.86)
k tan(Ka) − K tan(ka)
K + k tan(ka) tan(Ka) . (8.87)
For k → 0 we observe that tanδ 0 becomes proportial to k. The limit
a s = − lim
k→0
tanδ 0 (k)
k
(8.88)