With just theℓ= 0 term, the differential scattering cross section is.
dσ
dΩ→
sin^2 (δℓ)
k^2The cross section will have zeros when
k′
k= cot(ka) tan(k′a)
k′cot(k′a) =kcot(ka).There will be many solutions to this and the cross section will look like diffraction.
there will be many solutions to this
log σ
E
diffractive scattering
15.8 The Radial Equation foru(r) =rR(r)*.
It is sometimes useful to use
unℓ(r) =rRnℓ(r)
to solve a radial equation problem. We can rewrite the equation foru.
(
d^2
dr^2
+
2
rd
dr)
u(r)
r=
d
dr(
1
rdu
dr−
u
r^2)
+
2
r^2du
dr−
2 u
r^3=1
rd^2 u
dr^2−
1
r^2du
dr−
1
r^2du
dr+
2 u
r^3+
2
r^2du
dr−
2 u
r^3=
1
rd^2 u
dr^2
1
rd^2 u(r)
dr^2+
2 μ
̄h^2[
E−V(r)−ℓ(ℓ+ 1) ̄h^2
2 μr^2]
u(r)
r= 0
d^2 u(r)
dr^2+
2 μ
̄h^2[
E−V(r)−
ℓ(ℓ+ 1) ̄h^2
2 μr^2]
u(r) = 0This now looks just like the one dimensional equation except the pseudo potential due to angular
momentum has been added.
We do get the additional condition that
u(0) = 0
to keepRnormalizable.