Computational Physics

2.3 Calculation of scattering cross sections 25

*2.3 Calculation of scattering cross sections

In this section we derive Eqs. (2.7) and (2.8). At a large distance from the scattering
centre we can make anAnsatzfor the wave function. This consists of the incoming
beam and a scattered wave:

ψ(r)∝eik·r+f(θ )

eikr

r

. (2.19)

Here,θis the angle between the incoming beam and the line passing throughr
and the scattering centre. The functionfdoes not depend on the azimuthal angleφ
because the incoming wave has azimuthal symmetry, and the spherically symmetric
potential will not generatem=0 contributions to the scattered wave.f(θ )is called
the scattering amplitude. From theAnsatzit follows that the differential cross section
is given directly by the square of this amplitude:

dσ

d


=|f(θ )|^2 (2.20)

withtheappropriatenormalisation(seeforexampleRef.[1]).
Beyondrmax, the solution can also be written in the form(2.2)leaving out all
m=0 contributions because of the azimuthal symmetry:

ψ(r)=

∑∞

l= 0

Al

ul(r)

r

Pl(cosθ) (2.21)

where we have used the fact thatY 0 l(θ,φ)is proportional toPl(cosθ). Because the
potential vanishes in the regionr>rmax, the solutionul(r)/ris given by the linear
combination of the regular and irregular spherical Bessel functions, and as we have
seen this reduces for largerto

ul(r)≈sin

(

kr−

lπ

2

+δl

)

. (2.22)

We want to derive the scattering amplitudef(θ )by equating the expressions (2.19)
and(2.21)for the wave function. For largerwe obtain, using(2.22):

∑∞

l= 0

Al

[

sin(kr−lπ/ 2 +δl)

kr

]

Pl(cosθ)=eik·r+f(θ )

eikr

r

. (2.23)

We write the right hand side of this equation as an expansion similar to that in the
left hand side, using the following expression for a plane wave [4]

eik·r=

∑∞

l= 0

( 2 l+ 1 )iljl(kr)Pl(cosθ). (2.24)