The document discusses scattering cross sections and scattering amplitudes. It defines differential and total scattering cross sections, relating them to the scattering amplitude. The scattering amplitude f(θ,φ) is derived using Born approximation by solving the Schrodinger equation for scattering. The scattering amplitude is expressed as an integral involving the scattering potential V(r) and is related to the experimentally observable differential scattering cross section.

1. UIII-PAN 1
Scattering Cross Section:
Consider beam of particles of mass m travelling along z-axis with velocity v, scattered by scattering
potential (target) centered at origin.
Let N be the number of incident particles crossing unit area normal to the incident beam in unit time and n
be the number of particles scattered into solid angle dΩ in the direction (θ,φ) in unit time. Then
differential cross section is defined as
/
( , )
n d
N
  


where θ is called angle of scattering.
The solid angle d in the direction of ( , )  is given by
2
2
sin
sin
r d d
d d d
r
  
    
The total scattering cross section is given by
2
0 0
( , ) ( , )sind d d
 
             
For spherically symmetric potential ( , )   = ( )  ,
0
2 ( )sin d

       
Both ( , )   &  have the dimensions of area. Scattering cross sections are usually measured in barns.
1barn= 10-24
cm2
Scattering Amplitude:
Consider the scattering of two particles. Let the scattering potential V, depends only on relative distance r
between two particles. The Schrodinger equation for their relative motion in centre of mass system is
written as
2
2
( )
2
V r E  

   
...... (1)
Where
Mm
M m
 

, M is the mass of target and m is the mass of incident particle.
At the large distance from the scatterer (target), effect of potential is negligible (i.e. V(r) →0) then beam
of incident particles can be represented by the plane wave
ik r
i r
Ae 


Let the incident beam of particles is along z-axis,
ikz
i r
Ae 

.... (2)
If the detector is far away from the scatterer and we are interested only in asymptotic form of scattered
wave. As the scattered intensity depends on θ&φ, the spherically diverging wave can be represented as
( , )
ikr
s r
e
Af
r
  

.... (3)

2. UIII-PAN 2
Where f(θ,φ)is called scattering amplitude. For spherically symmetric potential f(θ,φ) depends only on θ.
i.e. ( )
ikr
s r
e
Af
r
 

.... (4)
Thus
ikz
e
and
( )
ikr
e
f
r

are the two eigen functions of particles and most general asymptotic solution can
be written as
( , )
ikr
ikz
s r
e
A e f
r
  
 
  
 
We have the expression for the probability current density
* *
( )
2
j
im
      
Therefore probability current density for incident wave
 * *
2 2 2
( )
2
ikz ikz ikz ikz
i
i
j A e ikAe Ae ik A e
im
k p
j A A v A
m m
 
  
   
Similarly probability current density for scattered wave
* * * *
2 2 2 2 2 2
2 2 2
( ) ( ) ( ) ( )
( )
2
1 1 1
( ) ( ) ( )
ikr ikr ikr ikr
s
s
A f ikAf Af A f
j e e e ik e
im r r r r
k p
j A f A f v A f
m r m r r
   
  
  
   
 
   
As 1/r2
is solid angle subtended by unit area of the detector at the scattering centre and also
Probability current density of scattered wave per unit solid angle
( )
Probability current density of incident wave
  
2 2 2
2
2
1
( )
( )
v A f r
r
v A

 

 
2
( ) ( )f   
The scattering amplitude is thus related to experimentally observable differential scattering cross section.
Since differential scattering cross section has the dimensions of area, scattering amplitude has the
dimensions of length.
Scattering Amplitude: Born Approximation
Consider the scattering of two particles. Let the scattering potential V, depends only on relative distance r
between two particles. The Schrodinger equation for their relative motion in centre of mass system is
written as
2
2
( ) ( ) ( )
2
r V r E r  

   
2
2 2
2 2
( ) ( ) ( ) ( )r E r V r r
 
    
2 2
2
2
( ) ( ) ( ) ( )k r V r r

    2
2
2
where, k E


……. (1)

3. UIII-PAN 3
The general solution of this equation is given by
3
2
2
( ) ( ') ( ') ( ') 'ir G r r V r r d r

     ……. (2)
It consists of two components:
1. The general solution of above equation for r ∞, i.e. incident plane wave i = ik r
e 
and
2. The particular solution of above equation expressed in terms of Green’s function corresponding
to the operator on the left hand side of equation (1) such that,
2 2
( ) ( ') ( ')k G r r r r     ……. (3)
Where ( ') and ( ')G r r r r  are given by their Fourier transforms as follows:
( ') 3
3
( ') 3
3
1
( ') ( )
(2 )
1
( ')
(2 )
iq r r
iq r r
G r r e G q d q
r r e d q





 
 


Put in equation (3) we have
2 2 ( ') 3 ( ') 3
3 3
1 1
( ) ( )
(2 ) (2 )
iq r r iq r r
k e G q d q e d q
 
 
   
2 2 ( ') 3 ( ') 3
( ) ( )iq r r iq r r
k e G q d q e d q 
   
2 2
( ) ( ) 1q k G q   2 2
1
( )
( )
G q
k q
 

( ')
3
3 2 2
1
( ')
(2 ) ( )
iq r r
e
G r r d q
k q

  
 ……. (4)
As is volume element in q-space, 3 2
sind q q dq d d   and ( ') ' cosq r r q r r   
2 ' cos
2
3 2 2
0 0 0
1
( ') sin
(2 ) ( )
iq r r
e
G r r q dq d d
k q
  
  

 
  
 
22
' cos
3 2 2
0 0 0
1
( ') sin
(2 ) ( )
iq r rq
G r r e d d
k q
 

  



  
   ……. (5)
2
0
1
1 '
' cos '
0 1 1
' '
' cos
0
2
sin
'
sin
'
iq r r x
iq r r iq r r x
iq r r iq r r
iq r r
d
e
e d e dx
iq r r
e e
e d
iq r r





 
 
 

 
 
  


 
   
 
 
   
 

 

' '2
2 2 2
0
1
( ')
(2 ) ( ) '
iq r r iq r r
q e e
G r r dq
k q iq r r
   
 
    
  


4. UIII-PAN 4
 ' '
2 2 2
0
1
( ')
4 ' ( )
iq r r iq r r
q e e
G r r dq
i r r k q
  
 
  
 
'
2 2 2
1
( ')
4 ' ( )
iq r r
qe
G r r dq
i r r k q
 

  
 
We may integrate this integral by the method of residues i.e. it is equal to 2πi times residue of the
integrand the poles. Since there are two poles, q k  , the integral has two possible values
'
'
( ')
4 '
( ')
4 '
ik r r
ik r r
e
G r r
r r
e
G r r
r r




 

  

  

Since ( ')G r r  represents outgoing spherical wave from r’, it is of our interest.
Therefore from equation (2) total scattered wave function:
'
3
2
( ) ( ') ( ') '
2 '
ik r r
i
e
r V r r d r
r r

  


 
 ……. (6)
The asymptotic value of ( )r can be obtained as follows:
2 2
1 1 1
' ' 2 '
' '
r r rr r r r
k r r kr k r
 
   
   
'
3
2
( ) ( ') ( ') '
2
ikr ik r
i
e e
r V r r d r
r

  

 
    ……. (7)
The asymptotic value of scattered wave function ( )r is equated to
( ) ( , )
ikr
i
e
r f
r
    
then the scattering amplitude is given by
' 3
2
( , ) ( ') ( ') '
2
ik r
f e V r r d r

  

 
   ……. (8)
and the differential scattering cross section is given by
2 22 ' 3
2 4
( , ) ( , ) ( ') ( ') '
4
ik r
f e V r r d r

     

 
   ……. (9)
Replacing ' '
( ') by eik r
ir  
 in equation (8) we get
' ' ' 3
2
( ' ) ' 3
2
( , ) ( ') '
2
( , ) ( ') '
2
ik r ik r
i k k r
f e V r e d r
f e V r d r

 


 

  
 
 
 



5. UIII-PAN 5
' 3
2
( , ) ( ') '
2
iq r
f e V r d r

 


  
2
2 'cos '
2
0 0 0
( , ) ( ') ' ' sin ' ' '
2
iqr
f V r r dr e d d
 

    


    
2 'cos '
2
0 0
( ) ( ') ' ' sin ' 'iqr
f V r r dr e d


  

   
Put cos θ’= y, we have
1
2 'y
2
0 1
( ) ( ') ' ' iqr
f V r r dr e dy




   
1 ' '
'y
1
2sin( ')
' '
iqr iqr
iqr e e qr
e dy
iqr qr



 
2
0
2
( ) 'sin( ') ( ') 'f r qr V r dr
q



   
2
2
2
2 4
0
4
( ) ( ) 'sin( ') ( ') 'f r qr V r dr
q

  

   
This is called first Born approximation.
Where
2 2
2 2 2 2 2 2
' ' 2 ' cos
cos 2 cos 2(1 cos )
2 sin
2
q k k k k k k
q k k k k
q k

  

    
    
 
  
 
Validity of first Born approximation:
In Born approximation ' '
( ') was replaced by eik r
ir  
 which is valid only if the scattered wave
function given by second term in equation (6), i.e.
'
3
2
( ) ( ') ( ') '
2 '
ik r r
i
e
r V r r d r
r r

  


 

2
'
2' ' 3
2
( ')e '
2 '
ik r r
ik r
i
e
V r d r
r r







'
' ' 3
2
( ')e ' 1
2 '
ik r r
ik re
V r d r
r r






In elastic scattering k’ = k and assuming that the scattering potential is largest near r = 0, we have
2
' 'cos '
2
0 0 0
' ( ') ' e sin ' ' ' 1
2
ikr ikr
e r V r dr d d
 

  


  
' 'cos '
2
0 0
' ( ') ' e sin ' ' 1ikr ikr
e r V r dr d


 

  

6. UIII-PAN 6
Put cos θ’= y, the second integral is written as
1 ' '
'y
1
'
ikr ikr
ikr e e
e dy
ikr




'
2
0
(2 1) ( ') ' 1ikr
e V r dr
k
 
  
Since the energy of the incident particle is proportional to k2
, the Born approximation infers that it is
valid for large incident energies and weak scattering potential. That is, when the average interaction
energy between the incident particle and scattering potential is much smaller than the incident particle’s
kinetic energy, the scattered wave can be considered to be a plane wave.
Partial Waves:
A plane wave eikz
can be expanded as a linear combination of spherical waves as:
0
(2 1) ( ) (cos )ikz l
l l
l
e i l j kr P 


 
( )lj kr : is spherical Bessel’s function of order l & (cos )lP  Legendre’s polynomial.
Each term in the sum of above equation represents spherical wave. The plane wave eikz
thus equivalent to
the superposition of an infinite number of spherical waves, these spherical waves are called partial waves.
The waves with l=0,1,2.... are respectively called s-wave, p-wave, d-wave and so on.
Asymptotically
1
( ) sin
2
l
l
j kr kr
kr
 
  
 
As
exp exp
2 2
sin
2 2
il il
ikr ikr
l
kr
i
 

   
      
       
 
0
(2 1) 1
(cos ) exp exp
2 2 2
l
ikz
l
l
i l il il
e P ikr ikr
ik r
 



     
        
    

This form shows that each partial wave can be represented as the sum of an incoming and an outgoing
spherical wave.
Partial Wave Analysis:
A partial wave analysis is precise method for the analysis of elastic scattering. It is done in two steps.
I. A wave function  which represents the sum of the incident and scattered waves is obtained in terms of
the partial waves.
II. The asymptotic value of this is equated to the ( )
ikr
ikz e
e f
r
   .
In scattering problems, the first few spherical waves are the most important. The results of extremely low
energy scattering can be analyzed satisfactorily with s-wave alone. If the energy is slightly higher, needs
s-wave and p-wave for analysis of scattering and so on.

7. UIII-PAN 7
Significant Number of Partial Waves:
In partial wave analysis of scattering at sufficiently low energies, S wave (l=0) will be important. As
energy increases, contributions from the higher waves become important. Then question arises is that up
to what partial wave one has to consider for given energy.
For spherically symmetric potentials, the influence of the potential felt only in the region of sphere of
radius r0 around scattering centre. For a particle with impact parameter b, linear momentum p= k and
angular momentum L= l ,
kb l
l kb

 
When impact parameter b>r0 , the particle will not see the potential region and therefore particle do not
get scattered, if l>kr0. Hence we need to consider partial waves up to l=kr0. The waves corresponding to
value of l up to l=kr0 are called significant number of partial waves.
Scattering by Central Potential (using partial wave analysis):
The Schrödinger equation that describes scattering is given by
2
2
( , ) ( ) ( , ) ( , )
2
r V r r E r     

    .... (1)
The wave function  has to be independent of  , as the incident particles are along z-axis.
We have the solution of above equation is
( , ) ( ) (cos )l lr R r P   , l = 0, 1, 2, ....... .... (2)
Where ( )lR r satisfies radial equation:
2
2 2 2 2
1 2 2 ( 1)
0l
l
dRd E V l l
r R
r dr dr r
     
        
.... (3)
Outside the range of potential (r>>r0), this equation reduces to
2 2 2
2 2 2
1 ( 1) 2
0 & 0l
l
dRd l l E
r k R k V
r dr dr r
   
         
.... (4)
The general solution of this equation is
( ) ' ( ) ' ( )l l lR r A j kr B n kr  .... (5)
Where ( )lj kr is spherical Bessel’s function and ( )ln kr is spherical Neumann’s function. The asymptotic
solution of equation (4)
' '
( ) sin( ) cos( )
2 2
l
A l B l
R r kr kr
kr kr
 
    .... (6)
Substitute
' cos & ' sinl l l lA A B A   .... (7)
1
( ) cos sin( ) sin cos( )
2 2
l l l l l
l l
R r A kr A kr
kr
 
 
 
     
( ) sin( )
2
l
l l
A l
R r kr
kr

   .... (8)
l ’s are called phase shifts for l = 0, 1, 2, ........
The phase shift measures amount by which the phase of the radial function for angular momentum
quantum number l differs from the corresponding one for the V=0 case.

8. UIII-PAN 8
The most general asymptotic solution of eqn (1) is then
0
( , ) sin( ) (cos )
2
l
l l
l
A l
r kr P
kr

   


   .... (9)
To obtain the scattering amplitude equate this asymptotic form with wave function ( )
ikz
ikz e
e f
r
   .
0
( ) sin( ) (cos )
2
ikr
ikz l
l l
l
Ae l
e f kr P
r kr

  


   
0 0
(2 1)
sin( ) (cos ) ( ) sin( ) (cos )
2 2
l ikr
l
l l l
l l
Ai l l e l
kr P f kr P
kr r kr
 
   
 
 

     
   
       
0
0
(2 1)
exp exp exp exp (cos ) ( )
2 2 2
exp exp exp exp exp exp (cos )
2 2 2
l ikr
l
l
l
l l l
l
i l il il e
ikr ikr P f
ikr r
A il il
ikr i ikr i P
ikr
 
 
 
  




     
       
    
    
        
    


Equating coefficients of ikr
e
on both sides
 
0 0
(2 1)exp (cos ) exp exp (cos )
2 2
l
l l l l
l l
il il
i l P A i P
 
  
 
 
   
     
   
 
Using the orthogonal property of Legendre’s polynomial, we have
 (2 1)expl
l lA i l i  .... (10)
Equating coefficients of ikr
e on both sides
 
0 0
(2 1)
exp (cos ) ( ) exp exp (cos )
2 2 2 2
l
l
l l l
l l
Ai l il il
P f i P
ik ik
 
   
 
 
    
      
   
 
From eqn (10)
 
0 0
(2 1) (2 1)
exp (cos ) ( ) exp exp 2 (cos )
2 2 2 2
l l
l l l
l l
i l il i l il
P f i P
ik ik
 
   
 
 
    
      
   
 
 
0
(2 1)
( ) exp exp 2 1 (cos )
2 2
l
l l
l
i l il
f i P
ik

  


  
      
 

As          exp( ) & exp 2 1 exp exp exp 2 exp sin
2
l
l l l l l l
l
i i i i i i i

                
 
0
(2 1)
( ) exp sin (cos )l l l
l
l
f i P
k
   



   ..... (11)
It is the basic result of partial wave analysis which gives the scattering amplitude as sum of contributions
from all partial waves.
The differential scattering cross section is given by
2
( ) ( )f  
From eqn (11)
 
2
2
0
1
( ) (2 1)exp sin (cos )l l l
l
l i P
k
    


  ..... (12)
0
0
(2 1)
exp exp (cos ) ( )
2 2 2
exp exp (cos )
2 2 2
l ikr
l
l
l
l l l
l
i l il il e
ikr ikr P f
ikr r
A il il
ikr i ikr i P
ikr
 
 
 
  




     
        
    
    
          
    



9. UIII-PAN 9
The total cross section
2
2
00
4
2 ( )sin (2 1)sin l
l
d l
k


      


   ..... (13)
Optical theorem:
For the case  = 0, we get from eqn (11)
 
0
1
(0) (2 1)exp sinl l
l
f l i
k
 


  
(cos )lP  = 1 for  = 0
The imaginary part of this scattering amplitude is given by
2
0
1
Im (0) (2 1)sin l
l
f l
k



 
From eqn (13), we have
4
Im (0)f
k

 
This theorem is known as optical theorem.
Phase Shift:
The phase shift measures amount by which the phase of the radial function for angular momentum
quantum number l differs from the corresponding one for the V=0 case. In Born approximation for phase
shifts, the phase shifts l is given by
2 2
2
0
= ( ) ( )l lV r j kr r dr





From this equation it can be seen that for attractive potentials (V(r) <0) phase shift is positive and for
repulsive potential (V(r) >0) phase shift is negative.
Ex: 1 Consider the scattering of a 5MeV alpha particle (i.e. helium nucleus with Z1=2 and A1=4) from a
gold nucleus (Z2=79 and A2=197). If the scattering angle of the alpha particle in the L-system is 600
, a)
find its scattering angle in C-system and b) give numerical estimation of Rutherford cross section using
formula
2 2 4 2
1 2
4 4 4
( )
4 sin ( )
2
Z Z e
k

 


Solution:
Since the mass ratio of alpha particle to gold nucleus is roughly equal to the ratio f their atomic masses,
m1/m2 = A1/A2= γ=4/197=0.0203.
Since L =600
sin
tan
cos
C
L
C


 
 

0 0sin
tan 60 61
0.0203 cos
C
C
C



   

We have

10. UIII-PAN 10
 
22 2 4 2 2 2 2 2
21 2 1 2
4 4 4 4 4 4
22 2 2
1 2
2
4
4
( )
4 sin ( ) 16 sin ( )
2 2
( )
16sin ( )
2
C C
C
Z Z e Z Z e
c
ck k
Z Z c
E
 
 
 

 

 
   
 
 
   
 
As α = (1/137) , c= 197.3 MeV fm and E = 8 MeV
2 28 2
( ) 30.87fm 0.31 10 m 0.31 barn  
    
Ex: 2 Calculate the differential scattering cross section in the Born approximation for Coulomb potential,
2
1 2
( )
Z Z e
V r
r
 , where 1Z e and 2Z e are the charges of the projectile and target nucleus, respectively.
Solution:
We shall consider the scattering of a particle having charge 1Z e by an atomic nucleus of charge 2Z e .
The interaction between these particles is screened by electrons surrounding the atomic nucleus. Hence
interaction potential is written as
2
1 2
( ) rZ Z e
V r e
r


where:  is the screening parameter
The first Born approximation gives the differential scattering cross section as
2
2 2 4 2
2 1 2
2 4
0
4
( ) ( ) sin( )rZ Z e
f e qr dr
q

  


  
0
1
sin( )r
e qr dr
q




22 2 4 2 2
1 2 1 2
2 4 2 2 2
4 21
( )
Z Z e Z Z e
q q q
 
 
 
   
 
As 2 sin
2
q k
 
  
 
2
2 2 2 4 2
1 2 1 2
2 2 2 4 4 4
( )
2 sin ( ) 4 sin ( )
2 2
Z Z e Z Z e
k k
 
 
 
 
 
   
 
 
It is called Rutherford’s scattering formula for scattering by a pure Coulomb potential.
Ex: 3 Calculate the differential scattering cross section and total scattering cross section in the Born
approximation for Yukawa potential
( )
0
( )
r R
V e
V r
r

 .
Solution:
The first Born approximation gives the differential scattering cross section as
2
2 2
2 ( / )0
2 4
0
4
( ) ( ) sin( )r RV
f e qr dr
q

  


  

11. UIII-PAN 11
( / ) ( / )
0 0
sin( )
2
iqr iqr
r R r R e e
e qr dr e dr
i
  
   
  
 
 
( / ) ( / ) ( / )
0 0 0
1
sin( )
2
r R r R iqr r R iqr
e qr dr e e dr e e dr
i
  
  
 
  
 
  
1 1
( / )
0 0 0
1
sin( )
2
iq r iq r
r R R R
e qr dr e dr e dr
i
               
 
  
 
 
  
( / )
0
1 1 1
sin( )
1 12
r R
e qr dr
i
iq iq
R R


 
 
  
         
    

( / )
2 20
2 2
1 2
sin( )
1 12
r R iq q
e qr dr
i
q q
R R


 
   
    
   

2 2
0
2
4 2
2
4
( )
1
V
q
R

  
 
 
 
As 2 sin
2
q k
 
  
 
2 2 2 2 4
0 0
2 2
4 2 2 4 2 2 2
2
4 4
( )
1
4 sin 1 4 sin
2 2
V V R
k k R
R
 
 
 
  
      
       
      
The total scattering cross section is given by
0
2 ( )sin d

      
2 2 4
0
24
0 2 2 2
8 1
sin
1 4 sin
2
V R
d
k R


  

 
  
   
  

2 2 2 2 2 2
2 2
2 2
Let 4 sin 2 (1 cos )
2
1
2 2 sin sin
x k R k R
xdx k R d d xdx
k R


   
 
   
 
   
for 0 0 and for 2x x kR       
 
22 2 4
0
24 2 2 2
0
8
1
kR
V R x
dx
k R x

 


2 2 4
0
4 2 2
16 1
1 4
V R
k R

 

Ex: 4 Consider the scattering of a particle of mass m from a hard sphere potential: V(r) = ∞ for r<a and
V(r) = 0 for r>a.
a. Calculate the total cross section in the low energy limit.
Find the numerical estimate for the cross section for the case of scattering 5 keV protons from a
hard sphere of radius a=6 fm.

12. UIII-PAN 12
b. Calculate the total cross section in the high energy limit.
Find the numerical estimate for the cross section for the case of 700 MeV protons from a hard
sphere of radius a=6 fm.
Solution:
The radial part of Schrödinger equation that describes scattering is given by
2
2 2 2 2
1 2 2 ( 1)
0l
l
dRd E V l l
r R
r dr dr r
     
        
Put ul(r) = rRl(r) in above equation we get
2
2 2 2 2
2 2 ( 1)
0l
l
d u E V l l
u
dr r
   
     
As the scattering is dominated at low energies by s-wave (l=0)
2
2 2 2
2 2
0l
l
d u E V
u
dr
  
     
2
2
2
0 forl
l
d u
k u r a
dr
    …………….. (1)
2
2
2
where
E
k


Solutions of this equation are
  1
2 0
( ) 0,
( ) sin( ),
l
u r r a
u r A k
u r
r r a
 
 
  
…………….. (2)
The wave function ul(r) is continuous at r=a
0 0
2 2
0
sin( ) 0 tan tan( )
sin sin ( )
A kr ka
ka
 

    
 
The lowest value of the phase shift 0 ka   ; it is negative , as it should be for a repulsive potential.
The total cross section is given by
2
2
0
4
(2 1)sin l
l
l
k

 


 
2 2
0 02 2
4 4
sin sin ka
k k
 
   
2 2
0 02 2
4 4
sin sin ka
k k
 
   
In the low energy limit ka<<1, then 2 2
sin ( )ka ka
2
0 4 a  
In the high energy limit ka>>1, the number of significant partial waves in this case are large. Then total
cross section is given by
max
2 2
2 2
0 0
4 4
(2 1)sin (2 1)sin
l
l
l l
l l ka
k k
 
 

 
     
max
2
max2 2
0
4 1 2
(2 1) ( 1)
2
l
l
l l
k k
 


    
Assuming maxl ka we have
2
2
2
( )ka
k

 
2
2 a  

13. UIII-PAN 13
Ex: 5 Calculate the total cross section for the low energy scattering of particle of mass m from attractive
square well potential: V(r) = -V0 for r<a and V(r) = 0 for r>a with V0>0.
Solution:
As the scattering is dominated at low energies by s-wave (l=0), the radial part of Schrödinger equation
(with ul =rRl ) that describes scattering is given by
2
2 2 2
2 2
0l
l
d u E V
u
dr
  
    
2
2
12
0 forl
l
d u
k u r a
dr
    …………….. (1)
2
2
22
0 forl
l
d u
k u r a
dr
    …………….. (2)
  2 2
0 1 22 2
2 2
where and
E
E V k k
 
  
Solutions of these equations are
  1 1
2 2 0
( ) sin( ),
( ) sin( ),
l
u r B k r r
u
a
u r B k r
r
r a
 
 
  
The wave function ul(r) and its first order derivative are continuous at r=a. i.e.
1 2
1 2and r a
r a
du du
u u
dr dr 

 
1 2 1 2 0sin( ) sin( )r a
u u k a k a 
    …………….. (3)
1 2
1 1 2 2 0cos( ) cos( )
r a
du du
k k a k k a
dr dr


    …………….. (4)
1 2 0
1 2
1 1
tan( ) tan( )k a k a
k k
  
2
2 0 1
1
tan( ) tan( )
k
k a k a
k
  
Also we have
2 0
2 0
2 0
tan( ) tan
tan( )
1 tan( )tan
k a
k a
k a




 

2 02
1
1 2 0
tan( ) tan
tan( )
1 tan( ) tan
k ak
k a
k k a



 

2 02
1
1 2 0
tan( ) tan
tan( )
1 tan( ) tan
k ak
k a
k k a



 

2 1 1 2
0
1 2 1 2
tan( ) tan( )
tan
tan( ) tan( )
k k a k k a
k k k a k a


 

Using the relation
12
2 1 2 1 2
0 02 2
1 1 2 1 1 2
tan( )tan( )4 4
sin 1
tan( ) tan( )
k k k a k a
k k k k a k k a
 
 

  
     
   
2
0 2
0
1
sin
1 (1/ tan )





14. UIII-PAN 14
Exercise examples:
1. Consider the scattering of a 5MeV alpha particle (i.e. helium nucleus with Z1=2 and A1=4) from
an aluminum nucleus (Z2=13 and A2=27). If the scattering angle of the alpha particle in the L-
system is 300
, a) find its scattering angle in C-system and b) give numerical estimation of
Rutherford cross section.
2. Find the differential scattering cross section and total scattering cross section in the Born
approximation for the following scattering potentials
a) An attractive square well potential: V(r) = -V0 for r<a and V(r) = 0 for r>a, with V0>0.
b) The delta potential V(r) = V0δ(r-a).
c) The potential V(r) = V0 e-r/a
.
3. Find the differential scattering cross section and total scattering cross section of slow particles
from spherical delta potential V(r) = V0δ(r-a) using partial wave analysis method.