Download Quantum Mechanical Cross Sections

Cross section for potential scattering
In a practical scattering situation we have a finite acceptance for a detector
with a solid angle DW. There is a range of momenta which are allowed by
kinematics which can contribute to the cross section. The cross section for
scattering into DW is then obtained as an integral over all the allowed
momenta for that solid angle. In the case of potential scattering we will
assume only elastic scattering is allowed. This means that there is a change
in momentum between the incoming particle and outgoing particle. We are
using periodic boundary conditions, or box normalization, ( see ref. 2) for
which the incoming and outgoing states are plane wave states. The
incoming wave |a> is
 


1
a (r )  r | a  3 / 2 exp( ik a  r ), (14a)
L
2nax
and k ax 
, nax  0,1,2,..., (15a)
L
and the same for the y and z components .
1
The outgoing state |b> is
 


1
b (r )  r | b  3 / 2 exp( ikb  r ), (14b)
L
2nbx
and k bx 
, nbx  0,1,2,..., (15b)
L
and the same for the y and z components .
All possible outgoing states |b> which can be accepted by the solid angle
must be included in the summation.
3
L
D  2
v
| b | V | a |
nx n y nz DW
2
2 (ba )
We next substitute the matrix element into the above equation. This will be an
integral over all relevant degrees of freedom needed to describe the states a
and b. For a position dependent interaction and the plane waves above, the
integration is over configuration space.
2




 b | V | a   d r ' d r  b | r  r | V | r '  r ' | a 
3
3
But we will consider only potentials diagonal ( local) in configuration space.
 
  
V (r , r ' )  V (r ) (r  r ' ), so

 


1
1
V(q )   dr 3 3 exp( iq  r )V (r )  3 f (q ), where
L
L





q  kb  k a , is the " momentum transfer" , (p  k)
Notice that the matrix element is proportional to the Fourier transform of the potential.

kb
q

ka

q
L3
D  2
v
 2
1
| f (q ) | 2 (ba ), (16)

6
nx n y nz DW L
3
1
D  2
v
 2
1 1 1
( )( )( ) | f (q ) | 2 (ba ), (17)

n x n y n z DW L L L
In eqn (17) we note the sum over possible n values. The smallest change
possible in n is dn = 1. We want to convert the sum in (17) into an integral over
wave numbers.
And from eqns (15) we realize that
dk x dk y dk z dnx dn y dnz 1 1 1


2 2 2
L L L
LLL
dk x dk y dk z
1 1 1
( )( )( )  
, (18)

2 2 2
n x n y n z DW L L L
k x k y k z DW
dk x dk y dk z k 2 dkdWk
dk 3


3
3
3
(2 )
(2 )
(2 )
4
We substitute (18) into (17) and obtain
1
D  2
v
 dW
DWk

k
k x k y k z DWk
 2
dkk 2
| f (q ) | 2 ( ba ), (19)
3
(2 )
1
and  ( ba )   ( ( E (kb )  E (k a ))), (20)

Note that in the delta function in (20) we have functions ( the energies) of the
momenta k, whereas the variable of integration is k. We use the following
property of functions of the delta function.
 ( f ( x))  
xi
 ( x  xi )
df
|
| xi
dx
, where x i are the zeroes of f(x)
In the case of potential scattering, or elastic scattering, the zeroes occur for
k = kb = ka . The derivative of the energy with respect to the wave number k is
5
d E (k ) 1 d
k p
2
2

(k )  m 
 v
dk 
 dk
E E
Integrating over k, the delta function selects the value k = ka which we will
simply call k, the magnitude of the initial wave number. Thus,
 2
2 DW k k 2
D  2 2
| f (q ) |
3
v  (2 )
 2
d
D
1
k2

 2 2
| f (q ) | , (21)
2
dW k DWk  v (2 )
We are now in a position to evaluate the differential elastic cross section using
eqn (21) for any potential V. We will need to find the Fourier transform of the
potential as a function of q. Bear in mind that (21) is an approximation using
only the first non diagonal element of the S-matrix.
6
Coulomb Scattering
Consider the Coulomb scattering cross section. The classical cross section was
calculated by Rutherford and was very important in determining the existence of the
atomic nucleus. If we perform the integration for f(q) given in eqn (21) using the
Coulomb potential we will end up with an indefinite result. In order to obtain a definite
result we use a screened potential with a parameter e to evaluate the integral and
then let e go to zero. It is plausible in a real scattering situation to use a screened
potential because in a typical scattering experiment the atomic electrons surrounding
the nucleus screen the Coulomb potential of the nucleus so that at large distance the
incoming projectile actually feels no force. The atom as a whole is electrically
neutral. Nevertheless, the real reason to use a screened potential is that it allows us
to calculate a quantum mechanical cross section.
z1 z2e 2
V (r ) 
exp( er ), where e is a small positive number.
r
 

2
3 exp( iq  r ) exp( er )
f (q )  z1 z 2 e  dr
r
d r  r drdW  r drdd cos(q )
3
2
2
In these expressions z1 and z2 are the atomic numbers of the incident projectile
and target, e is the charge on the electron. We can choose q as the z-axis and
then (q,) are the polar angles of r with respect to the q direction.
q
Let cos(q) = x, then
r
q

y

x
From the diagram on page 3,
2
a
2
b
1

2z1 z 2 e 2 2iq
f (q ) 
iq
q2  e 2

4z1 z 2 e 2
lim f (q ) 
e 0
q2
q  k  k  2k a kb cos(q )
2
2
2
2 exp( er )
f (q)  z1 z2e  drr
d  dx exp( iqrx )

r
0
0
1
q
But k a  kb  k , so, q 2  4k 2 sin 2 ( )
2


d
1
k 2  4z1 z 2 e 2 


 2 2
2
dW k  v (2 )  4k 2 sin 2 (q ) 


2 

2
d
1 k 2 ( z1 z2 e 2 ) 2
1 E 2 ( z1 z2 e 2 ) 2
 2 2

, (22)
4
dW k  v 4 k 4 sin 4 (q ) 4 p sin 4 (q )
2
2
p
where we used v  .
E
Equation (22) turns out to be correct for spin zero relativistic Coulomb scattering too.
The use of relativistic wave functions does not change the result. For the case of spin
½ particles there is an additional factor. Eqn, (22) reduces to the Rutherford cross
section in the non-relativistic limit.
Coulomb Scattering from an extended charge distribution
Equation (22) gives us the Coulomb scattering from an infinitely massive point
charge. In a real situation, such as scattering of an electron off a nuclear charge
distribution, we must consider how the charge is distributed to arrive at the cross
section. Let us consider the case of a spherically symmetric charge distribution
of density r(r) for a nucleus with total charge Q2. The potential V(r)
Is given by ( see ref. 3 )
rdR3 =dQ
V (r )  Q1  dR
R
s
r
3
r ( R)
|rR|
  
s r R
Q1
3
 
r (r )
f (q )  Q1  d r exp( iq  r )  d R  
|r R|
3
 
3
 
3 exp( iq  s )
f (q)  Q1Q2  d s
d Rr 0 (r ) exp( iq  R)

s
The first integral was evaluated earlier for a point charge target. We call this
integral f0(q). The second integral over the charge density is called the charge
form factor F(q).
 
F (q)   d Rr 0 ( R) exp( iq  R)
3
f (q)  f 0 (q) F (q)
The charge density is normalized so that
r ( R)  Q2 r 0 ( R), with  d 3 Rr 0 ( R)  1
Let’s consider the case of a uniformly charged sphere. This is a rough
approximation to nuclei. It fails in the details but gives us insight into important
qualitative features of scattering.
3
r0 
for 0  r  R
3
4R
r0  0 for r  R
This is a sphere of radius R
and total charge Q2 .
3
F (q) 
(sin( qR)  qR cos( qR))
3
(qR)
d
d
2
 ( ) point | F (q) | , (23)
dW
dW
Equation (23) describes Coulomb scattering from an extended charge distribution.
We see that |F(q)|<= 1 and reaches the value of 1 at q=0, F(0)=1. It also introduces
an additional fall off with respect to q by a factor of about 1/q2 for large q. This is a
general feature we expect from quantum mechanics. As the region of interaction
becomes larger the ability of the interaction to impart large momentum transfers
decreases. Large momenta are associated with small distances, small momenta
are associated with large distances. If we compare, for example, elastic scattering
of electrons from 3He with elastic scattering from 12C we see that there is a steeper
q dependence for the larger nucleus. We also notice from F(q) that there are
values of q for which F(q) = 0. This produces an oscillatory behavior which is
absent in the point charge cross section.
As a second example consider an exponential charge dependence,
3
r (r )  Q2 r0 exp( R), r0 
8
The charge form factor for this distribution is
F (q) 
(1 
1
q2
2
)
2

A commonly employed model for the proton charge distribution uses the
exponential dependence with  = 4.35 F-1 . So far we have considered
spinless projectiles. The case of an electron on a spinless point target includes
the additional factor below. The point cross section is the same as in eqn 22.
d
d
cos 2 (q / 2)
( ) D  ( )0
, where mtgt is the target mass.
dW
dW 1  2 E sin 2 (q / 2)
mtgt
d
d
(
) D | FC (q) |2
dW
dW
We have neglected magnetic interactions in the above. If the target
has spin then the incoming electron can interact magnetically with
the magnetization density in the target. This introduces another form
factor, the magnetic form factor.
(1) “Modern Quantum Mechanics – Revised Edition”, J. J. Sakurai,
Addison-Wesley, 1994
(2) “Relativistic Quantum Mechanics and Field Theory”, Franz Gross,
John Wiley & Sons, 1993
(3) “Nuclei and Particles”, second edition, Emilio Segre’,W. A.
Benjamin, 1977, chapters 6, 18