Transcript electron scattering (2)

16.451 Lecture 6: Cross Section for Electron Scattering
23/9/2003
Recall from last time:
e
Vn
 d ( ) 

  if
c d
 d 
e

where Vn is the normalization volume for the plane wave electron states, and if
is the transition rate from the initial to final state, which we calculate using a
standard result from quantum mechanics known as “Fermi’s Golden Rule:”
2
if 
M if

2
f

M if   *f V (r )  i d 3r

 f  dn/ dE f
(transitions / sec)
We will first calculate the matrix element Mif and then the density of states f
1
Matrix Element for Scattering, Mif
M if 

 *f
2
( Nonrelativistic QM treatment!!! )

V (r )  i d 3 r
e
e

pi

pf


pR
3 – momenta: pi , pf , pR
Use plane wave states to represent the incoming and outgoing electrons, and
let pi = ħ ki , pf = ħ kf , pR = ħ q, and normalization volume = Vn

 i (r ) 
1
Vn
e
 
ik i  r

 f (r ) 
1
e
 
ik f  r
Vn
(Note: slight change of notation here from last class to make sure we don’t miss
any factors of ħ. The recoil momentum of the proton in MeV/c is pR ;
the momentum transfer in fm -1 is q = pR / ħ )
3
Matrix element, continued...
M if


V (r )  i d 3 r
 

i (ki  k f )  r
 3
1
1

e
V (r ) d r 
Vn
Vn

 *f

Insight #1:
e
 
i qr
 3
V (r ) d r
RHS is the Fourier transform of the scattering potential V(r) ,
and it only depends on the momentum transfer q !
Next, proceed with caution:
V(r) is the Coulomb potential of the extended charge distribution of
the target atom that our electron is scattering from ...
• at large distances, the atom is electrically neutral, so V(r)  0 faster than 1/r
• at short distances, we have to keep track of geometry carefully, accounting for
the details of the nuclear charge distribution....
4
Matrix element, continued...
M if


V (r )  i d 3 r
 

i (ki  k f )  r
 3
1
1

e
V (r ) d r 
Vn
Vn

 *f

Insight #1:
e
 
i qr
 3
V (r ) d r
RHS is the Fourier transform of the scattering potential V(r) ,
and it only depends on the momentum transfer q !
Next, proceed with caution:
V(r) is the Coulomb potential of the extended charge distribution of
the target atom that our electron is scattering from ...
• at large distances, the atom is electrically neutral, so V(r)  0 faster than 1/r
• at short distances, we have to keep track of geometry carefully, accounting for
the details of the nuclear charge distribution....
5
Screened Coulomb potential:
V (r )  
Z e2
4  o r
For the atom, where Z is the atomic number, and
 is a distance scale of order Å, the atomic radius.
e r / 
But the electron interacts with charge elements dQ inside the nucleus:
 

Rs  r
incoming e 
x

s

r
 dQ
R
z
y
 3
dQ   Ze e ( R) d R
proton / nucleus
Bottom line:
V (r )   e

nucleus
dQ e  r /
4  o s
Now do the integral...
 r /

V (r )   e
6
nucleus
dQ e
4  o s
 3
dQ   Ze e ( R) d R

r  electron coordinate

R  coordinate of dQ

s  displacement of electron from dQ
normalization: [] = m-3
 dQ   Ze
substitute
for dQ:
Z e2
V (r )  
4  o

nucleus
 e ( R ) e  r /
s
d 3R
Finally, for the matrix element:
M if
1

Vn

e
 
i qr
 
 3
 e ( R) e r /
1  Z e 2 
iq  r
3
V (r ) d r 

d r e


Vn  4  o 
s
all space
nucleus


d 3R
7
How to deal with the variables:
M if
1

Vn
 
 Z e2 

( R ) e  r /
i
q
r
3
e


d r e
 4  
s
o  all space

nucleus


d 3R
problem: r, R and s in here!
Solution:
1. inside the nucleus, where  (R) differs from zero,
e r /   1  e  s / 
(where the screening factor really matters is at large r, and there r  s
to an even better approximation!)
2. there is a one to one mapping between all
electron positions r and all displacements
from the charge element dQ, so:
M if
1

Vn
 Z e2 


 4  
o 

 e
 
iq  r

all space
d 3r 

d 3s
all space
s / 
e
 e ( R) d 3 R
d 3s
s
( this expression can be factored into 2 parts ... )
8
Finishing the integral for Mif:
M if
1

Vn
 Z e2 


 4  
o 

 e
use the relation:
M if
1  Z e 2 



Vn  4  o 

e
 
iq  r
e s /  3
 e ( R) d R
d s
s
  
r Rs
 
iq  R
3
to simplify...
 e ( R) d 3 R 
Fourier transform of the
nuclear charge density  F(q2)
(Eureka!)

M if
1

Vn
 Z e2 


 4  
o 


  s / 
iq  s e
e
d 3s
s
Exact integral:
4
q 2   2
 4


 F (q 2 )
 q 2   2 


9
What does this mean?
M if  (constants)  (exact integral)  (Fourier transformof  (R))
M if
Consider:
1

Vn
 Z e2 


 4  
o 

F (q 2 ) 

e
 
iq  r
 4


 F (q 2 )
 q 2   2 


 e (r ) d 3 r
all space
If the scattering object is a point charge,

e (r )   (r ), i.e. the normalized
3
charge density is a Dirac delta function, with the property:


 3 (r ) d 3 r  1
all space
F(q2) = 1 for a point charge
(Really important result!)
Finally, work out the density of states factor:
from slide 1 ...
2 Vn
 d ( ) 

 
 c d
 d 
M if
2
10
M if 
f

 *f

V (r )  i d 3 r
 f  dn/ dE f
for the cross-section:
2 1  Z
 d ( ) 


 
 4  
 c Vn 
o
 d 
e2
2
2
2
 4  
dn
2
 2  2   F (q ) 
 dE f d
q   
All we have left to calculate is the “density of states” factor,
where Ef is the total energy in the final state when the electron
dn
dE f d
scatters at angle , and this factor accounts for the number of ways it can do that.
Consider the total final state energy:
dn
dn  dp f

dE f d
dp f d  dE f
11




e
e


q
E f  E   ER
(electron)

kf
(recoil)
E f  (cp f  mc 2 )  ( Mc 2  K )
(being careful with the factor of c !)
dE f  c dp f
dn
dn

dE f d
cdp f d
This is useful because the momentum
states are quantized – we have our
electrons in a normalization volume, and
we can “count the states” inside ...
12
Counting the scattered electron momentum states:
Recall the wave function:

 f (r ) 
1
e
 
ik f  r
with p f  k f
Vn
The normalization volume is arbitrary, but we have to be consistent ....
let Vn = L3, i.e. the electron wave function is contained in a cubical box,
so its wave function must be identically zero on all 6 faces of the cube.
z
L
L
x
Since:
L
 
k f  r  kx x  k y y  kz z
Then it follows that:
y
 f ( x, y, z ) 
1
L3
e
ik x x ik y y ik z z
e
e
k x L  nx 2 , etc. ...
13
So, momentum is quantized on a 3-d lattice:


pf   kf  
 n
2
L
x
iˆ  n y ˆj  n z kˆ

n x   (1, 2, 3 ...) etc.
pz
ħ 2/L
py
px
For a relativistic electron beam, the
quantum numbers nx etc. are very large,
but finite.
We use the quantization relation not to
calculate the allowed momentum, but
rather to calculate the density of states!
Allowed states are dots, 1 per cube of volume p = (2ħ/L)3
dn
1 state

d p
(2  / L) 3
14
Finally, consider the scattered momentum into d at  :
pz

shaded area:

pf
dAp  p 2f d
py
px
number of momentum points in the shaded ring:
dn 
Vn
(2  ) 3
p 2f dp f d


dn   dn   dAp dp f
 d p 


15
End of the calculation:
dn 
Vn
(2  ) 3
p 2f dp f d
We want the density of states factor:
p 2f
Vn
dn
dn


dE f d
cdp f d
(2  ) 3 c
FINALLY, from slide 9:
2 1  Z
 d ( ) 


 
 4  
 c Vn 
o
 d 
e2
2
2 
2
 4  
Vn
2

 2 2   F (q )  
  (2 ) 3
q   

 point charge cross  section   F (q 2 ) 


2
p 2f 
c 
