14. Scattering
14A. Cross Section and Differential Cross Sec.
Cross Section
• Imagine many particles fired at and scattering from a small target
• We want to predict the rate  at which particles are scattered
• It makes sense that it would be proportional to the
number density n of the projectiles
out
• It makes sense that it would be proportional to the speed
|v| of the projectiles
• Call the remaining proportionality constant    n v
in
• The cross-section  has units of area
• Classically, it is just the size of the target as viewed from the direction the
projectiles are approaching
• This formula is the definition of , and hence is correct by definition
• |v| is the difference in speeds, even if everything is relativistic
Flux and Differential Cross-Section
• The flux (in or out) is the number density times the relative velocity   n v
• Imagine we are far from a small target
  in
– It makes sense that scattered particles will be moving radially outwards
• Imagine a spherical detector radius r covering all possible angles
• The total rate for scattering will then be:
2


lim
r
 out  r  d  
  lim   out  r  dA


r  
r 
• The cross-section is then:

1

lim  r 2   out  rrˆ  d  

 in  in r 
• If detector only covers some angles, restrict integral
• Useful to define the differential cross-section:
d
1

lim  r 2  out  r  
d   in r 
 d 
  
 d
 d 
Lab Frame and Center of Mass Frame
There are two types of collision experiments
v'
that commonly occur:
L
v
• The lab frame, where the
target is stationary and the
projectile collides with it
v'cm
• The center of mass frame,
vcm

where the incoming particles
have net zero momentum
• Usually easier to calculate in the cm frame
• But then we must relate these two frames
m
• Let the mass of the projectile be m and the target M

• Commonly write things in terms of the ratio:
M
• The velocity of the center of mass compared to the lab frame can be found:
u cm
Ptot
m


v
mtot
mM
u cm 

1 
v
Choice of Axis and Description of Angles
• Denote the velocities beforehand by v and vcm
• Denote the velocities after by v' and v'cm
• We will assume that the initial direction is
in the z-direction
v  vzˆ
v'
v
L
v'cm
• Then the scattering angles  and L are the
vcm

polar angle of the direction of the outgoing
particles in the two frames
• The azimuthal angle  and L will be the same
  L
• However, the polar angles  and L will not be
equal, because of the change in velocities between the two frames
• The velocities in the two frames are related by the relative speed of the frames,
ucm:
v  vcm  ucm
and v  vcm  ucm
Comparing Angles in the Two Frames
v  v cm  u cm

1
ucm
u

v
v

v

cm
v  vcm  u cm cm 1  
1 

v'
v
• It follows that
• From conservation of momentum in the cm frame, it is
v'cm
clear that the final particles are back to back and
vcm

their momenta must be equal and opposite
• In cm frame, conservation of energy implies
• Now, let’s do some geometry:
  vcm
vcm
 – L
• The sum of two interior angles of a
v'cm
v'
triangle equals the other exterior angle:
L 
• Now use the law of sines:
ucm
sin    L  sin  L sin  L  sin  L




vcm
ucm
ucm
vcm
sin   L    sin  L
L
An Equivalent Formula
sin   L    sin  L
• Algebra can make this into a more useful formula
sin  cos  L  cos  sin  L   sin  L
sin  cos L   cos    sin  L
sin  cos  L   cos    1  cos2  L 
2
2
2
sin 2    cos    2  cos2  L   cos    2


1  2 cos    cos   cos   
2
2
L
cos  L 
cos   
1  2 cos    2
2
Differential Cross-Section In the Two Frames
  n v
d
 
d
d
cos  L 
cos   
1  2 cos    2
• The cross-section and the differential
cross-section can be computed in either the cm or the lab frame
– But it is usually easier in the cm frame
– We’ll do all calculations in the cm frame
• Because the density n, the rate , and the relative velocity v are all the
same in both frames, the total cross-section  will be the same*  cm   L
• But because the angles are different, the differential cross-section will not be
d
 d 
the same
d  
 d   d   d cos  
 d L
d


 

 d  L
 d   L  d    d cos  L 
d
 d 
 d  d cos      dL  d cos L 
2 3/2
d
 d  L
 d   d  1  2 cos    

 

1   cos 
 d  L  d  
*True even relativistically
Quantum Mechanics and Cross-Section
•
•
•
•
•
d
1

lim  r 2  out  r  
d   in r 
We need to figure out how to calculate these expression in QM
Typically, we don’t want to use multiple particles at once
Fortunately, even for one particle, we have a pretty
good idea of what the density and the flux are:
2
– Number density corresponds to probability density: n  
– Flux corresponds to probability current:
*
 j
j

Im

 

Our formula for differential
m
cross section becomes:
d
1
Note that since j appears in numerator

lim  r 2 jout  r  
d  jin R
and denominator, no need to normalize 
Solution in the Asymptotic Region
2
• We now need to start solving
E  r   
2  r   V  r   r 
2
Schrödinger’s Equation:
2
2 E
• First define the asymptotic wave
2
U
r

  2 V r 
k  2
number k and the scaled potential U by
• Schrödinger’s equation is now:
 k 2  2   r   U  r   r 
• Assume potential vanishes (sufficiently quickly) at infinity  k 2   2   r   0
– Roughly speaking, it must vanish faster than 1/r
• For the incoming wave, we expect a plane wave in the +z direction  in  eikz
• For the outgoing wave, makes sense to work in spherical coordinates
• Since we are taking r  , drop any terms with 1/r2:
  r     r , ,  
2
2

1

2
r



k
k 2 
r


0


 r 


2
2
r
r r
r out  r, ,    f k  ,   eikr  gk  ,   eikr
Asymptotic Form and Probability Currents
 in  e
ikz
eikr
eikr
 out  r , ,    f k  ,  
 g k  ,  
r
r
 out
eikr
 f k  ,  
r
• The gk corresponds to waves coming in from infinity
– Physically wrong, so ignore it
eikr
ikz
• The wave at infinity is therefore
lim   r    e 
f k  ,  
r 
r
*
• We will need
j  Im   
m
probability currents:
k

ikz
ikz
zˆ
• Probability current in: j  Im e  ikz eikz  Im  e zˆ ike  

 m
m
• Probability current out: in m
 * e  ikr 

eikr
 * e  ikr  eikr  
2 
fk  O  r  
jout  Im  f k

f k    Im  f k
 ikrˆ
r 
r
m 
r

 r
 m 

k fk
mr 2
2
rˆ  O  r 3 
Cross Section from Asymptotic Form
ikr
e
lim   r    eikz 
f k  ,  
r 
r
k
jin 
zˆ
m
jout 
k fk
mr 2
2
rˆ  O  r 3 
• Now we can do the differential cross-section:
d
1

lim  r 2 jout
d  jin r 
2



m
m k fk
2  k fk
3 
 
lim  r 
 O  r   

2
k r   
mr
k
m


 
• And then we can get the total cross-section
d
2
 
d    fk d 
d
• Of course, we don’t yet have a clue what fk is
2
d
 fk
d
2
14B. The Born Approximation
The Basic Idea
2
• We are trying to
2
2
k

k     r   U  r   r 

solve Schrodinger:
• We know what in is, so write   r   eikz    r 
• Substitute in:
 k 2  2  eikz    r   U  r   r 
k
2
2 E
2
U r  
2
2
V r 
  2    r   U  r   r 
• Now, imagine that we knew the right-hand side
• We want to find a straightforward way to find  if we know the right side
• We will use superposition
– Treat right hand side as a superposition of point sources
• First step – replace right side by a point source at the origin and solve equation
The Green’s Function
2
2
k


• Replace right side by a point source

  r   U  r   r 
– To simplify, place it at the origin
2
2
3
k


G
r




r 


– Call the result the Green’s function G(r):
• Spherically symmetric problem should
G r   G  r 
have a spherically symmetric solution:
2
• Write this out in
1
d
2
3
k
G
r

rG
r








r 
spherical coordinates
2 

r dr
2
• Away from the
d
2
rG
r


k


 rG  r  


origin, this is
2 

dr
• This has solutions:
rG  r    eikr   eikr
 ikr
G r   e
• But we are looking for outgoing waves, not incoming, so
r
• We still have to make sure it works at the origin
– This will tell us what  is
Getting G to Work at the Origin
2
2
3
• It is hard to check this equation at the origin, because
k


G
r



   r 
everything is infinite
 ikr
• To avoid this problem, integrate over a sphere of tiny radius R
G r   e
r
1    3  r  d 3r    k 2   2  G  r  d 3r  k 2  G  r  d 3r   rˆ G  r  dA
• Take the limit R  0:
R
 4 k  
0

1  ikr 
2  ik
1  lim  4 R   2  e
 4

R
R 0
r r 


• We now know that
2
ikr
eikr 2
2 d e
r dr  4 R
r
dr r
1 ikr
G r  
e
4 r
ikr

e
 k 2  2  4 r   3  r 
• We could easily have put the source
delta-function anywhere, so we generalize
• Laplacian is understood to affect r, not r'
r R
ik r r 
e
2
2
3
k





 4 r  r r  r
Solving the Actual Problem
• We solved our equation for a
“source” that is an arbitrary point
• We now use this to solve the actual problem
• Multiply top equation by U(r')(r')
ik r r 
e
2
2
 k    4 r  r   3 r  r
2
2
k



  r   U  r   r 
ik r r
e
2
2
3


k


U
r

r






 r  rU  r  r

 4 r  r
ik r r
e
2
2
3
3
3






k


U
r

r
d
r


r

r
U
r

r
d
r











  4 r  r

 U  r   r 
ik r r 

e
• We therefore see that
 r   
U  r   r  d 3r
• Substitute back in to
4 r  r
  r   eikz    r 
ik r r 
e
U  r 
ikz
3
  r   e   d r
  r 
• We therefore have:
4 r  r
• Integrate
over r':
The Born Approximation
ik r r 
• This looks useless
e
U  r 
ikz
3
– You can find  if you know 
  r   e   d r
  r 
4 r  r
• Substitute this equation into itself repeatedly:
ik r r 



U
r
e
U  r 


ikz
3
ikz 
3
  r   e   d r
  r  
e   d r
4 r  r 
4 r  r

e
ik r r 
ik r r 
ik r  r 


U
r
e
U
r
e
U  r  ikz




ikz
3
ikz 
3
3
 e   d r
e   d r
d r
e

4 r  r
4 r  r
4 r  r
e
ik r r 
ik r  r 
ik r  r 


U
r
e
U
r
e
U  r  ikz




3
3
3
  d r
d r
d r
e 


4 r  r
4 r  r
4 r  r
• This is a perturbative expansion in U
• If you keep just the first integral, it is called the first Born approximation, or
sometimes, the Born approximation
– We won’t go past the first Born approximation
e
ik r r 
Asymptotic Behavoir
• We need behavior at large r
to calculate cross-section
U  r  ikz
  r   e   d r
e
4 r  r
ikz
e
3
ik r r 
1/2
 2r  r r  
1 2r  r

2 

r
1






r
1


O
r
r  r  r  2r  r  r



2
2 


2
r
r
2
r




1
 r  rˆ  r  O  r 
ikr
1
e
ik  r rˆ r  
ikz
3
ikz 
ikz
3
ikz ikrˆ r 


 r   e 
d
r
e
U
r
e





e

d
r
U
r
e
 

4 r 
4 r
2
2
2
• Define the change in wave number
K  krˆ  kzˆ
ikr
e
  r   eikz 
f k  ,  
r
• Compare with the general asymptotic form:
• We therefore have:
1 3
 iK r 


f k  ,   
d
r
U
r
e


4 
Differential and Total Cross-Section
K  krˆ  kzˆ
f k  ,   
1 3
 iK r 


d
r
U
r
e


4 
• We are now ready to get the cross section using
2
2
d
1
3
 iK r
 f k  ,   
d rU  r  e
2 
d
16
• You then get the total cross section from
U r  
d
2
 2
d  4
2
2
V r 
 d rV  r  e
3
4
d
 
d
d
• Often need to write K in Cartesian coordinates
K  k  sin  cos  ,sin  sin  ,cos 1
• Also handy to work out K2:
2
2
2
2
2
ˆ
ˆ
ˆ
ˆ

k

k
r

z

2
r

z
ˆ
ˆ

 1  1  2cos 
K  k r  z 
2
2
K 2  2k 2 1  cos  
 iK r
2
Sample Problem
d
2
 2
d  4
Calculate the differential and total cross-section
r
V
r


e
r


for scattering from the potential
 d rV  r  e
3
4
 iK r
2
• Work on the Fourier transform of the potential  d 3rV  r  e  iK r
• Since potential is spherically symmetric, we can pretend K is any direction
– For convenience, pick it in the z-direction for this integration K  r  Kr cos


2
1
i  iKr cos cos 1
r
3
 iKr
2
 iKr cos
e
d cos   2 0 re dr
 d rV r  e  0 V  r  r dr 0 d 1 e
cos 1
Kr
2 i    r iKr   r iKr
2 i  1
1 
4

e
e
dr 

 2





0
K
K    iK   iK    K 2
– Done thinking of K in z-direction
• Now get the differential cross-section:
• Recall that K 2  2k 2 1  cos  
d
 2  4 



d  4 2 4   2  K 2 
d

d
4

2
4 2  2
2
 2k  2k cos  
2
2
2
Sample Problem (2)
Calculate the differential and total cross-section d
r

V
r


e
r


for scattering from the potential
d
4

4 2  2
2
 2k  2k cos  
2
2
• Now get the total cross-section
d
1
   d

1
d
4

8 2  2 d cos 
2
 2k  2k cos  
2
2
2
8 2  2
1

2k 2 4  2  2k 2  2k 2 cos 
4 2  2  1
1

 2 4  2 2
2 
k



4
k


16 2  2
 4 2 2
    4k 2 
cos 1
cos 1
2
The Coulomb Potential
V r  

r
e
r
d

d
4 
2
4

2
2
 2k  2k cos  
2
2
• Consider the Coulomb potential, given by
• We immediately get the differential cross section:
• Rewrite in terms of the energy E = 2k2/2
d C
ke2 q 2Q 2
ke2 q 2Q 2
4 1


csc

2
2 
2
2
d  4 E 1  cos  
16 E
2
16 2  2
 4 2 2
    4k 2 
ke qQ
V r  
r
d C
 2 ke2 q 2Q 2
 4 4
2
d
k 1  cos  
Why this is a cheat:
• We assumed potential falls off faster than 1/r, for  = 0 it does not
• Turns out this only causes a shift in the phase of  at large r, so answer is right
• Total cross-section is infinite
• Comes from small angles
• Experimentally, there is a minimum angle where we can tell if it scattered
14C. Method of Partial Waves
Spherically Symmetric Potentials
• Suppose our potential is not small, but is spherically symmetric
• We generally know how to find some
m

r

R
r
Y




l
l  ,  
solutions to Schrödinger’s equation:
V r   V  r 
1 d 2
l2  l 
• The radial function then satisfies ERl  
rRl   2 Rl   V  r  Rl

2 
2  r dr
r

• Which we rewrite as:
2
 2 l2  l

1 d2
rRl    k  2  U  r  Rl
2 
r dr
r


• Second order differential equation has two linearly independent solutions
Small r Behavior
 2 l2  l

1 d2
rRl    k  2  U  r  Rl
2 
r dr
r


• For small r, the 1/r2 term tends
1 d2
l2  l
rRl   2 Rl

2
2
to dominate the U and k terms
r dr
r
• Two linearly independent solutions, that roughly go like:
Rl  r l
Rl  r  l 1
• The latter one is unacceptable, because we want  finite
• This means for each l, there is only one acceptable solution, up to normalization
• We can always find these solutions, numerically if necessary
– Assume we have done so
• For large l, solution Rl tends to be very small near the origin
– Roughly, it vanishes if kr << l
• If U(r) vanishes or is negligible for r > r0, you can ignore U(r) if l > kr0
Large r Behavior
 2 l2  l

1 d2
rRl    k  2  U  r  Rl
2 
r dr
r


• Assume the potential falls off quickly at large r – then ignore U(r)
• Define x = kr, then

1 d2
l2  l 
xRl    1  2  Rl
2 
x dx
x 

• This equation has two known exact linearly independent solutions, called
spherical Bessel functions:
l
l
l  1 d  sin x
l  1 d  cos x
jl  x     x  
nl  x      x  


 x dx  x
x
dx

 x
• Most general solution, at large r, is therefore a linear combination of these
Rl  r    jl  kr    nl  kr 
Spherical Bessel Functions
l
 1 d  sin x
jl  x     x  

 x dx  x
l
l
l  1 d  cos x
nl  x      x  

x
dx

 x
• The spherical Bessel functions are closely related to regular Bessel functions
x
x
jl  x  
J l  1  x  , nl  x  
Yl  1  x 
2
2
2 2
• The jl’s are small at x = 0, and
the nl’s diverge
• We most want their asymptotic
behavior at large x:
lim jl  x  
sin  x  12  l 
x 
lim nl  x   
x 
x
cos  x  12  l 
x
l=0
l=1
l=2
l=3
jl(r)
nl(r)
Asymptotic Solution
• We assume we have the radial wave functions Rl, up to normalization
• At large r, we know
lim  Rl  r     jl  kr    nl  kr 
r 
it takes the form
  A cos  l ,   A sin  l
• Write these constants in the form
• The phase shift l is determined by the behavior of the Rl’s
• The phase shift is what you need to finish this analysis
– It vanishes for large l, because U(r) is irrelevant and nl badly behaved
• The amplitude A is arbitrary
• For large r, we now know what our solution looks like:
lim   r   lim  Rl  r  Yl m  ,    AYl m  ,    jl  kr  cos  l  nl  kr  sin  l 
r 
r 
• Use asymptotic form of spherical Bessel functions
A
lim   r    Yl m  ,   sin  kr  12  l  cos  l  cos  kr  12  l  sin  l 
r 
kr
A m
 Yl  ,   sin  kr  12  l   l 
kr
Our Remaining Goal
•
•
•
•
A m
lim   r    Yl  ,   sin  kr  12  l   l 
r 
kr
Write this in terms of exponentials
A
lim   r   
exp i  kr  12  l   l    exp  i  kr  12  l   l  Yl m  ,  
r 
2ikr
This expression has waves coming in and going out in all directions
ikr
e
We want a wave that looks like
lim   r    eikz 
f  ,  
r 
r
Most general solution will be a linear combination




Alm
lim   r    
exp i  kr  12  l   l   exp  i  kr  12  l   l  Yl m  ,  
r 
l , m 2ikr
• Want to find a combination to make it look like what we want
• First step: write eikz in terms of spherical harmonics at large r
ikz
e
in Spherical Harmonics at Large r
eikz   clm  r  Yl m  ,  
• Since spherical harmonics are complete, everything
l ,m
can be written in terms of them, including eikz:
• The functions clm can be found using orthogonality of the Ylm’s:
1
2
1
0
clm  r    d Yl m*eikr cos   d cos   eikr cos Yl m*  ,   d
1
m
• The Ylm’s go like eim, so
cl  r   2 m0  Yl 0*  ,   eikr cos d cos 
1
the  integral is easy:
1 
2  0
d

• Integrate by 0
 ikr cos
ikr cos cos 1
0
cl  r  
Y

e

Y

e
d
cos

 l  

l  


cos


1

1
parts on
ikr 
 d cos 


2 0
cos
2 0
ikr cos   0
ikr
0
 ikr

Yl   e
 O  r 2  


Y
0
e

Y

e




l
l




repeatedly
ikr
ikr
• We know these 0
2 2l  1  ikr
l ikr

c
r

e


1
e




l
functions, so

ikr 4 
1
l  ikr
ikz
ikr
0


lim e 
4

2
l

1
e


1
e
Y





l  


r 
2ikr l
Announcements
ASSIGNMENTS
Day
Read
Homework
Today
none
14.3
Wednesday none
none
Friday
no class
no class
Test Thursday
1:45, Olin 102
Covers through 13
Reviews on
Monday and Wed.
Homework Solutions
Chapter 12
3/2
Putting it Together . . .
•
•
•
•
•
•


Alm i kr  12  l l  i kr  12  l l  m
lim   r    
e
e
Yl  ,  
We have
r 
l , m 2ikr
ikr
e
ikz
We want lim   r    e 
f  ,  


r 
r
1
l  ikr
ikr
0


Where
lim eikz 
4

2
l

1
e


1
e
Y
 , 






l


r 
2ikr l
i  12  l  l 
We have to get the e-ikr
m
Al   m0 4  2l  1e
terms to match, so try
i  kr  2 l 
 ikr  i l
0
Substitute this in: lim   r    1
4

2
l

1
e

e
e
Y



l  


r 
2ikr l
1
Compare to eikz:
2i l
ikz
ikr
0
lim   r   e  
4

2
l

1
e
e

1
Y



l  
r 
2ikr l
• We therefore have:



f  ,   


1
2 i l
0
4

2
l

1
e

1
Y



l  
2ik l

The Differential and Total Cross-Section
1
2
2 i l
0
d
f

,


4

2
l

1
e

1
Y




 f k  ,  

k
l  
2ik l
d
i l
 i l
e

e
• Some algebra: 1 2il
 ei l sin  l
e  1  eil
2i
2i



d 4
 2
d k
• The differential
cross section:
• The total cross-section:
4
 2
k
4
 2
k
i l i l
e
sin  l sin  l 

4
 2
k
i l i l
e
sin  l sin  l 

l
l
l
l


2l  1eil sin  lYl 0  
2
l

2
2l  1eil sin  lYl 0   d 
l
 2l  1 2l   1  Yl 0   Yl0*   d 
 2l  1 2l   1 l ,l
4
 2
k
2
2
l

1
sin
l



l
The Procedure
• Find U(r): U  r   2V  r  2
Now, for each value of l:
– Find appropriate inner boundary conditions (typically Rl(0) = 0)
– Solve numerically or analytically:
 2 l2  l

1 d2
rRl    k  2  U  r  Rl
– At large r, match your solution to:
2 
r dr
r


Rl  r   l jl  kr   l nl  kr 
l  tan 1  l l 
– Deduce phase shift:
– Stop when l gets small, or l >> kr0, where r0 is where the potential gets
small
2
• Differential cross-section: d 4
 2  2l  1eil sin  lYl 0  
d k l
• Total cross-section:
4
 2
k
2
2
l

1
sin
l



l
Sample Problem
Calculate the differential and total crosssection from a hard sphere of radius a, with
potential as given at right, where ka is small
•
•
•
•
 r  a
V r   
0 r  a
Perturbation theory cannot work, because the potential is infinite
U(r) = V(r), because it is zero or infinity
Boundary condition must be that Rl(a) = 0
2
 2 l2  l

2
We need to solve outside: 1 d

l
rRl    k  2  U  r  Rl  k 2   l  R
2 

2  l
r dr
r


r


• Solution of this is known:
Rl  r   l jl  kr   l nl  kr 
• Boundary condition:
 l jl  ka 

0  Rl  a   l jl  ka   l nl  ka 
 l nl  ka 
• Now we attempt to find phase shift:
 jl  ka  



1

 l  tan 

 l  tan 1  l 
n
ka
 l  
 l 
Sample Problem (2)
Calculate the differential and total crosssection from a hard sphere of radius a, with
potential as given at right, where ka is small
•
•
•
•
•
•
 r  a
V r   
0 r  a
 jl  ka  
1
 l  tan 

 nl  ka  
Now use approximation ka small
This means we need to keep up to l ~ ka <<1
Just keep l = 0
sin  kr 
cos  kr 
j0  kr  
, n0  kr   
Look up spherical Bessel functions:
kr
kr
Get the phase shift:
 sin  ka  
1
 0  tan 
Now find differential
  tan 1   tan  ka    ka
  cos  ka  
cross section:
2
2
sin
ka 
2

4
d 4
i 0
0
i l 0
 2  2l  1sin  l e Yl    2 2  0  1e sin  0Y0   
2
k
k
d k l
• We assumed ka small, so
d
 a2
d
  4 a 2
Comments on Cross Sections
• Interestingly, the cross-section is equal to the surface area (NOT the
  4 a 2
silhouette area)
• Generally described as diffraction allows particles to scatter off of all sides
• Note that whenever the scale of the potential is small, scattering is dominated by
l = 1 (s-wave scattering)
• All angles are scattered equally
Limits on Total Cross Sections
4
 2
k
2
2
l

1
sin
l



l
• Note that if a particular value of l dominates the
cross-section, then there is a limit on the cross section:
4
  2  2l  1
k
• Before the discovery of the Higgs boson, it was pointed out that without
the Higgs boson, the cross-section for WW scattering was predicted   k 2
to be dominated by l = 0, and it grows as k2
• Higgs boson cancels part of amplitude, and suppresses the cross section, once
you get at or near the Higgs mass
• Predicted Higgs, or something had to be lighter than 1000 GeV/c2
– No lose theorem
• Higgs discovered in 2012 at 126 GeV/c2