• 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• It makes sense that it would be proportional to the speed

|v| of the projectiles• Call the remaining proportionality constant • 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

14. Scattering

Cross Section 14A. Cross Section and Differential Cross Sec.

in

out

n v

Flux and Differential Cross-Section• The flux (in or out) is the number density times the relative velocity• Imagine we are far from a small target

– 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:

• The cross-section is then:

• If detector only covers some angles, restrict integral• Useful to define the differential cross-section:

in n v

outlimr

dA

r 2outlim

rr d

r

in

2out

in

1ˆlim

rr r d

r

2out

in

1limr

dr

d

r d

dd

Lab Frame and Center of Mass FrameThere are two types of collision experimentsthat commonly occur:•The lab frame, where thetarget is stationary and theprojectile collides with it•The center of mass frame,where the incoming particleshave net zero momentum•Usually easier to calculate in the cm frame•But then we must relate these two frames•Let the mass of the projectile be m and the target M•Commonly write things in terms of the ratio:•The velocity of the center of mass compared to the lab frame can be found:

v L

vcm

totcm

totm

Pu

m

m M

v

m

M

1cm

u v

v'

v'cm

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 isin the z-direction

• Then the scattering angles and L are thepolar angle of the direction of the outgoingparticles in the two frames

• The azimuthal angle and L will be the same• 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 L

vcm

ˆvv z

v'

v'cm

L

andcm cm cm cm v v u v v u

Comparing Angles in the Two Frames

• It follows that • From conservation of momentum in the cm frame, it is

clear that the final particles are back to back andtheir momenta must be equal and opposite

• In cm frame, conservation of energy implies• Now, let’s do some geometry:• The sum of two interior angles of a

triangle equals the other exterior angle:• Now use the law of sines:

v L

vcm

cm cm

cm cm

v v u

v v u

cm cmv v

v'

v'cm

1cm

u v1

1cm

v v

v' v'cm

ucm

L

– L

sin sinL L

cm cmu v

sin L

cmv

cm

u

sin L

cmu

sin sinL L

An Equivalent Formula

• Algebra can make this into a more useful formula

sin sinL L

sin cos cos sin sinL L L

sin cos cos sinL L

22 2 2sin cos cos 1 cosL L

2 22 2sin cos cos cosL

22 21 2 cos cos cosL

2

coscos

1 2 cosL

Differential Cross-Section In the Two Frames

• The cross-section and the differentialcross-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*

• But because the angles are different, the differential cross-section will not be the same

2

coscos

1 2 cosL

n v d

dd

cm L

LL

d dd d

d d

cos cosL LL

d dd d d d

d d

cos

cosL L

d d d

d d d

3/221 2 cos

1 cosL

d d

d d

*True even relativistically

Quantum Mechanics and Cross-Section

• 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:– Number density corresponds to probability density:– Flux corresponds to probability current:

• Our formula for differentialcross section becomes:

• Note that since j appears in numeratorand denominator, no need to normalize

2n

2out

in

1limr

dr

d

r

j *Imm

j

2out

in

1limR

dr

d

j r

j

• We now need to start solvingSchrödinger’s Equation:

• First define the asymptotic wavenumber k and the scaled potential U by

• Schrödinger’s equation is now:

• Assume potential vanishes (sufficiently quickly) at infinity– Roughly speaking, it must vanish faster than 1/r

• For the incoming wave, we expect a plane wave in the +z direction• For the outgoing wave, makes sense to work in spherical coordinates• Since we are taking r , drop any terms with 1/r2:

inikze

2

2

2E V

r r r r

2 2k U r r r

22

2 Ek

2

2U V

r r

2 2 0k r

2

22

10k r

r r

2

22

r k rr

out , , , ,ikr ikrk kr r f e g e

Solution in the Asymptotic Region

, ,r r

• The gk corresponds to waves coming in from infinity– Physically wrong, so ignore it

• The wave at infinity is therefore• We will need

probability currents:• Probability current in:• Probability current out:

lim ,ikr

ikzkr

ee f

r

r

Asymptotic Form and Probability Currents

inikze out , , , ,

ikr ikr

k k

e er f g

r r

out ,ikr

k

ef

r

*Imm

j

in Im ikz ikze em

j ˆIm ikz ikze ike

m z

ˆk

m z

*out Im

ikr ikr

k k

e ef f

m r r

j

* 2ˆImikr ikr

k k

e ef ik f O r

m r r

r

2

32

ˆkk fO r

mr r

• Now we can do the differential cross-section:

• And then we can get the total cross-section

• Of course, we don’t yet have a clue what fk is

lim ,ikr

ikzkr

ee f

r

r

Cross Section from Asymptotic Form

in ˆk

mj z

2

3out 2

ˆkk fO r

mr j r

2out

in

1limr

dr

d

j

j

2

2 32

lim k

r

k fmr O r

k mr

2

kk fm

k m

2

k

df

d

2

k

dd f d

d

• We are trying tosolve Schrodinger:

• We know what in is, so write• Substitute in:

• 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 Basic Idea 14B. The Born Approximation

2 2k U r r r

ikze r r

22

2 Ek

2

2U V

r r

2 2 ikzk e U r r r

2 2k U r r r

• Replace right side by a point source– To simplify, place it at the origin– Call the result the Green’s function G(r):

• Spherically symmetric problem shouldhave a spherically symmetric solution:

• Write this out inspherical coordinates

• Away from theorigin, this is

• This has solutions:

• But we are looking for outgoing waves, not incoming, so

• We still have to make sure it works at the origin – This will tell us what is

The Green’s Function 2 2k U r r r

2 2 3k G r r

G G rr

2

2 32

1 dk G r rG r

r dr r

2

22

drG r k rG r

dr

ikr ikrrG r e e ikrG r e

r

• It is hard to check this equation at the origin, becauseeverything is infinite

• To avoid this problem, integrate over a sphere of tiny radius R

• Take the limit R 0:

• We now know that

• We could easily have put the sourcedelta-function anywhere, so we generalize

• Laplacian is understood to affect r, not r'

Getting G to Work at the Origin 2 2 3k G r r

ikrG r er

3 31 d r r 2 2 3k G r d r 2 3 ˆk G r d G r dA r r

2 2 2

04 4

ikr ikrR

r R

e d ek r dr R

r dr r

2

20

11 lim 4 ikr

RR

ikR e

r r

1

4ikrG r e

r

2 2 3

4

ikrek

r

r

2 2 3

4

ikek

r r

r rr r

4

• 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')

• Integrateover r':

• We therefore see that• Substitute back in to

• We therefore have:

Solving the Actual Problem

U r r

2 2 3

4

ikek

r r

r rr r

2 2k U r r r

2 2 3

4

ikek U U

r r

r r r r r rr r

2 2 3 3 3

4

ikek U d U d

r r

r r r r r r r rr r

3

4

ikeU d

r r

r r r rr r

ikze r r

3

4

ikikz e U

e d

r r r

r r rr r

• This looks useless– You can find if you know

• Substitute this equation into itself repeatedly:

• 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

The Born Approximation

3

4

ikikz e U

e d

r r r

r r rr r

3 3

4 4

ik ikikz ikze U e U

e d e d

r r r rr rr r r r

r r r r

3 3 3

3 3 3

4 4 4

4 4 4

ik ik ikikz ikz ikz

ik ik ikikz

e U e U e Ue d e d d e

e U e U e Ud d d e

r r r r r r

r r r r r r

r r rr r r

r r r r r r

r r rr r r

r r r r r r

• We need behavior at large rto calculate cross-section

• Define the change in wave number

• Compare with the general asymptotic form:• We therefore have:

Asymptotic Behavoir

3

4

ikikz ikze U

e d e

r r rr r

r r

2 22 r r r r r r

1/22

2 2

21

rr

r r

r r 22

1 21

2r O r

r

r r

1ˆr O r r r

ˆ31

4ik rikz ikze d e U e

r

r rr r r ˆ3

4

ikrikz ikz ike

e d U er

r rr r

ˆ ˆk k K r z

,ikr

ikzk

ee f

r r

31,

4i

kf d U e

K rr r

• We are now ready to get the cross section using

• You then get the total cross section from

• Often need to write K in Cartesian coordinates

• Also handy to work out K2:

Differential and Total Cross-Section

ˆ ˆk k K r z

sin cos ,sin sin ,cos 1k K

31,

4i

kf d U e

K rr r

2,k

df

d

2

32

1

16id U e

K rr r

2 23

2 44id

d V ed

K rr r

2

2U V

r r

dd

d

22 2 ˆ ˆk K r z 2 2 2ˆ ˆˆ ˆ2k r z r z 2 1 1 2cosk

2 22 1 cosk K

Sample ProblemCalculate the differential and total cross-section for scattering from the potential rV r e r

2 23

2 44id

d V ed

K rr r

• Work on the Fourier transform of the potential• Since potential is spherically symmetric, we can pretend K is any direction

– For convenience, pick it in the z-direction for this integration

– Done thinking of K in z-direction• Now get the differential cross-section:• Recall that

3 id V e K rr r

cosKr K r

2 13 2 cos

0 0 1cosi iKrd V e V r r dr d e d

K rr r

cos 1cos

cos 102 r iKri

re dr eKr

0

2 r iKr r iKrie e dr

K

2 1 1i

K iK iK

2 2

4

K

22

2 4 2 2

4

4

d

d K

2 2

24 2 2 2

4

2 2 cos

d

d k k

2 22 1 cosk K

Sample Problem (2)Calculate the differential and total cross-section for scattering from the potential rV r e r

• Now get the total cross-section

dd

d

2 21

21 4 2 2 2

8 cos

2 2 cos

d

k k

2 2

24 2 2 2

4

2 2 cos

d

d k k

cos 12 2

2 4 2 2 2cos 1

8 1

2 2 2 cosk k k

2 2

2 4 2 2 2

4 1 1

4k k

2 2

4 2 2 2

16

4k

• Consider the Coulomb potential, given by • We immediately get the differential cross section:• Rewrite in terms of the energy E = 2k2/2

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

The Coulomb Potential

2 2

24 2 2 2

4

2 2 cos

d

d k k

2 2

4 2 2 2

16

4k

rV r e

r

ek qQV r

r

2 2 2

224 1 cosC ed k q Q

d E

2 2 2

4 122

csc16ek q Q

E

2 2 2 2

24 4 1 cosC ed k q Q

d k

• Suppose our potential is not small, but is spherically symmetric• We generally know how to find some

solutions to Schrödinger’s equation:

• The radial function then satisfies• Which we rewrite as:

• Second order differential equation has two linearly independent solutions

Spherically Symmetric Potentials 14C. Method of Partial Waves

V V rr

,ml lR r Y r

2 2 2

2 2

1

2l l l l

d l lER rR R V r R

r dr r

2 2

22 2

1l l

d l lrR k U r R

r dr r

• For small r, the 1/r2 term tendsto dominate the U and k2 terms

• Two linearly independent solutions, that roughly go like:

• 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

Small r Behavior

2 2

2 2

1l l

d l lrR R

r dr r

1l ll lR r R r

2 2

22 2

1l l

d l lrR k U r R

r dr r

• Assume the potential falls off quickly at large r – then ignore U(r)• Define x = kr, then

• This equation has two known exact linearly independent solutions, called spherical Bessel functions:

• Most general solution, at large r, is therefore a linear combination of these

Large r Behavior

2 2

22 2

1l l

d l lrR k U r R

r dr r

2 2

2 2

11l l

d l lxR R

x dx x

1 sinl

l

l

d xj x x

x dx x 1 cos

ll

l

d xn x x

x dx x

l l lR r j kr n kr

l = 0l = 1l = 2l = 3

jl(r)nl(r)

• The spherical Bessel functions are closely related to regular Bessel functions

• The jl’s are small at x = 0, andthe nl’s diverge

• We most want their asymptoticbehavior at large x:

Spherical Bessel Functions

1 sinl

l

l

d xj x x

x dx x

1 cosl

l

l

d xn x x

x dx x

1 12 2

,2 2l ll l

x xj x J x n x Y x

12

12

sinlim

coslim

lx

lx

x lj x

xx l

n xx

• We assume we have the radial wave functions Rl, up to normalization• At large r, we know

it takes the form• 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:

• Use asymptotic form of spherical Bessel functions

Asymptotic Solution

lim l l lrR r j kr n kr

cos , sinl lA A

lim lim ,ml lr r

R r Y

r , cos sinml l l l lAY j kr n kr

1 12 2lim , sin cos cos sinm

l l lr

AY kr l kr l

kr

r

12, sinm

l l

AY kr l

kr

• Write this in terms of exponentials

• This expression has waves coming in and going out in all directions• We want a wave that looks like

• Most general solution will be a linear combination

• 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

Our Remaining Goal

12lim , sinm

l lr

AY kr l

kr

r

1 12 2lim exp exp ,

2m

l l lr

Ai kr l i kr l Y

ikr

r

lim ,ikr

ikz

r

ee f

r

r

1 12 2

,

lim exp exp ,2

mml

l l lrl m

Ai kr l i kr l Y

ikr

r

• Since spherical harmonics are complete, everythingcan be written in terms of them, including eikz:

• The functions clm

can be found using orthogonality of the Ylm’s:

• The Ylm’s go like eim, so

the integral is easy:• Integrate by

parts oncosrepeatedly

• We know thesefunctions, so

eikz in Spherical Harmonics at Large r

,

,ikz m ml l

l m

e c r Y

* cosm m ikrl lc r d Y e

1 2 cos *

1 0cos ,ikr m

ld e Y d

1 0* cos

0 12 , cosm ikr

l m lc r Y e d

1cos 10 0 cos 0 cos

cos 1 1

2cos

cosikr ikr

l l l

dc r Y e Y e d

ikr d

00 cos 22 ikr

lY e O rikr

0 02

0 ikr ikrl lY e Y e

ikr

0 2 2 11

4likr ikr

l

lc r e e

ikr

01lim 4 2 1 1

2likz ikr ikr

lrl

e l e e Yikr

Announcements

3/2

ASSIGNMENTSDay Read HomeworkToday none 14.3Wednesday none noneFriday no class no class

Test Thursday1:45, Olin 102

Covers through 13Reviews on

Monday and Wed.

Homework Solutions Chapter 12

• We have

• We want

• Where

• We have to get the e-ikr

terms to match, so try• Substitute this in:

• Compare to eikz:

• We therefore have:

Putting it Together . . .

01lim 4 2 1 1 ,

2likz ikr ikr

lr

l

e l e e Yikr

lim ,ikr

ikz

r

ee f

r

r

1 12 2

,

lim ,2

l l

mi kr l i kr l ml

lrl m

Ae e Y

ikr

r

12

0 4 2 1 li lml mA l e

2 01lim 4 2 1

2li kr ikr i l

lrl

l e e e Yikr

r

2 01lim 4 2 1 1

2liikz ikr

lrl

e l e e Yikr

r

2 01, 4 2 1 1

2li

ll

f l e Yik

• Some algebra:

• The differentialcross section:

• The total cross-section:

The Differential and Total Cross-Section

211

2 2

l l

l l

i ii i e e

e ei i

2,k

df

d

2 01, 4 2 1 1

2li

k ll

f l e Yik

sinli

le

2

02

42 1 sinli

l ll

dl e Y

d k

2

02

42 1 sinli

l ll

l e Y dk

0 0*

2

4sin sin 2 1 2 1l li i

l l l ll l

e l l Y Y dk

,2

4sin sin 2 1 2 1l li i

l l l ll l

e l lk

22

42 1 sin l

l

lk

• Find U(r):Now, for each value of l:

– Find appropriate inner boundary conditions (typically Rl(0) = 0)– Solve numerically or analytically:– At large r, match your solution to:

– Deduce phase shift:– Stop when l gets small, or l >> kr0, where r0 is where the potential gets small

• Differential cross-section:

• Total cross-section:

The Procedure 22U r V r

22

42 1 sin l

l

lk

2 2

22 2

1l l

d l lrR k U r R

r dr r

l l l l lR r j kr n kr

1tanl l l

2

02

42 1 sinli

l ll

dl e Y

d k

Sample ProblemCalculate the differential and total cross-section from a hard sphere of radius a, with potential as given at right, where ka is small

0

r aV r

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• We need to solve outside:

• Solution of this is known:• Boundary condition:

• Now we attempt to find phase shift:

2 2

22 2

1l l

d l lrR k U r R

r dr r

22

2 l

l lk R

r

l l l l lR r j kr n kr

0 l l l l lR a j ka n ka

ll

l l

j ka

n ka

1tan ll

l

1tan ll

l

j ka

n ka

Sample Problem (2)Calculate the differential and total cross-section from a hard sphere of radius a, with potential as given at right, where ka is small

0

r aV r

r a

• Now use approximation ka small• This means we need to keep up to l ~ ka <<1• Just keep l = 0• Look up spherical Bessel functions:• Get the phase shift:• Now find differential

cross section:

• We assumed ka small, so

0 0

sin cos,

kr krj kr n kr

kr kr

2

02

42 1sin li

l ll

dl e Y

d k

1tan ll

l

j ka

n ka

02

00 02

42 0 1 sinie Y

k

2

2

sin ka

k

10

sintan

cos

ka

ka

1tan tan ka ka

2da

d

24 a

• Interestingly, the cross-section is equal to the surface area (NOT thesilhouette 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

Comments on Cross Sections24 a

Limits on Total Cross Sections

• Note that if a particular value of l dominates thecross-section, then there is a limit on the cross section:

• Before the discovery of the Higgs boson, it was pointed out that withoutthe Higgs boson, the cross-section for WW scattering was predictedto 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

2

42 1l

k

2k

22

42 1 sin l

l

lk