Scattering of particles - topic 1 - june 2007

• Particle Scattering:– Differential cross section– Trajectories and currents– Mean free path

• Quantal Scattering:

– Currents– Differential cross section

– Integral Equation

• Classical Scattering:– Trajectories; impact parameter– Differential cross section– Total cross section– Example: Hard sphere scattering

Cross section - mean free path - macroscopic cross section

ddd

dd

d

d)sin(

0

2

0

Number of scattered particles into :

Differential Cross Section:

Total Cross Section:

innj

dd

The Scattering Cross Section

Nout

Nout

Number of particles

Number of particles : seen as ”particles” in a current, or probability density current

Number of particles : seen as ”particles” in a number of possible trajectories (impact parameters

In both cases: DIFFERENTIAL CROSS SECTION

In both cases: probabilistic formulations

Example - Classical scattering:

d

b

bdbdd

ddd

dd )sin(

d

dbb

d

d

sin

b R

2

2cos

R

b

2sin

2

R

d

db

4

2R

d

d

2R

Hard Sphere scattering:

Independent of angles!= Geometrical Cross sectional area of sphere!

Quantal Scattering - No Trajectory! (A plane wave hits some object and a spherical wave emerges)

innj

dd

r

eCf

ikr

scattered ),(

scatteredinn

rkiinn Ce

• Solve the time independent Schrödinger equation• Approximate the solution to one which is valid far away from the scattering center• Write the solution as a sum of an incoming plane wave and an outgoing spherical wave.• Must find a relation between the wavefunction and the current densities that defines the

cross section.

Procedure:

scj

Current Density:

imm

ij *** Re)(

2

Incomming current density:

2C

m

kjin

Outgoing spherical current density:

2

22),(mr

kfCjsc

r

eCf

ikr

sc ),(

ikzrkiin CeCe

zyx ez

fey

fex

frf

)(

eeer

fr

.......

2),(),(

rOr

eikCf

r

eCfr

ikrikr

The Schrödinger equation - scattering form:

)( )( )(2

22

rErrVm

)( )()( 22 rrUrk

:get we)(2

)( and 2

with 2

22

rVm

rUm

kE

Now we must define the current densities from the wave function…

The final expression:

2

2

22

2

22

),(),(

f

mk

Cd

drfmrk

C

jd

drj

d

d

in

sc

2),(

fd

d

Summary

Then we have:

2),(

fd

d

…. Now we can start to work

)( )()( 22 rrUrk

Write the Schrödinger equation as:

)),(( r

efeC

ikrrki

Asymptotics:

Integral equation

)( )()( 22 rrUrk

)()(G ''22 rrrrk

With the rewritten Schrödinger equation we can introducea Greens function, which (almost) solves the problem for a delta-function potential:

Then a solution of:

can be written:

rdrrUrrGrr 30 )()()()()(

where we require: 0)( 022 rk

because….

rdrrUrrGkrkrk 3220

2222 )()()()()()()()(

This term is 0 This equals )( rr

Integration over the delta function gives result:

)()()()( 22 rrUrk

rdrrUrrGrr 30 )()()()()(

Formal solution:

Transforms formally differential equation to integral equation

)()(G ''22 rrrrk

Green’s function

)()2(

1)()( 3

322 rrderGk rsi

r

erG

ikr

4)(

”Proof”:

The result is well known (function of scalar distance only!):

Easier to solve for r’ = 0 ( | r | instead of | r - r’ | )

rdrrUrr

eer

rrikrki

3

01 )()(4

1)(

One obtains:

rdrrUrr

eer

rrikrki

3)()(

4

1)(

The Born series (first Born approximation usually used:

rkier )(0

rdrrUrr

eer

rrikrki

3

12 )()(4

1)(

)( )()( 22 rrUrk

Schrödinger equation as:

)),(( r

efeC

ikrrki

Asymptotics:

SUMMARY

)1(

)21(

2

2

2/12

2

2

22

r

rrr

r

r

r

rrr

rrrrrr

The potential is assumed to have short range, i.e. Active only for small r’ :

rr

rikikrrrik eee

1)

rrr

11

ff kp

r

rk

Asymptotics - Detector is at near infinite r

2)

),(

3)()(4

1)(

f

rkiikr

rki rdrrUer

eer f

Asymptotic excact result:

Still formal expression

The Born approximation:

rdrVem

r

eer rkki

ikrrki f

3)(

2)(

4

2)(

rdrVem

f rkki f 3)(

2)(

2),(

The scattering amplitude is then:

Suitable for simple evaluations:

Fourier transform of the potential for the value of the momentum change or ”momentum transfer”

Use incomming wave instead of )'(r Under integration sign:

rdrVem

f rkkiB

q

f 3)(

2)(

4

2),(

fk

k

q

2sin2

kq

2

00

cos'

0

22

sin)(4

2)( ddedrrrV

mf iqrB

2cos

2sin2sin

dkdq

2cos

0

2sin)(

2drqrrV

q

m

Spherically Symmetric potentials - typical evaluation:

Total Cross Section:

ddd

dd )sin(

qdqqfk

dfk

BB

22

02

0

2)(

2)sin()(2

momentum transfer

A feature - 1’st. Born Approximation:

rdrVem

f rkkiB

q

f 3)(

2)(

4

2),(

qdqqfk

kB

22

02

)(2

2),( Bf

d

d

fkkq

Because

scattering angle is related to MOMENTUM TRANSFER

fk

k

q

INTEGRAL OVER MOMENTUM TRANSFER

Example - Hard sphere

b R

2

4

2R

d

d

2R

Classical Hard Sphere scatteringDifferenial cross section constant, no angular dependence!

Homework discussion

Homework discussion