Lecture 17:Scattering in One Dimension

Part 2

Phy851 Fall 2009

Transfer Matrix

• For the purpose of propagating a wavethrough multiple elements, it is moreconvenient to relate the amplitudes on theright-side of the boundary to those on theleft-side via the ‘Transfer Matrix’:

=

b

aM

d

c

dcba +=+

( ) ( )dckbak −=− 21

−=

− b

a

kkd

c

kk 1122

1111

−

−=

−

b

a

kkkkd

c

11

1

22

1111

−

−=

b

a

kkk

k

kd

c

112

2

2

11

1

1

2

1

( )

+−

−+=

1212

1212

221 2

1,

kkkk

kkkk

kkkM

Definition oftransfer

matrix, M:Boundaryconditionequations

b.c. eqs inmatrix form

Solve for theright amplitudesin terms of the

left

Calculate theinverse

Multiplymatrices to

get:

We don’t use T,because the T-

matrix is somethingelse

Extracting r and t from M

• Consider case a=1, b=r, c=t, and d=0:

=

rM

t 1

0

02221

1211

=+

=+

rMM

trMM

22

21

M

Mr −=

22

]det[

M

Mt =

22

211211 M

MMMt −=

A Generalized Transfer Matrixapproach to complex scattering

potentials

• How would you go about computing r and tfor a complicated structure?

Systematic approach:• compute the Transfer Matrix for each element,

then multiply them all together to get a fullTransfer Matrix for the object– For a problem with only one boundary, it will

always save time to plug and chug.– For 2 or more boundaries, the Transfer Matrix

approach should save time

• Then compute probabilities via:

x5x4

x3x2

x1

1 2 34

5E

tr1

2

22

12

M

MR = RT −=1

6

Definition of the 2-vectorrepresentation of the wavefunction

• At any point, x, along the wave, the state ofthe system, ψ(x), will be represented by a twovector:

• Example: consider a free particle:

→+=

)(

)()()()(

x

xxxx

L

RLR ψ

ψψψψ

b

a

d

c

x

1x 2xbax +=)( 1ψ dcx +=)( 2ψ

ikxikx BeAex −+=)(ψBut also we must have:

1ikxAea =This meansthat: 1ikxBeb −=

2ikxAec =2ikxBed −=

( )12 xxikaec −=( )12 xxikbed −−=

Thus wehave:

So we have:matrixform:

=

− b

a

e

e

d

cikL

ikL

0

0

12 xxL −=

The Transfer-matrix for free propagation

• Thus to propagate a two-vector over adistance L, with wavevector k, we have:

• A potential step can be characterized by k1and k2. We already determined that the M-matrix is:

• Now we are ready to go to work on somescattering problems

=

−ikL

ikL

freee

ekLM

0

0)(

+−

−+=

1212

1212

212 2

1),(

kkkk

kkkk

kkkMstep

`Free Propagator’

LR

• Example: Scattering through two steppotentials:

• Thus the full M-matrix for this scatterer is:

x1=0

Transfer-Matrix Approach to Scatteringthrough Multiple Elements

x

x2=L

I II III

1

r

a

b

c

d

tk1 k2k3

( )

=

rkkM

b

astep

1, 12

( )

=

b

aLkM

d

cfree 2 ( ) ( )

=

rkkMLkM

d

cstepfree

1, 122

( )

=

d

ckkM

tstep 23,0

( ) ( ) ( )

=

rkkMLkMkkM

tstepfreestep

1,,

0 12223

2

22

12

M

MR = RT −=1

( ) ( ) ( )12223 ,, kkMLkMkkMM stepfreestep=

• Example: Scattering through square barrier

• The full M-matrix for this scatterer is:

Step potential result

( ) ( ) ( )kKMKLMKkMM stepfreestep ,,=

€

Mstep k,K( ) =12k

k + K k −Kk −K k + K

€

M =cos[KL]+ i

k 2 + K 2( )2kK

sin[KL] −i k 2 −K 2( )2kK

sin[KL]

+i k 2 −K 2( )2kK

sin[KL] cos[KL]− ik 2 + K 2( )2kK

sin[KL]

€

Mstep K,k( ) =12K

K + k K − kK − k K + k

x1=0

x

x2=L

I II III

1

r

a

b

c

d

tk1=k k2=K k3=k

E

( )

=

−iKL

iKL

freee

eKLM

0

0LR

LR

LR

R

R

Example Continued:

• The reflection Amplitude is then:

• So the reflection probability is

• This has a minimum of R =0 wheneverKL=nπ

– These are the Transmission Resonances– This describes a frequency filter– Choose L = (n/2)λ to transmit wavelength λ– For KL=(n+1/2)π, transmission is reduced to:

€

r = −M21

M22

= −k 2 −K 2( )sin[KL]

K 2 + k 2( )sin[KL]+ 2ikK cos[KL]

€

R =| r |2=k 2 −K 2( )

2

k 2 + K 2( )2 + 4k 2K 2 cot 2[KL]

€

T =1−k 2 −K 2( )

2

k 2 + K 2( )2=

4K 2k 2

k 2 + K 2( )2≈ 4 K

2

k 2For K << k

can be very small

€

M =cos[KL]+ i

k 2 + K 2( )2kK

sin[KL] −i k 2 −K 2( )2kK

sin[KL]

+i k 2 −K 2( )2kK

sin[KL] cos[KL]− ik 2 + K 2( )2kK

sin[KL]

( ) ( )∞=

−==

0

1

)sin(

)cos(cot

n

n

nn

ππ

π 11 →−= RT

( )( ) ( )( ) ( )

01

0

)sin(

)cos(cot

21

21

21 =

−=

+

+=+ nn

nn

ππ

π

Transmission Resonance• Analogy: light passing through a high-index

medium

• Interference due to multiple TransmissionPathways– Constructive Interference when:

Incident

R T

€

K 2L( ) = n 2π( )

Round-tripphase shift

€

T =1− R =1−kL( )2 − KL( )2( )

2

kL( )2 + KL( )2( )2

+ 4 kL( )2 KL( )2 cot2[KL]

0

2222

22V

m

K

m

k+=

hh 20

2 kKk +=

Depends only on KL and α( ) 22 α+= KLkL

LmV

Lk20

0

2

h==α

α is a measure of barrier `area’

K is wavevector inside material

Resonance Transmission Profiles

α = 100

α = 10

•Common Approach:•To increase α, increase V0

•Then use L to tune wavelength

Plots made with Mathematica

Generalized Approach

• With this Method we can calculate thereflection and transmission amplitudes forany sequence of potential steps:

]1,2[]2[]2,3[]3[]3,4[]4[]4,5[]5[]5,6[sfsfsfsfs MMMMMMMMMM =

22

21

M

Mr −=

2rR = RT −=1

x5x4

x3x2

x1

1 2 34

5

E

rt1

6

[ ]( )

( )

=

−

−

−−

−

1

1

0

0mmm

mmm

xxik

xxikmf

e

eM

[ ]

+−

−+=

mnmn

mnmn

n

mns kkkk

kkkk

kM

2

1, ( )h

nn

VEmk

−=2