lecture 17: scattering in one dimension part 2 · lecture 17: scattering in one dimension part 2...
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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