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1ECE 407 Spring 2009 Farhan Rana Cornell University
Handout 18
Phonons in 2D Crystals: Monoatomic Basis and Diatomic Basis
In this lecture you will learn:
Phonons in a 2D crystal with a monoatomic basis Phonons in a 2D crystal with a diatomic basis Dispersion of phonons LA and TA acoustic phonons LO and TO optical phonons
ECE 407 Spring 2009 Farhan Rana Cornell University
1a
x
21 amanRnm
Phonons in a 2D Crystal with a Monoatomic Basis
y
2a
yanxanyanxan
43
21
General lattice vector:
Nearest-neighbor vectors:
Atomic displacement vectors:
tRutRu
tRunmy
nmxnm ,
,,
Atoms, can move in 2D therefore atomic displacements are given by a vector:
yaxapyaxapyaxapyaxap
43
21
Next nearest-neighbor vectors:
2ECE 407 Spring 2009 Farhan Rana Cornell University
Vector Dynamical Equations
mmm
RRm
12
tRu ,1
tmRutRu ,, 12
mmtRutmRummtRutRudt
tRudM .,,.,,, 111221
2
Vector dynamical equation:
If the nearest-neighbor vectors are known then the dynamical equations can be written easily.
Component dynamical equation:To find the equations for the x and y-components of the atomic displacement, take the dot-products of the above equation on both sides with and , respectively:x
xmmtRutmRudt
tRudM x ..,,, 1121
2
y
ymmtRutmRudt
tRudM y ..,,
,112
12
ECE 407 Spring 2009 Farhan Rana Cornell University
Vector Dynamical Equations for a 2D Crystal
1a
x
y
2a
21 amanRnm
yanxanyanxan
43
21
General lattice vector:
Nearest-neighbor vectors:
yaxapyaxapyaxapyaxap
43
21
Next nearest-neighbor vectors:
4,3,2,12
4,3,2,112
2
.,,
.,,,
jjjnmjnm
jjjnmjnm
nm
pptRutpRu
nntRutnRudt
tRudM
summation over 4 nn
summation over 4 next nn
12
3ECE 407 Spring 2009 Farhan Rana Cornell University
,,2
,,2
,,2
,,2
,,2
,,2
,,2
,,2
,,,,,
42
42
32
32
22
22
12
12
31112
2
tpRutRutpRutRu
tpRutRutpRutRu
tpRutRutpRutRu
tpRutRutpRutRu
tnRutRutnRutRudt
tRudM
nmynmynmxnmx
nmynmynmxnmx
nmynmynmxnmx
nmynmynmxnmx
nmxnmxnmxnmxnmx
Dynamical Equations
4,3,2,12
4,3,2,112
2
.,,
.,,,
jjjnmjnm
jjjnmjnm
nm
pptRutpRu
nntRutnRudt
tRudM
If we take the dot-product of the above equation with we get:x
ECE 407 Spring 2009 Farhan Rana Cornell University
,,2
,,2
,,2
,,2
,,2
,,2
,,2
,,2
,,,,,
42
42
32
32
22
22
12
12
41212
2
tpRutRutpRutRu
tpRutRutpRutRu
tpRutRutpRutRu
tpRutRutpRutRu
tnRutRutnRutRudt
tRudM
nmxnmxnmynmy
nmxnmxnmynmy
nmxnmxnmynmy
nmxnmxnmynmy
nmynmynmynmynmy
Dynamical Equations
4,3,2,12
4,3,2,112
2
.,,
.,,,
jjjnmjnm
jjjnmjnm
nm
pptRutpRu
nntRutnRudt
tRudM
If we take the dot-product of the above equation with we get:y
4ECE 407 Spring 2009 Farhan Rana Cornell University
Solution of the Dynamical Equations
Assume a wave-like solution of the form:
Then:
tiRqiyxnmy nmxnm eeququ
tRutRu
tRu nm
.
,,
,
tRueeequ
que
eeququ
tnRutnRu
tnRu
nmnqi
tiRqi
y
xnqi
tinRqi
y
x
jnmy
jnmxjnm
j
nmj
jnm
,
,,
,
.
..
.
We take the above solution form and plug it into the dynamical equations
ECE 407 Spring 2009 Farhan Rana Cornell University
Dynamical Matrix and Phonon Bands
ququ
MM
ququ
qDy
x
y
x
0
02Compare with the standard form:
Solutions:
X M
FBZ
a2
a2
kg 2x10
N/m 100N/m 200
26-2
1
M
coscos122
sin4sinsin2
sinsin2coscos122
sin42
22
12
222
1
ququ
Mququ
aqaqaq
aqaq
aqaqaqaqaq
y
x
y
x
yxy
yx
yxyxx
5ECE 407 Spring 2009 Farhan Rana Cornell University
Transverse (TA) and Longitudinal (LA) Acoustic Phonons
2
42
22
42
222
2
12
222
2
2
1
ququ
Mququ
aqaqaq
aqaq
aqaqaqaqaq
y
x
y
x
yxy
yx
yxyxx
Case I: 0,0 yx qq
0121 Aqu
quaq
Mq
xy
xxxxLA
Longitudinal acoustic phonons: atomic motion in the direction of wave propagation
Transverse acoustic phonons: atomic motion in the direction perpendicular to wave propagation
102 Aqu
quaq
Mq
xy
xxxxTA
TA
LA
X
M
FBZ
ECE 407 Spring 2009 Farhan Rana Cornell University
Transverse (TA) and Longitudinal (LA) Acoustic Phonons
2
42
22
42
222
2
12
222
2
2
1
ququ
Mququ
aqaqaq
aqaq
aqaqaqaqaq
y
x
y
x
yxy
yx
yxyxx
Case II: qqqqq yxyx 0,0
114 21 Aqu
quaq
Mq
y
xLA
Longitudinal acoustic phonons: atomic motion in the direction of wave propagation
Transverse acoustic phonons: atomic motion in the direction perpendicular to wave propagation
111 Aqu
quaq
Mq
y
xTA
TA
LA
X
M
FBZ
TA
LATA
LA
6ECE 407 Spring 2009 Farhan Rana Cornell University
Transverse (TA) and Longitudinal (LA) Acoustic Phonons
TA
LATA
LATA
LAIn general for longitudinal acoustic phonons near the zone center:
y
x
y
xqq
qA
ququ
And for transverse acoustic phonons near the zone center:
x
y
y
xqq
qA
ququ
In general, away from the zone center, the LA phonons are not entirely longitudinal and neither the TA phonons are entirely transverse
ECE 407 Spring 2009 Farhan Rana Cornell University
Transverse (TA) and Longitudinal (LA) Acoustic Phonons
LATA
TA and LA
7ECE 407 Spring 2009 Farhan Rana Cornell University
Periodic Boundary Conditions in 2D
tiRqiyxnmy nmxnm eeququ
tRutRu
tRu nm
.
,,
,
Our solution was:
Periodic boundary conditions for a lattice of N1xN2 primitive cells imply:
22
where
21
21 where22
integer an is where2.1
,,
11
1
11
1111
1111
.
..11
11
11
NmN-Nm
-mN
mmaNqe
eeququ
tRueeququ
taNRu
aNqi
tiRqi
y
xnm
tiaNRqi
y
xnm nmnm
21 amanRnm
General lattice vector: 21,21 212211 bbq
General reciprocal lattice vector inside FBZ:
Similarly:
22 where 222
2
22
NmN-Nm
FBZ
1b2b
ECE 407 Spring 2009 Farhan Rana Cornell University
Counting Degrees of FreedomIn the solution the values of the phonon wavevector are dictated by the periodic boundary conditions:
2211 bbq
22 where 11
1
11
NmN-Nm
22 where 222
2
22
NmN-Nm
There are N1N2 allowed wavevectors in the FBZ(There are also N1N2 primitive cells in the crystals)
There are N1N2 phonon modes per phonon bandCounting degrees of freedom:
There are 2N1N2 degrees of freedom corresponding to the motion in 2D of N1N2 atoms
The total number of different phonon modes in the two bands is also 2N1N2
FBZ
8ECE 407 Spring 2009 Farhan Rana Cornell University
Phonons in a 2D Crystal with a Diatomic Basis
Atomic displacement vectors:
tdRutdRutdRutdRu
tdRutdRu
nmy
nmx
nmy
nmx
nm
nm
,,,,
,,
22
22
11
11
22
11
The two atoms in a primitive cell can move in 2D therefore atomic displacements are given by a four-component column vector:
1a
x
y
2a
1
2
21 amanRnm
2
2
2
2
43
21
yaxahyaxah
yaxahyaxah
1st nearest-neighbor vectors (red to blue):
yanxanyanxan
43
21
2nd nearest-neighbor vectors (red to red):
3rd nearest-neighbor vectors (red to red):
yaxapyaxapyaxapyaxap
43
21
ECE 407 Spring 2009 Farhan Rana Cornell University
Diatomic Basis: Force Constants
plus
1a
x
y
2a
1
The force constants between the 1st2nd and 3rd nearest-neighbors need to be included (at least)
1a
x
y
2a
23
9ECE 407 Spring 2009 Farhan Rana Cornell University
4,3,2,111113
4,3,2,111112
4,3,2,1111212
112
1
.,,
.,,
.,,,
jjjnmjnm
jjjnmjnm
jjjnmjnm
nm
pptdRutpdRu
nntdRutndRu
hhtdRuthdRudt
tdRudM
summation over 4 1st nn
summation over 4 2nd nn
Diatomic Basis: Dynamical Equations
4,3,2,122223
4,3,2,122222
4,3,2,1222212
222
2
.,,
.,,
.,,,
jjjnmjnm
jjjnmjnm
jjjnmjnm
nm
pptdRutpdRu
nntdRutndRu
hhtdRuthdRudt
tdRudM
summation over 4 1st nn
summation over 4 2nd nn
Dynamical equation for the red(1) atom:
Dynamical equation for the blue(2) atom:
summation over 4 3rd nn
summation over 4 3rd nn
ECE 407 Spring 2009 Farhan Rana Cornell University
Diatomic Basis: Dynamical Equations
tiRqi
dqiy
dqix
dqiy
dqix
nmy
nmx
nmy
nmx
nm
nm ee
equequ
equequ
tdRutdRutdRutdRu
tdRutdRu nm
.
.2
.2
.1
.1
22
22
11
11
22
11
2
2
1
1
,,,,
,,
Assume a solution of the form:
ququququ
MM
MM
ququququ
qD
y
xy
x
y
xy
x
2
21
1
2
2
1
1
2
2
21
1
000000000000
To get a matrix equation of the form:
10
ECE 407 Spring 2009 Farhan Rana Cornell University
The Dynamical Matrix
aqaqaq
yx
x
coscos122
sin42
3
221
2cos
2cos2 1
aqaq yx
2sin
2sin2 1
aqaq yx
2cos
2cos2 1
aqaq yx
2sin
2sin2 1
aqaq yx
2cos
2cos2 1
aqaq yx
2sin
2sin2 1
aqaq yx
2sin
2sin2 1
aqaq yx
2cos
2cos2 1
aqaq yx
ququququ
MM
MM
ququququ
qD
y
xy
x
y
xy
x
2
21
1
2
2
1
1
2
2
21
1
000000000000
aqaqaq
yx
y
coscos122
sin42
3
221
aqaqaq
yx
x
coscos122
sin42
3
221
aqaqaq
yx
y
coscos122
sin42
3
221
aqaq yx sinsin2 3
aqaq yx sinsin2 3
aqaq yx sinsin2 3
aqaq yx sinsin2 3
qD The matrix is:
ECE 407 Spring 2009 Farhan Rana Cornell University
Diatomic Basis: Solution and Phonon Bands
N/m100N/m200N/m300
kg1042
3
2
1
2621
MM
For calculations: Opticalbands
Acousticbands
One obtains:
- 2 optical phonon bands (that have a non-zero frequency at the zone center)
- 2 acoustic phonon bands (that have zero frequency at the zone center)
TA
LA
LO
TO
TA
LA
LO
TO
X M
FBZ
a2
a2
X
TA
LA
TOLO
11
ECE 407 Spring 2009 Farhan Rana Cornell University
Longitudinal (LO) and Transverse (TO) Optical Phonons
Case I: 0,0 yx qq
0
01
2021
2
21
1
1MMA
ququququ
Mq
xy
xxxy
xx
rxLO
212
21
1
1010
20
MM
A
ququququ
Mq
xy
xxxy
xx
rxTO
Longitudinal optical phonons: atomic motion in the direction of wave propagation and basis atoms move out of phase
Transverse optical phonons: atomic motion in the direction perpendicular to wave propagation and basis atoms move out of phase
X
M
FBZ
21
111MMMr
ECE 407 Spring 2009 Farhan Rana Cornell University
Longitudinal (LA) and Transverse (TA) Acoustic Phonons
Case I: 0,0 yx qq
0101
?0
2
21
1
A
ququququ
q
xy
xxxy
xx
xLA
1010
?0
2
21
1
A
ququququ
q
xy
xxxy
xx
xTO
Longitudinal acoustic phonons: atomic motion in the direction of wave propagation and basis atoms move in phase
Transverse acoustic phonons: atomic motion in the direction perpendicular to wave propagation and basis atoms move in phase
X
M
FBZ
12
ECE 407 Spring 2009 Farhan Rana Cornell University
Counting Degrees of Freedom and the Number of Phonon Bands
2211 bbq
22 where 11
1
11
NmN-Nm
22 where 222
2
22
NmN-Nm
There are N1N2 allowed wavevectors in the FBZThere are N1N2 phonon modes per phonon bandCounting degrees of freedom:
There are 4N1N2 degrees of freedom corresponding to the motion in 2D of 2N1N2 atoms (2 atoms in each primitive cell)
The total number of different phonon modes in the four bands is also 4N1N2
Periodic boundary conditions for a lattice of N1xN2 primitive cells imply: