lattice waves (phonons) in 3d crystals group iv …1 ece 407 – spring 2009 – farhan rana –...
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Handout 19
Lattice Waves (Phonons) in 3D CrystalsGroup IV and Group III-V Semiconductors
LO and TO Phonons in Polar Crystals and
Macroscopic Models of Acoustic Phonons in Solids
In this lecture you will learn:
• Lattice waves (phonons) in 3D crystals• Phonon bands in group IV and group III-V Semiconductors
• Macroscopic description of acoustic phonons from elasticity theory• Stress, strain, and Hooke’s law
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Counting the Number of Phonon bands in 3D Crystals
332211 bbbq
22 where
22 where
22 where
333333
222222
111111
NmN-Nm
NmN-Nm
NmN-Nm
There are N1N2N3 allowed wavevectors in the FBZThere are N1N2N3 phonon modes per phonon band
Counting degrees of freedom and the number of phonon bands: Monoatomic Basis
• There are 3N1N2N3 degrees of freedom corresponding to the motion in 3D of N1N2N3
atomsThe number of phonon bands must be 3 (two TA bands and one LA band)
Counting degrees of freedom and the number of phonon bands: Diatomic Basis
• There are 6N1N2N3 degrees of freedom corresponding to the motion in 3D of 2N1N2N3 atomsThe number of phonon bands must be 6 (two TA bands and one LA band for acoustic phonons and two TO bands and one LO band for optical phonons)
Periodic boundary conditions for a lattice of N1xN2xN3 primitive cells imply:
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Phonon Bands in Silicon
qz
qy
qx
qz
qy
qx
Silicon has a FCC lattice with two basis atoms in one primitive cell
FBZ of Silicon
The number of phonon bands must be 6; two TA bands and one LA band for acoustic phonons and two TO bands and one LO band for optical phonons
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Phonon Bands in Silicon
qz
qy
qx
qz
qy
qx
Calculations
Data (Neutron scattering)
meV6400 xTOxLO qq
X
Fre
qu
ency
(T
Hz)
3
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Phonon Bands in Diamond
Fre
qu
ency
(cm
-1)
X
Fre
qu
ency
(cm
-1)
L
LO
TO/TO
LA
TA/TA
LO
TO/TO
LA
TA/TA
Calculations
Data (Neutron scattering)
qz
qy
qx
qz
qy
qx
meV16500 xTOxLO qq
1-cm 13302
02
0
c
qc
q xTOxLO
Phonon frequencies are also expressed in units of equivalent photon wavelength inverse:
Large!!
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Phonon Bands in GaAs
qz
qy
qx
qz
qy
qx
GaAs has a FCC lattice with two basis atoms in one primitive cell
FBZ of GaAs
The number of phonon bands must be 6; two TA bands and one LA band for acoustic phonons and two TO bands and one LO band for optical phonons
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Calculations
Data (Neutron scattering)
qz
qy
qx
qz
qy
qx
meV 330
meV360
xTO
xLO
q
q
Fre
qu
ency
(T
Hz)
LO
TO
LA
TA
TOTO
LO
LA
TA
TA
TA
LA
TO
X K L
Phonon Bands in GaAs
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Optical Phonons in Polar CrystalsConsider a crystal, like GaAs, made up of two different kind of atoms with a polar covalent bond
When the atoms move, an oscillating charge dipole is created with a dipole moment given by:
tdRu ,11
tndRu j ,12
tdRutndRuftRp jj ,,, 1112
jn
The material polarization, or the dipole moment density, is then:
j
jj
j tdRutndRuZnf
tRpZn
tRP ,,,, 1112
where:
3
1n Number of primitive cells per unit volume
A non-zero polarization means an electric field!
Z Number of nearest neighbors
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Optical Phonons in Polar Crystals: D-Field and E-Field
tRu ,1
tnRu ,12
A non-zero polarization means an electric field! How do we find it? tnRu ,22
tnRu ,32
tnRu ,32
The divergence of the D-field is zero inside the crystal:
0. uD
But inside the crystal:
PE
PED
..
Since:
Therefore:
tRP
tRE,.
,.
j
jj
j tdRutndRuZnf
tRpZn
tRP ,,,, 1112
We must also have:
0, tRE
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Optical Phonons in Polar Crystals: Dynamical Equations
tREMf
nntndRutdRuMdt
tdRud
tREMf
nntdRutndRuMdt
tdRud
jjjj
jjjj
,ˆˆ.,,,
,ˆˆ.,,,
22122
22
222
11112
12
112
Dynamical equations (assuming only nearest neighbor interactions):
Suppose:
tiRqi
tiRqitiRqi
dqi
dqi
eqPtRP
eqEtREeequ
equ
tdRu
tdRu
.
..
.2
.1
22
11
,
,
,
,
2
1
We have:
00, qEqtRE
The above two imply that the E-field has non-zero component only in the direction parallel to given by: q
q
qqPqE ˆ
ˆ.
qqP
qEqtRP
tREˆ.
.ˆ,.
,.
We also have:
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
qEMf
nnququM
ququrj
jjr
ˆˆ.1212
2
Optical Phonons in Polar Crystals: TO Phonons
Subtract the two equations and take the limit q0 to get:
Take the cross-product of both sides with to get: q̂
qqEMf
qnnququM
qququrj
jjr
ˆˆˆˆ.ˆ 12122
r
TO
r
r
Mb
q
Mb
qququMb
qququ
0
ˆˆ 12122
qAbqnnAj
jj ˆˆˆˆ.
Transverse Optical Phonons:
bnnj
jj ˆˆ
1
1
1
3
11n
1
1
1
3
12n
1
1
1
3
13n
1
1
1
3
14n
For example in GaAs:
34
100
010
001
34
ˆˆ
j
jj nn
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Optical Phonons in Polar Crystals: LO Phonons
qEMf
nnququM
ququrj
jjr
ˆˆ.1212
2
Again start from:
Take the dot-product of both sides with to get:
Longitudinal Optical Phonons:
q̂
qqEMf
qnnququM
qququrj
jjr
ˆ.ˆ.ˆˆ.ˆ. 12122
rrLO
rr
Mnf
Mb
q
qququM
nfqququ
Mb
qququ
2
12
2
12122
0
ˆ.ˆ.ˆ.
qAbqnnAj
jj ˆ.ˆ.ˆˆ.
rTOLO
rTOLO
Mnf
Mnf
222
222 00
bnnj
jj ˆˆ
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Optical Phonons in Polar Crystals: Dielectric ConstantConsider the response of polar optical phonons to an externally applied E-fieldThe total electric field (external plus internal) is:
22
2
22
2
TO
r
TO
r
Mnf
qEMnf
qD
qEqPqEqD
2212
12122 ˆˆ.
TO
r
rjjj
r
qEMf
ququ
qEMf
nnququM
ququ
tiRqieqEtRE .,
22
2
12TO
rqE
Mnf
ququnfqP
We have:
The D-field is:
bnnj
jj ˆˆ
0q
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Optical Phonons in Polar Crystals: Lydanne-Sachs-Teller Relation
0
0
22
2
2
22
2
TOr
TO
r
TO
r
Mnf
Mnf
Mnf
The LO-TO phonon frequency splitting was given by:
0
0
22
22
22
TOLO
TOr
TOLO Mnf
The above relationship is called the Lydanne-Sachs-Teller relationThe above relation does not change if more than nearest-neighbor interactions are also included in the analysisOne can also write:
We have:
Low frequency dielectric constant
22
2 0
TO
TO
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Vector Dynamical Equations: Bond-Stretching and Bond-Bending
mm
m
RRm
ˆ
121R
2R
• In general, atomic displacements can cause both bond-stretching and bond-bending
• Both bond-stretching and bond-bending give rise to restoring forces
Bond-stretching component
Bond-bending component
mmtRutmRudt
tRudM ˆˆ.,,
,112
12
Bond-stretching contribution:
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Vector Dynamical Equations: Bond-Stretching and Bond-Bending
mm
m
RRm
ˆ
121R
2R
Bond-stretching component
Bond-bending component
Bond-stretching and bond-bending contributions:
2211
1111
1121
2
ˆˆ.,,
ˆˆ.,,
ˆˆ.,,,
nntRutmRu
nntRutmRu
mmtRutmRudt
tRudM
First find two mutually orthogonal unit vectors that are also perpendicular to m̂
Let these be: 1n
2n̂and
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Macroscopic Description of Acoustic Phonons in Solids
Acoustic phonons can also be described using a macroscopic formalism based on the theory of elasticity
Let the local displacement of a solid from its equilibrium position be given by the vector
ru
ru
ru
ru
z
y
x
ru
Strain Tensor:
Consider a stretched rubber band:
0 L xL+L
L
There is a uniform strain given by:
LL
xxu
e xxx
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Stress and Strain
Stress Tensor:
Stress is the force acting per unit area on any plane of the solidIt is a tensor with 9 components (as shown)
x
y
z
yY
xXzZ
xY
xZ
zYzX
yXyZ
yXFor example, is the force acting per unit area in the x-direction on a plane that has a normal vector pointing in the y-direction
Strain Tensor:
The strain tensor is defined by its 6 components:e
z
rux
rue
yru
z
rue
x
ru
yru
e
zru
ey
rue
xru
e
xzzx
zyyz
yxxy
zzz
yyy
xxx
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Hooke’s LawStress Tensor:
In solids with cubic symmetry, if the stress tensor produces no torque (and no angular acceleration) then one must have:
zxyzxy XZZYYX So there are only 6 independent stress tensor components:
yxzzyx XZYZYX
Hooke’s Law:
A fundamental theorem in the theory of elasticity is Hooke’s law that says that strain is proportional to the stress and vice versa. Mathematically, the 6 stress tensor components are related to the 6 strain tensor components by a matrix:
xy
zx
yz
zz
yy
xx
y
x
z
z
y
x
e
e
e
e
e
e
ccc
c
cc
cccc
X
Z
Y
Z
Y
X
666261
31
2221
16131211
...
......
......
.....
....
..
Elastic stiffness constants
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Hooke’s Law for Cubic Materials
In solids with cubic symmetry (SC, FCC, BCC) the matrix of elastic constants have only three independent components:
xy
zx
yz
zz
yy
xx
y
x
z
z
y
x
e
e
e
e
e
e
c
c
c
ccc
ccc
ccc
X
Z
Y
Z
Y
X
44
44
44
111212
121112
121211
00000
00000
00000
000
000
000
Elastic energy:
The elastic energy per unit volume of a strained cubic material is:
2224412
222112
1xyzxyzxxzzzzyyyyxxzzyyxx eeeceeeeeeceeecV
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Wave Equation for Acoustic Phonons in Cubic SolidsConsider a solid with density Consider a small volume of this solid that is in motion, as shown
x
We want to write Newton’s second law for its motion in the x-direction First consider only the force due to the stress tensor component Xx
r
x
rX
t
tru
xrX
zyxxx
rXxx
rXzyt
truzyx
xx
xxx
x
2
2
2
2
,
ˆ2
ˆ2
,
xx
rX x ˆ2
xx
rX x ˆ2
y
z
z
rXy
rX
xrX
t
tru zyxx
2
2 ,
Now add the contribution of all forces acting in the x-direction:
yy
rX y ˆ2
yy
rX y ˆ2
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
z
rXy
rX
xrX
t
tru zyxx
2
2 ,We have:
Similarly for acceleration in the y- and z-directions we get:
z
rYy
rY
xrY
t
tru zyxy
2
2 ,
zrZ
y
rZ
xrZ
t
tru zyxz
2
2 ,
Using the Hooke’s law relation, the above equation for motion in the x-direction can be written as:
zxru
yx
rucc
z
ru
y
ruc
x
ruc
zre
y
rec
xre
x
rec
xre
ct
tru
zyxxx
zxxyzzyyxxx
22
44122
2
2
2
442
2
11
4412112
2
,
Wave Equation for Acoustic Phonons in Cubic Solids
Wave equation for acoustic phonons
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
zxru
yx
rucc
z
ru
y
ruc
x
ruc
t
tru zyxxxx 22
44122
2
2
2
442
2
112
2 ,
LA phonons:
Consider a LA phonon wave propagating in the x-direction:
tixqix eeAtru x ,
Plug the assumed solution in the wave equation to get:
xqc
11 velocity of wave = 11c
TA phonons:
Consider a TA phonon wave propagating in the y-direction:
tiyqix eeAtru y ,
Plug the assumed solution in the wave equation to get:
yqc
44 velocity of wave = 44c
Wave Equation for Acoustic Phonons in Cubic Solids
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Acoustic Phonons in Silicon
qz
qy
qx
qz
qy
qx
In Silicon:
3
21144
21112
21111
mkg2330
mN1080.0
mN1064.0
mN1066.1
c
c
c
For LA phonons propagating in the -X direction:
velocity of wave = seckm 44.811
c
For TA phonons propagating in the -X direction:
velocity of wave = seckm 86.544
c
Results from elasticity theory
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
zxru
yx
rucc
z
ru
y
ruc
x
ruc
t
tru zyxxxx 22
44122
2
2
2
442
2
112
2 ,
Wave Equation for Acoustic Phonons in Cubic Solids
yzru
yxru
ccx
ru
z
ruc
y
ruc
t
tru zxyyyy 22
44122
2
2
2
442
2
112
2 ,
Consider a phonon wave propagating in the direction:
tirqi
y
x
y
xee
qu
qu
tru
tru
.
,
,
Plug the assumed solution in the wave equation to get two coupled equations:
2
ˆˆ yx2
ˆˆ yxqq
qu
qu
qu
qu
ccq
ccq
ccq
ccq
y
x
y
x
2
4411
2
4412
2
4412
2
4411
2
22
22
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Wave Equation for Acoustic Phonons in Cubic Solids
qu
qu
qu
qu
ccq
ccq
ccq
ccq
y
x
y
x
2
4411
2
4412
2
4412
2
4411
2
22
22
The two solutions are as follows:
LA phonon:
TA phonon:
1
1
22 441211 A
qu
quq
ccc
y
x
1
1
21211 A
qu
quq
cc
y
x