1
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Handout 17
Lattice Waves (Phonons) in 1D Crystals: Monoatomic Basis and Diatomic Basis
In this lecture you will learn:
• Equilibrium bond lengths• Atomic motion in lattices• Lattice waves (phonons) in a 1D crystal with a monoatomic basis• Lattice waves (phonons) in a 1D crystal with a diatomic basis• Dispersion of lattice waves• Acoustic and optical phonons
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
The Hydrogen Molecule: Equilibrium Bond Length
0 x
rV
d-d
BE
AE
sE1:2 AE:1
BE:1ssV2
The equilibrium distance between the two hydrogen atoms in a hydrogen molecule is set by the balance among several different competing factors:
• The reduction in electronic energy due to co-valent bonding is 2Vss . If the atoms are too far apart, Vss becomes to small
• If the atoms are too close, the positively charged nuclei (protons) will repel each other and this leads to an increase in the system energy
• Electron-electron repulsion also plays a role
sssB VEE 1
sssA VEE 1
ssss VxdrHxdr ˆˆˆ 11
2
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
A Mass Attached to a Spring: A Simple Harmonic Oscillator
x
0
x
0
u
Equilibrium position Stretched position
Potential Energy:
2
21
ukuVPE
Kinetic Energy:2
2
dtduM
KE
M M
spring constant = k(units: Newton/meter)
Dynamical Equation (Newton’s Second Law):
ukdudV
dt
udM 2
2 Restoring force varies linearly with the displacement “u ” of the mass from its equilibrium position
Solution:
tBtAtu oo sincos where:Mk
o
PE varies quadratically with the displacement “u ” of the mass from the equilibrium position
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
A 1D Crystal: Potential Energy
Consider a 1D lattice of N atoms:
xaa ˆ1
x
1anRn
• The potential energy of the entire crystal can be expressed in terms of the positions of the atoms. The potential energy will be minimum when all the atoms are at their equilibrium positions.
• Let the displacement of the atom at the lattice site given by from its equilibrium position be
• One can Taylor expand the potential energy of the entire crystal around its minimum equilibrium value:
k jkj
EQkj
jj EQj
EQN
RuRuRuRu
V
RuRuV
VRuRuRuRuV
2
321
21
............,,
nR
nRu
0
Potential energy varies quadratically with the displacements of the atoms from their equilibrium positions
Atoms can move only in the x-direction
3
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
A 1D Crystal: Potential and Kinetic Energies
A1D lattice of N atoms:
xaa ˆ1
x
1anRn
k jkjkjEQ
k jkj
EQkjEQ
tRutRuRRKV
tRutRuRuRu
VVV
,,,21
,,21 2
Potential Energy:
j
j
dt
tRduMKE
2,
2
Kinetic Energy:
• The kinetic energy of all the atoms is the sum of their individual kinetic energies
EQkj
kj RuRu
VRRK
2
,
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
A 1D Crystal: Dynamical Equation
A1D lattice of N atoms:
xaa ˆ1
x
1anRn
j
jjnn
n tRuRRKRu
V
dt
tRudM ,,
,2
2
Write Newtons law for the atom sitting at the site :nR
EQkj
kj RuRu
VRRK
2
,
Remember that:
The restoring forces on the atoms vary linearly with the displacement of the atoms from their equilibrium positions
4
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Dynamical Equation for Nearest-Neighbor Interactions
A1D lattice of N atoms:
j
jjnn
n tRuRRKRu
V
dt
tRudM ,,
,2
2
Assume nearest-neighbor interactions:
njnjnjjn RRK ,1,1, 2,
tRutRutRutRudt
tRudM nnnn
n ,,,,,
112
2
This gives:
The constants “” provide restoring forces as if the atoms were connected together with springs of spring constant “”
We have N linear coupled differential equations for N unknowns
1.......2,1,0, NntRu n
The constant is called “force constant” (not spring constant) in solid state physics
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Solution of the Dynamical Equation: Lattice Waves (Phonons)
A1D lattice of N atoms:
tRutRutRutRudt
tRudM nnnn
n ,,,,,
112
2
tiRqin eequtRu n
.Re,
Assume a solution of the form:
Represents a wave with wavevector , frequency , and amplitude
q
qu
tiRqiaqi
tiaRqitiRqin
eeque
eequeequtRu
n
nn
..
..1
1
11
,
Note that:
tiRqin eequtRu n
.,Or: Slight abuse of notation
xaa ˆ1
1anRn
tiRqiaqi
tiaRqitiRqin
eeque
eequeequtRu
n
nn
..
..1
1
11
,
5
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
A1D lattice of N atoms:
tRutRutRutRudt
tRudM nnnn
n ,,,,,
112
2
Plug in the assumed solution:
To get:
queququeququM aqiaqi 11 ..2
Which simplifies to:
2.
sin4
.cos12
12
12
aqM
aqM
Or:
2.
sin4 1aqM
xaa ˆ1
1anRn
xqq x ˆ tiRqin eequtRu n
.,
Solution of the Dynamical Equation: Lattice Waves (Phonons)
Since is always positive, the negative sign is chosen when the sine term is negative
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
A1D lattice of N atoms: xaa ˆ1
1anRn
Solution is: and tiRqin eequtRu n
.,
2.
sin4 1aqM
Solution of the Dynamical Equation: Lattice Waves (Phonons)
x
x
a nRu
• The lattice waves are like the compressional sound waves in the air
6
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
A1D lattice of N atoms: xaa ˆ1
1anRn
Solution is: and
The relation:
represents the dispersion of the lattice waves or phonons
2sin
42.
sin4 1 aq
Maq
Mx
xq
tiRqin eequtRu n
.,
2.
sin4 1aqM
a
a
Solution of the Dynamical Equation: Lattice Waves (Phonons)
x
a nRu
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Solution is: and
For the shortest wavelength:
The largest wavevector is then:
The wavevector values can be restricted to the First BZ
• No new solutions are found for values of the wavevector outside the first BZ
a2
xq
tiRqin eequtRu n
.,
2.
sin4 1aqM
a
a
Lattice Waves and the First BZ
x
Question: What is the shortest wavelength (or the largest wavevector) the lattice waves can have?
aqx
2
First BZ
nRu = 2a
7
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
xq
2sin
4 aqM
x
a
a
Phase and Group Velocities
First BZ
Case I: For qx ≈ 0 (i.e. qx a << ):
aqM x
Linear dispersion
Phase velocity and group velocity of lattice waves are defined as:
xdqd
qqv
xq
qqq
qv
x
Dqg
x
Dp
ˆ
ˆˆ
1
1
gp vaM
v
Case II: For qx a = :
M 4
0
4
g
p
v
aM
v
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Periodic Boundary Conditions
In the solution:
allowed values of the wavevector depend on the boundary conditions
tiRqin eequtRu n
.,
A1D lattice of N atoms: xaa ˆ1
1anRn
Periodic Boundary Condition:
0 1N-1
The N-th atom is the same as the 0-th atom This implies:
22
integer2
1
1
,,
0
0
.
.0
.
Nm
N
m
aNm
q
e
e
eequtRueequtRu
x
aNqi
RRqi
tiRqitiRqiN
x
N
N
xq
a
a
First BZ
aN2
8
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Counting and Conserving Degrees of FreedomA1D lattice of N atoms: xaa ˆ1
1anRn
• We started with N degrees of freedom which were related to the motion in 1D of Ndifferent atoms• The dynamical variables were the amplitudes of the displacements of N different atoms
1.......2,1,0, NntRu n
We then ended up with lattice waves:
• There are N different lattice wave modes corresponding to the N different possible wavevector values in the first BZ• The dynamical variables are the amplitudes of the N different lattice wave modes
22
and2
,N
mN
aNm
qtqu x
The number of degrees of freedom are the same before and after – as they should be!
xq
a
a
First BZ
aN2
nn RqitqiRqin etqueequtRu
.. ,ReRe,
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Counting and Conserving Degrees of Freedom
The atomic displacements,
taken together provide a complete description of the motion of all the atoms in the crystal
In general, one can expand the atomic displacements in terms of all the lattice wave modes (resembles a Fourier series expansion):
1.......2,1,0, NntRu n
FBZ in
.*
.
FBZ in
.*
.
FBZ in
.
2,
2,
22
Re,
q
RqiRqi
q
tqiRqitqiRqi
q
tqiRqin
nn
nn
n
etqu
etqu
eequ
eequ
eequtRu
Therefore, the lattice wave amplitudes also provide a complete description of the motion of all the atoms in the crystal
9
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
A1D lattice of N red atoms and N blue atoms and N primitive cells:
xaa ˆ1
x
1anRn
Lattice Waves (Phonons) in a 1D crystal: Diatomic Basis
Unit cell
1M 2M
1n
2n
The nearest neighbor vectors are:
1d
2d
The basis vectors are:
1n
2n
The spring model for nearest neighbor interactions:
x
Let the displacement of the red atom in the n-th cell be: tdRu n ,11
Let the displacement of the blue atom in the n-th cell be: tdRu n ,22
1d
2d
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
tdRutdRutdRutdRudt
tdRudM nnnn
n ,,,,,
22111212112211
2
1
tdRutdRutdRutdRudt
tdRudM nnnn
n ,,,,,
11122211221222
2
2
Write the dynamical equations for both the atoms assuming nearest neighbor interactions
Assume a solution of the form:
tidRqin
tidRqin
eequtdRu
eequtdRu
n
n
2
1
.222
.111
,
,
This is again a wave-like solution with a wavevector and frequency and different amplitudes for the two atoms
q
Notice the phases
x
1M 2M
1n
2n
12
The constants 1 and 2 are called “force constants” (not spring constants) in solid state physics
Lattice Waves (Phonons) in a 1D crystal: Diatomic Basis
10
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Plug the solutions in the dynamical equations to get:
queququeququM nqinqi
2.
112.
12112 12
queququeququM nqinqi
1.
221.
21222 21
Write the equations in a matrix form:
qu
qu
M
M
qu
qu
ee
eenqinqi
nqinqi
2
1
2
12
2
1
21.
2.
1
.2
.121
0
021
21
x
1M 2M
1n
2n
12
This is a 2x2 matrix eigenvalue equation that needs to be solved for each value of the wavevector to get the dispersion of the lattice waves
Lattice Waves (Phonons) in a 1D crystal: Diatomic Basis
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
x
1M 2M
1n
2n
12
The Dynamical Matrix
Or:
qu
quM
qu
quqD
2
12
2
1
• The matrix is called the dynamical matrix of the medium
• For any medium, in any dimension, the dispersion relations for the lattice waves (phonons) are obtained by solving a similar matrix eigenvalue equation
qD
qu
qu
M
M
qu
qu
ee
eenqinqi
nqinqi
2
1
2
12
2
1
21.
2.
1
.2
.121
0
021
21
Or:
qu
qu
qu
quqDM
2
12
2
11
11
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Optical and Acoustic Phonons
x
1M 2M
1n
2n
12
The frequency eigenvalues are:
2.
sin1621
212
21
212
221212 aq
MMMMq
rr
21
111MMMr
xaa ˆ1
xq
a
a
• The two frequency eigenvalues for each wavevector value in the FBZ give two phonon bands
• The higher frequency band is called the optical phonon band
• The lower frequency band is called the acoustic phonon band
Optical
Acoustic
FBZ
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Optical and Acoustic Phonons: Special Cases
x
1M 2M 12
xaa ˆ1
Case I: 0q
aqMM
q x2121
210
Acoustic band:
1
1
0
0
2
1 Aqu
qu
tidRqin
tidRqin
eeAtdRu
eeAtdRu
n
n
2
1
.22
.11
,
,
rM
q 210
Optical band:
212
1 1
0
0
MMA
qu
qu
tidRqin
tidRqin
eeMM
AtdRu
eeAtdRu
n
n
2
1
.
2
122
.11
,
,
xq
a
a
Optical
Acoustic
12
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Optical and Acoustic Phonons: Special Cases
x
1M 2M 12
xaa ˆ1
Case I: Acoustic Mode0q
Acoustic band (basis atoms move in-phase):
tidRqin
tidRqin
eeAtdRu
eeAtdRu
n
n
2
1
.22
.11
,
,
x
x
a
u
xq
a
a
Optical
Acoustic
0q
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Optical and Acoustic Phonons: Special Cases
x
1M 2M 12
xaa ˆ1
Case I: Optical Mode0q
Optical band (basis atoms move 180o out-of-phase):
x
x
a
u
xq
a
a
Optical
Acoustic
tidRqin
tidRqin
eeMM
AtdRu
eeAtdRu
n
n
2
1
.
2
122
.11
,
,
0q
13
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Optical and Acoustic Phonons: Special Cases
x
1M 2M 12
xaa ˆ1
Case II: Acoustic Mode 2121 ,,ˆ MMMx
aq
xq
a
a
Optical
Acoustic
Mx
a22
ˆ
1.
2
1 1
ˆ
ˆ
nqieA
xa
u
xa
u
tidRqi
tidRqinqin
tidRqin
eeA
eeeAtdRu
eeAtdRu
n
n
n
1
21
1
.
..22
.11
,
,
x
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Optical and Acoustic Phonons: Special Cases
x
1M 2M 12
xaa ˆ1
Case II: Optical Mode2121 ,,ˆ MMMx
aq
xq
a
a
Optical
AcousticM
xa
12ˆ
1.
2
1 1
ˆ
ˆ
nqieA
xa
u
xa
u
x
tidRqi
tidRqinqin
tidRqin
eeA
eeeAtdRu
eeAtdRu
n
n
n
1
21
1
.
..22
.11
,
,
14
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Counting and Conserving Degrees of Freedom
• We started with 2N degrees of freedom which were related to the motion in 1D of 2Ndifferent atoms• The dynamical variables were the amplitudes of the displacements of 2N different atoms
1.......2,1,0,, 2211 NntdRutdRu nn
We then ended up with lattice waves:
• There are N different modes per phonon band corresponding to the N different possible wavevector values in the first BZ• There are 2 phonon bands and therefore a total of 2Ndifferent phonon modes • The dynamical variables are the amplitudes of the 2Ndifferent phonon modes
The number of degrees of freedom are the same before and after – as they should be!
xq
a
a
FBZ
aN2
x
1M 2M 12
xaa ˆ1
Optical
Acoustic
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Counting and Conserving Degrees of Freedom
The atomic displacements,
taken together provide a complete description of the motion of all the 2N atoms in the crystal
In general, one can expand the atomic displacements in terms of all the lattice wave modes – all wavevectors and all bands:
OA q
Rqidqi
dqiRqi
dqi
dqi
OA q
tqiRqidqi
dqitqiRqi
dqi
dqi
OA q
tqiRqidqi
dqi
n
n
nn
nn
n
eetqu
etque
etqu
etqu
eeequ
equee
equ
equ
eeequ
equ
tdRu
tdRu
, FBZ in
..*
2
.*1.
.2
.1
, FBZ in
..*
2
.*1.
.2
.1
, FBZ in
..
2
.1
22
11
2
1
2
1
2
1
2
1
2
1
,
,
21
,
,
21
21
21
Re,
,
Therefore, the lattice wave amplitudes also provide a complete description of the motion of all the atoms in the crystal
1.......2,1,0,, 2211 NntdRutdRu nn
Top Related