Download LECTURE 5

Crystal Vibration
Equilibrium
Hot
Cold
Energetic atomic vibrations
Conduction of heat in insulators involves the generation and
propogation of atomic vibrations through the bonds that couple the
atoms. (An intuitive figure.)
Crystal Vibration
Interatomic Bonding
Spring constant (C)
Energy
Parabolic Potential of
Harmonic Oscillator
ro
Eb
Distance
s-1
s
s+1
x
Mass (M)
Transverse wave:
3
ur
ur1
ur+1
x
a
Atoms executing longitudinal vibrations parallel to x.
Crystal Vibration of a Monoatomic Linear Chain
Longitudinal wave of a 1-D Array of Spring Mass System
Spring constant, g
M
Mass, m
Equilibrium
Position
a
Deformed
Position
xun-1
s-1
usx n
x n+1
us+1
us: displacement of the sth atom from its equilibrium position
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6
Solution of Lattice Dynamics
s-1
s
s+1
Same M
Wave solution:
u(x,t) ~ uexp(-iwt+iKx)
Time dep.:
w: frequency
K: wavelength
= uexp(-iwt)exp(isKa)exp(iKa)
cancel
Identity:
Trig:
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w-K Relation: Dispersion Relation
K = 2/l
lmin = 2a
Kmax = /a
-/a<K< /a
l: wavelength
2a
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Polarization and Velocity
w 2 M = C2  exp  iKa   exp iKa  = 2C 1  cos Ka 
1
2C
1  cos Ka 2
w=
M
Group Velocity:
Frequency, w
dw
vg =
dK
Speed of Sound:
dw
vs = lim
K 0 dK
0
Wave vector, K
/a
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Two Atoms Per Unit Cell
M2
Lattice Constant, a
M1
yn-1
xn
2
M1
M2
d xn
dt
2
d 2 yn
dt
2
yn
= f  yn  yn1  2 xn 
xn+1
f: spring constant
= f  xn1  xn  2 yn 
Solution:
Ka
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Acoustic and Optical Branches
Ka
1/µ = 1/M1 + 1/M2
What is the group velocity of the optical branch?
What if M1 = M2 ?
K
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Polarization
TA & TO
Lattice Constant, a
LA & LO
yn-1
xn
xn+1
yn
Total 6 polarizations
Optical
Vibrational
Modes
Frequency, w
LO
TO
LA
0
TA
Wave vector, K /a
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Dispersion in Si
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Dispersion in GaAs (3D)
LO
Frequency (1012 Hz)
8
LO
TO
TO
6
LA
LA
4
2
0
L
TA
0.4
0.2
0
(111) Direction 
TA
0.2
Ka/
0.4
0.6
(100) Direction
0.8
1.0
X
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Allowed Wavevectors (K)
A linear chain of N=10 atoms
with two ends jointed
a
Solution: us ~uK(0)exp(-iwt)sin(Kx), x =sa
B.C.:
us=0 = us=N=10
x
K=2n/(Na), n = 1, 2, …,N
Na = L
Only N wavevectors (K) are allowed (one per mobile atom):
K=
-8/L
-6/L -4/L -2/L
0 2/L
4/L
6/L 8/L /a=N/L
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Allowed Wave Vectors in 3D
2 4
N
K x , K y , K z = 0;
;
;...;
L
L
L
N3: # of atoms
Kz
Ky
Kx
2/L
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Phonon
Energy
•The linear atom chain can only have
N discrete K  w is also discrete
Distance
• The energy of a lattice vibration mode at
frequency w was found to be
1

u =  n  w
2

hw
• where ħw can be thought as the energy of a
particle called phonon, as an analogue to photon
• n can be thought as the total number of
phonons with a frequency w, and follows the
Bose-Einstein statistics:
1
n =
 w 
  1
exp 
 k BT 
Equilibrium distribution
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Total Energy of Lattice Vibration
El = 
p


1

  n w K , p  2 w K , p
K
p: polarization(LA,TA, LO, TO)
K: wave vector
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Reflected waves
Backward wave
-k
k
B'
A'
C'
Forward wave
x
A
B
C
a
An electron wave propagation through a linear lattice. For certain k values
the reflected waves at successive atomic planes reinforce each other to
give rise to a reflected wave travelling in the backward direction. The
electron then cannot propagate through the crystal.
c
2
k=±/a
Energy = Ec
s2
Energy = Es
Forward and backward waves in the crystal with k = ± /a give rise to
two possible standing waves, c and s. Their probability density
distributions, c2 and s2 , have maxima either at the ions or
between the ions.