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3
3
Crystal Vibrations
Vibrations of atoms in a crystal give us waves (sound waves). This lecture studies these sound waves.
Model: consider atoms as points (sites) and consider a crystal as points connected by springs.
3.1. Vibrations of crystals with monotonic basis
Fig. 1. a 1D crystal
3.1.1. Displacement and Force
Consider a 1D crystal with one atom per unit cell. Here we label the sites (atoms) using an integer s = 1, 2, ..., N
Displacement of site s: how far the site moves away from its equilibrium position: us = Rs - Rs H0L where Rs is the position of the site s and Rs H0L is
the equilibrium positions.
Force on site s: FS
FS = C Hus+1 - us L + CHus-1 - us L
(3.1)
Here we assume each spring follows the Hooke’s law and C is the spring constant F = C DL (assuming higher order terms are small)
3.1.2. Equation of motion
m
â2 us
â t2
= FS = C Hus+1 - us L + CHus-1 - us L = CHus+1 + us-1 - 2 us L
(3.2)
3.1.3. Sound waves
To solve this equation, we consider a plane wave:
us = u expHä K s aL expH-ä Ω tL
(3.3)
This us is complex and the real part of it is the true displacement.
Here the amplitude of this wave is u. The wavevector is K = 2 Π  Λ. a is the lattice constant. The frequency of this wave is Ω. For this wave, the
equation of motion turns into
-m Ω2 u expHä K s aL expH-ä Ω tL = C@expHä K aL + expH-ä K aL - 2D u expHä K s aL expH-ä Ω tL
(3.4)
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Here we used the following relations:
us = u expHä K s aL expH-ä Ω tL
(3.5)
us+1 = u exp@ä K Hs + 1L aD expH-ä Ω tL = u expHä K s aL expH-ä Ω tL expHä K aL
us-1 = u exp@ä K Hs - 1L aD expH-ä Ω tL = u expHä K s aL expH-ä Ω tL expH-ä K aL
â 2 us
â t2
â2
= u expHä K s aL
â t2
expH-ä Ω tL = -Ω2 u expHä K s aL expH-ä Ω tL
(3.6)
(3.7)
(3.8)
We can simplify the EOM:
-m Ω2 = C@expHä K aL + expH-ä K aL - 2D = 2 C@cosHK aL - 1D
(3.9)
So
Ω2 = 2
C
m
and thus
@1 - cosHK aLD
C
1 - cosHK aL
m
2
(3.10)
(3.11)
Ω=2
Using the trigonometric identity
C
Ω=2
1-cos Α
2
= sin
Α
2
Ka
(3.12)
sin
m
2
Fig. 2. Ω as a function of K.
3.1.4. the long wave length limit
At small momentum (very small K, K a << 1), sin
C
Ω»a
Ka
2
=
Ka
2
+ OI
M , so
Ka 2
2
K
m
This limit is known as “the long wavelength limit”, becomes small K means very large wavelength (Λ = 2 Π  K ® ¥ when K ® 0)
(3.13)
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5
Here, the frequncy is a linear function of the wavevector and at K = 0, Ω=0. This type of sound waves with Ω µ K are known as “acoustic sounds”.
For an acoustic sound mode, the relation between Ω and K is very similar to the corresponding relation for light. For light, we have Ω = c K where
c is the speed of light. Here, we have Ω = v K where v = a
3.1.5. the continuum limit
C  m is the speed of this sound wave.
If we set the lattice spacing to be very small Ha ® 0L, when this lattice model should recover the continuum limit. Here, the continuum limit is the
same as the long wave length limit, because both these two limit have k a << 1, so
C
Ka
Ω=2
C
sin
m
»a
2
(3.14)
K
m
3.1.6. Brillouin zone boundary K = Π  a
At K = Π  a
us = u expHä K s aL expH-ä Ω tL = u expHä Π sL expH-ä Ω tL = H-1Ls u expH-ä Ω tL
(3.15)
This is a standing wave, where even sites and odd sites move in the opposite way.
3.2. Quantization of sound waves
Quantum mechanics tells us that for a wave with frequency Ω and wavevector k, we can consider it as a beam of particles with energy E = ÑΩ and
momentum P = Ñ K.
For EM waves (light), the corresponding particles are photons.
For sound waves, the corresponding particles are called "phonons".
The energy of a phone is E=ÑΩ and we know that
C
Pa
E =2Ñ
(3.16)
sin
m
2Ñ
At small momentum (very small P, P a  2 Ñ << 1), we have
C
E=a
P
(3.17)
m
3.3. Sound velocities: phase velocity and group velocity
3.3.1. Phase velocity
The phase velocity of a wave is
Ω
vp =
(3.18)
K
C
m
For this sound wave, Ω = 2
C
Ω
vp =
sin
m
Ka
2
, so we have
Ka
=2
K
sin
2
K
at small K (the long wavelength limit),
(3.19)
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vp =
sin
C
Ω
Ka
=2
K
m
2
C
Ka
2
»2
K
m
C
(3.20)
a
=
K
m
3.3.2. Group velocity
The group velocity measures the velocity of a wave packet. It is defined as:
âΩ
vg =
(3.21)
âK
C
m
For this sound wave, Ω = 2
C
âΩ
vg =
sin
âK
Ka
2
Ka
â
=2
sin
m âK
. For the first Brillouin zone (-Π  a < K < Π  a), Ω = 2
C
=a
2
C
m
Ka
cos
m
sin
Ka
2
, because in this region sin
Ka
2
³0
(3.22)
2
At small K (the long wavelength limit), vg » v p
C
vg = a
Ka
cos
m
C
»a
2
(3.23)
m
which is the same as v p at small K
At the edge of the Brillouin zone, K = Π  a,
C
vg = a
Π
cos
m
=0
(3.24)
2
Zero group velocity means that there is no energy flow. As discussed above, K = Π  a has a standing wave, which indeed has no energy flow.
3.4. two atoms per primitive basis
Fig. 3. a 1D crystal formed by two different types of atoms
3.4.1. Equations of motion
Two types of atoms. ABABABABAB...
Ÿ Use an integer s to label each unit cell (each unit cell contains two atoms: 1 A tome and 1 B atom)
Ÿ Use us to describe the displacement of A atoms
Ÿ Use vs to describe the displacement of B atoms
The equations of motion is
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M1
M2
â2 us
â t2
â2 vs
â t2
= FS = C Hvs - us L + CHvs-1 - us L = CHvs + vs-1 - 2 us L
= FS = C Hus+1 - vs L + CHus - vs L = CHus+1 + us - 2 vs L
7
(3.25)
(3.26)
Here M1 and M2 are the masses for atoms A and B respectively.
So we have
M1
M2
â2 us
â t2
â2 vs
â t2
= CHvs + vs-1 - 2 us L
(3.27)
= CHus+1 + us - 2 vs L
(3.28)
3.4.2. Sound wave solutions
Let’s consider sound waves
us = u expHä K a sL expH-ä Ω tL
(3.29)
vs = v expHä K a sL expH-ä Ω tL
(3.30)
Here, a is the lattice constant (the size of a unit cell. NOT the distance between neighboring A and B atoms).
Now, the EOM will take the form
-M1 Ω2 u expHä K a sL expH-ä Ω tL = CIvs + vs ã-ä k a - 2 uM expHä K a sL expH-ä Ω tL
(3.31)
-M2 Ω2 v expHä K a sL expH-ä Ω tL = CIus+1 ãä k a + us - 2 vM expHä K a sL expH-ä Ω tL
(3.32)
M1 Ω2 u = -CIv + v ã-ä K a - 2 uM
(3.33)
M2 Ω2 v = -CIu ãä K a + u - 2 vM
(3.34)
So
So
IM1 Ω2 - 2 CM u + C I1 + ã-ä K a M v = 0
C Iã
äKa
+ 1M u + IM2 Ω - 2 CM v = 0
2
(3.35)
(3.36)
We can write these two equations into a matrix form
u
0
M1 Ω2 - 2 C CH1 + ã-ä K a L
K O=K O
v
0
CHãä K a + 1L M2 Ω2 - 2 C
(3.37)
where u and v are the two unknowns for these two equations.
For this type of linear equations, to have nontrivial solution (u and v being nonzero), we must have
det
M1 Ω2 - 2 C CH1 + ã-ä K a L
CHãä K a + 1L M2 Ω2 - 2 C
=0
(3.38)
This condition give us a relation between Ω and K
det
M1 Ω2 - 2 C CH1 + ã-ä K a L
CHãä K a + 1L M2 Ω2 - 2 C
= IM1 Ω2 - 2 CM IM2 Ω2 - 2 CM - C2 I1 + ã-ä K a M Iãä K a + 1M =
IM1 M2 Ω4 + 4 C2 - 2 C M1 Ω2 - 2 C M2 Ω2 M - C2 I2 + ã-ä K a + ãä K a M =
So
IM1 M2 Ω4 + 4 C2 - 2 C M1 Ω2 - 2 C M2 Ω2 M - 2 C2 H1 + cos K aL = 0
(3.39)
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M1 M2 Ω4 - H2 C M1 + 2 C M2 L Ω2 + 2 C2 H1 - cos K aL = 0
(3.40)
The solution to this equation is:
Ω2 =
H2 C M1 + 2 C M2 L2 - 8 M1 M2 C2 H1 - cos K aL
2 C M1 + 2 C M2 ±
=
2 M1 M2
(3.41)
HC M1 + C M2 L2 - 2 M1 M2 C2 H1 - cos K aL
C M1 + C M2 ±
M1 M2
where the “+”solution is known as the optical branch and the “-”one is known as the acoustic branch
Fig. 4. Ω as a function of K. Notice that there are two branches of solutions. The low branch is called acoustic branch and the upper one is the optical branch.
3.4.3. the long wave length limit (K » 0)
In the long wave length limit (or the continuum limit), where K a << 1, we can set cos K a = 1 for the optical branch
Ω2 =
C M1 + C M2 +
HC M1 + C M2 L2
=
2 C M1 + 2 C M2
M1 M2
1
=2C
M1 M2
1
(3.42)
+
M1
M2
For the acoustic branch, setting cos K a = 1 will give us Ω2 = 0, which is not good enough. So we need to keep higher order terms here and set
cos K a » 1 - HK aL2  2. Then we have,
Ω2 =
C M1 + C M2 -
HC M1 + C M2 L2 - M1 M2 C2 K 2 a2
=
M1 M2
C M1 + C M2
1-
M1 M2
So
1
C
Ω=
1-
M1 M2 C2 K 2
HC M1 + C M2 L2
a2
»
(3.43)
C M1 + C M2 1
M1 M2
M1 M2 C2 K 2
2 HC M1 + C M2 L2
a2 =
a2 K
2 M1 + M2
The acoustic mode has Ω µ K, so we have Ω = 0 at K = 0. But the optical mode has finite Ω at K = 0.
1
C K2
a2
2 M1 + M2
(3.44)
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3.4.4. the edge of the zone K = Π  a
At zone edge K = Π  a, cos K a = -1
Ω2 =
C M1 + C M2 ±
HC M1 + C M2 L2 - 4 M1 M2 C2
=
C M1 + C M2 ±
M1 M2
HC M1 - C M2 L2
M1 M2
=
C M1 + C M2 ± HC M1 - C M2 L
(3.45)
M1 M2
So,
Ω2 = 2
C
or Ω2 = 2
M1
C
(3.46)
M2
Here, one of the modes only involves the motion of A atoms (with mass M1 ) and the other mode only involves the motion of the B atoms (with
mass M2 ).
3.4.5. Energy gap
Notice that between the acoustic and optical branches, there is a range of frequency which cannot be reached, regardless of the momentum K (the
equation has no solution). This range is called an “energy gap”.
3.4.6. Questions: Why at K=Π/a, the frequency of the accoustic phonon mode only relies on M1 , but is
independent of M2 . For the optical phonon model, why it frequency only depends on M2 but not M1 at K=Π/a?
Answer: at K = Π  a, for the accoustic mode, only atoms with mass M1 moves. For the optical modes only atoms with mass M2 moves.
3.5. Longitudinal and transverse phonon modes
®
®
®
In higher dimensions, the deformation us becomes a vector us and the wavevector K becomes a vector K . Depending on the angle between u and
®
K , we can distinguish two different types of sound waves:
®
®
Longitudinal modes: u is parallel to K
®
®
Transverse modes: u is perpendicular to K
In a D-dimensional space, there will be 1 longitudinal acoustic branch and D-1 transverse acoustic branches.
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Fig. 5. Longitudinal and transverse modes. LA: longitudinal acoustic modes. TA: transverse acoustic modes. LO: longitudinal optical modes and TO; transverse optical
modes.
Q: Why LA has a higher velocity than TA?
A: Will be ansewred in the next chapter.
3.6. Number of phonon modes
For a lattice with n atoms per unit cell, there are Hn - 1L d optical phonons and d accoustic phonon modes.
For these d accoustic phonon modes, one of them is longidudinal and the other d - 1 modes are transverse accoustic modes.
In each unit cell (n atoms per unit cell), there are n d degrees of freedom, this number matches the number of phonon modes (accoustic+optical).