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October 19th, 2010
Class 12
Crystal Vibrations: Monoatomic basis
We found that the dispersion relation for monoatomic basis was
4C
4C
 Ka 
 Ka 
sin 
2 
sin 2 

 what leads to the dispersion relation  K  
M
M
 2 
 2 
That for the long wavelength limit (small K) leads to,
cosKa  1 

 

2C
C
1
Ka2   and  2  1  cos Ka  Ka 2
2
M
M
Ca 2
or  
K
M
and for K=±/a leads to stationary waves (the derivative is zero at those points). These are the
limits of the first Brillouin zone.
for Kmax=/a  2/min=/a or min=2a and
 

s
u s  u exp  is a   u exp is   u  1
a 

Thus, consecutive planes move in opposite phase.
All the physics of the problem is already in the first Brillouin zone, second and others are
redundant and do not produce new information.
To understand this, notice that us are actually the amplitude of the oscillation at each atom, thus
any wave for which the amplitude at each atomic position is the same and NOT a new solution
since there are no atoms to see intermediate amplitudes. Notice in Figure 5 (page 93) that the two
waves predict the same amplitude at each point.
Notice that the ratio of the amplitude of the atomic displacement between consecutive planes is
u s 1 u expi s  1Ka 

 expiKa 
us
u expi sKa 
All independent values of this ratio are obtained for Ka between - and . Given any value of K
outside the mentioned [- ,] interval, there exist another wavevector K’=K-2n/a (with n and
integer) that it is in the interval. For this wavevector K’ in the first Brillouin zone
 
us 1
2n  
 2n 
 exp iKa   exp  i  K '
a   exp iKa exp  i
a   exp iK ' a 

us
a  
 a 
 
Thus the relative amplitude is exactly the same for K as it is for a K’ in the first Brillouin zone.
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Group Velocity
When a packet of waves with different wave lengths (and frequency) travel together, it is not the
speed of each of the components what matters, but the velocity of the group (imagine a pulse
sent through space). In non-dispersive mediums, frequency and wavevector are proportional and,
as we will see, phase velocity (the one of individual components) is equal to group velocity (like
light in vacuum), all the components travel at the same speed. But in general that is not the case,
each component travels at a different speed, and the actual package travels at a speed giving by:
d
dK
To see this, consider two waves with a slightly different frequency and wavevector (and just for
illustration purpose, the same amplitude too). The composed waves will be:
vg 
ue i t  Kx   ue i ' t  K ' x 
This equation can be re-arranged to give
K 
 
2ue i  t  Kx  cos 
t
x  Where  and K are the average frequency and wave vector.
2 
 2
That is a high frequency wave with a low frequency envelope. The velocity at which the
package moves is the velocity of the envelope, thus
vg 

thus the vg indeed give the speed of the package.
K
If for a particular medium, the frequency and the wavevector are proportional, then group and
phase velocity are the same, and equal to the speed of each component. In this case the medium
is non-dispersive and all the components move at the same speed, thus that package does not
change shape. If instead different components travel at different speeds, the package will spread,
or disperse as it moves on (thus the relationship between  and K is known as the dispersion
relationship)
d
For a continuum of frequencies, v g 
dK
In the case of phonons, using the dispersion relation obtained above.
 Ca 2   Ka 
cos
0  Ka  


d  M   2 
vg 

dK  Ca 2   Ka 

cos
    Ka  0

M   2 

vg is discontinuous for K=0 where it goes from -1
to 1 in going from negative values of K to positive values of K. Which is not surprising, it only
means that the wave now moves in the opposite direction.
In the limits of the Brillouin zone, Ka=± and vg=0, consistent with standing waves.
2
As we saw before, for the long wavelength limit (small K),

Ca 2
K
M
Thus the group velocity is equal to /K. This can be understood by realizing that in the long
wave length limit K~0, >>a and the interaction with the lattice is minimal leading to a nondispersive behavior.
Force constant from the interaction with non-consecutive planes
If we consider now the effect of a plane p places from the reference plane s then


us p a   u exp i s  p Ka

from where the dispersion condition can be easily shown to be
2 
2
C p 1  cos  pKa  and considering all the planes
M
2 
2
M
 C 1  cos pKa 
p 0
p
Where Cp, the force constant associated with the interaction between plane s and p is given by:
(See derivations provided)
 /a
Ma
Cp  
 2 cos pKa dK
2 / a
More than one type of atoms per primitive basis
When the basis consist on one atom, there are three independent vibration modes, 1 longitudinal
(the atoms in the plane oscillate in a direction perpendicular to the plane) and 2 transversal where
the atoms oscillates parallel to the plane (the two modes come from the two dimensions of the
plane). When the basis have more than one type of atom, then each type of atom give rise to
three independent vibration modes, thus in a crystal with a p-atoms basis, there are 3p
independent modes. Out of the 3p modes, 3 and called acoustical modes and 3p-3 are called
optical modes for a reason that will be evident soon. Also both optical and acoustical modes
have transversal and longitudinal in the rate of 2-to-1.
In what follows we will consider a cubic crystal with a two atoms base with alternate planes of
atoms 1 (with mass M1) and atoms 2 (with mass M2).
d 2u s
 C vs  vs1  2us 
dt 2
d 2v
Atom 2 : M 2 2s  C us  us 1  2vs 
dt
Atom1 : M 1
3
Solutions to the above equations can be proposed to be of the form
u s  u exp  it  sKa  and v s  v exp  i t  sKa 
Where a is the distance between plane of atoms of the same kind
Using these solutions in the equation above,
 M1 2u  C v  v exp iKa   2u 
 M 2 2v  C u  u expiKa   2v 
What leads to a set of two homogeneous equations with two unknowns,
2C  M  u  C1  e   0
 C 1  e u  2C  M    0
 iKa
2
1
iKa
2
1
This set of equations has non-trivial solution only if the following determinant is zero
2C  M1 2
 C1  exp  iKa 


 C1  exp iKa 
2C  M  
2
2
what leads to
M 1 M 2 4  2C M 1  M 2  2  2C 2 1  cos Ka   0
with solutions
2C M 1  M 2   4C 2 M 1  M 2   8M 1 M 2C 2 1  cos Ka 
 
2M 1M 2
2
2
For the limit Ka << 1 cos(Ka)~1- ½ (Ka)2 or 1- cos(Ka)~ ½ (Ka)2
2C M 1  M 2   4C 2 M 1  M 2   4 M 1 M 2 C 2 Ka 
 
2M 1M 2
2
2
2
and for small values of Ka we can expand the square root in series
values of x
2

4M1M 2C 2 Ka  


2C M1  M 2    2C M1  M 2  



4
C
M

M
1
2


2 
2 M 1M 2
what provide the two solutions for small Ka
4
a 2  bx ~ a 
bx
for small
2a
4M 1M 2C 2 Ka 
 1
4C M 1  M 2 
1 
4C M 1  M 2 

2 

 2C 

2 M 1M 2
2 M 1M 2
M
M
2 
 1
To the first order in Ka
and
2
4C M 1  M 2  
4 M 1 M 2C 2 Ka 
4C M 1  M 2 
C
Ka 2
2 

2M 1M 2
2M 1  M 2 
2
By using these solutions in the set of equations for u and v and solving for them, it can be seen
M
u
that for the first of the solutions in the K~0 limit,   2 , thus the consecutive planes vibrate
v
M1
in opposite direction keeping the center of mass fixed. If the two atoms are ions of opposite sign,
this mode can be excited by an electromagnetic field, thus this mode is called optical mode. The
other solution for the K~0 leads to u=v thus the planes move in the same direction as it is for
long wavelengths sound waves, thus this mode is known as acoustical mode.
For the other limit, Ka=± cos(Ka)=-1 and thus the solutions are :
2C M 1  M 2   4C 2 M 1  M 2   16 M 1 M 2 C 2 2C M 1  M 2   2C M 1  M 2 
 

2M 1M 2
2M 1M 2
2
2
Or  2 
2C
M2
and  2 
2C
M1
This leads to a solution where each type of atom vibrates independently from the other. One
solution correspond to one type of atom vibrating while the other is at rest and viceversa
There is a gap between optical and acoustical mode at the boundary of the first Brillouin zone
where frequencies between (2C/M2)2 and (2C/M1)2 are not allowed.
Figure 8 shows both modes. Also in Figure 8, it can be seen that a 3-D crystal, longitudinal and
transversal vibrations split in frequency.
In a crystal with a p atom base there are 3 acoustical modes (1 longitudinal and 2 transversal) and
3p-3 optical modes.
5
Quantization of Elastic Waves
By describing vibration in the lattice by a quantum harmonic oscillator model, it is observed that
the lattice energy is quantized, in other words, the energy of a lattice vibrating with a single
frequency  is:


1
2
   n   
That means that no just any value of energy is allowed but the ones in that equation. For n=0 the
lowest energy state is obtained and it is known as zero point energy. Other values of n represent
exited states of the crystal vibration where n photons populate the state. Phonons are not
subjected to the Pauli Exclusion Principle (as photons are not) thus any number of phonons can
occupy a given state, however you add phonons one by one, a fraction of a phonon would not do.
The quantization of the energy has the consequence to produce quantization on the amplitude of
the oscillation.
To see this consider a vibration described by
u  uo cos Kx cos t
As for a harmonic oscillator, the temporal average of the energy is half kinetic and half potential.
1  u 
Lets consider the kinetic energy density K      , where  is the mass density
2  t 
2
The volume integral of the kinetic energy density gives (notice that the integral of the cos2 is Lx/2
while the integral in y and z gives Ly and Lz), thus the volume integral is V/2
¼ V2uo2sin2t
with a time average of 1 V2uo2 thus (since <sin2t>=1/2). Making this average kinetic
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energy equal to ½ of the total energy
1 V2uo2= ½(n+½) ħ
8
we obtain that
uo2= 4(n+½) ħ/V= 4(n+½) ħ/M
Thus the vibration amplitude is related to the occupancy n of the mode.
The solution of the motion’s equation for atoms in the crystal gives 2 so, what is the sign of ?
Conventionally  is assumed positive. For unstable structures, 2 is negative and  is
imaginary.
6
Phonon Momentum
A phonon of wavevector K can interact with photons, neutrons, electrons, and other particles as
if they had a momentum ħK. Looking at this in a different way, if a photon (or other quantum
particle) collides with the crystal, without losing energy, the selection rule for the incoming
particle wave vector imposes that
k’=k+G
No energy is transferred to the crystal but the crystal as a whole recoils with a momentum -ħG.
If inelastic scattering occurs, some energy is indeed lost by the incoming particle to the crystal
that will excite phonon modes, besides the fact that the transferred energy should be at least ħ,
the energy of a phonon of frequency, there is a generalized selection rule that can be written as
k’+K=k+G
where K is the phonon’s wave vector, thus the wavevector of the scattered particle is conditioned
to the wavevector of the created phonon such that the above equation is satisfied. In the same
way, if a phonon is annihilated in the interaction, the equivalent selection rule is:
k’=K+k+G
Neutrons are often used to determine the dispersion relations for the phonon. After scattering,
the energy and direction of scattered neutrons is recorded thus k and k’ are determined.
The kinetic energy of the neutrons is p2/2Mn and the momentum is p= ħk, so conservation of
energy imposes
 2 k 2  2 k' 2

  Where ħ is the energy of the phonon created (+) or absorbed (-) from
2M n 2M n
where  is obtained.
By knowing the crystal lattice (G) and from the formula k’±K=k+G, K is determined, where G
is chosen such that K is in the first Brillouin zone.
The ± sign applies to phonon created or absorbed.
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