Download Phys 446: Solid State Physics / Optical Properties Lattice vibrations

Phys 446:
Solid State Physics / Optical
Properties
Lattice vibrations:
Thermal, acoustic, and optical properties
Fall 2015
Lecture 4
Andrei Sirenko, NJIT
1
Solid State Physics
Last weeks:
Lecture 4
(Ch. 3)
• Diffraction from crystals
• Scattering factors and selection rules for diffraction
Today:
• Lattice vibrations: Thermal, acoustic, and optical
properties
This Week:
• Start with crystal lattice vibrations.
• Elastic constants. Elastic waves.
• Simple model of lattice vibrations – linear atomic chain
• HW1 and HW2 discussion
2
1
Material to be included in the 1st QZ
• Crystalline structures. Diamond structure. Packing ratio
7 crystal systems and 14 Bravais lattices
• Crystallographic directions
and Miller indices
d hkl 
• Definition of reciprocal lattice vectors:
• What is Brillouin zone
• Bragg formula:
2d·sinθ = mλ
;
n
12
 h2 k 2 l 2 
 2  2  2 
b
c 
a
k = G
3
• Factors affecting the diffraction amplitude:
Atomic scattering factor (form factor):
reflects distribution of electronic cloud.
In case of spherical distribution
f a   n(r )e ik rl d 3 r
r0
f a   4r 2 n(r )
0
• Structure factor
F   f aj e
sin Δk  r 
dr
Δk  r
2i ( hu j  kv j  lw j )
j
where the summation is over all atoms in unit cell
• Be able to obtain scattering wave vector or frequency from
geometry and data for incident beam (x-rays, neutrons or light)
4
2
Material to be included in the 2nd QZ
TBD
Elastic stiffness and compliance. Strain and stress: definitions and
relation between them in a linear regime (Hooke's law):
 ij   Sijkl kl
 ij   Cijkl  kl
kl
kl
• Elastic wave equation:
 u Ceff  u x

t 2
 x 2
2
2
Ceff
sound velocity v 

5
• Lattice vibrations: acoustic and optical branches
In three-dimensional lattice with s atoms per unit cell there are
3s phonon branches: 3 acoustic, 3s - 3 optical
• Phonon - the quantum of lattice vibration.
Energy ħω; momentum ħq
• Concept of the phonon density of states
• Einstein and Debye models for lattice heat capacity.
Debye temperature
Low and high temperatures limits of Debye and Einstein models
•
Formula for thermal conductivity
1
K  Cvl
3
• Be able to obtain scattering wave vector or frequency from
geometry and data for incident beam (x-rays, neutrons or light)
6
3
7
Elastic properties
Elastic properties are determined by forces
acting on atoms when they are displaced from
the equilibrium positions
Taylor series expansion of the energy near
the minimum (equilibrium position):
U ( R)  U 0 
U
R
( R  R0 ) 
R0
1  2U
2 R 2
( R  R0 )  ...
R0
For small displacements, neglect higher terms. At equilibrium,
So,
ku 2
U ( R)  U 0 
2
where
k
 2U
R 2
U
R
0
R0
R0
u = R - R0 - displacement of an atom from equilibrium position
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4
F 
force F acting on an atom:
U
 ku
R
k - interatomic force constant. This is Hooke's law in simplest form.
Valid only for small displacements. Characterizes a linear region in which the
restoring force is linear with respect to the displacement of atoms.
Elastic properties are described by considering a crystal as a homogeneous
continuum medium rather than a periodic array of atoms
In a general case the problem is formulated as follows:
• Applied forces are described in terms of stress ,
• Displacements of atoms are described in terms of strain .
• Elastic constants C relate stress  and strain , so that  = C.
In a general case of a 3D crystal the stress and the strain are tensors
9
Stress has the meaning of local applied “pressure”.
Applied force F(Fx, Fy, Fz) 
Stress components ij (i,j = 1, 2, 3)
General case for stress: i.e ij
Fj 
 Fi dV  
V
V
 ij
x j
x  1, y  2, z  3
 ij
x j
dV    ij dS j
S
Shear forces must come in pairs: ij = ji (no angular acceleration)
 stress tensor is diagonal, generally has 6 components
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5
Stress has the meaning of local applied “pressure”.
Applied force F(Fx, Fy, Fz) 
Stress components ij (i,j = 1, 2, 3)
General case for stress: i.e ij
Fj 
 ij
x  1, y  2, z  3
x j
Hydrostatic pressure – stress tensor is equivalent to a
scalar: i.e xx =yy =zz
 p 0 0 
ˆ  [ ij ]  0  p 0 


0 0  p 
Stress tensor is a “field tensor” that can have any symmetry not
related to the crystal symmetry. Stress tensor can change the
crystal symmetry
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Stress has the meaning of local applied “pressure”.
Applied force F(Fx, Fy, Fz) 
Stress components ij (i,j = 1, 2, 3)
Compression stress: i = j, i.e xx , yy , zz
 xx 
x  1, y  2, z  3
Fx
Ax
Shear stress: i ≠ j, i.e xy , yx , xz zx , yz , zy
 yx 
Fy
Ax
Shear forces must come in pairs: ij = ji (no angular acceleration)
 stress tensor is diagonal, generally has 6 components
12
6
Strain tensor
3x3
 ij 
ui
x j
In 3D case, introduce the
displacement vector as
u = uxx + uyy + uzz
Strain tensor components are defined as
 xx 0 0


ˆ  [ ij ]  0  yy 0 
0 0  
zz 

V ' dV
  xx   yy   zz  Tr (ˆ )
dV
ˆ
u x
x
u
 xy  x
y
 xx 
Share deformations:
 xx   yy   zz  Tr (ˆ )  0
Can be diagonalized in x-y-z coordinates at a
certain point of space
In other points the tensor is not necessarily
diagonal
Strain tensor components are defined as
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 ij 
ui
x j
 xx 
u x
x
 xy 
u x
y
Since ij and ji always applied together, we can
define shear strains symmetrically:
1  ui u j 


2  x j xi 
 ij   ji  

So, the strain tensor is also diagonal and has 6 components
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7
Elastic stiffness (C) and compliance (S) constants
relate the strain and the stress in a linear fashion:
This is a general form of the Hooke’s law.
6 components ij, 6 ij  36 elastic constants
 ij   Cijkl  kl
kl
 ij   Sijkl kl
kl
Notations: Cmn where 1 = xx, 2 = yy, 3 = zz, 4 = yz, 5 = zx, 6 = xy
For example, C11 Cxxxx , C12 Cxxyy , C44 Cyzyz
Therefore, the general form of the Hooke’s law is given by
15
Elastic constants in cubic crystals
Due to the symmetry (x, y, and z axes are equivalent) C11 = C22 = C33 ;
C12 = C21 = C13 = C31 = C23 = C32 ; C44 = C55 = C66
Also, the off diagonal shear components are zero:
C45 = C54 = C46 = C64 = C56 = C65
and mixed compression/shear coupling does not occur:
C45 = C54 = C46 = C64 = C56 = C65
 the cubic elastic stiffness
tensor has the form:
only 3 independent constants
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8
Elastic constants in cubic crystals
Longitudinal compression
(Young’s modulus):
C11 
 xx F A

 xx u L
L
Transverse expansion:
Shear modulus:
C12 
C44 
 xx
 yy
 xy F A

 xy

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Uniaxial pressure setup
for optical characterization of correlated oxides
Pressure control
•Variables:
Uniaxial pressure
Temperature
External magnetic field
Measured sample properties:
Far-IR Transmission / Reflection
Raman scattering
Optical
cryostat
sample
Low T
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9
Elastic waves
Considering lattice vibrations three major approximations are made:
• atomic displacements are small: u << a , where a is a lattice
parameter
• forces acting on atoms are assumed to be harmonic, i.e.
proportional to the displacements: F = - Cu
(same approximation used to describe a harmonic oscillator)
• adiabatic approximation is valid – electrons follow atoms, so that
the nature of bond is not affected by vibrations
The discreteness of the lattice must be taken into account
For long waves  >> a, one may disregard the atomic nature –
solid is treated as a continuous medium.
Such vibrations are referred to as elastic (or acoustic) waves.
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Elastic waves
First, consider a longitudinal wave of compression/expansion
mass density 
segment of width dx at
the point x;
elastic displacement u



 2u 1 F  xx


t 2 A x
x
where F/A = xx
Assuming that the wave propagates along the
 xx  C11 xx
[100] direction, can write the Hooke’s law in the form
Since
u
 xx  x
x
 2 u C11  2 u x

get wave equation:
t 2
 x 2
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10
A solution of the wave equation - longitudinal plane wave
u ( x, t )  Aei ( qx t )
vL 
C11

where q - wave vector; frequency ω = vLq
- longitudinal sound velocity
Now consider a transverse wave which is controlled by shear stress
and strain:
In this case

 2u  xy

t 2
x
where  xy  C44 xy
 wave equation is  2u
C44  2u x

t 2
 x 2
 xy 
and
vT 
u
x
- transverse
sound velocity
C44

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Two independent transverse modes: displacements along y and z
For q in the [100] direction in cubic crystals, by symmetry the
velocities of these modes are the same - modes are degenerate
Normally C11 > C44  vL > vT
We considered wave along [100].
In other directions, the sound velocity depends
on combinations of elastic constants:
v
Ceff

Ceff - an effective elastic constant. For cubic crystals:
Relation between ω and q - dispersion relation. For sound ω = vq
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11
Model of lattice vibrations
one-dimensional lattice: linear chain of atoms
harmonic approximation:
force acting on the nth atom is
equation of motion (nearest neighbors interaction only):
 2u
M 2  Fn  C (un 1  un )  C (un 1  un )  C (2un  un 1  un 1 )
t
M is the atomic mass, C – force constant
Now look for a solution of the form
u ( x, t )  Aei ( qxn t )
where xn is the equilibrium position of the n-th atom  xn = na
obtain
23

 the dispersion relation is
Note: we change q  q + 2/a the atomic displacements and
frequency ω do not change  these solutions are physically identical
 can consider only
i.e. q within the first Brillouin zone
The maximum frequency is 2 C
M
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12
25
At the boundaries of the Brillouin
zone q = /a  un  A(1) n e  it
standing wave
Phase and group velocity
phase velocity is defined as v p 
group velocity
vg 

q
d
dq
vg  a
C
qa
cos
M
2
vg = 0 at the boundaries of the Brillouin zone (q = /a) 
no energy transfer – standing wave
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13
27
Long wavelength limit:  >> a ; q = 2/ << 2/a  qa << 1
small q - close to the center of Brillouin zone

4C
qa
sin

M
2
v p  vg  a
C
M
C
qa
M
- linear dispersion
- sound velocity for the one dimensional lattice
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14
Diatomic lattice
one-dimensional linear chain, atoms of two types: M1 and M2
Optical Phonons
can interact with light
For diamond
Optical phonon
frequency is  1300 cm-1
 7700 nm
(far-IR)
29
Model of diatomic lattice
one-dimensional linear chain, atoms of two types: M1 and M2
Treat in similar way, but
need two equations of
motion:
Again, look for a solution of the form
u n  A1e i ( qna t )
u n 1  A2 e i ( q ( n 1) a t )
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Substitute this solution into equations of motion
15
get system of two linear equations for the unknowns A1 and A2
In matrix form:
determinant of the matrix must be zero 
Solve this quadratic equation, get dispersion relation:
Depending on sign in this formula there are two different
solutions corresponding to two different dispersion curves
31
Note: the first Brillouin zone is now
from -/2a to +/2a
The lower curve - acoustic branch, the upper curve - optical branch.
at q = 0 for acoustic branch ω0 = 0; A1 = A2
 the two atoms in the cell have the same amplitude and the phase
dispersion is linear for small q
for optical branch
at q = 0
 1
1 

0  2C 

 M1 M 2 
M1A1 +M2A2 = 0
 the center of mass of the atoms remains fixed. The two atoms move out of
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phase. Frequency is in infrared – that's why called optical
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q
q
34
17
35
36
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Summary
 Elastic properties – crystal is considered as continuous anisotropic
medium
 Elastic stiffness and compliance tensors relate the strain and the
stress in a linear region (small displacements, harmonic potential)
Hooke's law:
 ij   Cijkl  kl
kl
 Elastic waves
 u Ceff  u x

t 2
 x 2
2
2
 ij   Sijkl kl
kl
sound velocity
v
Ceff

 Model of one-dimensional lattice: linear chain of atoms
 More than one atom in a unit cell – acoustic and optical branches
 All crystal vibrational waves can be described by wave vectors
within the first Brillouin zone in reciprocal space
What do we need? 3D case consideration
Phonons. Density of states
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19
HW1
Compute the theoretical density of NaCl based on its crystal structure.
For NaCl structure, the
crystal lattice parameter is
a= 2 ( r Na+ + r Cl -),
where r is ionic radius.
Cl-
Na+
a
a


M 4(A Na  A Cl )

V
a3 NA
4 ions (22.99  35.45) g/mol
2[(0.102x10  0.181x10  7 )]cm3 (6.023x10 23 )ions/mol
7
 2.14 g/cm 3
(actual  2.16g/cm 3 )
39
Vibrations in three-dimensional lattice.
Phonons
Phonon Density of states
Specific heat
(Ch. 3.3-3.9)
40
20
Three-dimensional lattice
In simplest 1D case with only nearest-neighbor interactions we had
M
equation of motion
solution
u
 Fn  C (un 1  un )  C (un 1  un )
t 2
2
u ( x, t )  Aei ( qxn t )
In general 3D case the equations of motion are:
M
 2 u n
  Fnm
2
t
m ,
N unit cells, s atoms in each 
3N’s equations
Fortunately, have 3D periodicity 
Forces depend only on difference m-n
Write displacements as
u n i ( x , t ) 
1
ui (q)ei ( qrn t )
41
M
substitute into equation of motion, get
  2ui (q)  
1
Fnmij e iq(rm rn )u j (q)  0
M M

, j m
Dij ( q ) - dynamical matrix
  2ui (q)   Dij (q)u j (q)  0
,j



phonon dispersion
curves in Ge
Det Dij (q)   2 1  0
- dispersion relation
3s solutions – dispersion branches
3 acoustic, 3s - 3 optical
direction of u determines polarization
(longitudinal, transverse or mixed)
Can be degenerate because of symmetry
42
21
Phonons
• Quantum mechanics: energy levels of the harmonic oscillator are quantized
• Similarly the energy levels of lattice vibrations are quantized.
• The quantum of vibration is called a phonon
(in analogy with the photon - the quantum of the electromagnetic wave)
Allowed energy levels of the harmonic oscillator:
where n is the quantum number
A normal vibration mode of frequency ω is given by
u  Aei ( qr t )
mode is occupied by n phonons of energy ħ momentum p = ħq
Number of phonons is given by Planck function:
(T – temperature)
n

1
e  kT  1
The total vibrational energy of the crystal is the sum of the energies of the
individual phonons:
(p denotes particular
43
phonon branch)
44
22
45
46
23
Density of states
Consider 1D longitudinal waves. Atomic displacements are given by:
u  Aeiqx
Boundary conditions: external constraints applied to the ends
Periodic boundary condition:
Then
eiqL  1
 condition on the admissible values of q:
2
n where n  0 ,  1,  2 , ...
L
regularly spaced points, spacing 2π/L
q
Number of modes in the
interval dq in q-space :
dω
L
dq
2
Number of modes in the frequency
L
range (ω, ω + dω):
D( )d 
D(ω) - density of states
2
ω
dq
dq
q
47
determined by dispersion ω = ω(q)
Density of states in 3D case
Now have
Periodic boundary condition:
eiq x L  e

iq y L
 eiq z L  1
l, m, n - integers
Plot these values in a q-space,
obtain a 3D cubic mesh
number of modes in the spherical
shell between the radii q and q + dq:
V = L3 – volume of the sample
 Density of states
48
24
Few notes:
• Equation we obtained is valid only for an isotropic solid,
(vibrational frequency does not depend on the direction of q)
• We have associated a single mode with each value of q.
This is not quite true for the 3D case: for each q there are 3 different
modes, one longitudinal and two transverse.
• In the case of lattice with basis the number of modes is 3s,
where s is the number of non-equivalent atoms.
They have different dispersion relations. This should be taken into
account by index p =1…3s in the density of states.
49
Lattice specific heat (heat capacity)
dQ
Defined as (per mole) C 
If constant volume V
dT
The total energy of the phonons at temperature T in a crystal:
E   nqp  p (q)  0
q, p
n 
1
e  kT  1
(the zero-point energy is chosen as the
origin of the energy).
- Planck distribution
Then
replace the summation over q by an integral over frequency:
Then the lattice heat capacity is:
Central problem is to find the density of states
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25
Debye model
• assumes that the acoustic modes give the dominant contribution
to the heat capacity
• Within the Debye approximation the velocity of sound is taken a
constant independent of polarization (as in a classical elastic
continuum)
The dispersion relation: ω = vq,
v is the velocity of sound.
In this approximation the density of states is given by:
Vq 2
1
Vq 2 1 V 2
D( ) 


2 2 d dq 2 2 v 2 2 v 3
Need to know the limits of integration over ω. The lower limit is 0.
How about the upper limit ? Assume N unit cells is the crystal, only
one atom in per cell  the total number of phonon modes is 3N 
13
Debye
 6 2 v 3 N 
13
frequency
  v 6 2 n
  D  

V



51
The cutoff wave vector which corresponds to this frequency is
modes of wave vector larger than qD are not
allowed - number of modes with q ≤qD
exhausts the number of degrees of freedom
Then the thermal energy is
where is "3" from ?

where x ≡ ħω/kBT and xD ≡ ħωD/kBT ≡ θD/T
Debye temperature:
52
26
The total phonon energy is then
where N is the number of atoms in the crystal and xD ≡ θD/T
To find heat capacity, differentiate
So,
In the limit T >>θD, x << 1,  Cv = 3NkB - Dulong-Petit law
53
Opposite limit, T <<θD : let the upper limit in the integral xD  
Get

within the Debye model at low
temperatures Cv  T3
The Debye temperature is
normally determined by
fitting experimental data.
Curve Cv(T/θ) is universal
– it is the same for
different substances
54
27
Einstein model
The density of states is approximated by a δ-function at some ωE :
D(E) = Nδ(ω – ωE) where N is the total number of atoms –
simple model for optical phonons
Then the thermal energy is
The heat capacity is then
The high temperature limit is the same as that for the Debye model:
Cv = 3NkB - the Dulong-Petit law
At low temperatures Cv ~ e-ħω/kBT - different from Debye T3 law
Reason: at low T acoustic phonons are much more populated  the
55
Debye model is much better approximation that the Einstein model
Real density of vibrational states is much more complicated than those
described by the Debye and Einstein models.
This density of states must be taken into account in order to obtain
quantitative description of experimental data.
The density of states for Cu.
The dashed line is the Debye
approximation.
The Einstein approximation
would give a delta peak at
some frequency.
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28
Summary
 In three-dimensional lattice with s atoms per unit cell there are
3s phonon branches: 3 acoustic, 3s - 3 optical
 Phonon - the quantum of lattice vibration.
Energy ħω; momentum ħq
 Density of states is important characteristic of lattice vibrations;
It is related to the dispersion ω = ω(q).
2
Simplest case of isotropic solid, for one branch: D ( ) 
Vq
1
2
2 d dq
 Heat capacity is related to the density of states.
 Debye model – good when acoustic phonon contribution dominates.
At low temperatures gives Cv  T3
 Einstein model - simple model for optical phonons (ω(q) is constant)
 At high T both models lead to the Dulong-Petit law: Cv = 3NkB
 Real density of vibrational states is more complicated
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