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Lecture 4 Andrei Sirenko, NJIT 1

Phys 446:

Solid State Physics / Optical Properties

Lattice vibrations: Thermal, acoustic, and optical properties

Fall 2015

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Solid State Physics Lecture 4

(Ch. 3)Last weeks:

• 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

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Material to be included in the 1st QZ

• Crystalline structures. Diamond structure. Packing ratio7 crystal systems and 14 Bravais lattices

• Crystallographic directions and Miller indices

• Definition of reciprocal lattice vectors:

• What is Brillouin zone

• Bragg formula: 2d·sinθ = mλ ; k = G

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2

2

2

2

2

2

cl

bk

ah

ndhkl

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•Factors affecting the diffraction amplitude:

Atomic scattering factor (form factor): reflects distribution of electronic cloud.

In case of spherical distribution

•Structure factor

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)

rdenf lia

3)( rkr

0

0

2

Δ

Δsin)(4

r

a drrk

rkrnrf

j

lwkvhuiaj

jjjefF )(2

3

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Elastic stiffness and compliance. Strain and stress: definitions and relation between them in a linear regime (Hooke's law):

•Elastic wave equation:

klkl

ijklij C klkl

ijklij S

2

2

2

2

x

uC

t

u xeff

sound velocity

effC

v

Material to be included in the 2nd QZTBD

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• 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

• Be able to obtain scattering wave vector or frequency from geometry and data for incident beam (x-rays, neutrons or light)

CvlK3

1

4

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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):

...)(2

1)()( 02

2

00

00

RRR

URR

R

UURU

RR

For small displacements, neglect higher terms. At equilibrium, 00

RR

U

So,

2)(

2

0

kuURU where

0

2

2

RR

Uk

u = R - R0 - displacement of an atom from equilibrium position

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force F acting on an atom: kuR

UF

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

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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)

x 1, y 2, z 3General case for stress: i.e ij

ijj

j

Fx

Shear forces must come in pairs: ij = ji (no angular acceleration)

stress tensor is diagonal, generally has 6 components

iji ij j

jV V S

FdV dV dSx

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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)

x 1, y 2, z 3General case for stress: i.e ij ij

jj

Fx

Hydrostatic pressure – stress tensor is equivalent to a scalar: i.e xx =yy =zz

0 0

ˆ [ ] 0 0

0 0 ij

p

p

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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x

yyx A

F

Stress has the meaning of local applied “pressure”.

Applied force F(Fx, Fy, Fz) Stress components ij (i,j = 1, 2, 3)

x 1, y 2, z 3Compression stress: i = j, i.e xx , yy , zz

x

xxx A

F

Shear stress: i ≠ j, i.e xy , yx , xz zx , yz , zy

Shear forces must come in pairs: ij = ji (no angular acceleration)

stress tensor is diagonal, generally has 6 components

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Strain tensor3x3

In 3D case, introduce the displacement vector asu = uxx + uyy + uzzStrain tensor components are defined as

j

iij x

u

y

uxxy

x

uxxx

0 0

ˆ [ ] 0 0

0 0

'ˆ( )

xx

ij yy

zz

xx yy zz

V dVTr

dV

Can be diagonalized in x-y-z coordinates at a certain point of space

In other points the tensor is not necessarily

diagonal

Share deformations:

ˆ( ) 0xx yy zz Tr

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Strain tensor components are defined as

j

iij x

u

y

uxxy

x

uxxx

Since ij and ji always applied together, we can define shear strains symmetrically:

1

2ji

ij jij i

uu

x x

So, the strain tensor is also diagonal and has 6 components

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Elastic stiffness (C) and compliance (S) constants

klkl

ijklij C 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

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

klkl

ijklij S

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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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Elastic constants in cubic crystals

Longitudinal compression

(Young’s modulus): Lu

AFC

xx

xx

11

Transverse expansion:

L

yy

xxC

12

AF

Cxy

xy 44

Shear modulus:

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Uniaxial pressure setupfor optical characterization of correlated oxides

sample

•Variables:

Uniaxial pressureTemperatureExternal magnetic field

Measured sample properties:

Far-IR Transmission / ReflectionRaman scatteringOptical

cryostat

Pressure control

Low T

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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

xx

F

At

u xx

1

2

2

where F/A = xx

Assuming that the wave propagates along the [100] direction, can write the Hooke’s law in the form xxxx C 11

Since x

uxxx

get wave equation: 2

211

2

2

x

uC

t

u x

11

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A solution of the wave equation - longitudinal plane wave)(),( tqxiAetxu where q - wave vector; frequency ω = vLq

11C

vL - longitudinal sound velocity

Now consider a transverse wave which is controlled by shear stress and strain:

In this case

xt

u xy

2

2

where xyxy C 44 andx

uxy

wave equation is

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2

2

x

uC

t

u x

44C

vT - transverse sound velocity

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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:

effCv

Ceff - an effective elastic constant. For cubic crystals:

Relation between ω and q - dispersion relation. For sound ω = vq

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Model of lattice vibrations

one-dimensional lattice: linear chain of atoms

harmonic approximation: force acting on the nth atom is

)2()()( 11112

2

nnnnnnnn uuuCuuCuuCFt

uM

equation of motion (nearest neighbors interaction only):

M is the atomic mass, C – force constant

Now look for a solution of the form )(),( tqxi nAetxu

where xn is the equilibrium position of the n-th atom xn = na

obtain

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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 CM

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At the boundaries of the Brillouin

zone q = /a standing wave

tinn eAu )1(

Phase and group velocity

phase velocity is defined as

group velocity

qvp

dq

dvg

2cos

qa

M

Cavg

vg = 0 at the boundaries of the Brillouin zone (q = /a) no energy transfer – standing wave

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Long wavelength limit: >> a ; q = 2/ << 2/a qa << 1

qaM

Cqa

M

C

2sin

4 - linear dispersion

small q - close to the center of Brillouin zone

M

Cavv gp - sound velocity for the one dimensional lattice

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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)

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Model of diatomic lattice

one-dimensional linear chain, atoms of two types: M1 and M2

Again, look for a solution of the form)(

1tqnai

n eAu

Treat in similar way, but need two equations of motion:

))1((21

tanqin eAu

Substitute this solution into equations of motion

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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

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Note: the first Brillouin zone is nowfrom -/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

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112

MMC M1A1 +M2A2 = 0

the center of mass of the atoms remains fixed. The two atoms move out of phase. Frequency is in infrared – that's why called optical

for optical branchat q = 0

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q

q

18

35

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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:

Elastic waves sound velocity

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 considerationPhonons. Density of states

klkl

ijklij C klkl

ijklij S

2

2

2

2

x

uC

t

u xeff

effC

v

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HW1

a

a

Cl-

Na+

For NaCl structure, the crystal lattice parameter is a= 2 ( r Na+ + r Cl -), where r is ionic radius.

Compute the theoretical density of NaCl based on its crystal structure.

)2.16g/cm(actual

g/cm142

)ions/mol(6.023x10)]cm0.181x1002[(0.102x1

g/mol 35.45)(22.99ions4

Na

)AA(4

V

M

3

3

23377

A3

ClNa

.

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Vibrations in three-dimensional lattice.

Phonons

Phonon Density of states

Specific heat

(Ch. 3.3-3.9)

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Three-dimensional lattice

In general 3D case the equations of motion are:

)()( 112

2

nnnnn uuCuuCFt

uM

In simplest 1D case with only nearest-neighbor interactions we had

equation of motion solution

)(),( tqxi nAetxu

,2

2

m

mn

nuF

tM

N unit cells, s atoms in each 3N’s equations

)()(1

),( tiiin

neuM

txu

rqq

Fortunately, have 3D periodicity Forces depend only on difference m-n

Write displacements as

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0)(1

)(,

matrix dynamical - )(

)(2 qq

q

rrqmn

m

nmj

j

D

ijii ueF

MMu

ji

substitute into equation of motion, get

0)()()(,

2 qqq jj

jii uDu

0)( 2 1qDet

jiD

- dispersion relation3s solutions – dispersion branches

3 acoustic, 3s - 3 optical

direction of u determines polarization(longitudinal, transverse or mixed)

Can be degenerate because of symmetry

phonon dispersion curves in Ge

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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

mode is occupied by n phonons of energy ħ momentum p = ħq

Number of phonons is given by Planck function:(T – temperature)

The total vibrational energy of the crystal is the sum of the energies of the individual phonons:(p denotes particular

phonon branch)

)( tie rqAu

1

1

kTen

44

23

45

46

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Density of states

Consider 1D longitudinal waves. Atomic displacements are given by: iqxAeu

Boundary conditions: external constraints applied to the ends

Periodic boundary condition:

1iqLeThen condition on the admissible values of q:

..210 where2

, ., , n nL

q

regularly spaced points, spacing 2π/L

dq

dω

ω

q

Number of modes in the interval dq in q-space :

dqL

2

Number of modes in the frequency range (ω, ω + dω):

D(ω) - density of statesdetermined by dispersion ω = ω(q)

dqL

dD

2

)(

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Density of states in 3D case

Now have

Periodic boundary condition: 1 LiqLiqLiq zyx eee

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

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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.

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Lattice specific heat (heat capacity)

dT

dQC Defined as (per mole) If constant volume V

0)(,

qq

q pp

pnE

The total energy of the phonons at temperature T in a crystal:

(the zero-point energy is chosen as the origin of the energy).

1

1

kTen

- 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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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:

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2

2

2

2

2

2

1

2

1

2)(

v

V

v

Vq

dqd

VqD

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

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3132

66

nvV

NvD

Debye frequency

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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:

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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

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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

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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 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 Debye model is much better approximation that the Einstein model

The heat capacity is then

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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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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).

Simplest case of isotropic solid, for one branch:

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

dqd

VqD

1

2)(

2

2