Lattice Waves (Phonons) in 3D Crystals Group IV …1 ECE 407 – Spring 2009 – Farhan Rana – Cornell University Handout 19 Lattice Waves (Phonons) in 3D Crystals Group IV and Group

1

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Handout 19

Lattice Waves (Phonons) in 3D CrystalsGroup IV and Group III-V Semiconductors

LO and TO Phonons in Polar Crystals and

Macroscopic Models of Acoustic Phonons in Solids

In this lecture you will learn:

• Lattice waves (phonons) in 3D crystals• Phonon bands in group IV and group III-V Semiconductors

• Macroscopic description of acoustic phonons from elasticity theory• Stress, strain, and Hooke’s law


Counting the Number of Phonon bands in 3D Crystals

332211 bbbq

22 where

22 where

22 where

333333

222222

111111

NmN-Nm

NmN-Nm

NmN-Nm

There are N1N2N3 allowed wavevectors in the FBZThere are N1N2N3 phonon modes per phonon band

Counting degrees of freedom and the number of phonon bands: Monoatomic Basis

• There are 3N1N2N3 degrees of freedom corresponding to the motion in 3D of N1N2N3

atomsThe number of phonon bands must be 3 (two TA bands and one LA band)

Counting degrees of freedom and the number of phonon bands: Diatomic Basis

• There are 6N1N2N3 degrees of freedom corresponding to the motion in 3D of 2N1N2N3 atomsThe number of phonon bands must be 6 (two TA bands and one LA band for acoustic phonons and two TO bands and one LO band for optical phonons)

Periodic boundary conditions for a lattice of N1xN2xN3 primitive cells imply:

2


Phonon Bands in Silicon

qz

qy

qx

qz

qy

qx

Silicon has a FCC lattice with two basis atoms in one primitive cell

FBZ of Silicon

The number of phonon bands must be 6; two TA bands and one LA band for acoustic phonons and two TO bands and one LO band for optical phonons


Phonon Bands in Silicon

qz

qy

qx

qz

qy

qx

Calculations

Data (Neutron scattering)

meV6400 xTOxLO qq

X

Fre

qu

ency

(T

Hz)

3


Phonon Bands in Diamond

Fre

qu

ency

(cm

-1)

X

Fre

qu

ency

(cm

-1)

L

LO

TO/TO

LA

TA/TA

LO

TO/TO

LA

TA/TA

Calculations


qz

qy

qx

qz

qy

qx

meV16500 xTOxLO qq

1-cm 13302

02

0

c

qc

q xTOxLO

Phonon frequencies are also expressed in units of equivalent photon wavelength inverse:

Large!!


Phonon Bands in GaAs

qz

qy

qx

qz

qy

qx

GaAs has a FCC lattice with two basis atoms in one primitive cell

FBZ of GaAs

The number of phonon bands must be 6; two TA bands and one LA band for acoustic phonons and two TO bands and one LO band for optical phonons

4


Calculations


qz

qy

qx

qz

qy

qx

meV 330

meV360

xTO

xLO

q

q

Fre

qu

ency

(T

Hz)

LO

TO

LA

TA

TOTO

LO

LA

TA

TA

TA

LA

TO

X K L

Phonon Bands in GaAs


Optical Phonons in Polar CrystalsConsider a crystal, like GaAs, made up of two different kind of atoms with a polar covalent bond

When the atoms move, an oscillating charge dipole is created with a dipole moment given by:

tdRu ,11

tndRu j ,12

tdRutndRuftRp jj ,,, 1112

jn

The material polarization, or the dipole moment density, is then:

j

jj

j tdRutndRuZnf

tRpZn

tRP ,,,, 1112

where:

3

1n Number of primitive cells per unit volume

A non-zero polarization means an electric field!

Z Number of nearest neighbors

5


Optical Phonons in Polar Crystals: D-Field and E-Field

tRu ,1

tnRu ,12

A non-zero polarization means an electric field! How do we find it? tnRu ,22

tnRu ,32

tnRu ,32

The divergence of the D-field is zero inside the crystal:

0. uD

But inside the crystal:

PE

PED

..

Since:

Therefore:

tRP

tRE,.

,.

j

jj

j tdRutndRuZnf

tRpZn

tRP ,,,, 1112

We must also have:

0, tRE


Optical Phonons in Polar Crystals: Dynamical Equations

tREMf

nntndRutdRuMdt

tdRud

tREMf

nntdRutndRuMdt

tdRud

jjjj

jjjj

,ˆˆ.,,,

,ˆˆ.,,,

22122

22

222

11112

12

112

Dynamical equations (assuming only nearest neighbor interactions):

Suppose:

tiRqi

tiRqitiRqi

dqi

dqi

eqPtRP

eqEtREeequ

equ

tdRu

tdRu

.

..

.2

.1

22

11

,

,

,

,

2

1

We have:

00, qEqtRE

The above two imply that the E-field has non-zero component only in the direction parallel to given by: q

q

qqPqE ˆ

ˆ.

qqP

qEqtRP

tREˆ.

.ˆ,.

,.

We also have:

6


qEMf

nnququM

ququrj

jjr

ˆˆ.1212

2

Optical Phonons in Polar Crystals: TO Phonons

Subtract the two equations and take the limit q0 to get:

Take the cross-product of both sides with to get: q̂

qqEMf

qnnququM

qququrj

jjr

ˆˆˆˆ.ˆ 12122

r

TO

r

r

Mb

q

Mb

qququMb

qququ

0

ˆˆ 12122

qAbqnnAj

jj ˆˆˆˆ.

Transverse Optical Phonons:

bnnj

jj ˆˆ

1

1

1

3

11n

1

1

1

3

12n

1

1

1

3

13n

1

1

1

3

14n

For example in GaAs:

34

100

010

001

34

ˆˆ

j

jj nn


Optical Phonons in Polar Crystals: LO Phonons

qEMf

nnququM

ququrj

jjr

ˆˆ.1212

2

Again start from:

Take the dot-product of both sides with to get:

Longitudinal Optical Phonons:

q̂

qqEMf

qnnququM

qququrj

jjr

ˆ.ˆ.ˆˆ.ˆ. 12122

rrLO

rr

Mnf

Mb

q

qququM

nfqququ

Mb

qququ

2

12

2

12122

0

ˆ.ˆ.ˆ.

qAbqnnAj

jj ˆ.ˆ.ˆˆ.

rTOLO

rTOLO

Mnf

Mnf

qq

222

222 00

bnnj

jj ˆˆ

7


Optical Phonons in Polar Crystals: Dielectric ConstantConsider the response of polar optical phonons to an externally applied E-fieldThe total electric field (external plus internal) is:

22

2

22

2

TO

r

TO

r

Mnf

qEMnf

qD

qEqPqEqD

2212

12122 ˆˆ.

TO

r

rjjj

r

qEMf

ququ

qEMf

nnququM

ququ

tiRqieqEtRE .,

22

2

12TO

rqE

Mnf

ququnfqP

We have:

The D-field is:

bnnj

jj ˆˆ

0q


Optical Phonons in Polar Crystals: Lydanne-Sachs-Teller Relation

0

0

22

2

2

22

2

TOr

TO

r

TO

r

Mnf

Mnf

Mnf

The LO-TO phonon frequency splitting was given by:

0

0

22

22

22

TOLO

TOr

TOLO Mnf

The above relationship is called the Lydanne-Sachs-Teller relationThe above relation does not change if more than nearest-neighbor interactions are also included in the analysisOne can also write:

We have:

Low frequency dielectric constant

22

2 0

TO

TO

8


Vector Dynamical Equations: Bond-Stretching and Bond-Bending

mm

m

RRm

ˆ

121R

2R

• In general, atomic displacements can cause both bond-stretching and bond-bending

• Both bond-stretching and bond-bending give rise to restoring forces

Bond-stretching component

Bond-bending component

mmtRutmRudt

tRudM ˆˆ.,,

,112

12

Bond-stretching contribution:


Vector Dynamical Equations: Bond-Stretching and Bond-Bending

mm

m

RRm

ˆ

121R

2R

Bond-stretching component

Bond-bending component

Bond-stretching and bond-bending contributions:

2211

1111

1121

2

ˆˆ.,,

ˆˆ.,,

ˆˆ.,,,

nntRutmRu

nntRutmRu

mmtRutmRudt

tRudM

First find two mutually orthogonal unit vectors that are also perpendicular to m̂

Let these be: 1n

2n̂and

9


Macroscopic Description of Acoustic Phonons in Solids

Acoustic phonons can also be described using a macroscopic formalism based on the theory of elasticity

Let the local displacement of a solid from its equilibrium position be given by the vector

ru

ru

ru

ru

z

y

x

ru

Strain Tensor:

Consider a stretched rubber band:

0 L xL+L

L

There is a uniform strain given by:

LL

xxu

e xxx


Stress and Strain

Stress Tensor:

Stress is the force acting per unit area on any plane of the solidIt is a tensor with 9 components (as shown)

x

y

z

yY

xXzZ

xY

xZ

zYzX

yXyZ

yXFor example, is the force acting per unit area in the x-direction on a plane that has a normal vector pointing in the y-direction

Strain Tensor:

The strain tensor is defined by its 6 components:e

z

rux

rue

yru

z

rue

x

ru

yru

e

zru

ey

rue

xru

e

xzzx

zyyz

yxxy

zzz

yyy

xxx

10


Hooke’s LawStress Tensor:

In solids with cubic symmetry, if the stress tensor produces no torque (and no angular acceleration) then one must have:

zxyzxy XZZYYX So there are only 6 independent stress tensor components:

yxzzyx XZYZYX

Hooke’s Law:

A fundamental theorem in the theory of elasticity is Hooke’s law that says that strain is proportional to the stress and vice versa. Mathematically, the 6 stress tensor components are related to the 6 strain tensor components by a matrix:

xy

zx

yz

zz

yy

xx

y

x

z

z

y

x

e

e

e

e

e

e

ccc

c

cc

cccc

X

Z

Y

Z

Y

X

666261

31

2221

16131211

...

......

......

.....

....

..

Elastic stiffness constants


Hooke’s Law for Cubic Materials

In solids with cubic symmetry (SC, FCC, BCC) the matrix of elastic constants have only three independent components:

xy

zx

yz

zz

yy

xx

y

x

z

z

y

x

e

e

e

e

e

e

c

c

c

ccc

ccc

ccc

X

Z

Y

Z

Y

X

44

44

44

111212

121112

121211

00000

00000

00000

000

000

000

Elastic energy:

The elastic energy per unit volume of a strained cubic material is:

2224412

222112

1xyzxyzxxzzzzyyyyxxzzyyxx eeeceeeeeeceeecV

11


Wave Equation for Acoustic Phonons in Cubic SolidsConsider a solid with density Consider a small volume of this solid that is in motion, as shown

x

We want to write Newton’s second law for its motion in the x-direction First consider only the force due to the stress tensor component Xx

r

x

rX

t

tru

xrX

zyxxx

rXxx

rXzyt

truzyx

xx

xxx

x

2

2

2

2

,

ˆ2

ˆ2

,

xx

rX x ˆ2

xx

rX x ˆ2

y

z

z

rXy

rX

xrX

t

tru zyxx

2

2 ,

Now add the contribution of all forces acting in the x-direction:

yy

rX y ˆ2

yy

rX y ˆ2


z

rXy

rX

xrX

t

tru zyxx

2

2 ,We have:

Similarly for acceleration in the y- and z-directions we get:

z

rYy

rY

xrY

t

tru zyxy

2

2 ,

zrZ

y

rZ

xrZ

t

tru zyxz

2

2 ,

Using the Hooke’s law relation, the above equation for motion in the x-direction can be written as:

zxru

yx

rucc

z

ru

y

ruc

x

ruc

zre

y

rec

xre

x

rec

xre

ct

tru

zyxxx

zxxyzzyyxxx

22

44122

2

2

2

442

2

11

4412112

2

,

Wave Equation for Acoustic Phonons in Cubic Solids

Wave equation for acoustic phonons

12


zxru

yx

rucc

z

ru

y

ruc

x

ruc

t

tru zyxxxx 22

44122

2

2

2

442

2

112

2 ,

LA phonons:

Consider a LA phonon wave propagating in the x-direction:

tixqix eeAtru x ,

Plug the assumed solution in the wave equation to get:

xqc

11 velocity of wave = 11c

TA phonons:

Consider a TA phonon wave propagating in the y-direction:

tiyqix eeAtru y ,

Plug the assumed solution in the wave equation to get:

yqc

44 velocity of wave = 44c



Acoustic Phonons in Silicon

qz

qy

qx

qz

qy

qx

In Silicon:

3

21144

21112

21111

mkg2330

mN1080.0

mN1064.0

mN1066.1

c

c

c

For LA phonons propagating in the -X direction:

velocity of wave = seckm 44.811

c

For TA phonons propagating in the -X direction:

velocity of wave = seckm 86.544

c

Results from elasticity theory

13


zxru

yx

rucc

z

ru

y

ruc

x

ruc

t

tru zyxxxx 22

44122

2

2

2

442

2

112

2 ,


yzru

yxru

ccx

ru

z

ruc

y

ruc

t

tru zxyyyy 22

44122

2

2

2

442

2

112

2 ,

Consider a phonon wave propagating in the direction:

tirqi

y

x

y

xee

qu

qu

tru

tru

.

,

,

Plug the assumed solution in the wave equation to get two coupled equations:

2

ˆˆ yx2

ˆˆ yxqq

qu

qu

qu

qu

ccq

ccq

ccq

ccq

y

x

y

x

2

4411

2

4412

2

4412

2

4411

2

22

22



qu

qu

qu

qu

ccq

ccq

ccq

ccq

y

x

y

x

2

4411

2

4412

2

4412

2

4411

2

22

22

The two solutions are as follows:

LA phonon:

TA phonon:

1

1

22 441211 A

qu

quq

ccc

y

x

1

1

21211 A

qu

quq

cc

y

x

Documents

Lattice Waves (Phonons) in 3D Crystals Group IV …1 ECE 407 – Spring 2009 – Farhan Rana – Cornell University Handout 19 Lattice Waves (Phonons) in 3D Crystals Group IV and Group