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1ECE 407 Spring 2009 Farhan Rana Cornell University

Handout 18

Phonons in 2D Crystals: Monoatomic Basis and Diatomic Basis

In this lecture you will learn:

Phonons in a 2D crystal with a monoatomic basis Phonons in a 2D crystal with a diatomic basis Dispersion of phonons LA and TA acoustic phonons LO and TO optical phonons

ECE 407 Spring 2009 Farhan Rana Cornell University

1a

x

21 amanRnm

Phonons in a 2D Crystal with a Monoatomic Basis

y

2a

yanxanyanxan

43

21

General lattice vector:

Nearest-neighbor vectors:

Atomic displacement vectors:

tRutRu

tRunmy

nmxnm ,

,,

Atoms, can move in 2D therefore atomic displacements are given by a vector:

yaxapyaxapyaxapyaxap

43

21

Next nearest-neighbor vectors:


Vector Dynamical Equations

mmm

RRm

12

tRu ,1

tmRutRu ,, 12

mmtRutmRummtRutRudt

tRudM .,,.,,, 111221

2

Vector dynamical equation:

If the nearest-neighbor vectors are known then the dynamical equations can be written easily.

Component dynamical equation:To find the equations for the x and y-components of the atomic displacement, take the dot-products of the above equation on both sides with and , respectively:x

xmmtRutmRudt

tRudM x ..,,, 1121

2

y

ymmtRutmRudt

tRudM y ..,,

,112

12


Vector Dynamical Equations for a 2D Crystal

1a

x

y

2a

21 amanRnm

yanxanyanxan

43

21

General lattice vector:

Nearest-neighbor vectors:


43

21

Next nearest-neighbor vectors:

4,3,2,12

4,3,2,112

2

.,,

.,,,

jjjnmjnm

jjjnmjnm

nm

pptRutpRu

nntRutnRudt

tRudM

summation over 4 nn

summation over 4 next nn

12


,,2

,,2

,,2

,,2

,,2

,,2

,,2

,,2

,,,,,

42

42

32

32

22

22

12

12

31112

2

tpRutRutpRutRu

tpRutRutpRutRu

tpRutRutpRutRu

tpRutRutpRutRu

tnRutRutnRutRudt

tRudM

nmynmynmxnmx

nmynmynmxnmx

nmynmynmxnmx

nmynmynmxnmx

nmxnmxnmxnmxnmx

Dynamical Equations

4,3,2,12

4,3,2,112

2

.,,

.,,,

jjjnmjnm

jjjnmjnm

nm

pptRutpRu

nntRutnRudt

tRudM

If we take the dot-product of the above equation with we get:x


,,2

,,2

,,2

,,2

,,2

,,2

,,2

,,2

,,,,,

42

42

32

32

22

22

12

12

41212

2

tpRutRutpRutRu

tpRutRutpRutRu

tpRutRutpRutRu

tpRutRutpRutRu

tnRutRutnRutRudt

tRudM

nmxnmxnmynmy

nmxnmxnmynmy

nmxnmxnmynmy

nmxnmxnmynmy

nmynmynmynmynmy

Dynamical Equations

4,3,2,12

4,3,2,112

2

.,,

.,,,

jjjnmjnm

jjjnmjnm

nm

pptRutpRu

nntRutnRudt

tRudM

If we take the dot-product of the above equation with we get:y


Solution of the Dynamical Equations

Assume a wave-like solution of the form:

Then:

tiRqiyxnmy nmxnm eeququ

tRutRu

tRu nm

.

,,

,

tRueeequ

que

eeququ

tnRutnRu

tnRu

nmnqi

tiRqi

y

xnqi

tinRqi

y

x

jnmy

jnmxjnm

j

nmj

jnm

,

,,

,

.

..

.

We take the above solution form and plug it into the dynamical equations


Dynamical Matrix and Phonon Bands

ququ

MM

ququ

qDy

x

y

x

0

02Compare with the standard form:

Solutions:

X M

FBZ

a2

a2

kg 2x10

N/m 100N/m 200

26-2

1

M

coscos122

sin4sinsin2

sinsin2coscos122

sin42

22

12

222

1

ququ

Mququ

aqaqaq

aqaq

aqaqaqaqaq

y

x

y

x

yxy

yx

yxyxx


Transverse (TA) and Longitudinal (LA) Acoustic Phonons

2

42

22

42

222

2

12

222

2

2

1

ququ

Mququ

aqaqaq

aqaq

aqaqaqaqaq

y

x

y

x

yxy

yx

yxyxx

Case I: 0,0 yx qq

0121 Aqu

quaq

Mq

xy

xxxxLA

Longitudinal acoustic phonons: atomic motion in the direction of wave propagation

Transverse acoustic phonons: atomic motion in the direction perpendicular to wave propagation

102 Aqu

quaq

Mq

xy

xxxxTA

TA

LA

X

M

FBZ



2

42

22

42

222

2

12

222

2

2

1

ququ

Mququ

aqaqaq

aqaq

aqaqaqaqaq

y

x

y

x

yxy

yx

yxyxx

Case II: qqqqq yxyx 0,0

114 21 Aqu

quaq

Mq

y

xLA

Longitudinal acoustic phonons: atomic motion in the direction of wave propagation

Transverse acoustic phonons: atomic motion in the direction perpendicular to wave propagation

111 Aqu

quaq

Mq

y

xTA

TA

LA

X

M

FBZ

TA

LATA

LA



TA

LATA

LATA

LAIn general for longitudinal acoustic phonons near the zone center:

y

x

y

xqq

qA

ququ

And for transverse acoustic phonons near the zone center:

x

y

y

xqq

qA

ququ

In general, away from the zone center, the LA phonons are not entirely longitudinal and neither the TA phonons are entirely transverse



LATA

TA and LA


Periodic Boundary Conditions in 2D

tiRqiyxnmy nmxnm eeququ

tRutRu

tRu nm

.

,,

,

Our solution was:

Periodic boundary conditions for a lattice of N1xN2 primitive cells imply:

22

where

21

21 where22

integer an is where2.1

,,

11

1

11

1111

1111

.

..11

11

11

NmN-Nm

-mN

mmaNqe

eeququ

tRueeququ

taNRu

aNqi

tiRqi

y

xnm

tiaNRqi

y

xnm nmnm

21 amanRnm

General lattice vector: 21,21 212211 bbq

General reciprocal lattice vector inside FBZ:

Similarly:

22 where 222

2

22

NmN-Nm

FBZ

1b2b


Counting Degrees of FreedomIn the solution the values of the phonon wavevector are dictated by the periodic boundary conditions:

2211 bbq

22 where 11

1

11

NmN-Nm

22 where 222

2

22

NmN-Nm

There are N1N2 allowed wavevectors in the FBZ(There are also N1N2 primitive cells in the crystals)

There are N1N2 phonon modes per phonon bandCounting degrees of freedom:

There are 2N1N2 degrees of freedom corresponding to the motion in 2D of N1N2 atoms

The total number of different phonon modes in the two bands is also 2N1N2

FBZ


Phonons in a 2D Crystal with a Diatomic Basis

Atomic displacement vectors:

tdRutdRutdRutdRu

tdRutdRu

nmy

nmx

nmy

nmx

nm

nm

,,,,

,,

22

22

11

11

22

11

The two atoms in a primitive cell can move in 2D therefore atomic displacements are given by a four-component column vector:

1a

x

y

2a

1

2

21 amanRnm

2

2

2

2

43

21

yaxahyaxah

yaxahyaxah

1st nearest-neighbor vectors (red to blue):

yanxanyanxan

43

21

2nd nearest-neighbor vectors (red to red):

3rd nearest-neighbor vectors (red to red):


43

21


Diatomic Basis: Force Constants

plus

1a

x

y

2a

1

The force constants between the 1st2nd and 3rd nearest-neighbors need to be included (at least)

1a

x

y

2a

23


4,3,2,111113

4,3,2,111112

4,3,2,1111212

112

1

.,,

.,,

.,,,

jjjnmjnm

jjjnmjnm

jjjnmjnm

nm

pptdRutpdRu

nntdRutndRu

hhtdRuthdRudt

tdRudM

summation over 4 1st nn

summation over 4 2nd nn

Diatomic Basis: Dynamical Equations

4,3,2,122223

4,3,2,122222

4,3,2,1222212

222

2

.,,

.,,

.,,,

jjjnmjnm

jjjnmjnm

jjjnmjnm

nm

pptdRutpdRu

nntdRutndRu

hhtdRuthdRudt

tdRudM

summation over 4 1st nn

summation over 4 2nd nn

Dynamical equation for the red(1) atom:

Dynamical equation for the blue(2) atom:

summation over 4 3rd nn

summation over 4 3rd nn


Diatomic Basis: Dynamical Equations

tiRqi

dqiy

dqix

dqiy

dqix

nmy

nmx

nmy

nmx

nm

nm ee

equequ

equequ

tdRutdRutdRutdRu

tdRutdRu nm

.

.2

.2

.1

.1

22

22

11

11

22

11

2

2

1

1

,,,,

,,

Assume a solution of the form:

ququququ

MM

MM

ququququ

qD

y

xy

x

y

xy

x

2

21

1

2

2

1

1

2

2

21

1

000000000000

To get a matrix equation of the form:

10


The Dynamical Matrix

aqaqaq

yx

x

coscos122

sin42

3

221

2cos

2cos2 1

aqaq yx

2sin

2sin2 1

aqaq yx

2cos

2cos2 1

aqaq yx

2sin

2sin2 1

aqaq yx

2cos

2cos2 1

aqaq yx

2sin

2sin2 1

aqaq yx

2sin

2sin2 1

aqaq yx

2cos

2cos2 1

aqaq yx

ququququ

MM

MM

ququququ

qD

y

xy

x

y

xy

x

2

21

1

2

2

1

1

2

2

21

1

000000000000

aqaqaq

yx

y

coscos122

sin42

3

221

aqaqaq

yx

x

coscos122

sin42

3

221

aqaqaq

yx

y

coscos122

sin42

3

221

aqaq yx sinsin2 3

aqaq yx sinsin2 3

aqaq yx sinsin2 3

aqaq yx sinsin2 3

qD The matrix is:


Diatomic Basis: Solution and Phonon Bands

N/m100N/m200N/m300

kg1042

3

2

1

2621

MM

For calculations: Opticalbands

Acousticbands

One obtains:

- 2 optical phonon bands (that have a non-zero frequency at the zone center)

- 2 acoustic phonon bands (that have zero frequency at the zone center)

TA

LA

LO

TO

TA

LA

LO

TO

X M

FBZ

a2

a2

X

TA

LA

TOLO

11


Longitudinal (LO) and Transverse (TO) Optical Phonons

Case I: 0,0 yx qq

0

01

2021

2

21

1

1MMA

ququququ

Mq

xy

xxxy

xx

rxLO

212

21

1

1010

20

MM

A

ququququ

Mq

xy

xxxy

xx

rxTO

Longitudinal optical phonons: atomic motion in the direction of wave propagation and basis atoms move out of phase

Transverse optical phonons: atomic motion in the direction perpendicular to wave propagation and basis atoms move out of phase

X

M

FBZ

21

111MMMr


Longitudinal (LA) and Transverse (TA) Acoustic Phonons

Case I: 0,0 yx qq

0101

?0

2

21

1

A

ququququ

q

xy

xxxy

xx

xLA

1010

?0

2

21

1

A

ququququ

q

xy

xxxy

xx

xTO

Longitudinal acoustic phonons: atomic motion in the direction of wave propagation and basis atoms move in phase

Transverse acoustic phonons: atomic motion in the direction perpendicular to wave propagation and basis atoms move in phase

X

M

FBZ

12


Counting Degrees of Freedom and the Number of Phonon Bands

2211 bbq

22 where 11

1

11

NmN-Nm

22 where 222

2

22

NmN-Nm

There are N1N2 allowed wavevectors in the FBZThere are N1N2 phonon modes per phonon bandCounting degrees of freedom:

There are 4N1N2 degrees of freedom corresponding to the motion in 2D of 2N1N2 atoms (2 atoms in each primitive cell)

The total number of different phonon modes in the four bands is also 4N1N2

Periodic boundary conditions for a lattice of N1xN2 primitive cells imply:

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Documents

nearestneighbor vectors

monoatomic basis phonons

ta acoustic phonons

component dynamical

atomic displacement

dynamical equationsassume

dynamical equationsece

d crystals