handout 18 phonons in 2d crystals: monoatomic basis and ... · phonons in 2d crystals: monoatomic...

1

ECE 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


1a

x

21 amanRnm

Phonons in a 2D Crystal with a Monoatomic Basis

y

2a

yanxan

yanxan

ˆˆ

ˆˆ

43

21

General lattice vector:

Nearest-neighbor vectors:

Atomic displacement vectors:

tRu

tRutRu

nmy

nmxnm ,

,,

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

yaxapyaxap

yaxapyaxap

ˆˆˆˆ

ˆˆˆˆ

43

21

Next nearest-neighbor vectors:

2


Vector Dynamical Equations

mm

m

RRm

ˆ

12

tRu ,1

tmRutRu ,, 12

mmtRutmRummtRutRudt

tRudM ˆˆ.,,ˆˆ.,,

,11122

12

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 ˆ.ˆˆ.,,

,112

12

y

ymmtRutmRudt

tRudM y ˆ.ˆˆ.,,

,112

12


Vector Dynamical Equations for a 2D Crystal

1a

x

y

2a

21 amanRnm

yanxan

yanxan

ˆˆ

ˆˆ

43

21


Nearest-neighbor vectors:

yaxapyaxap

yaxapyaxap

ˆˆˆˆ

ˆˆˆˆ

43

21

Next nearest-neighbor vectors:

4,3,2,12

4,3,2,112

2

ˆˆ.,,

ˆˆ.,,,

jjjnmjnm

jjjnmjnm

nm

pptRutpRu

nntRutnRudt

tRudM

summation over 4 nn

summation over 4 next nn

1

2

3


,,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

ˆˆ.,,

ˆˆ.,,,

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

ˆˆ.,,

ˆˆ.,,,

jjjnmjnm

jjjnmjnm

nm

pptRutpRu

nntRutnRudt

tRudM

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

4


Solution of the Dynamical Equations

Assume a wave-like solution of the form:

Then:

tiRqi

y

x

nmy

nmxnm ee

qu

qu

tRu

tRutRu nm

.

,

,,

tRue

eequ

que

eequ

qu

tnRu

tnRutnRu

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

qu

qu

M

M

qu

quqD

y

x

y

x

0

02

Compare with the standard form:

Solutions:

X M

FBZ

a2

a2

kg 2x10

N/m 100

N/m 200

26-

2

1

M

coscos122

sin4sinsin2

sinsin2coscos122

sin42

22

12

222

1

qu

quM

qu

qu

aqaqaq

aqaq

aqaqaqaqaq

y

x

y

x

yxy

yx

yxyxx

5


Transverse (TA) and Longitudinal (LA) Acoustic Phonons

242

22

42

222

2

12

222

2

2

1

qu

quM

qu

qu

aqaqaq

aqaq

aqaqaqaqaq

y

x

y

x

yxy

yx

yxyxx

Case I: 0,0 yx qq

0

121 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

1

02 Aqu

quaq

Mq

xy

xxxxTA

TA

LA

X

M

FBZ



242

22

42

222

2

12

222

2

2

1

qu

quM

qu

qu

aqaqaq

aqaq

aqaqaqaqaq

y

x

y

x

yxy

yx

yxyxx

Case II: qqqqq yxyx 0,0

1

14 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

1

11 Aqu

quaq

Mq

y

xTA

TA

LA

X

M

FBZ

TA

LATA

LA

6



TA

LATA

LATA

LAIn general for longitudinal acoustic phonons near the zone center:

y

x

y

x

q

q

qA

qu

qu

And for transverse acoustic phonons near the zone center:

x

y

y

x

q

q

q

Aqu

qu

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

7


Periodic Boundary Conditions in 2D

tiRqi

y

x

nmy

nmxnm ee

qu

qu

tRu

tRutRu 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

N

mN

-Nm

-mN

mmaNq

e

eequ

qutRuee

qu

qutaNRu

aNqi

tiRqi

y

xnm

tiaNRqi

y

xnm

nmnm

21 amanRnm


21,21 212211 bbq

General reciprocal lattice vector inside FBZ:

Similarly:

22 where 2

22

2

22

Nm

N-

Nm

FBZ

1b2b


Counting Degrees of Freedom

In the solution the values of the phonon wavevector are dictated by the periodic boundary conditions:

2211 bbq

22 where 11

1

11

Nm

N-

Nm

22 where 2

22

2

22

Nm

N-

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 band

Counting 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

8


Phonons in a 2D Crystal with a Diatomic Basis

Atomic displacement vectors:

tdRu

tdRu

tdRu

tdRu

tdRu

tdRu

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

ˆˆ

2

ˆˆ2

ˆˆ

2

ˆˆ

43

21

yaxah

yaxah

yaxah

yaxah

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

yanxan

yanxan

ˆˆ

ˆˆ

43

21

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

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

yaxapyaxap

yaxapyaxap

ˆˆˆˆ

ˆˆˆˆ

43

21


Diatomic Basis: Force Constants

plus

1a

x

y

2a

1

The force constants between the 1st

2nd and 3rd nearest-neighbors need to be included (at least)

1a

x

y

2a

2

3

9


4,3,2,111113

4,3,2,111112

4,3,2,1111212

112

1

ˆˆ.,,

ˆˆ.,,

ˆˆ.,,,

jjjnmjnm

jjjnmjnm

jjjnmjnm

nm

pptdRutpdRu

nntdRutndRu

hhtdRuthdRudt

tdRudM

summation over 4 1st nnsummation over 4 2nd nn

Diatomic Basis: Dynamical Equations

4,3,2,122223

4,3,2,122222

4,3,2,1222212

222

2

ˆˆ.,,

ˆˆ.,,

ˆˆ.,,,

jjjnmjnm

jjjnmjnm

jjjnmjnm

nm

pptdRutpdRu

nntdRutndRu

hhtdRuthdRudt

tdRudM

summation over 4 1st nnsummation 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

equ

equ

equ

equ

tdRu

tdRu

tdRu

tdRu

tdRu

tdRunm

.

.2

.2

.1

.1

22

22

11

11

22

11

2

2

1

1

,

,

,

,

,

,

Assume a solution of the form:

qu

ququ

qu

M

M

M

M

qu

ququ

qu

qD

y

x

y

x

y

x

y

x

2

2

1

1

2

2

1

1

2

2

2

1

1

000

000

000

000

To get a matrix equation of the form:

10


The Dynamical Matrix

aqaq

aq

yx

x

coscos12

2sin42

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

qu

ququ

qu

M

M

M

M

qu

ququ

qu

qD

y

x

y

x

y

x

y

x

2

2

1

1

2

2

1

1

2

2

2

1

1

000

000

000

000

aqaq

aq

yx

y

coscos12

2sin42

3

221

aqaq

aq

yx

x

coscos12

2sin42

3

221

aqaq

aq

yx

y

coscos12

2sin42

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/m100

N/m200

N/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

0

1

20

21

2

2

1

1

1MM

A

qu

ququ

qu

Mq

xy

xx

xy

xx

rxLO

212

2

1

1

101

0

20

MM

A

qu

ququ

qu

Mq

xy

xx

xy

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

0

10

1

?0

2

2

1

1

A

qu

ququ

qu

q

xy

xx

xy

xx

xLA

1

01

0

?0

2

2

1

1

A

qu

ququ

qu

q

xy

xx

xy

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

Nm

N-

Nm

22 where 2

22

2

22

Nm

N-

Nm

There are N1N2 allowed wavevectors in the FBZThere are N1N2 phonon modes per phonon band

Counting 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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