handout 17 lattice waves (phonons) in 1d crystals ...handout 17 lattice waves (phonons) in 1d...

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ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Handout 17

Lattice Waves (Phonons) in 1D Crystals: Monoatomic Basis and Diatomic Basis

In this lecture you will learn:

• Equilibrium bond lengths• Atomic motion in lattices• Lattice waves (phonons) in a 1D crystal with a monoatomic basis• Lattice waves (phonons) in a 1D crystal with a diatomic basis• Dispersion of lattice waves• Acoustic and optical phonons


The Hydrogen Molecule: Equilibrium Bond Length

0 x

rV

d-d

BE

AE

sE1:2 AE:1

BE:1ssV2

The equilibrium distance between the two hydrogen atoms in a hydrogen molecule is set by the balance among several different competing factors:

• The reduction in electronic energy due to co-valent bonding is 2Vss . If the atoms are too far apart, Vss becomes to small

• If the atoms are too close, the positively charged nuclei (protons) will repel each other and this leads to an increase in the system energy

• Electron-electron repulsion also plays a role

sssB VEE 1

sssA VEE 1

ssss VxdrHxdr ˆˆˆ 11

2


A Mass Attached to a Spring: A Simple Harmonic Oscillator

x

0

x

0

u

Equilibrium position Stretched position

Potential Energy:

2

21

ukuVPE

Kinetic Energy:2

2

dtduM

KE

M M

spring constant = k(units: Newton/meter)

Dynamical Equation (Newton’s Second Law):

ukdudV

dt

udM 2

2 Restoring force varies linearly with the displacement “u ” of the mass from its equilibrium position

Solution:

tBtAtu oo sincos where:Mk

o

PE varies quadratically with the displacement “u ” of the mass from the equilibrium position


A 1D Crystal: Potential Energy

Consider a 1D lattice of N atoms:

xaa ˆ1

x

1anRn

• The potential energy of the entire crystal can be expressed in terms of the positions of the atoms. The potential energy will be minimum when all the atoms are at their equilibrium positions.

• Let the displacement of the atom at the lattice site given by from its equilibrium position be

• One can Taylor expand the potential energy of the entire crystal around its minimum equilibrium value:

k jkj

EQkj

jj EQj

EQN

RuRuRuRu

V

RuRuV

VRuRuRuRuV

2

321

21

............,,

nR

nRu

0

Potential energy varies quadratically with the displacements of the atoms from their equilibrium positions

Atoms can move only in the x-direction

3


A 1D Crystal: Potential and Kinetic Energies

A1D lattice of N atoms:

xaa ˆ1

x

1anRn

k jkjkjEQ

k jkj

EQkjEQ

tRutRuRRKV

tRutRuRuRu

VVV

,,,21

,,21 2

Potential Energy:

j

j

dt

tRduMKE

2,

2

Kinetic Energy:

• The kinetic energy of all the atoms is the sum of their individual kinetic energies

EQkj

kj RuRu

VRRK

2

,


A 1D Crystal: Dynamical Equation


xaa ˆ1

x

1anRn

j

jjnn

n tRuRRKRu

V

dt

tRudM ,,

,2

2

Write Newtons law for the atom sitting at the site :nR

EQkj

kj RuRu

VRRK

2

,

Remember that:

The restoring forces on the atoms vary linearly with the displacement of the atoms from their equilibrium positions

4


Dynamical Equation for Nearest-Neighbor Interactions


j

jjnn

n tRuRRKRu

V

dt

tRudM ,,

,2

2

Assume nearest-neighbor interactions:

njnjnjjn RRK ,1,1, 2,

tRutRutRutRudt

tRudM nnnn

n ,,,,,

112

2

This gives:

The constants “” provide restoring forces as if the atoms were connected together with springs of spring constant “”

We have N linear coupled differential equations for N unknowns

1.......2,1,0, NntRu n

The constant is called “force constant” (not spring constant) in solid state physics


Solution of the Dynamical Equation: Lattice Waves (Phonons)


tRutRutRutRudt

tRudM nnnn

n ,,,,,

112

2

tiRqin eequtRu n

.Re,

Assume a solution of the form:

Represents a wave with wavevector , frequency , and amplitude

q

qu

tiRqiaqi

tiaRqitiRqin

eeque

eequeequtRu

n

nn

..

..1

1

11

,

Note that:

tiRqin eequtRu n

.,Or: Slight abuse of notation

xaa ˆ1

1anRn

tiRqiaqi

tiaRqitiRqin

eeque

eequeequtRu

n

nn

..

..1

1

11

,

5



tRutRutRutRudt

tRudM nnnn

n ,,,,,

112

2

Plug in the assumed solution:

To get:

queququeququM aqiaqi 11 ..2

Which simplifies to:

2.

sin4

.cos12

12

12

aqM

aqM

Or:

2.

sin4 1aqM

xaa ˆ1

1anRn

xqq x ˆ tiRqin eequtRu n

.,


Since is always positive, the negative sign is chosen when the sine term is negative


A1D lattice of N atoms: xaa ˆ1

1anRn

Solution is: and tiRqin eequtRu n

.,

2.

sin4 1aqM


x

x

a nRu

• The lattice waves are like the compressional sound waves in the air

6



1anRn

Solution is: and

The relation:

represents the dispersion of the lattice waves or phonons

2sin

42.

sin4 1 aq

Maq

Mx

xq

tiRqin eequtRu n

.,

2.

sin4 1aqM

a

a


x

a nRu


Solution is: and

For the shortest wavelength:

The largest wavevector is then:

The wavevector values can be restricted to the First BZ

• No new solutions are found for values of the wavevector outside the first BZ

a2

xq

tiRqin eequtRu n

.,

2.

sin4 1aqM

a

a

Lattice Waves and the First BZ

x

Question: What is the shortest wavelength (or the largest wavevector) the lattice waves can have?

aqx

2

First BZ

nRu = 2a

7


xq

2sin

4 aqM

x

a

a

Phase and Group Velocities

First BZ

Case I: For qx ≈ 0 (i.e. qx a << ):

aqM x

Linear dispersion

Phase velocity and group velocity of lattice waves are defined as:

xdqd

qqv

xq

qqq

qv

x

Dqg

x

Dp

ˆ

ˆˆ

1

1

gp vaM

v

Case II: For qx a = :

M 4

0

4

g

p

v

aM

v


Periodic Boundary Conditions

In the solution:

allowed values of the wavevector depend on the boundary conditions

tiRqin eequtRu n

.,


1anRn

Periodic Boundary Condition:

0 1N-1

The N-th atom is the same as the 0-th atom This implies:

22

integer2

1

1

,,

0

0

.

.0

.

Nm

N

m

aNm

q

e

e

eequtRueequtRu

x

aNqi

RRqi

tiRqitiRqiN

x

N

N

xq

a

a

First BZ

aN2

8


Counting and Conserving Degrees of FreedomA1D lattice of N atoms: xaa ˆ1

1anRn

• We started with N degrees of freedom which were related to the motion in 1D of Ndifferent atoms• The dynamical variables were the amplitudes of the displacements of N different atoms

1.......2,1,0, NntRu n

We then ended up with lattice waves:

• There are N different lattice wave modes corresponding to the N different possible wavevector values in the first BZ• The dynamical variables are the amplitudes of the N different lattice wave modes

22

and2

,N

mN

aNm

qtqu x

The number of degrees of freedom are the same before and after – as they should be!

xq

a

a

First BZ

aN2

nn RqitqiRqin etqueequtRu

.. ,ReRe,


Counting and Conserving Degrees of Freedom

The atomic displacements,

taken together provide a complete description of the motion of all the atoms in the crystal

In general, one can expand the atomic displacements in terms of all the lattice wave modes (resembles a Fourier series expansion):

1.......2,1,0, NntRu n

FBZ in

.*

.

FBZ in

.*

.

FBZ in

.

2,

2,

22

Re,

q

RqiRqi

q

tqiRqitqiRqi

q

tqiRqin

nn

nn

n

etqu

etqu

eequ

eequ

eequtRu

Therefore, the lattice wave amplitudes also provide a complete description of the motion of all the atoms in the crystal

9


A1D lattice of N red atoms and N blue atoms and N primitive cells:

xaa ˆ1

x

1anRn

Lattice Waves (Phonons) in a 1D crystal: Diatomic Basis

Unit cell

1M 2M

1n

2n

The nearest neighbor vectors are:

1d

2d

The basis vectors are:

1n

2n

The spring model for nearest neighbor interactions:

x

Let the displacement of the red atom in the n-th cell be: tdRu n ,11

Let the displacement of the blue atom in the n-th cell be: tdRu n ,22

1d

2d


tdRutdRutdRutdRudt

tdRudM nnnn

n ,,,,,

22111212112211

2

1

tdRutdRutdRutdRudt

tdRudM nnnn

n ,,,,,

11122211221222

2

2

Write the dynamical equations for both the atoms assuming nearest neighbor interactions

Assume a solution of the form:

tidRqin

tidRqin

eequtdRu

eequtdRu

n

n

2

1

.222

.111

,

,

This is again a wave-like solution with a wavevector and frequency and different amplitudes for the two atoms

q

Notice the phases

x

1M 2M

1n

2n

12

The constants 1 and 2 are called “force constants” (not spring constants) in solid state physics


10


Plug the solutions in the dynamical equations to get:

queququeququM nqinqi

2.

112.

12112 12

queququeququM nqinqi

1.

221.

21222 21

Write the equations in a matrix form:

qu

qu

M

M

qu

qu

ee

eenqinqi

nqinqi

2

1

2

12

2

1

21.

2.

1

.2

.121

0

021

21

x

1M 2M

1n

2n

12

This is a 2x2 matrix eigenvalue equation that needs to be solved for each value of the wavevector to get the dispersion of the lattice waves



x

1M 2M

1n

2n

12

The Dynamical Matrix

Or:

qu

quM

qu

quqD

2

12

2

1

• The matrix is called the dynamical matrix of the medium

• For any medium, in any dimension, the dispersion relations for the lattice waves (phonons) are obtained by solving a similar matrix eigenvalue equation

qD

qu

qu

M

M

qu

qu

ee

eenqinqi

nqinqi

2

1

2

12

2

1

21.

2.

1

.2

.121

0

021

21

Or:

qu

qu

qu

quqDM

2

12

2

11

11


Optical and Acoustic Phonons

x

1M 2M

1n

2n

12

The frequency eigenvalues are:

2.

sin1621

212

21

212

221212 aq

MMMMq

rr

21

111MMMr

xaa ˆ1

xq

a

a

• The two frequency eigenvalues for each wavevector value in the FBZ give two phonon bands

• The higher frequency band is called the optical phonon band

• The lower frequency band is called the acoustic phonon band

Optical

Acoustic

FBZ


Optical and Acoustic Phonons: Special Cases

x

1M 2M 12

xaa ˆ1

Case I: 0q

aqMM

q x2121

210

Acoustic band:

1

1

0

0

2

1 Aqu

qu

tidRqin

tidRqin

eeAtdRu

eeAtdRu

n

n

2

1

.22

.11

,

,

rM

q 210

Optical band:

212

1 1

0

0

MMA

qu

qu

tidRqin

tidRqin

eeMM

AtdRu

eeAtdRu

n

n

2

1

.

2

122

.11

,

,

xq

a

a

Optical

Acoustic

12



x

1M 2M 12

xaa ˆ1

Case I: Acoustic Mode0q

Acoustic band (basis atoms move in-phase):

tidRqin

tidRqin

eeAtdRu

eeAtdRu

n

n

2

1

.22

.11

,

,

x

x

a

u

xq

a

a

Optical

Acoustic

0q



x

1M 2M 12

xaa ˆ1

Case I: Optical Mode0q

Optical band (basis atoms move 180o out-of-phase):

x

x

a

u

xq

a

a

Optical

Acoustic

tidRqin

tidRqin

eeMM

AtdRu

eeAtdRu

n

n

2

1

.

2

122

.11

,

,

0q

13



x

1M 2M 12

xaa ˆ1

Case II: Acoustic Mode 2121 ,,ˆ MMMx

aq

xq

a

a

Optical

Acoustic

Mx

a22

ˆ

1.

2

1 1

ˆ

ˆ

nqieA

xa

u

xa

u

tidRqi

tidRqinqin

tidRqin

eeA

eeeAtdRu

eeAtdRu

n

n

n

1

21

1

.

..22

.11

,

,

x



x

1M 2M 12

xaa ˆ1

Case II: Optical Mode2121 ,,ˆ MMMx

aq

xq

a

a

Optical

AcousticM

xa

12ˆ

1.

2

1 1

ˆ

ˆ

nqieA

xa

u

xa

u

x

tidRqi

tidRqinqin

tidRqin

eeA

eeeAtdRu

eeAtdRu

n

n

n

1

21

1

.

..22

.11

,

,

14



• We started with 2N degrees of freedom which were related to the motion in 1D of 2Ndifferent atoms• The dynamical variables were the amplitudes of the displacements of 2N different atoms

1.......2,1,0,, 2211 NntdRutdRu nn

We then ended up with lattice waves:

• There are N different modes per phonon band corresponding to the N different possible wavevector values in the first BZ• There are 2 phonon bands and therefore a total of 2Ndifferent phonon modes • The dynamical variables are the amplitudes of the 2Ndifferent phonon modes

The number of degrees of freedom are the same before and after – as they should be!

xq

a

a

FBZ

aN2

x

1M 2M 12

xaa ˆ1

Optical

Acoustic



The atomic displacements,

taken together provide a complete description of the motion of all the 2N atoms in the crystal

In general, one can expand the atomic displacements in terms of all the lattice wave modes – all wavevectors and all bands:

OA q

Rqidqi

dqiRqi

dqi

dqi

OA q

tqiRqidqi

dqitqiRqi

dqi

dqi

OA q

tqiRqidqi

dqi

n

n

nn

nn

n

eetqu

etque

etqu

etqu

eeequ

equee

equ

equ

eeequ

equ

tdRu

tdRu

, FBZ in

..*

2

.*1.

.2

.1

, FBZ in

..*

2

.*1.

.2

.1

, FBZ in

..

2

.1

22

11

2

1

2

1

2

1

2

1

2

1

,

,

21

,

,

21

21

21

Re,

,

Therefore, the lattice wave amplitudes also provide a complete description of the motion of all the atoms in the crystal

1.......2,1,0,, 2211 NntdRutdRu nn

handout 17 lattice waves (phonons) in 1d crystals ...handout 17 lattice waves (phonons) in 1d...

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