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M.P. Vaughan Introduction to Condensed Matter Physics Phonons I Normal Modes

Transcript of Introduction to Condensed Matter · PDF filePY3105 Introduction to Condensed Matter Physics...

M.P. Vaughan

Introduction to Condensed

Matter Physics

Phonons I – Normal Modes

Overview

Coláiste na hOllscoile Corcaigh, Éire

University College Cork, Ireland ROINN NA FISICE Department of Physics

PY3105 Introduction to

Condensed Matter Physics

Overview of phonons I

• The simple harmonic oscillator

• Anharmonicity

• Diatomic linear chain

• Dispersion relations

• Acoustic and optical modes

The Simple Harmonic

Oscillator

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Classical simple harmonic oscillator

A mass m is attached to an ideal spring with spring constant

kS (gravity is neglected). The displacement of the mass is

denoted by u0 = z – z0.

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PY3105 Introduction to

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Classical simple harmonic oscillator

The restoring force exerted on the mass by the spring is

given by Hooke’s Law

.0ukF S

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PY3105 Introduction to

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Classical simple harmonic oscillator

.00 ukum S

By Newton’s Second Law, we therefore have

This has the general solution

,00

0

titiBeAeu

where the resonant frequency 0 is given by

.

2/1

0

m

kS

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Particular solution for SHM

A particular solution may be obtained by putting u0(0) = 0.

Defining the amplitude of u0 to be A0 = 2A, this gives

.sin 000 tAu

Hence we have simple harmonic motion (SHM).

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Potential energy for SHM

The potential energy U(z) stored in the spring is obtained

from

.2

021

00

zzkFdzzUzU S

z

z

Since U(z0) is an arbitrary constant, we are free to set it to

zero, so that we have simply

.2

021 zzkzU S

Note that

.2

2

Skdz

Ud

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Anharmonicity

In reality, the behaviour of real springs will only approximate

Hooke’s Law for small displacements. More generally, we

may characterise a real spring in terms of its potential

energy. Expanding this as a Taylor series (in 1D), we have

2

02

2

00!2

1zz

dz

Udzz

dz

dUzUzU

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PY3105 Introduction to

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Anharmonicity

At z0, U(z) passes through a minimum, so we can set the

first derivative to zero, giving

,!2

1 3

0

2

02

2

0 zzzzdz

UdzUzU O

where O[xn] means ‘terms of order xn and above’. These

terms are called the anharmonic terms, whilst the first two

terms are just those for a simple harmonic oscillator

.2

1 2

02

2

0 zzdz

UdzUzU

Normal Modes

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Normal modes

We now consider the case of two masses m1 and m2

coupled together by ideal springs of spring constant kS as

shown above.

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PY3105 Introduction to

Condensed Matter Physics

Normal modes

Using Hooke’s Law, the forces on each mass are

.

,

212222

121111

uukukumF

uukukumF

SS

SS

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PY3105 Introduction to

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Normal modes

Defining

.2

,2

2

2

21

2

22

2

2

11

2

11

uuu

uuu

,and2

2

2

1

2

1m

k

m

k SS

we can write the coupled differential equations as

These equations must now be solved simultaneously.

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Normal modes – eigenvalues

We look for solutions such that the coupled system oscillates

with a single angular frequency . That is, we require

.2

,2

2

2

2

2

21

2

2

1

2

2

2

11

2

1

uuu

uuu

Hence, we have

.

,

222

111

titi

titi

eBeAu

eBeAu

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PY3105 Introduction to

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Normal modes – eigenvalues

In matrix form, we may write this as

Thus, we must solve the characteristic equation

.2

2

2

12

2

1

2

2

2

2

2

1

2

1

u

u

u

u

.02

22

2

22

2

2

1

2

1

2

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PY3105 Introduction to

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Normal modes – eigenvalues

The solution of the characteristic equation is

.2/14

2

2

2

2

1

4

1

2

2

2

1

2

When m1 = m2 = m, this reduces to

.3,

,2

2

0

2

0

2

0

2

0

2

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University College Cork, Ireland ROINN NA FISICE Department of Physics

PY3105 Introduction to

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Normal modes – eigenvectors

Putting

,2/14

2

2

2

2

1

4

1

2

the eigenvalue solutions may be written

.22

2

2

1

2

This may then be substituted into the matrix equation to find

the eigenvectors.

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PY3105 Introduction to

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Normal modes – eigenvectors

We shall simplify the problem by assuming solutions of the

form

.12

1

22

2

2

12 uu

The displacements are related to one another by

.cos

,cos

22

11

tAu

tAu

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Normal modes – eigenvectors

Thus, we may put ,11 Au etc.

In the case the eigenvectors reduce to ,21 mm

12 uu

and

.12 uu

The first of these corresponds to acoustic phonons in a

crystal, the second to optical phonons.

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Normal modes – ‘acoustic’ mode

.2/14

2

2

2

2

1

4

1

2

2

2

1

2

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Normal modes – ‘optical’ mode

.2/14

2

2

2

2

1

4

1

2

2

2

1

2

The Diatomic Linear

Chain

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Diatomic linear chain

Let us extend our analysis to a linear chain of alternating

masses m1 and m2 connected by ideal springs with spring

constant kS. The distance between identical masses is taken

to be a and the equilibrium positions and displacements are

labelled as above.

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Diatomic linear chain

We may view the distance between identical masses, a, as

the lattice constant of a 1D crystal.

Typically, we impose periodic (Born-von Karman)

boundary conditions on this 1D chain, visualised by joining

the ends of the chain into a ring.

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Diatomic linear chain

If the total length of the chain is L and there are N primitive

cells (of length a) in the chain, then

LNa

and periodic boundary conditions require that, for any

function of the distance z around the ring

.zfLzf

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Diatomic linear chain

The boundary conditions are satisfied for a function

,ziqnAezf

if the wavevector qn takes the values

.2

N

n

aqn

For large values of N, qn is taken to be continuous and the n

subscript may be dropped.

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Diatomic linear chain

Using Hooke’s Law, the acceleration of mass m1 about

position z1,n is given by

.2

,

1,2,2,1

,1,21,2,1,11

nnSnS

nnSnnSn

uukuk

uukuukum

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PY3105 Introduction to

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Diatomic linear chain

Similarly, the acceleration of mass m2 about position z2,n is

given by

.2

,

1,1,1,2

,21,1,1,2,22

nnSnS

nnSnnSn

uukuk

uukuukum

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Diatomic linear chain

We look for solutions of the form

,

,

2,2

1,1

tqnai

n

tqnai

n

eAu

eAu

and again, for convenience, set

.and2

2

2

1

2

1m

k

m

k SS

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Diatomic linear chain

The first of the differential equations is then

,2 1,2,2

2

1,1

2

1,1

2

nnnn uuuu

.12 2

2

11

2

11

2 iqaeAAA

which, after dividing out the common exponential terms,

becomes

Similarly, the second equation becomes

.12 1

2

22

2

22

2 iqaeAAA

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Diatomic linear chain

We therefore obtain the characteristic equation

,021

122

2

22

2

2

1

2

1

2

iqa

iqa

e

e

which yields the quadratic equation

.0cos1242 2

2

2

1

2

2

2

1

22

2

2

1

4 qa

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PY3105 Introduction to

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Diatomic linear chain

This has the solution

.

cos1211

2/1

22

2

2

1

2

2

2

12

2

2

1

2

qa

Taking the boundaries of the 1D Brillouin zone to be – and

, we have therefore found the normal modes of the diatomic

linear chain as a function of wavevector q.

The graphs of against q are known as the dispersion

relations. The modes of the crystal are called phonons.

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Dispersion relations

Optical modes

Acoustic modes

Linear approximation for

acoustic branch

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Dispersion relations

• The dispersion relations follow two

branches:

• Acoustic branch, so-called because in a

typical material the frequencies are of the

order of audible sound

• Optical branch, so-called because in a

typical material the frequencies are

comparable with visible light

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Approximation for acoustic modes

.11

2/1

22

2

2

1

222

2

2

12

2

2

1

2

aq

For wavevectors close to zero, we may approximate the

cosine function as

.2

1cos22aq

qa

Inserting this into the expression for the acoustic branch of

2 (subtraction of the square root term) gives

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Approximation for acoustic modes

.

2

2/1

2

2

2

1

2

2

2

1 qa

Using the approximation

2

112

2/12 xx

and taking the square root of the approximate expression for

2 gives

The acoustic mode dispersion is therefore approximately

linear near q = 0 (shown as dashed line on graph).

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Zone boundaries

At the zone boundaries

.2,2

,4

11

2

2

2

1

2/1

22

2

2

1

2

2

2

12

2

2

1

2

,qa

Inserting this into the expression for 2

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Zone boundaries

Hence, the gap between the acoustic and optical branches at

the boundaries is

.2 12

2/1

We see that if m1 = m2 = m (i.e. a monatomic linear chain)

the branches join up.