Yu-Jun Zhao

Department of Physics, SCUT

Lecture X

Phonons I. : Crystal Vibrations

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Vibrations of crystals with monatomic basis

Consider the elastic vibrations of a crystal with one atom in

the primitive cell.

Describe with a single coordinate us the displacement of the

plane s from its equilibrium position.

Three mode for each wavevector, one of longitudinal

polarization and two of transverse polarization.

longitudinal mode transverse mode

Note: for N atoms there are 3N mode, N L-modes and 2N T-modes.

The equation of the motion is:

)2( 112

2

ssss uuuC

dt

udM

a traveling wave solution:

us = u exp[i(sKa t)],

Then we have:

)cos1(2]2)exp()[exp(2 KaCiKaiKaCM

the dispersion relation (色散关系):

or

2sin

42/1

Ka

M

C

2sin

4)cos1)(/2( 22 Ka

M

CKaMC

Group velocity

The group velocity is the transmission velocity of a wave

packet, which is also the velocity of energy propagation

in the medium.

dimension onein /

dimension in three )(

dKdv

Kv

g

Kg

With the particular dispersion relation

2sin

42/1

Ka

M

C

2cos

2/12 Ka

M

Cavg

Q: Does the velocity of “sound” depend on its frequency?

Long wavelength limit

When >> a i.e. Ka << 1

22

22

2sin

4K

M

CaKa

M

C

KaM

C2/1

i.e.

aM

CdKdvg

2/1

/

In this limit, the velocity of sound is independent of frequency.

Derivation of force constants from experiment

In metals the effective forces may be of quite long range,

carried from ion to ion through the conduction electron sea.

Considering p nearest planes, the force on s plane

)(20

1

s

p

j

jsj uuCF

The equation of motion is

0

2

2

)(2p

j

sjsjs uuC

dt

udMThe dispersion relation

0

1

2 )cos1(2 p

j

j jKaCM

a

aKp pKadK

MaC

/

/

2 cos2

By multiplying both sides

by cos 𝑝𝐾𝑎 , and

integrating we may get:

Two atoms per primitive basis

Assume that each plane interacts only with the nearest-

neighbor planes. The equations of motion is

)2(

)2(

12

2

2

12

2

1

ssss

ssss

vuuCdt

vdM

uvvCdt

udM

the solution in a form of a traveling wave:

)](exp[

)](exp[

tsKaivv

tsKaiuu

s

s

substitute the solution in the equations

CviKaCuvM

CuiKaCvuM

2)]exp(1[

2)]exp(1[

2

2

1

2

The homogenous linear equations have a solution only if the

determinant of the coefficients of the unknown u, v vanishes.

02 )]exp(1[

)]exp(1[ 2

2

2

2

1

MCiKaC

iKaCMC

or 0)cos1(2)(2 22

21

4

21 KaCMMCMM

the dispersion relation

)cos1()(

211

cos2)(

2

21

21

21

21

21

2

2

2

121

21

2

KaMM

MM

MM

MMC

KaMMMMMMMM

C

at long wavelength limit (Ka << 1)

branch) l(acoustica )(2

branch) (optical 11

2

2

21

22

21

2

KMM

Ca

MMC

1

2

M

M

v

uFor the optical branch

The atoms vibrate against each other, but their center

of mass is fixed.

vu For the acoustical branch

The atoms and their center of mass move together.

at the 1st Brillouin zone boundary (Ka = )

2

2

1

2 /2 ;/2 MCMC

There is a frequency gap at boundary of the 1st Brillouin zone.

With p atoms in the primitive cell and N primitive cells,

there are pN atoms, 3pN degrees of freedom, N LA branches,

2N TA branches, (p1)N LO branches and (2p2)N TO

branches.

If there are p atoms in the primitive cell, there are 3p

branches to the dispersion relation: 3 acoustical branches

(1 LA and 2 TA) and 3p3 optical branches (p1 LO and

2p2 TO ).

The dispersion relation shows new features in crystals

with two or more atoms per primitive basis.

acoustical and optical branches (声学支和光学支)

Q: What is the vibration mode corresponding to𝐾 ≪

𝜋

𝑎, 𝐾 =

𝜋

𝑎?

)](exp[

)](exp[

tsKaivv

tsKaiuu

s

s

The dispersion relation shows new features in crystals

with two or more atoms per primitive basis.

acoustical and optical branches (声学支和光学支)

Q: What is the vibration mode corresponding to𝐾 ≪

𝜋

𝑎, 𝐾 =

𝜋

𝑎?

Quantization of the elastic waves

The energy of a lattice vibration is quantized. The quantum

of the energy is called phonon in analogy with the photon

of the electromagnetic wave.

Thermal vibrations in crystals are thermally excited phonons.

The energy of the elastic mode

)2

1( n

this mode occupied by n phonons with a frequency

the zero point energy 2

1

The amplitude of the elastic vibration

)/()2

1(42

0 Vnu

Consider a standing wave mode

tKxuu coscos0

The time average kinetic energy is

)2

1(

2

1

8

1 2

0

2 nuVE

tKxuMt

uM 222

0

2

2

sincos2

1

2

1

The kinetic energy of an atom is

What is the sign of ?

The energy of a phonon must be positive.

It is conventional and suitable to view as positive.

A mode with imaginary (negative 2) will be unstable.

Phonon momentum

A phonon of wavevector K will interact with particles

such as photons, neutrons, and electrons as if it has a

momentum .K

However a phonon does not carry physical momentum.

For most practical purpose a phonon acts as if its momentum

were , which is called the crystal momentum.K

In crystal there exists wavevector selection rules for allowed

transition between quantum states.

The true momentum of the whole system always is

rigorously conserved.

For the elastic scattering of a photon by a crystal, the

wavevector selection rule is

Gkk

'

GkKk

'

If the scattering of the photon is inelastic, with the creation

of a phonon , the wavevector selection rule becomesK

If a phonon is absorbed in the inelastic scattering, the

wavevector selection rule becomes

K

GKkk

'

Inelastic scattering by phonons

Phonon dispersion relations (K) are most often determined

by inelastic scattering of neutrons with the emission or

absorption of a phonon.

A neutron sees the crystal lattice chiefly by interaction with

the nuclei of the atoms.

The wavelength and the energy of the slow neutron is just

within the order of phonons.

The general wavevector selection rule is:

KkGk

'

The statement of conservation of energy is:

nn M

k

M

k

2

'

2

2222

Where are the wavevector and the energy of a

phonon created (+) or absorbed () in the process.

and K

22

3D Example: Normal modes of Silicon vs. Lead

L — longitudinal

T — transverse

O — optical

A — acoustic

8.828THz

2.245THz

r

Si

Si

M

M

Si

LOA(X) LTO(Gamma)

Pb

Physical processes related to neutron scattering

A new neutron source (CSNC) will be set up at DongGuan in

two years.

• Repetition rate ： 25 Hz

•Average proton current： 62.5 A

• Proton kinetic energy： 1.6 GeV

•Average beam power： 100 kW

• Target: Tungsten

• Moderators: H2O,LCH4 ,LH2

• Spectrometers:

HRPD

HIPD

Reflectometer,

SANS

Chopper spectrometer

Homework

2018/4/7

1. Problem 4.1 of textbook.

2. Problem 4.2 of textbook

3. Problem 4.5 of textbook.

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