Main Topics from Chapters 3-5 Due to time, not all topics will be on test. Some problems ask to discuss the meaning or implication. Lattice Dynamics (Monatomic,

Main Topics from Chapters 3-5Due to time, not all topics will be on test.

Some problems ask to discuss the meaning or implication.

Lattice Dynamics (Monatomic, Diatomic, Mass Defect, 2D Lattices)

Strain (compliance, reduced notation, tensors)Harmonic Oscillator (Destruction/Creation, Hamiltonian &

Number Operators, Expectation Values)Energy Density and Heat Capacity (phonons, electrons

and photons)Quasiparticle Interactions (e-e, e-phonon, e-photon,

defect interations)Electrical and Thermal Conductivity

Lattice Vibrations

Longitudinal Waves

Transverse Waves

When a wave propagates along one direction, 1D problem.Use harmonic oscillator approx., meaning amplitude vibration small.

The vibrations take the form of collective modes which propagate. Phonons are quanta of lattice vibrations.

The force on the nth atom;

)( 1 nn uuK

• The force to the right;

• The force to the left;

)( 1 nn uuK

The total force = Force to the right – Force to the left

a a

Un-1 Un Un+1

Eqn’s of motion of all atoms are of this form, only the value of ‘n’ varies

Monatomic Linear Chain

)2( 11

..

nnnn uuuKum

Thus, Newton’s equation for the nth atom is

11 nnnnn uuKuuKum

0expn nu A i kx t ..

2n nu u

Brillouin Zones of the Reciprocal Lattice

1st Brillouin Zone (BZ=WS)

k

a

a

2a

a

2 0

M

K4

a

3a

4a

3

a

4

2nd Brillouin Zone

3rd Brillouin Zone

Each BZ contains identical

information about the lattice

2p/a

Reciprocal Space Lattice:

There is no point in saying that 2 adjacent atoms are out of phase by more than (e.g., 1.2 =-0.8 )

Modes outside first Brillouin zone can be mapped to first BZ 2/sin2 0 kaq m

K0

Fig 4.43From Principles of Electronic Materials and

Devices, Third Edition, S.O. Kasap (© McGraw-Hill, 2005)

Four examples of standing waves in a linear crystal corresponding to q = 1, 2, 4, and N. q is maximum when alternating atoms are vibrating in

opposite directions. A portion from a very long crystal is shown.

Are These Waves Longitudinal or Transverse?

Diatomic Chain(2 atoms in primitive basis)2 different types of atoms of masses m1 and m2 are connected by identical springs

Un-2Un-1 Un Un+1 Un+2

K K K K

m1 m1m2 m2m a)

b)

(n-2) (n-1) (n) (n+1) (n+2)

a

Since a is the repeat distance, the nearest neighbors separations is a/2

Two equations of motion must be written; One for mass m1, and One for mass m2.

2,12111 2 nnnn uuuKum 1,11222 2 nnnn uuuKum

•As there are two values of ω for each value of k, the dispersion relation is said to have two branches

Upper branch is due to thepositive sign of the root.

Negative sign: k for small k. Dispersion-free propagation of sound waves

Optical Branch

Acoustical Branch

• This result remains valid for a chain containing an arbitrary number of atoms per unit cell.

0 л/a 2л/a–л/a k

wA

BC

2/1 22

221

222

21

22

21

2 2/sin4 qa

A when the two atoms oscillate in antiphase

• At C, M oscillates and m is at rest.• At B, m oscillates and M is at rest.

Number and Type of Branches• Every crystal has 3 acoustic

branches, 1 longitudinal and 2 transverse

• Every additional atom in the primitive basis contributes 3 further optical branches (again 2 transverse and 1 longitudinal)

2D Lattice

K

Ulm Ul+1,m

Ul,m-1

Ul,m+1

Ul-1,m

)( ,1 lmml uuK )( ,1 lmml uuK

)( 1, lmml uuK

)( 1, lmml uuK

• Write down the equation(s) of motion

)2()2( 1,1,,1,1

..

lmmlmllmmlmllm uuuKuuuKuM

What if I asked you to include second nearest neighbors with a different spring constant?

)( tamkalkiolm

yxeuu

2D Lattice

C

Ulm Ul+1,m

Ul,m-1

Ul,m+1

Ul-1,m

)2()2( 1,1,,1,1

..

lmmlmllmmlmllm uuuKuuuKuM

)coscos2(22 akakM

Kyx

Similar to the electronic bands on the test, plot w vs k for the [10] and

[11] directions.

Identify the values of at k=0 and at the edges.

Specific Heat or Heat Capacity

The heat energy required to raise the temperature a certain amount

The thermal energy is the dominant contribution to the heat capacity in most solids. In non-magnetic insulators, it is the only contribution.

Classical Picture of Heat CapacityWhen the solid is heated, the atoms vibrate around their sites like a set of harmonic oscillators.

Therefore, the average energy per atom, regarded as a 3D oscillator, is 3kT, and consequently the energy per mole is = 3 3BNk T RT

v

dC

dT

Dulong-Petit law: states that specific heat of any solid is independent of temperature and the same result(3R~6cal/K-mole) for all materials!

nn

nn

n

P

P _

Average energy of a harmonic oscillator and hence of a lattice mode at temperature T

Energy of oscillator

1

2n n

The probability of the oscillator being in this level as given by the Boltzman factor

exp( / )n Bk T

Thermal Energy & Heat Capacity Einstein Model

_0

0

1 1exp /

2 2

1exp /

2

Bn

Bn

n n k T

n k T

_

/

1

2 1Bk Te

Mean energy of a harmonic oscillator

Low Temperature LimitT

2

1

TkB

TkB

12

1_

TBke

2

1_

Zero Point

Energy

exponential term gets

bigger

..........!2

12

x

xex

Tke

B

TBk

1

112

1_

TkB

_ 1

2 Bk T

_

Bk T

High Temperature Limit

Bk T

is independent of frequency of oscillation. This is a classical limit because the energy steps are now small compared with thermal/vibrational energy

<<

Heat Capacity C (Einstein)Heat capacity found by differentiating average phonon energy

2

2

1

k TB

k TB

B

Bv

ke

k TdC

dTe

k

2

2

1

T

T

v B

eC k

T e

where

T(K)

Area =2

Bk

Bk

vC

The difference between classical and Einstein models comes from zero point energy.

Points:ExperimentCurve: EinsteinPrediction

The Einstein model near T= 0 did not agree with experiment, but was better

than classical model.Taking into account the distribution of

vibration frequencies in a solid this discrepancy can be accounted for.

1. Approx. dispersion relation of any branch by a linear extrapolation2. Ensure correct number of modes by imposing a cut-off frequency ,

above which there are no modes. The cut-off freqency is chosen to make the total number of lattice modes correct. Since there are 3N lattice vibration modes in a crystal having N atoms, we choose so that:

Debye approximation to the dispersion

vk

Debye approximation has two main steps

Einstein approximation to the dispersion

D

D

23

9( )

D

Ng

2 3 3 3 3

1 2 3 9( ) 3

2 L T D D

V N N

v v

0

( ) 3D

g d N

2

2 3 30

1 2( ) 3

2

D

L T

Vd N

v v

3

2 3 3

1 2( ) 3

6 DL T

VN

v v

2( ) /g

Density of states (DOS) per unit frequency range g()

• The number of modes/states with frequencies and +d will be g()d.

( ) ( )Sdn k dk g d # modes with wavenumber from k to k+dk=

dk

ddk

d( ) ( )Sg k

d

dk

2cos

2 2

a K ka

m2 sin

2

K ka

m

for 1D monoatomic lattice

1

cos2

K kaa

m

( ) ( )Sg k

( )gL

2 2max

2 1

a

2 2 2sin cos 1 cos 1 sinx x x x

The energy of lattice vibrations will then be found by integrating the energy of single oscillator over the distribution of vibration frequencies. Thus

/0

1

2 1kTg d

e

1/ 22 2max

2N

Mean energy of a harmonic oscillator

for 1D

It would be better to find 3D DOS in order to compare the results with experiment.

Debye Model adjusts Einstein Model

3D Example: The number of allowed states per unit energy range for free electron?

• Each k state represents two possible electron states, one for spin up, the other is spin down.

( ) 2 ( )g E dE g k dk ( ) 2 ( )dk

g E g kdE

2 2

2

kE

m 2dE k

dk m

2

2mEk

( )g E 2 ( )g kdk

dE2

22

V

kk

2

2mE

2

m

k

3/ 2 1/ 22 3

(2 )2

( )V

m Eg E

2

2( ) ,

2

Vkg k dk dk

L

L

L

Octant of the crystal:

kx,ky,kz(all have positive values)

The number of standing waves;

3

3 3 33s

L Vk d k d k d k

/L

k

dk

zk

yk

xk

214

8k dk

3 23

14

8s

Vk d k k dk

2

322s

Vkk d k dk

2

22S

Vkk

The Heat Capacity of a Cold Fermi Gas (Metal)

Close to EF, we can ignore the variation in the density of states: g() g(EF).

TkEgdT

TdUC BFe

22

3

dTfgTU ,0

By heating up a metal (kBT << EF), we take a group of electrons at the energy - (with respect to EF), and “lift them up” to . The number of electrons in this group g(EF)f()d and each electron increased its energy by 2 :

dTfEgUU F ,20

0

F

F E

nEg

2

3

F

BBe E

TkNkC

2

2

The small heat capacity of metals is a direct consequence of the Pauli principle. Most of the electrons cannot change their energy.

kBT

Bam!

Random Collisions

On average, I go about

seconds betweencollisions

with phonons and impurities

electron

phonon

Otherwise metals would have infinite conductivity

Electrons colliding with phonons (T > 0)

Electrons colliding with impurities

0Tphimp is independent of T

The thermal vibration of the lattice (phonons) will prevent the atoms from ever all being on their correct sites at the same time.The presence of impurity atoms and other point defects will upset the lattice periodicity

Fermi’s Golden Rule

fpfi iHfW 22

Transition rate:

i

ff ,

Quantum levels of the non-perturbed

system

Perturbation is applied

Transition is induced

r(E) is the ‘density of states available at energy E’.

• See Fermi‘s Golden Rule paper in Additional Material on the course homepage

Absorption

When the ground state finds itself in the presence of a photon of the appropriate frequency, the perturbing field can induce

the necessary oscillations, causing the mix to occur.

This leads to the promotion of the system to the upper energy state and the annihilation of the photon.

This process is stimulated absorption (or simply absorption).

Einstein pointed out that the Fermi Golden Rule correctly

describes the absorption process.

i

f

fi

iN

fif - degeneracy of state f

Quantum Oscillator

)(2

aam

x

0)(02

22 aa

mx

000000002

22 aaaaaa

m

Atoms still have energy at T=0.

What is <x2> for the ground state of the quantum harmonic oscillator?

mm 2

10002

(1D Case)

For 3D quantum oscillator, the result is multiplied by 3:

mx

2

32

M

GII

2exp

2

0⇒

• These quantized normal modes of vibration are called

PHONONS• PHONONS are massless quantum mechanical particles which

have no classical analogue.– They behave like particles in momentum space or k space.

• Phonons are one example of many like this in many different areas of physics. Such quantum mechanical particles are often called

“Quasiparticles”

Examples of other Quasiparticles:Photons: Quantized Normal Modes of electromagnetic waves.

Magnons: Quantized Normal Modes of magnetic excitations in magnetic solidsExcitons: Quantized Normal Modes of electron-hole pairs

Phonon spectroscopy =

Constraints:Conservation laws of

Momentum Energy

Conditions for: elastic scattering in

In all interactions involving phonons, energy must be conserved and crystal momentum must be conserved to within a reciprocal lattice vector.

Elastic anisotropy

klijklij c

Stress tensor , Compliance C, Stiffness S Message: Wave propagation in anisotropic media is quite

different from isotropic media:• There are in general 21 independent elastic constants (instead of 2 in the isotropic case), which can be reduced still further by considering the symmetry conditions found in different crystal structures. • There is shear wave splitting (analogous to optical birefringence, different polarizations in diff. directions)• Waves travel at different speeds depending in the direction of propagation

birefringenceDeformation tensor

)(2

1

i

j

j

iij x

u

x

u

o

cba

90

x=(a-b)/2 or

The cubic axes are equivalent, so the diagonal components for normal and shear distortions must

be equal.

And cubic is not elastically isotropic because a deformation along a cubic axis differs from the stress

arising from a deformation along the diagonal.e.g., [100] vs. [111]

2CC

C 121144

x

C

CC

CA 44

1211

442

Zener Anisotropy Ratio: