Chapter 3. Lattice Waves

Lattice waves• Lattice waves

: Vibrational motion of the atoms in a crystalline solid in terms of a wave passing through the atoms of the crystal as they are displaced by their thermal energy from their rest positions.

• The thermal properties of solids are strongly related to the lattice waves

• The movement of electrons (mobility) are hindered due to scattering by lattice waves.

• Lattice waves have their particle-like counterpart, called phonons: quanta of energy ħωn, ωn: normal vibrational modes

• Energy exchanging interactions with lattice waves occur in integral multiple of ħωn.

Lattice waves• Two examples:

1. Vibrations associated with a one-dimensional crystal in which all the atoms have the same mass and the same atomic spacing.

“Acoustic modes” : long wavelength longitudinal vibration corresponds to the sound wave.

2. Vibrations with two or more different kinds of atoms in a one-dimensional crystals.① Two different masses with a common atomic spacing

② Two different atomic spacings for atoms with the same mass

𝑚

𝑎

𝜉𝑟-1 𝜉𝑟 𝜉𝑟+1

𝑟-1 𝑎 𝑟𝑎

displacement

position𝑟+1 𝑎

“Optical modes” : long wavelength transverse vibrations characterized by neighboring atoms being displaced in opposite directions. The long wavelength vibration can be excited by interaction with light if the material is at least partially ionic.

• Transverse waves in a one dimensional infinite string

𝜉: displacement away from the x-axis

Transverse waves in a 1-D infinite lattice

-11up

+11downward

Force at

: ~

: ~

r rrr

r rrr

x ra

F F Fa

F F Fa

𝑚

𝑎

𝜉𝑟-1 𝜉𝑟 𝜉𝑟+1

𝑟-1 𝑎 𝑟𝑎

displacement

position𝑟+1 𝑎

Assumption 1. Restrict the forces between nearest neighbor atoms 2. The force is an attractive force 𝐹 3. 𝐹 is constant and in the direction of the nearest neighbor atoms

(Assumption: 𝜉 ≪ 𝑎

Transverse waves in a 1-D infinite lattice

• The harmonic solution

: mathematical wave passing through the displaced atoms.

Such a wave has physical reality only at the locations of atoms, i.e., only at x=ra. Then,

𝜉 𝑥, 𝑡 𝐴 exp 𝑖 𝑘𝑥 𝜔𝑡

𝜉 𝑟𝑎, 𝑡 𝐴 exp 𝑖 𝑘𝑟𝑎 𝜔𝑡

O

∴ The net upward force on the atom at 𝑥 𝑟𝑎

𝐹 𝐹 𝑚𝑑 𝜉𝑑𝑡 ⇒

𝑑 𝜉𝑑𝑡 𝜂𝜉 2𝜂𝜉 𝜂𝜉 where 𝜂 𝐹/𝑚𝑎

𝜉 𝐴𝑒 𝜉𝜉 𝐴𝑒 𝜉

-11 ~ r rrrF F

a

+11 ~ r rrrF F

a

• Dispersion relationship

Lattice wave has a dispersive system: The velocity varies with frequency and wave length Reducible to 0 ≤ k ≤ π/a (The first Brillouin zone)

𝜔 4𝜂 sin 𝑘 𝑎/2 ⇒ 𝜔 2𝜂 / sin 𝑘 𝑎/2

Transverse waves in a 1-D infinite lattice

𝜔 2𝜂 𝜂 𝑒 𝑒

2𝜂 1 cos 𝑘 𝑎

The shortest wavelength

2, (1 cos 2 ) 2sincf

/F ma

Dispersion relation between 𝜔 and 𝑘

Transverse waves in a 1-D infinite lattice

𝜉′ 𝐴𝑒

𝐴𝑒 𝐴𝑒 / ⋅ 𝑒 𝜉𝑒 ⋅ / ⋅ 𝜉 exp 𝑖𝑛2𝜋𝑟 𝜉

𝜔 2𝜂 / sin 𝑘 𝑎/2

sin 𝑘 𝑎/2 → 𝑘𝑎/2

𝑣~

𝑣~

𝜔𝑘 𝜂 / 𝑎 𝐹𝑎/𝑚 /

small 𝑘 region velocity becomes constant

cf) Displacement is identical for any 𝑘 and 𝑘 𝑘

For small 𝑘 (long wavelength)

Transverse waves in a 1-D infinite lattice• Note:

① 𝑘 space is the reciprocal lattice②

𝜔max 2 𝐹/𝑚𝑎Debye frequency

→ 2 a

Infinite wavelength

All atoms displaced by the same amount in the same direction.

• Neighboring atoms are displaced by the same distance in opposite directions.

• The shortest wavelength; • The dashed wave (n>2) has shorter wavelength. However it does not

give any new information on the position of atoms.• Equivalent to the Bragg reflection condition

• Cannot propagate: group velocity at k=/a is equivalent to zero

Acoustical Branch

𝜆 2𝑎

𝑛𝜆 2𝑑 sin 𝜃 𝜆 2𝑎, 𝑛 1, 𝑑 𝑎

(∵ 𝜔 2𝜂 / sin 𝑘 𝑎/2 )

Transverse waves in a 1-D finite lattice• General solution for transverse waves

1, 2, , ( 1)mk m nL

𝜔 𝜔 sin𝑚𝜋𝑎

2𝐿

𝜔 sin𝑚𝜋𝑎

2 𝑛 1 𝑎 𝑛: # of moving atoms

∴ 𝜔 𝜔 sin𝑚𝜋

2 𝑛 1 : normal modes

sin( / ( 1))The general solution for the atom at is

i trm m

r m rmm

A m r n ex ra

A

Finite set of discrete (,k) values

𝜉 𝑥, 𝑡 𝐴 exp 𝑖 𝑘𝑥 𝜔𝑡 𝐵 exp 𝑖 𝑘𝑥 𝜔𝑡

𝑛 2 : total number of atoms

(boundary conditions)

(allowed frequencies)

(displacement at 𝑥 𝑟𝑎 and for 𝜔 )

𝜉 0 𝜉 𝐿 0

Longitudinal waves in a 1-D infinite lattice

• The restoring force for longitudinal displacement depends on the spatial variation of the force (F) between atoms

• Let F(a) represent the force between atoms when separated by a normal lattice spacing (a), then the net force on the rth

atom is

𝑟-1 𝑎 𝑟𝑎 𝑟+1 𝑎

𝜉𝑟-1 𝜉𝑟 𝜉𝑟+1

𝐹 𝐹 𝑎 𝜉 𝜉 𝐹 𝑎 𝜉 𝜉

𝐹 𝑎 𝜉 𝜉 𝐹 𝑎 𝜉 𝜉𝑑𝐹𝑑𝜉 ⋯

𝐹 𝑎 𝜉 𝜉 𝐹 𝑎 𝜉 𝜉𝑑𝐹𝑑𝜉 ⋯

For very small displacement

∴ 𝐹 𝜉 2𝜉 𝜉 1𝑑𝐹𝑑𝜉 𝑚

𝑑 𝜉𝑑𝑡

Longitudinal waves in a 1-D infinite lattice

𝐹 𝜉 2𝜉 𝜉𝑑𝐹𝑑𝜉 𝑚

𝑑 𝜉𝑑𝑡

Let 𝜂′1𝑚

𝑑𝐹𝑑𝜉

Then 𝑑 𝜉𝑑𝑡 𝜂′𝜉 2𝜂′𝜉 𝜂′𝜉

L: longitudinal acoustic waveT1, T2: transverse acoustic wavesT1=T2 for isotropic crystal structure

long λ longitudinal wave ≡ sound waves

3D

Crystallographic direction

Freq

uenc

y (T

Hz)

Q: Why are the frequencies for L greater than T?

𝑑 𝜉𝑑𝑡 𝜂𝜉 2𝜂𝜉 𝜂𝜉

simiar to transverse waves except for 𝜂 → 𝜂′

Longitudinal waves in a 1-D infinite lattice

• Long wavelength longitudinal wave ≡ sound waves• Velocity is given by the slope at k=0

𝑣𝜔𝑘 ~

𝜂 / 𝑎

𝑣𝜔𝑘 ~

𝜂′ / 𝑎

𝐹 ∝ 𝑟

𝑣 𝑛𝜂 𝑎 𝑛 𝑣

𝜂1𝑚

𝑑𝐹𝑑𝜉 ,

The longitudinal waves are 𝑛 times faster than transverse waves.

𝜂′ 𝑛𝜂𝜂𝐹

𝑚𝑎 ,

Density of states for lattice waves• Density of states: # of allowed vibrational modes, N(), per

unit frequency interval, d.

max

max

1/22

max

max

sin2( +1)

cos2( +1) 2( +1)

1 : frequency spacing2( +1)

mn

d mdm n n

n

1/22

max max

2( +1)( ) 1nN d d

In a frequency interval d, there are d/ states. Therefore,

N(v) starts with a value of [2(n+1)/vmax] at v=0 and then increases with increasing v to a large value as v approaches vmax.

cf) cos 𝜃 = 1 sin 𝜃

between allowed modes

From 𝜔 2𝜂 / sin

Lattice waves for two kinds of atoms

•① Lattice parameter is the same (a)② Masses are different (m and M)

ex) compound

•① Lattice parameters are different (a and b)② Mass is the same (m)

ex) more atoms in a unit cell

m M

a b

The first case (different masses & same spacing)

• Follow the same procedure of transverse wave except that near atoms are different kinds.

m M

Zr-1 Zr Zr+1𝜉𝑟-1 𝜉𝑟 𝜉𝑟+1

(2r-1)a 2ra (2r+1)a r=0, 1, 2, ⋯

Ⅰ. 𝑑 𝜉𝑑𝑡 𝜂 𝑍 2𝜂 𝜉 𝜂 𝑍 where 𝜂

𝐹𝑀𝑎

Ⅱ. 𝑑 𝑍𝑑𝑡 𝜂 𝜉 2𝜂 𝑍 𝜂 𝜉 where 𝜂

𝐹𝑚𝑎

𝜉 𝐴𝑒 for 𝑥 2𝑟 1 𝑎𝑍 𝐵𝑒 for 𝑥 2𝑟𝑎

• Assume that harmonic waves with the same values of 𝑘 and 𝜔 for both types of atoms

M M M

M M M

2 2

2

2 02 0

ika ika ika

ika ika

e A e A B e BA A e B e B

𝜉 𝐴𝑒 𝑒 𝐴𝑒

M M

m m

2

2

( 2 ) 2 cos 02 cos ( 2 ) 0

A B kaA ka B

M m M m M m

M m M m

4 2 2

4 2 2

(2 2 ) 4 4 cos 02( ) 4 sin 0

kaka

𝜔 𝜂 𝜂 𝜂 𝜂 4𝜂 𝜂 sin 𝑘 𝑎 (dispersion relations)

M M

m m

2

2

2 2 cos0

2 cos 2ka

ka

: two separate branches in the vibration spectrum

𝐴𝐵

2𝜂 cos 𝑘 𝑎𝜔 2𝜂

for 𝑥 2𝑟 1 𝑎

for 𝑥 2𝑟𝑎𝑍 𝐵𝑒

𝜉 𝐴𝐵 𝑒 𝑍

𝑑 𝜉𝑑𝑡 𝜂 𝑍 2𝜂 𝜉 𝜂 𝑍

𝑑 𝑍𝑑𝑡 𝜂 𝜉 2𝜂 𝑍 𝜂 𝜉

The first case (different masses & same spacing)

1) For k = 0

< optical mode >

< acoustic mode >

ω+

ω-

The first case (different masses & same spacing)

𝐴𝐵

2𝜂 cos 𝑘 𝑎𝜔 2𝜂

2𝜂2𝜂 1

∵ 1 𝑥 / 1 𝑥/2 for 𝑥<<1)

<<1

From previous solution,

𝜔 𝜂 𝜂 𝜂 𝜂 14𝜂 𝜂 𝑘 𝑎

𝜂 𝜂

/

𝜔 𝜂 𝜂 𝜂 𝜂 12𝜂 𝜂 𝑘 𝑎

𝜂 𝜂

𝜂 𝜂 𝜂 𝜂

2𝜂 𝜂 𝑘 𝑎

𝜂 𝜂

∵ cos 𝑘𝑎 1 and 𝜔 0 at k = 0

:Equal displacement of neighboring atoms

𝐴𝐵

𝜂𝜂

𝜔 2 𝜂 𝜂

a)

b)

:Opposite displacement: In the long wavelength mode(k=0), neighboring atoms are displaced in opposite directions.

2) Near k = π / 2a, sin2ka → 1

𝜂 𝜂 𝜂 𝜂 4𝜂 𝜂𝜂 𝜂 𝜂 𝜂

① 𝜔 2𝜂

𝐴𝐵 0

② 𝜔 2𝜂

𝐴𝐵 ∞

gap opens for m≠M

𝐴𝐵

2𝜂 cos 𝑘 𝑎𝜔 2𝜂

The first case (different masses & same spacing)

Large mass

Small mass

AB

AB

𝜔 𝜂 𝜂 𝜂 𝜂 4𝜂 𝜂 sin 𝑘 𝑎

𝜔2𝜂 𝜂 𝑘 𝑎

𝜂 𝜂

𝜔 2 𝜂 𝜂 𝜔 2𝜂

𝜔 2𝜂

k = π / 2ak = 0

Vibration spectrum pf CdTe

LO: Longitudinal optical vibrationTO: Transverse optical vibration

LA: Longitudinal acoustic vibrationTA: Transverse acoustic vibration

An EM wave that propagates the lattice displaces the oppositely charged ions in opposite directions and forces them to vibrate at the frequency of the wave. Most of the energy is then absorbed from the EM wave and converted to lattice vibrational energy (heat).

Reststrahlen absorption: (German: residual rays) • long wavelength transverse modes in partially ionic crystals could be

directly excited by light of a suitable.• Strong interaction between a light wave and a lattice wave under the

unusual conditions for resonance.extinction coefficient K versus wavelength