Solid State Physics - جامعة ذي قارsci.utq.edu.iq/images/pdf/dr1.pdf · monoatomic linear chain with only nearest neighbour interaction a a U n-1 U n U n+1 Monoatomic Chain

Solid State Physics

Lecturer: Dr. Lafy Faraj

CHAPTER 1

Phonons and Lattice vibration

Crystal Dynamics

• Atoms vibrate about their equilibrium position at absolute zero.

• The amplitude of the motion increases as the atoms

gain more thermal energy at higher temperatures.

• In this chapter we discuss the nature of atomic motions, sometimes referred to as lattice vibrations.

• Atomic motions are governed by the forces exerted on atoms when they are displaced from their equilibrium positions.

3 Dr. Lafy Fraj AL-Badry

Hooke's Law

• One of the properties of elasticity is that it takes about twice as much force to stretch a spring twice as far. That linear dependence of displacement upon stretching force is called Hooke's law.

FSpring constant K

It takes twice as much force to stretch a spring twice as far.

F2

𝐹𝑠𝑝𝑟𝑖𝑛𝑔 = −𝐾. 𝑥


Monoatomic Chain

• The simplest crystal is the one dimensional chain of identical atoms.

• Chain consists of a very large number of identical atoms with identical masses.

• Atoms are separated by a distance of “a”. • Atoms move only in a direction parallel to the chain. • Only nearest neighbours interact (short-range forces).

a a a a a a

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


• Start with the simplest case of monoatomic linear chain with only nearest neighbour interaction

a a

Un-1 Un Un+1

Monoatomic Chain

The force on the nth atom;

)( 1 nn uuK

•The force to the right;

•The force to the left;

)( 1 nn uuKHooks law F=-Kx

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


• All atoms oscillate with a same amplitude A and frequency ω. Then we can offer a solution;

Monoatomic Chain

0expn nu A i kx t

..2

n nu u

naxn 0

nn unax Undisplaced

position

Displaced

position

0)2( 11

..

nnnn uuuKum

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

H.W

since 𝐹 = 𝑚𝑑2𝑢

𝑑𝑡2= 𝑚𝑢 𝑛 Where (m) is mass of atom

Where (k) is wavevector


Equation of motion for nth atom

By using We get

By using

We get

𝑐𝑜𝑠𝑥 =𝑒𝑖𝑥 + 𝑒−𝑖𝑥

2 𝑚𝜔2 = 2𝐾(1 − 𝑐𝑜𝑠𝑘𝑎)

1 − 𝑐𝑜𝑠𝑥 = 2𝑠𝑖𝑛2𝑥

2

𝜔 =4𝐾

𝑚𝑠𝑖𝑛

𝑘𝑎

2

Maximum value of it is 1

𝜔𝑚𝑎𝑥 =4𝐾

𝑚 • The maximum allowed frequency is:

𝑚𝑢 𝑛 = 𝐾(𝑢𝑛+1 − 2𝑢𝑛 + 𝑢𝑛−1)

−𝑚𝜔2𝐴𝑒𝑖 𝑘𝑛𝑎−𝜔𝑡 = 𝐾 𝐴𝑒𝑖 𝑘(𝑛+1)𝑎−𝜔𝑡 − 2𝐴𝑒𝑖 𝑘𝑛𝑎−𝜔𝑡 + 𝐴𝑒𝑖 𝑘(𝑛−1)𝑎−𝜔𝑡

By substituting the solution in eq. of motion

When 𝑘𝑎

2=

𝜋

2 𝑘 = ±

𝜋

𝑎


“Phonon Dispersion Relations” or Normal Mode Frequencies or ω versus k relation for the monatomic chain.

max 2

/s

K

m

V k

0 л / a 2 л / a –л / a k

Because of Brillouin zone periodicity with a period of 2π/a,

only the first BZ is needed. Points A, B & C correspond to the

same frequency, so they all have the same instantaneous atomic

displacements.

k

C A B

0


• Briefly look in more detail at the group velocity, vg.

• The dispersion relation is:

• So, the group velocity is:

vg (dω/dk) = a(K/m)½cos(½ka) (H.W)

vg = 0 at the BZ edge [k = (π/a)]

– This tells us that a wave with λ corresponding to a zone

edge wavenumber k = (π/a) will not propagate.

That is, it must be a standing wave!

4sin

2

K ka

m

1st BZ

Edge

𝑣 𝑔

𝐾𝑎2/𝑚

1/2

𝑘 10

Dr. Lafy Fraj AL-Badry

First Brillouin Zone

What range of k is physically significant for elastic waves?

The ratio of the displacements of two successive planes is given by

𝑢𝑛+1𝑢𝑛

=𝑢𝑒𝑖 𝑛+1 𝑘𝑎

𝑢𝑒𝑖𝑛𝑘𝑎= 𝑒𝑖𝑘𝑎

The range -π → +π for (𝑘𝑎) covers all independent values of 𝑒𝑖𝑘𝑎

−𝜋

𝑎< 𝑘 ≤

𝜋

𝑎 First Brillouin Zone

If k outside these limits, we use formula: 𝑘′ = 𝑘 − 2𝜋𝑛/𝑎

𝑢𝑛+1

𝑢𝑛= 𝑒𝑖𝑘

′𝑎 𝑤ℎ𝑒𝑟𝑒 (𝑒𝑖2𝜋𝑛 = 1) Then displacement ratio H.W

2𝜋𝑛/𝑎 is reciprocal lattice vector 11


Since there is only one possible propagation direction and one polarization direction, the 1D crystal has only one sound velocity.

In this calculation we only take nearest neighbor interaction although this is a good approximation for the inert-gas solids, its not a good assumption for many solids.

If we use a model in which each atom is attached by springs of different spring constant to neighbors at different distances many of the features in above calculation are preserved.

• Wave equation solution still satisfies.

• The detailed form of the dispersion relation is changed but ω is still periodic function of k with period 2π/a

• Group velocity vanishes at k=(±)π/a

• There are still N distinct normal modes

• Furthermore the motion at long wavelengths corresponds to sound waves with a velocity given by (velocity formulü)

Monoatomic Chain


Chain of two types of atom

• Two different types of atoms of masses M and m are connected by identical springs of spring constant K;


K K K K

M M m M m a)

b)

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

a

• This is the simplest possible model of an ionic crystal.

• Since a is the repeat distance, the nearest neighbors

separations is a/2


• We will consider only the first neighbour interaction although it is a poor approximation in ionic crystals because there is a long range interaction between the ions.

• The model is complicated due to the presence of two different types of atoms which move in opposite directions.

Our aim is to obtain ω-k relation for diatomic lattice


Two equations of motion must be written;

One for mass M, and

One for mass m.



M m M m M


𝑀𝑢 𝑛 = 𝐾 𝑢𝑛+1 − 𝑢𝑛 − 𝐾 𝑢𝑛 − 𝑢𝑛−1

𝑀𝑢 𝑛 = 𝐾 𝑢𝑛+1 − 2𝑢𝑛 + 𝑢𝑛−1

𝑚𝑢 𝑛−1 = 𝐾 𝑢𝑛 − 𝑢𝑛−1 − 𝐾 𝑢𝑛−1 − 𝑢𝑛−2

𝑚𝑢 𝑛−1 = 𝐾 𝑢𝑛 − 2𝑢𝑛−1 + 𝑢𝑛−2

Equation of motion for mass M (nth):

mass x acceleration = restoring force

Equation of motion for mass m (n-1)th:



M m M m M


0 / 2nx na

0expn nu A i kx t

Offer a solution for the mass M

For the mass m;

α : complex number which determines the relative amplitude and phase of the vibrational wave.

0expn nu A i kx t

𝑢𝑛 = 𝐴𝑒𝑥𝑝 𝑖 𝑘𝑥𝑛 −𝜔𝑡 𝑥𝑛= 𝑛𝑎/2

𝑢𝑛−1 = 𝛼𝐴𝑒𝑥𝑝 𝑖 𝑘𝑥𝑛 − 𝜔𝑡

𝑢 𝑛 = −𝜔2𝐴𝑒𝑥𝑝 𝑖 𝑘𝑥𝑛 −𝜔𝑡

𝑢 𝑛−1 = −𝜔2𝛼𝐴𝑒𝑥𝑝 𝑖 𝑘𝑥𝑛 −𝜔𝑡 16


Equation of motion for 𝑛𝑡ℎ atom (M):

𝑀𝑢 𝑛 = 𝐾 𝑢𝑛+1 − 2𝑢𝑛 + 𝑢𝑛−1

−𝜔2𝑀𝐴𝑒𝑖𝑘𝑛𝑎2 −𝜔𝑡

= 𝐾 𝛼𝐴𝑒𝑖𝑘(𝑛+1)𝑎

2 −𝜔𝑡− 2𝐴𝑒

𝑖𝑘𝑛𝑎2 −𝜔𝑡

+ 𝛼𝐴𝑒𝑖𝑘(𝑛−1)𝑎

2 −𝜔𝑡

𝜔2𝑀 = 2𝐾 1 − 𝛼𝑐𝑜𝑠𝑘𝑎

2

By substituting the solution:

Cancel the common terms

H.W

where cosx =𝑒𝑖𝑥 + 𝑒−𝑖𝑥

2


Equation of motion for the (𝑛 − 1)𝑡ℎ atom (m)

𝑚𝑢 𝑛−1 = 𝐾 𝑢𝑛 − 2𝑢𝑛−1 + 𝑢𝑛−2

−𝛼𝐴𝜔2𝑚𝑒𝑖𝑘 𝑛−1 𝑎

2 −𝜔𝑡= 𝐾 𝐴𝑒

𝑖𝑘𝑛𝑎2 −𝜔𝑡

− 2𝛼𝐴𝑒𝑖𝑘 𝑛−1 𝑎

2 −𝜔𝑡+ 𝐴𝑒

𝑖𝑘 𝑛−2 𝑎

2 −𝜔𝑡

𝛼𝜔2𝑚 = 2𝐾 𝛼 − 𝑐𝑜𝑠𝑘𝑎

2

By substituting the solution:

Cancel the common terms

H.W


• Now we have a pair of algebraic equations for α and ω as a function of k.


for m

for M

𝛼𝜔2𝑚 = 2𝐾 𝛼 − 𝑐𝑜𝑠𝑘𝑎

2

𝜔2𝑀 = 2𝐾 1 − 𝛼𝑐𝑜𝑠𝑘𝑎

2

𝛼 =2𝐾𝑐𝑜𝑠(𝑘𝑎 2)

2𝐾 − 𝜔2𝑚=

2𝐾 − 𝜔2𝑀

2𝐾𝑐𝑜𝑠(𝑘𝑎 2)

A quadratic equation for ω2 can be obtained by cross-multiplication

α can be found as:

𝜔4 − 2𝐾𝑚 +𝑀

𝑚𝑀𝜔2 + 4𝐾2

𝑠𝑖𝑛2(𝑘𝑎 2)

𝑚𝑀= 0

The two roots are:

𝜔2 = 𝐾𝑚 +𝑀

𝑚𝑀± 𝐾

𝑚 +𝑀

𝑚𝑀

2

− 4𝑠𝑖𝑛2(𝑘𝑎 2)

𝑚𝑀

1/2

𝑥1,2 =−𝑏 ± 𝑏2 − 4𝑎𝑐

2𝑎

We get:

H.W

Dispersion relation


• 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 the

+ve sign of the root.

Lower branch is due to the

-ve sign of the root.

Optical Branch

Acoustical Branch

• The dispersion relation is periodic in k with a period

2 π /a = 2 π /(unit cell length).

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

0 л / a 2 л / a –л / a k

A

B

C


Let’s examine the limiting solutions at 0, A, B and C.


𝑚𝑀±𝐾

𝑚 +𝑀

𝑚𝑀

2

− 4𝑠𝑖𝑛2(𝑘𝑎 2)

𝑚𝑀

1/2

𝜔2 ≈ 𝐾𝑚 +𝑀

𝑚𝑀±𝐾

𝑚+𝑀

𝑚𝑀

2

−(𝑘𝑎)2

𝑚𝑀

1/2


𝑚𝑀1 ± 1 −𝑚𝑀

𝑘𝑎

𝑚 +𝑀

2 1/2

Use Taylor expansion: (1 − 𝑥)1/2≈ 1 −1

2𝑥 for x<<1


𝑚𝑀1 ± 1 −

1

2𝑚𝑀

𝑘𝑎

𝑚 +𝑀

2

1- In long wavelength region

In this case (ka<<1); sin(ka/2)≈ ka/2


Taking +ve root; in case ka<<1

𝜔𝑜𝑝2 ≈ 2𝐾

𝑚 +𝑀

𝑚𝑀

(max value of optical branch)

(min value of acoustical brach)

Taking -ve root; in case ka<<1

𝜔𝑎𝑐2 ≈

1

2𝐾(𝑘𝑎)2

𝑚 +𝑀

By substituting these values of ω in α (relative amplitude)

equation and using cos(ka/2) ≈1 for ka<<1 we find the

corresponding values of α for acoustical branch as;

𝛼 =2𝐾𝑐𝑜𝑠(𝑘𝑎 2)

2𝐾 − 𝜔2𝑚

𝛼 = 1 acoustical branch H.W

H.W

H.W



k

A

B

C

Optical

Acoustical

0 π / a 2 π / a –π / a

The relative amplitude α=1 of acoustical branch represents long-wavelength sound

waves in the neighborhood of point 0 in the graph; the two types of atoms oscillate

with same amplitude and phase.

and the velocity of sound is

𝑣𝑠 =𝜔

𝑘 𝑣𝑠 = a

𝐾

2(𝑀 +𝑚)



0 π / a 2 π / a –π / a

k

A

B

C

Optical

Acoustical

The relative amplitude α=-M/m of optical branch, This solution corresponds to point

A in dispersion graph. This value of α shows that the two atoms oscillate in

antiphase (the two atoms are oscillating 180º out of phase) with their center of

mass at rest.

𝛼 = −𝑀

𝑚 Optical branch

For optical branch

We substitute 𝜔𝑜𝑝2 ≈ 2𝐾

𝑚 +𝑀

𝑚𝑀

in equation 𝛼 =2𝐾𝑐𝑜𝑠(𝑘𝑎 2)

2𝐾 − 𝜔2𝑚

We get H.W



2- At Brillouin Zone boundary

• At max.acoustical point C, M oscillates and m is at rest.

• At min.optical point B, m oscillates and M is at rest.


𝑚𝑀±𝐾

𝑚 +𝑀

𝑚𝑀

2

− 4𝑠𝑖𝑛2(𝑘𝑎 2)

𝑚𝑀

1/2


𝑚𝑀±𝐾

𝑚 +𝑀

𝑚𝑀

2

−4

𝑚𝑀

1/2

𝜔2 =𝐾

𝑚𝑀(𝑀 +𝑚) ± (𝑀 −𝑚)

𝜔𝑎𝑐2 =

2𝐾

𝑀 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 (𝐶) Taking -ve root; (max value of acoustical brach)

Taking +ve root; 𝜔𝑜𝑝2 =

2𝐾

𝑚 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 (𝐵) (min value of optical brach)

In this case ka= π , i.e sin(ka/2)=1.

0 π/a 2π/a–π/a

k

AB

C

Optical

Acoustical

𝜔𝑎𝑐 =2𝐾

𝑀

𝜔𝑜𝑝 =2𝐾

𝑚


Acoustic/Optical Branches

• The acoustic branch has this name because it gives rise to long wavelength vibrations - speed of sound.

• The optical branch is a higher energy vibration (the frequency is higher, and you need a certain amount of energy to excite this mode). The term “optical” comes from how these were discovered - notice that if atom 1 is +ve and atom 2 is -ve, that the charges are moving in opposite directions. You can excite these modes with electromagnetic radiation (ie. The oscillating electric fields generated by EM radiation)


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Solid State Physics - جامعة ذي قارsci.utq.edu.iq/images/pdf/dr1.pdf · monoatomic linear chain with only nearest neighbour interaction a a U n-1 U n U n+1 Monoatomic Chain