# Chapter 3 Atomic Vibration and Phonons

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Chapter 3 Atomic Vibration and Phonons

1. Monatomic Linear Chain

      x

0 a 2a 3a(N2)a (N1)a

     

u1 u2 u3 u4 uN1 uN

• A linear atomic chain contains N atoms, each with mass m, at x = 0, a, 2a, 3a …. (N1)a.

The displacement of the nth atom from its equilibrium position is un.

Considering nearest neighbor interaction only, the equation of motion of the nth atom is (except those near the boundary) :

m = ( un+1  un)  ( un  un1), (1)

where  is the force constant.

The number of equation is N ( 1023). The atoms near the edges would have an environment different from that at the interior. The equations of motion of these atoms would be different from (1). However, these atoms have a small population, they are assumed to obey Eq.(1) without affecting the results. This assumption can be realized by assuming that the original atomic chain is part of an infinitely long chain.

• Periodic boundary condition

The original crystal has N equations of motion for each atom. The extended imaginary crystal has more equations. We keep the number of equations to be N by introducing the periodic boundary condition:

un(t) = un+N(t)

The motion of an atom in the crystal is assumed to be exactly identical to another atom at the corresponding position of the added imaginary crystal.

(nN)th nth (n+N)th

…….    ….    ….    …. ……..

imaginary real imaginary

• To solve the equations of motion, use the trial solution :

un(t) = A exp i(qna  t), where A is the amplitude, q = 2/ = propagation constant and  = angular velocity. Equation (1) becomes :
 m 2 A exp i(qna  t)
= [Aexp i(q(n+1)at)  Aexp i(qnat)]
 [Aexp i(qnat)  Aexp i(q(n1)at)]

 m 2 =  [exp (iqa) + exp (iqa)  2]

m 2 = 2 (1  cos qa)

• (q) is periodic, because
(q+2/a) = (4/m)1/2 sin 

= (4/m)1/2 sin (qa/2+ )

= (q)

Period is 2/a.

• Allowed values of q's are discrete because:

un(t) = un+N(t),
exp i(qnat) = exp i[q(N+n)at),
1 = exp i qNa,
qNa = 2l, where l is an integer

An allowed q would give an . A q value defines a collective motion of the atoms, which defined as a normal mode of the crystal vibration.

• Total number of allowed q values is fixed, because un,q+2/a(t) = A exp [i(q+2/a)na  (q+2/a)t]

= A exp i(qna  (q)t) (for (q) is periodic)

= un,q(t).
Conclusion: The normal modes described by q and q+2/a are identical. The period of q is 2/a.

• The range [/a, /a] is selected to define the First Brillouin zone to include all the allowed q values.

Reciprocal lattice vector = 2/a

1st Brillouin zone : [/a, /a]

1st B.Z.

 

2/a /a 0/a 2/a

• The number of allowed q is ()/q = ()/() = N
= no. of atoms = no. of primitive unit cells
• Long wavelength limit refers to the modes with small q very close to 0 (or equivalently   ). Hence,
  = aq.
• Sound speed is defined at the long wavelength limit

Example : Prove that the modes described by q ( = 3.3a) and q+2/a (' = 0.76a) are equivalent.

Note: The physical meaning of this result is that atoms in a solid are discrete particles, while the material is not a continuous medium.

u

2.Diatomic linear chain :

A diatomic linear crystal is a 2-D crystal consisting of two kinds of atoms.

 ¤  ¤ ¤ x

M m M m M m

a/2 a/2

• Equations of motion for M and m in the nth cell are:

M = ( um n  uM n)  ( uM n  um n1) (2)

m = ( uM n+1  um n)  ( um n  uM n) (3)

Apply the trial solution:

uM n(t) = AM exp i(qna   t), and (4)

um n(t) = Am exp i[qa(n+1/2)  t]. (5)

 M2 AM = 2[ Am cos() AM], and (6)

 m2 Am = 2[ AM cos() Am]. (7)

(2  M2) AM  2 cos() Am = 0, and (8)

2 cos() AM + (2  m2) Am = 0. (9)

• Dispersion relationship

If Eq.(8) and (9) have a non-zero solution, then:

= 0.

By solving the equation, the dispersion relationship is :

2 =   [()2  sin2(qa/2)]1/2 (10)

Different from a monatomic chain, one q gives two 's.

 has two branches :

"+" for Optical branch

"" for Acoustic branch

optical branch

acoustic branch

q

-/a 0 /a
Example 1

Optical branch at q =/a,  has the minimum value of :

2 =   [()2  sin2(qa/2)]1/2 (10)

op, min = { +  [()2  ]1/2}1/2
= { +  [  ]1/2}1/2
= [ +  ]1/2 = (2/m)1/2

Example 2

Optical branch at q = 0,  has the maximum value :

2 =   [()2  sin2(qa/2)]1/2 (10)

op, max = { +  [()2  0]1/2}1/2
= (2 )1/2
Band width = op, max  op, min = (2 )1/2  (2/m)1/2

From (8) or (9)

(2  M2) AM  2 cos() Am = 0 (8)

for the optical mode at q = 0 :
(2  M op, max 2) AM  2 cos Am = 0

(2  M op, max 2) AM  2 Am = 0

= = =  < 0

AM M + Amm = 0 = (M + m) AC.G.

Since the C.G. of the unit cell does not move. An optical mode describes the relative motion of atoms within a basis.

An optical mode can be excited by EM wave.

The mode is termed "optical" since it can be excited by an EM wave.

positive charge

negative charge

Example 3

2 =   [()2  sin2(qa/2)]1/2 (10)

Acoustic branch at q  0,  has the minimum value :
ac, min2     [1  (qa/2)2]1/2
    [1  (qa)2]
=

Amplitude of atoms, from (8): (2  M2) AM2cos(qa/2) Am = 0
(2  Mac, min 2) AM  2 Am = 0

= 2/(2  Mac min2) =  1

 AM  Am, amplitudes of vibration of the two kinds of atoms in the same unit cell are about the same. The motions of the atoms are approximately in phase.

 q  0 is the condition of long wavelength limit,

ac, min  ()1/2q, sound speed = ()1/2

+ charge - charge

• Show that  is periodic. Since
2(q) =   [()2  sin2(qa/2)]1/2

2(q+2/a) = ….  {….  … sin2[(q+2/a)a/2]}1/2

= …. {….  … sin2 (qa/2)}1/2
= 2(q)

(q) has a period of 2/a.

• Allowed values of q are discrete, since periodic boundary condition ensures :
uM n = uM n+N
AM exp i(qna   t) = AM exp i(q[n+N]a   t)

 qNa = 2h
 allowed q of a normal mode has the form of
q = 2h/(Na) , where h is an integer.

q = 2/Na

• Total no. of allowed q is limited, since :

uM n(q+2/a) = AM exp i( [q+2/a] na  (q+2/a) t)
= AM exp i(qna  (q) t)
= uM n(q)

Therefore uM n(q+2/a) and uM n(q) represent the same normal mode.
Define the range [/a, /a] to be the 1st Brillouin zone.

No. of allowed q = ()/() = N = no. of unit cells

• No. of normal modes = (no. of q's) x (no. of branches) = no. of allowed  = 2N.

Note : In general, the no. of normal modes of a crystal = DnN,

D = the dimension of a crystal (i.e. 3 for 3-D crystal, 2 for planar, 3 for chain)

Dn = no. of branches

= no. of acoustical branches + no. of optical branches

= D + D(n  1)

N = no. of primitive unit cells

= no. of allowed q

Example : Dispersion relationship of diamond.

D = 3, n = 2 (primitive unit cell)

Three acoustic branches, three optical branches.

No. of normal modes = 6N

LO LO

TO TO
LA LA

TA TA

q

3. Vibration of 3-D crystal

• vectors are discrete, since from the periodic boundary condition :
q(, t) = q(+N1+ N2+ N3, t),

where is the lattice vector of the nth atom, N1, N2 and N3 are integers.

z

N3c

y

c a

x

Aqexp i( t)= Aqexp i([+N1+N2+N3]t),

 N1= 2n1

 N2 = 2n2, and

 N3 = 2n3,

where n1, n2, n3 are integers.

Proof : N1 = N1 (+ + )
= N1 (n1/N1) 2 + 0 + 0
= 2 n1

• No. of vectors are finite. If

= (n1/N1)+ (n2/N2)+ (n3/N3) and

' = [(n1+ N1)/N1]+ (n2/N2)+ (n3/N3)

= (n1/N1 + 1)+ (n2/N2)+ (n3/N3)

are two allowed wave vectors.

= Aqexp i(   t).

= Aqexp i(   t)

=Aqexp i{[(n1/N1+1)+(n2/N2)+(n3/N3)]t}

= Aqexp i(+ t)

(Note : = m+ n+ p)

= Aqexp i (+ m t)

= Aqexp i (+ 2m   t), and

= Aqexp i( t)

= q(, t)

Both and ' index the same normal mode.

Example : The first Brillouin zone of a 2-D crystal.

Exercise : With the aid of the diagram, show that for a vector in reciprocal lattice space, from the zone center to the zone boundary is

 2/2 = 0.

Examples : 1st B.Z. of S.C.

B.C.C. (reciprocal space is F.C.C.)

4. Number density of normal modes in a reciprocal lattice space

's are equally spaced in the 1st B.Z. bounded by /N1, /N2 and /N3. All the ’s are in the 1st Brillouin zone having a volume of:

()/(N1N2N3)

= ()()/(N1N2N3)

= [()()  ()()]/(N1N2N3)

= [()()  0]/(N1N2N3)

= (2)3/(N1N2N3)

= (2)3/Vc (Vc  volume of the specimen)

Number density of in a reciprocal space

= number density of normal modes = 1/[23/Vc] = Vc/(2)3.

5. Density of modes

At the long wavelength limit ( = vq) of an acoustic branch, a sphere in the reciprocal space with a radius  cuts some allowed ’s. They represent the normal modes having the frequency .

The number of modes with frequencies between  and +d correspond to all the ’s lying between the two spheres.

qy

        

         

Reciprocal lattice         

space          qx

        

        

        

• Density of normal modes g()

g() d = no. of modes between  and +d

= dq3 = 4q2 dq

= 4 d()

• For a 3-D crystal, there are three acoustic branches, so that
g() = 2

6. Debye Approximation

Since the number of acoustic modes is 3N (a finite number), there is a maximum  denoted as D, such that :

3N = g() d

= 2 d

=

Define Debye Temperature TD = D/kB,

kB is the Boltzmann's constant = 1.3810-23 J K-1

7. Phonons

• The energy of a simple harmonic oscillator is quantized in the form of E =  (n + 1/2), where n = 0, 1, 2, …
• The total energy of a normal mode is Ei,n=i(n+1/2). An energy quantum i is defined as a phonon. It is the energy quanta associated with a collective motion of all the atoms in one vibration mode.
• A normal mode has equally spaced energy levels.

3(1+1/2)

2(2+1/2)

1(4+1/2)

1(3+1/2) 2(1+1/2)

1(2+1/2) 3/2

1(1+1/2) 2/2

E = 1/2

Mode q1q2 ……….. qn

• A set of integers {n1, n2, n3, ….} is used to describe the overall vibration status of a solid.

8. Average phonon numbers in a mode is defined as <n>, so that the average vibration energy is <E> = (n +1/2).

<E> =

= (set x = /(kBT) )

=

=

=

= =

<E> =  ( + 1/2)

 Comparison with E =  (n + 1/2), one defines the average phonon no. <n> =

Exercise : Show that at high temperatures  < kBT,

(i) <n>  kBT/()

(ii) <E>  kBT (the classical result)

Exercise : Show that at low temperatures  > kBT,

(i) <n>  exp(/kBT)

(ii) <E>   exp(/kBT) + /2

11. Debye model of heat capacity

<E>all atoms= +

• High temperatures,  < kBT,

exp(/kBT)  1  1+/(kBT)1 = /(kBT)

<E>all atoms  +

Cv = 3 NkBT = constant (classical)

• Low temperatures,  > kBT,

Using TD = [1/3

g() = 2

= 9N( )2 = 2

<E> = +

if x = /(kBT) > 1, xmax = D(kBT)  .

<E>all atoms = 9NkBT4/TD3 +

= 9NkBT4/TD3 (4/15) +

= 34NkBT4/5TD3 +

Cv = 124NkBT3/5TD3  T3.

9. Thermal conductivity

• Consider the contribution of phonons only.
• Two regions in the solid are separated by a mean free path of a phonon l (distance that a phonon can travel without being scattered)
• The net energy flux passing through the interface is Je =

nv<E>1  nv<E>2. n is the phonon density. <E>1 and <E>2 are the phonon energies at points A and B. v is the drift velocity of a phonon.

• Energy current density is
Je = nv (<E>1  <E>2) = nv () c, where c is the specific heat capacity of a normal mode.
•  is the thermal conductivity, where Je =  . Therefore,   nclv or Clv, where C = specific heat per unit volume.

Interface

l

A B T T+

1