9. Lattice Dynamics

8/9/2019 9. Lattice Dynamics

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LATTICE VIBRATIONS PHONONS

Linear chain of identical particles

We first consider a linear chain of identical particles connected by springs of thesame strength so that only nearest neighbor atoms interact with each other; this isshown schematically in Fig. 1. The separation of the particles in equilibrium is equal

to a, the lattice parameter.

n-1 n n+1 n+2

un-1 u n u n+1 u n+2

Fig.1. Linear chain of identical particles connected by springs.

The force acting on the atom n arising from the springs connecting this atom with

atoms n-1 and n+1 is

Fn = !(u n+1 "u n ) " !(un "u n "1 ) = !(u n+1 +u n" 1 "2u n ) (PH1)

where !is the corresponding spring constant. The equation of motion of the nth atom

is then

md2u

n

dt2= !(u n+1 + un"1 " 2un )

(PH2)

where m is the mass of the particles.

In the continuum limit, when the separation, a, between the particles decreases to

zero, lim(a! 0)un+1

+ un"1" 2u

n

a2

=#

2u

#x2

, where x is the coordinate along the chain. If V is

the volume per particle ( a =V1/3) then in this limit equation (PH2) becomes

m

V2/3

!2u

!t2 = "

!2u

!x2

(PH3.1)

This is the wave equation that can be written in the familiar form


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!2u

!t2= c

2!2u

!x2

(PH3.2)

with c2=

!V2/3

m.

Solutions of (PH3.2) are harmonic waves

u=ukexp i(!t" kx)#$%

&'(+ complex conjugate (PH4)

where ukis the amplitude of the vibrations with the frequency "and the wave vector

k = ! c; the wavelength, #, is determined from the relation ! =2" k . The

relationship between the wave vector k and the frequency "is in this case linear and

thus the (group) velocity of the waves, defined as d! dk , is constant equal to

c = a!m

"

#

$%

&

'1/2

; such waves are called non-dispersive.

For the discrete system of particles, the vibrations of which are described by equation

(PH2), we seek, in analogy with (PH4), the solution in the form

un = ukexp i(!kt "kna)

#$%

&'(+ complex conjugate (PH5)

where the frequency of vibrations, !k , is generally a function of k. It is obvious from

(PH5) that if k changes by 2$/a then the displacement does not change since

exp(2!ni) = 1. Therefore, it is sufficient to consider !" a < k < " a . This range of k

defines the first Brillouin zoneof the wave vector. Inserting (PH5) into (PH2) yields

!" k2

m= #exp !ika[ ] + exp ika[ ]! 2( ) and, therefore,

!k

2=

2"

m1# cos(ka)[ ]=

4"

msin

2(ka 2) (PH6a)

The wavelength of the vibrations described by (PH5) is again ! = 2" k but the group

velocity of these vibrations is

c=d!kdk=a

"m

#

$

%%%%

&

'

(((((

1/2

cos(ka 2) (PH6b)

and thus these vibrations are dispersive. In the long wavelength limit when k! 0

c! a"m

#

$

%&

'

(1/2

as in the continuum case. The vibrational spectrum and/or dispersion


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3

relation is described by equation (PH6a) and in the present case it has the form shown

in Fig. 2.

!

k"/a-"/a

Fig, 2. Dispersion relation or vibrational spectrum for the linear chain of identical

particles connected by springs.

Definition of Phonons

The most general solution of equation (PH2) is a linear combination of waves withvarious k vectors

un = akexp !ikna( )+ akexp ikna( )"

#$

%

&'

k( (PH7)

where ak=u

kexp i! k t( ) . The energy of the vibrating system of the particles is

E=m

2

dun

dt

!

"

####

$

%

&&&&&n

'2

+(2

un+1

)un( )

2

n

' (PH8)

where the summation over n is over all the N particles of the chain. The first term is

the kinetic and the second term the potential energy. Inserting (PH7) into (PH8)

gives1

1 When evaluating the energy we use periodic boundary conditions for the chain containing N

atoms. In this case k =2!

Na! , where ! is an integer and exp ikna( )

n=1

N

! =0 for k " 0

N for k =0


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4

E = 2mN akak!k

2

k

" =m bkbk! k

2

k

" (PH9)

where we define bk= 2Na

k.

The motion of a harmonic oscillator (single vibrating particle) is described by the

equationm!!u=!"u

The solution is u = Asin!t where A is the amplitude of vibrations and !2= " m.

The energy of such harmonic oscillator is

E=1

2m!u

2+1

2!u

2=1

2mA

2"2 .

When the problem of the harmonic oscillator is solved quantum mechanically its

energy is

E = !!(n + 1/2) ,

where n=0,1,2,.... correspond to various quantum states.

Comparing the energy of the classical single particle harmonic oscillator with

equation (PH9), which gives the energy of the vibrating system of interacting

particles, it is seen that the latter can be formally regarded as the energy of non-

interacting quasi-particles each vibrating as a harmonic oscillator with frequency!k ; setting 2b

kbk= A

kA

k is the amplitude of vibrations . Each of these quasi-

particles, called phonons, is characterized by a value of the wave vector k. In thequantum mechanical framework the total energy of the vibrating system, i.e. of the

phonons present in this system, is then

E = Nk + 1/ 2( )k

! !"k (PH10)

where Nk is the number of phonons in the state k.

Linear chain of two particles of different types

We now consider a linear chain composed of two types of particles that have different

masses m1and m2 connected by different springs. The springs connecting differentparticles are generally different but only nearest neighbor atoms are assumed to

interact with each other (see Fig. 3). The separation of like particles is a, and the

separation of unlike particles is a1. The period of the system is equal to a.


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5

a

n - 11n +1

1n

1n - 1

2

a1

n + 12

n2

Fig. 3. . Linear chain of two types of particles connected by springs.

The force acting on the atom n1, arising from the springs connecting this atom with

atoms n2 (spring constant !1) and n2! 1 (spring constant !2), is

Fn1=!1(un

2"un

1)+!2 (un

2"1"un

1) ; the force acting on the atom n2 , arising from

the springs connecting this atom with atoms n1 and n1 +1, is

Fn2=!1(un1

"un2)+!2(un1+1

"un2) . Equations of motion for atoms n1and n2 are

then

m1

d2u

n1

dt2=F

n1

and m2

d2u

n2

dt2=F

n2

We again seek the solution in the form of plane waves analogous to (PH5)

un1 = u1k exp i(!(k) " t # kna)$% &'+ complex conjugate

un2 = u2k exp i(!(k) " t # kna)$% &'+ complex conjugate

(PH11)

Again, as in the previous case, if k is changed by 2$/a the displacement does not

change and, therefore, it is sufficient to consider !" a < k < " a ; this again defines

the first Brillouin zone of the vector k. Inserting (PH11) into equations of motion

yields

u1k !2 "#1 +#2

m1

$

%&&

'

())+u2

k #1 +#2exp("ika)m1

$

%&&

'

())= 0

u1k #1 +#2exp(+ika)

m2

$

%

&&'

(

))+u2k !2 "#1 +#2

m2

$

%

&&'

(

))= 0

(PH12)

These equations determine frequencies !k and amplitudes unk. The former are

eigenvalues of the matrix


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6

!1 + !2m1

" #2 "!1 + !2exp("ika)

m1

"!1 + !2exp(+ika)

m2

!1 + !2m

2

" #2%

&&&&

(

))))

and the latter are the corresponding eigenvectors. The equation for the eigenvalues isthe determinant of the above matrix equals zero and it is a simple quadratic equation

for!2

. The solutions are

!12=

1

2!0

2 1" 1"#2 sin2(ka 2)$

%&&&

'

())) !2

2=

1

2!0

2 1+ 1"#2 sin2(ka2)$

%&&&

'

())) (PH13)

where

! 02=

("1+"

2)(m

1+ m

2)

m1m2and

#

2= 16

"1"2

("1 +"2 )2m

1m

2

(m1 + m2 )2

$

%&

'

()

For long wavelengths, when k ! 0, "1 ! 0 and "2 ! "0 . At the boundary of the

Brillouin zone, when k ! "

a

, #12

!1

2#0

21$ 1$ %

2( ) and #22 !1

2# 0

21+ 1$ %

2( ) .The vibrational spectrum and/or dispersion relation has in the present case the form

shown in Fig. 4.

!

k"/a-"/a

!

OPTICAL

ACOUSTIC

Fig. 4. Dispersion relation, or vibrational spectrum, for the linear chain of two types

of particles connected by springs.


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7

It follows from equations (PH12) that the ratio of the amplitudes (eigenvectors) is

u1k

u2

k =

!1 + !2exp("ika)

!1 + !2 "m1#2 =

!1 + !2 "m2#2

!1 + !2exp(ika) (PH14)

Considering first the frequency !1

. When k! 0 (long wavelength) "1

! 0 and

thusu1

k

u2k!1. This means that the amplitudes of the neighboring atoms are the same.

This case is called acoustic and corresponding phonons are called acoustic phonons.

When considering the frequency !2

then when k! 0 "2!"

0 and thus

u1

k

u2

k=

!1+!

2

!1+!

2"m

1#02=

1

1"m1+m

2

m2

="m

2

m1

. This means that the amplitudes of

the neighboring atoms have opposite signs; this case is called optical andcorresponding phonons optical phonons.

It can be again shown that the energy of the system can be written as a sum of

energies of independent quasiparticles (phonons) and thus it is again given by

equation (PH10).

Three dimensional periodic structure

We now deal with a periodic system composed of N particles in the repeat cell

defined by three non-coplanar vectors a1,a2 and a3 . The position vectors of the

particles within the repeat cell are ri (i=1,....,N) and the position vectors of the origins

of the cells are Rn= N

iai

i=1

3

! where N i are integers. Let us assume that the potential

energy, U, of the system is a function of the positions of the atoms, ri+R

nwhich we

shall mark Ri,n . The atom at the position Ri,n is displaced by u Ri,n( ) and, assumingthat this displacement is small, we can expand the potential energy into the Taylor

series as follows

U= U0+ !U

!Ri,n" u" Ri,n( )

"=1

3

#i,n# +

1

2

!2U

!Ri,n" !R

j,m$ u" Ri,n( )u$ Rj,m( )

",$=1

3

#i,nj,m

# (PH15)


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where Ri,n! and u

!

are coordinates of the vectors Ri,n and u , respectively. In

equilibrium!U

!Ri,n"= 0 and thus the first order term in u vanishes. If particle at Ri,n

deviates by u! Ri,n( )from the equilibrium position, the force acting on this particle is

Fi,n! = "U"u! Ri,n( )

= Fi,n;j,m!# u# Rj,m( )

#=1

3

$j,m$

where

Fi,n; j,m!"

=# $2U

$Ri,n! $R

j,m"

(PH16)

is the force constant matrix. Equation of motion for the atom at Ri,n is then

mi

d2u! Ri,n( )dt2

= Fi,n;j,m!" u" Rj,m( )

"=1

3

#j,m# (PH17)

The force constant matrix satisfies the following symmetry relations:

(a) Fi,n; j,m!"

=Fj,m;i,n!"

(b) Fi,n;j,m!"

i,n;j,m# =0

The relationship (a) follows from the definition (PH16) and the relationship (b)

follows from the requirement of translational invariance that demands that if all

atoms are displaced identically (i. e. rigid body translation is imposed), the energy of

the system must not change. If the system considered possesses inversion symmetrythen

(c) Fi,n; j,m!"

=Fi,n;j,m"!

We assume again that the displacements are plane waves of the type

u Ri,n( )=Ui(k)exp i k!Rn" #(k)t( )$%&'

() (PH18)


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where k is the wave vector. If we add to the vector k a vector K such that

K!Rn=2"M , where M is an integer, then the displacement is not altered. Such

vector Kis called a reciprocal vector and it is generally given as

K = jibii=1

3

! (PH19)

where jiare integers and

b1 =2! a2 "a 3

a1# (a

2" a

3), b2 =2!

a3 " a1

a1# (a

2" a

3), b3 =2!

a1 " a2

a1# (a

2" a

3)

(PH20)

are the basis vectors of the reciprocal lattice. We can, therefore, limit k to a region

such that k ! K / 2 , where Kis any reciprocal lattice vector, i. e. to the first Brillouin

zone.

By inserting (PH18) into (PH17) we obtain

mi!2(k)U i

"(k)= Fi,n;j,m"# exp i k$(Rm%Rn )

&'(

)*+{ }Uj

#(k)#=1

3

,j,m, (PH21)

and owing to the translation symmetry this equation is the same for any value of n.

We can choose, therefore, n=0 and R0 = 0 . We now define the dynamic matrix for a

given vector kas follows

Di,j!"

(k) = Fi;j,m!"

exp ik #Rm{ }m$ (PH22)

where the summation extends over all repeat cells.

Equations of motion are now transformed into the system of 3N homogeneous linear

equations for U i!

(k)

D i,j!"#m i$

2(k)%!"% ij{ }Uj"(k) =0

"=1

3

&j

& (PH23)

Corresponding eigenvalues are the squares of the frequencies, !2(k) , and the

eigenvectors are the amplitudes U i(k) . For each value of kwe obtain 3N solutions

and in the following we mark the frequencies and amplitudes corresponding to thesedifferent solutions ! s (k) and U i,s (k) , respectively, where s=1,2,....,3N. The

subscript s numbers different branches (or bands) of the phonon spectrum.

Owing to the translational invariance, the potential energy U depends only on the

relative positions of the atoms forming the system, i. e. on the vectors


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Rk,p!R!,q =Rk,p;!,q ; this does not imply that U is only a function of the separation

of the atoms. Hence, the force constant matrix is

Fi;j,m!"=#

$2U

$Ri,0;k,p! $R

j,m;!,q"

k,p;!,q% (PH24.1)

For central forces, when U is only a function of the separation of atoms, i.e. of

Ri,n;j,m , but not necessarily described by a pair potential,

Fi;j,m!"

=# $2U

$Ri,0;k,p$Rj,m;!,q

Ri,0;k,p! R

j,m;!,q"

Ri,0;k,pRj,m;!,q

%

&

''''k,p;!,q

(

+ ()ij)m0)k!)pq#)jk)mp)i!)q0 ) $U

$Ri,0;k,p

)!"

Ri,0;k,p#

Ri,0;k,p! R

i,0;k,p"

Ri,0;k,p2

*

+

,,,,,,

-

.

///////

0

1

2222

(PH24.2)

Ideal body-centered-cubic lattice with first nearest neighbor interaction

The force constant matrix elements that need to be evaluated are F1;1,m!"

, where vectors

Rm

are

R1=

1

2111[ ] R

2=

1

21 11[ ] R3 =

1

211 1[ ] R4 =

1

21 1 1[ ]

R!1 =

1

21 1 1[ ] R

!2 =1

211 1[ ] R

!3 =1

21 11[ ] R

!4 =1

2111[ ]

Owing to the inversion symmetry F1;1,m!"

=F1;1,#m!"

and the invariance with respect to

the translations requires that F1;1,0!"

=# F1;1,k!"

k=#3k$ 0

3

% . Since there is only one atom per unit

cell the dynamic matrix has only one element:

D1,1

!"= 2F

1;1,1

!" coskx +ky +kz

2a #1$

%&

'

()+ 2F1;1,2

!" coskx#ky#kz

2a #1$

%&

'

()

*

+,,

+ 2F1;1,3

!" coskx#k

y+k

z

2a #1

$

%&

'

()+ 2F1;1,4

!" coskx+k

y#k

z

2a #1

$

%&

'

()-

.//

where kx , ky and k z are components of the kvector.


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11

Using the trigonometric relation cos! " 1= "2sin2(!/ 2)

D1,1!"

=# 2F1;1,1!" sin 2

k x +k y +k z

4a +2F1;1,2

!" sin 2k x# k y# k z

4a

%&

+

2F1;1,3!"

sin

2 kx# ky +k z4

a +

2F1;1,4!"

sin

2 kx +k y# k z4

a

'

()

In the following we write D1,1!"=D

!"

and (PH23) then becomes

D!"#m$

2(k)%!"{ }U"(k) =0

"=1

3

& (25)

There are three solutions for each

value of k. They corresponding to

three acoustic branches, as shown

schematically in the Fig. 5.

!

Brillouin zone boundary

k

Fig. 5 Dispersion relation, or vibrational

spectrum, for BCC. lattice.


12/16

12

Phonons at interfaces and free surfaces

The calculation of lattice vibrations is made using so called slab method. In this case

the unit cell (often called supercell) contains either two interfaces or two free

surfaces. The repeat cell containing two interfaces, for example grain boundaries, is

shown schematically in Fig. 6a. The structure studied is then a periodic arrangement

of such repeat cells.

INTERFACE

Fig. 6a. Repeat cell (supercell) containing two interfaces.

The repeat cell containing two surfaces is shown schematically in Fig. 6b. The

structure studied is then a periodic arrangement of such repeat cells.

SURFACE

VACUUM

VACUUM


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13

Fig. 6b. Repeat cell (supercell) containing two free surfaces.

For the case when the interface is the %= 5 (12 0) /[001] symmetrical tilt boundary

and the repeat cell contains 120 atoms, the corresponding phonon dispersion curves

are shown in Fig. 7; a many-body central-force potential for gold was used in this

calculation.

THz

20100

0

1

2

3

4

5

Y0

0-10-20

0

1

2

3

4

5

THz

X

Fig. 7. Phonon dispersion curves for the repeat cell contains 120 atoms and the S = 5 (12 0) /[001]

symmetrical tilt boundary. Peeled-off branches at low frequencies are acoustic interfacial waves and

branches peeled-off at high frequencies are optical branches associated with the interface.

The corresponding phonon dispersion curves for the ideal crystal, calculated using the

same type and size of the repeat cell as for the case of grain boundaries, are shown for

comparison in Fig. 8.


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14

THz

20100

0

1

2

3

4

5

Y0

0-10-20

0

1

2

3

4

5

THz

X

Fig. 8. Phonon dispersion curves for the ideal FCC. crystal when using the repeat cell containing 120 atoms.

Thermodynamic properties associated with phonons

In the quantum mechanical framework the total energy of the vibrating system, i.e. of

the phonons present in this system, is, similarly as in (PH10),

E= N(k,s)+1/ 2( )k,s! !"s(k) (PH25)

where N(k,s) = 0,1,2,... is the number of phonons in the state (k, s). The summation

extends over both the branches (bands), marked by s, and the wave vectors k.

The partition function for the phonon system is, using (PH25),


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15

Z= exp!E N(k,s)"

#$ %

&'

kBT

"

#

$$$$

%

&

''''N(k,s)

( = exp!N(k,s)+1/ 2( ) !)s(k)

k,s(

kBT

"

#

$$$$$$

%

&

''''''

N(k,s)( =

exp!N(k,s)+1/ 2( ) !)s(k)

kBT

"

#

$$

$$

%

&

''

''k,s*

N(k,s)( = exp!

N(k, s)+1/ 2( )!)s(k)

kBT

"

#

$$

$$

%

&

''

''N(k,s)(

k,s*

(PH26)

where the summation extends over all possible occupation numbers

N(k,s) = 0,1,2,...and multiplication includes all states (k, s). The summation over

N(k,s) can be written as

exp!!"s(k)

2kBT

#

$

%%

%%%

&

'

((((((

exp!N(k,s)!"s(k)

kBT

)

*

+

++

,

-

.

..N(k,s)/=

exp!!"s(k)2kBT

#

$

%%%%%

&

'

((((((

1!exp!!"s(k)kBT

#

$

%%%%%

&

'

((((((

since the last summation is a geometrical series with the quotient exp !!" s(k)k BT

#$%

&'(

.

Hence, the partition function is

Z =

exp !!"

s(k)

2kBT

#

$%

&

'(

1! exp ! !"s(k)kBT

#$% &

'(

k,s) =

1

2sinh ! !"s(k)2k

BT

#$% &

'(

k,s) (PH27)

Employing statistical physics:

Free energy

F= !kBTlnZ= k

BT ln 2sinh

!"s(k)2k

BT

#

$%%

&

'((

)

*

+++

,

-

.

.

.k,s/ (PH28)

Entropy

S=!"F"T

#

$

%%%%%

&

'

(((((V

=kB!)s(k)2kBT

coth !)s(k)

2kBT

#

$

%%%%%

&

'

((((((! ln 2sinh

!)s(k)2kBT

#

$

%%%%%

&

'

((((((

*

+

,,,,

-

.

////

0

1

2222

32222

4

5

2222

62222

k,s7 (PH29)


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16

Internal energy

E=F+TS= kBT !!s(k)

2kBT1+

2

exp !!s(k)kBT

"

#

$$$$

$

%

&

''''''

(1

)

*

++++++++

+

,

-

.

.

.

.

.

.

.

.

.

k,s/

E= !!s(k)12+

1

exp !!s(k)kBT

"

#

$$$$$

%

&

''''''(1

0

1

22222222

3

22222222

4

5

22222222

6

22222222

k,s/

(PH30)

Comparison with (PH25) reveals that the number of phonons in the state (k, s), at a

given temperature T, isN(k,s)=

1

exp!!s(k)kBT

"

#

$$$$$

%

&

''''''(1

(PH31)

i. e. it is given by the Bose-Einstein distribution since any state can be occupied by

any number of phonons.

Specific heat

CV =!E!T

"#$

%&'V

=k B

!( s ()2k

BT

*

+,

-

./

2

sinh 2 !( s ()2k

BT

"#$

%&'

,s

0 (PH32)

When the dispersion relation ! s ( ) , determined by solving equations (PH23), is

known, all the above-mentioned thermodynamical quantities associated with lattice

vibrations (phonons) can be evaluated.

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9. Lattice Dynamics

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