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8/9/2019 9. Lattice Dynamics
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1
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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2
!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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!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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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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9
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
8/9/2019 9. Lattice Dynamics
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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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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.
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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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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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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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Z= exp!E N(k,s)"
#$ %
&'
kBT
"
#
$$$$
%
&
''''N(k,s)
( = exp!N(k,s)+1/ 2( ) !)s(k)
k,s(
kBT
"
#
$$$$$$
%
&
''''''
N(k,s)( =
exp!N(k,s)+1/ 2( ) !)s(k)
kBT
"
#
$$
$$
%
&
''
''k,s*
N(k,s)( = exp!
N(k, s)+1/ 2( )!)s(k)
kBT
"
#
$$
$$
%
&
''
''N(k,s)(
k,s*
(PH26)
where the summation extends over all possible occupation numbers
N(k,s) = 0,1,2,...and multiplication includes all states (k, s). The summation over
N(k,s) can be written as
exp!!"s(k)
2kBT
#
$
%%
%%%
&
'
((((((
exp!N(k,s)!"s(k)
kBT
)
*
+
++
,
-
.
..N(k,s)/=
exp!!"s(k)2kBT
#
$
%%%%%
&
'
((((((
1!exp!!"s(k)kBT
#
$
%%%%%
&
'
((((((
since the last summation is a geometrical series with the quotient exp !!" s(k)k BT
#$%
&'(
.
Hence, the partition function is
Z =
exp !!"
s(k)
2kBT
#
$%
&
'(
1! exp ! !"s(k)kBT
#$% &
'(
k,s) =
1
2sinh ! !"s(k)2k
BT
#$% &
'(
k,s) (PH27)
Employing statistical physics:
Free energy
F= !kBTlnZ= k
BT ln 2sinh
!"s(k)2k
BT
#
$%%
&
'((
)
*
+++
,
-
.
.
.k,s/ (PH28)
Entropy
S=!"F"T
#
$
%%%%%
&
'
(((((V
=kB!)s(k)2kBT
coth !)s(k)
2kBT
#
$
%%%%%
&
'
((((((! ln 2sinh
!)s(k)2kBT
#
$
%%%%%
&
'
((((((
*
+
,,,,
-
.
////
0
1
2222
32222
4
5
2222
62222
k,s7 (PH29)
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Internal energy
E=F+TS= kBT !!s(k)
2kBT1+
2
exp !!s(k)kBT
"
#
$$$$
$
%
&
''''''
(1
)
*
++++++++
+
,
-
.
.
.
.
.
.
.
.
.
k,s/
E= !!s(k)12+
1
exp !!s(k)kBT
"
#
$$$$$
%
&
''''''(1
0
1
22222222
3
22222222
4
5
22222222
6
22222222
k,s/
(PH30)
Comparison with (PH25) reveals that the number of phonons in the state (k, s), at a
given temperature T, isN(k,s)=
1
exp!!s(k)kBT
"
#
$$$$$
%
&
''''''(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.