Download Part II - TTU Physics

Normal Modes of Vibration
One dimensional model # 1: The Monatomic Chain
• Consider a
Monatomic Chain of
Identical Atoms
with nearest-neighbor,
“Hooke’s Law”
type forces (F = - Kx)
between the atoms.
• This is equivalent to a forcespring model, with masses
m & spring constants K.
• To illustrate the procedure for treating the interatomic potential in
the harmonic approximation, consider just two neighboring atoms.
• Assume that they interact with a known potential V(r). See Figure.
Expand V(r) in a Taylor’s series in displacements about the
equilibrium separation, keeping only up through quadratic terms
in the displacements:
V(r)
Repulsive
0
a
Attractive
r2
r1
This potential is the same as
that associated with a spring
with spring constant K:
Force  K (r  a)
 d 2V 
K   2 
 dr r a
One Dimensional Model # 1:
The Monatomic Chain
• This is the simplest possible solid.
• Assume that the chain contains a very large number
(N  ) of atoms with identical masses m. Let the
atomic separation be a distance a.
• Assume that the atoms move only in a direction parallel
to the chain.
• Assume that only nearest-neighbors interact with each
other (the forces are short-ranged).
a
a
Un-2
a
Un-1
a
Un
a
Un+1
a
Un+2
• Consider the simple case
of a monatomic linear
chain with only nearestneighbor interactions.
• Expand the energy near
the equilibrium point for
the nth atom. Then, the
Newton’s 2nd Law
equation of motion
becomes: ..
a
a
0
Un-1
Un
l
Un+1
This can be seen as follows.
The total force on the nth
atom is the sum of 2 forces:
The force to the right is:
a
a
K (u n1  u n )
The force to the left is:
K (u n  u n1 )
Un-1
Un
Un+1
Total Force = Force to the right – Force to the left
..
The Equation of Motion of each atom is of this
form.Only the value of ‘n’ changes.
• Assume that all atoms oscillate with the same amplitude
A & the same frequency ω. Assume harmonic solutions
for the displacements un of the form:
un  A exp i  kx  t 
0
n
.
un 
d 2 un
2
u n  2   i   2 A exp i  kxn0  t  
dt
..
Undisplaced 0
xn  na
Position:
dun
 i A exp i  kxn0  t  
dt
..
u n   un
2
Displaced
xn  na  un
Position:
• Put all of this into the equation of motion:
..
• Now, carry out some simple math manipulation:
Equation of Motion for the nth Atom
..
mu n  K (un1  2un  un1 )
 m
2

Ae
i kxn0 t


K


Ae
i kxn10 t
  2A e 
k ( n  1) a
kna
i kxn0 t
 Ae 
i kxn10 t
k ( n  1) a
kna

i  kna  ka t 
i  kna t 
i  kna  ka t 
2 i  kna t 
 m A e
 K Ae
 2A e
 Ae
2
 m A e
i  kna t 

 K Ae
i  kna t 
e
ika
 2A e
i  kna t 
 Ae
i  kna t 
Cancel Common Terms & Get:
m 2  K  eika  2  eika 
e
 ika




• Mathematical Manipulation finally gives:
• After more manipulation, this simplifies to
Solution to the Normal
Mode Eigenvalue Problem
for the monatomic chain.
• The maximum allowed frequency is:
• The physical significance of these results is that, for the
monatomic chain, the only allowed vibrational frequencies ω must
be related to the wavenumber k = (2π/λ) or the wavelength λ in this way.
• This result is often called the “Phonon Dispersion Relation” for
the chain, even though these are classical lattice vibrations &
there are no (quantum mechanical) phonons in the classical theory.
“Phonon Dispersion Relations” or Normal Mode
Frequencies or ω versus k relation for the monatomic chain.

K
max  2
m
Vs   / k
C
–л/a
A
B
0
л/a
2л/a
k
Because of BZ 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.
Monatomic Chain
Dispersion Relation
4K
ka

sin
m
2
Some Physics Discussion
• We started from the Newton’s 2nd Law equations of motion for N
coupled harmonic oscillators. If one atom starts vibrating, it does not
continue with constant amplitude, but transfers energy to the others in a
complicated way. That is, the vibrations of individual atoms are not
simple harmonic because of this exchange of energy among them.
• On the other hand, our solutions represent the oscillations of
N UNCOUPLED simple harmonic oscillators.
• As we already said, these are called the Normal Modes of the
system. They are a collective property of the system as a whole
& not a property of any of the individual atoms. Each mode represented
by ω(k) oscillates independently of the other modes. Also, it can
be shown that the number of modes is the same as the original
number of equations N. Proof of this follows.
To establish which wavenumbers are possible for the
one-dimensional chain, reason as follows: Not all values
are allowed because of periodicity. In particular, the nth
atom is equivalent to the (N+n)th atom. This means that
the assumed solution for the displacements:
un  A exp i  kxn0  t  
must satisfy the periodic boundary condition: un  u N n
This, in turn requires that there are an integer number of
wavelengths in the chain. So, in the first BZ, there are only N
allowed values of k.
Na  p
Na 2
2
Na  p   

 Nk 
p
p
k
a
• The physical significance of wave numbers k outside of the
First Brillouin Zone [-(π/a)  k  (π/a)]?
• At the Brillouin Zone edge:
• This k value corresponds to the maximum frequency. A detailed
analysis of the displacements shows that, in that mode, every atom
is oscillating π radians out of phase with it’s 2 nearest neighbors.
That is, a wave at this value of k is A STANDING WAVE.
Black: k = π/a
or  = 2a
Green:
k = (0.85)π/a
or  = 2.35 a
x
Points A and C have the same frequency
& the same atomic displacements. It can
be shown that the group velocity
vg = (dω/dk) there is negative, so that a
wave at that ω & that k moves to the left.
The green curve (below) corresponds to
point B in the ω(k) diagram. It has the
same frequency & displacement as points
A and C, but vg = (dω/dk) there is
positive, so that a wave at that ω & that k
moves to the right.
u
n
x
u
a
 2m  4K sin2 ka
2

C
-π/a 00
K
m
Vs   / K
2
B
A
π/a 2π/a kk
ω(k) (dispersion relation)
Points A & C are symmetrically
equivalent; adding a multiple of 2π/a
to k does not change either ω or vg,
so point A contains no physical
information that is different from
point B.
T The points k = ± π/a have special
significance
n  n2  n 2n 
k
a
n
x
Bragg reflection occurs at k= ± nπ/a
The Monatomic Chain
k = (/a) = (2/);  = 2a
k  0;   
For visualization purposes, it is
sometimes useful to visualize a
plane of atoms, made up of a
large number of parallel chains
like the one we just analyzed.
See the next few slides:
• Briefly look in more detail at the group velocity, vg.
• The dispersion relation is:
4K
ka

sin
m
2
• So, the group velocity is:
vg  (dω/dk) = a(K/m)½cos(½ka)
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!
– At the BZ edge, the displacements have the form (for site n):
Un= Uoeinka = Uo ei(nπ/a) = Uo(-1)n
Group Velocity, vg in the 1st BZ
vg  (dω/dk)
= a(K/m)½cos(½ka)
At the 1st BZ Edge,
vg = 0
• This means that a wave
with λ corresponding to
a zone edge
wavenumber
k =  (π/a)
Will Not Propagate!
• That is, it must be a
1st BZ
Edge
Standing Wave!
One Dimensional Model # 2:
The Diatomic Chain
• Consider a Diatomic Chain of Two Different Atom Types with
nearest-neighbor, Hooke’s Law type forces (F = - kx) between the
atoms. This is equivalent to a force-spring model with two
different types of atoms of masses, M & m connected by identical
springs of spring constant K.
(n-2)
(n-1)
K
(n)
K
M
m
(n+1)
K
K
M
(n+2)
m
M
a)
a
b)
Un-2
Un-1
Un
Un+1
Un+2
• This is the simplest possible model of a diatomic crystal.
• a is the repeat distance, so, the nearest-neighbor separation is (½)a
• This model is complicated due to the presence of 2 different atom
types, which, in general, move in opposite directions.
M
m
Un-2
M
Un-1
Un
m
Un+1
M
Un+2
• The GOAL is to find the dispersion relation ω(k) for this model.
• There are 2 atom types, with masses M & m, so there will be 2
equations of motion, one for M & one for m.
..
M u n  K (un1  un )  K (un  un1 )
 K (un1  2un  un1 )
..
mu n .. K (un  un1 )  K (un1  un2 )
-1
mu n  K (un  2un1  un2 )
-1
Equation of Motion
for M
Equation of Motion
for m
• As before, assume harmonic (plane wave) solutions for the atomic
displacements Un:
M
Un-2
Un-1
un  A exp i  kxn0  t 
Un
xn0  na / 2
un   A exp i  kxn0  t 
M
m
M
m
Un+1
Un+2
Displacement
for M
Displacement
for m
α = complex number which determines the relative amplitude and phase of
the vibrational wave. Put all of this into the two equations of motion.
u n   2 A exp i  kxn0  t  
..
Carry out some simple math
manipulation as follows:
Equation of Motion for the nth Atom (M)
..
M u n  K (un1  2un  un1 )
 2 MAe
 kna

i
t 
 2

 2 MAe
 kna

i
t 
 2

 k  n 1 a

 k  n 1 a

 kna


i
t 
i
t  
i
t 
2
2


 K   Ae 
 2 Ae  2    Ae 




 kna

 kna

 kna


i
t  i ka
i
t 
i
t   i ka 
 K   Ae  2  e 2  2 Ae  2    Ae  2  e 2 




Cancel Common Terms
ka
ka
i
i


2
2
2
 M  K   e  2   e 


ka 

 2 M  2 K 1   cos 
2 

eix  eix  2cos x
Equation of Motion for the (n-1)th Atom (m)
..
mu n1  K (un  2un1  un2 )
 A 2 me
 2 mAe
 k  n 1 a

i
t 
2


 kna

i
t   i ka
 2

2
e
 k  n 1 a

 k  n  2 a

 i kna t 
i
t 
i
t  
2
2


 K  Ae  2   2 Ae 
 Ae 





 kna

 kna

 i kna
t 
i
t   i ka
i
t   i 2 ka 
 K  Ae  2   2 Ae  2  e 2  Ae  2  e 2 




Cancel Common Terms
ka
i

2
 ika 
2
 me
 K 1  2 e  e 


ka
ka
i
i


2
2
2
 m  K  e  2  e 


eix  eix  2cos x
i
ka
2
ka


 m  2 K  cos   
2


2
• The Equation for M becomes:
(1)
• The Equation for m becomes:
(2)
• (1) & (2) are two coupled, homogeneous, linear
algebraic equations in the 2 unknowns α & ω as
functions of k.
2 K cos(ka / 2)
2K   2 M
• More algebra gives:  

2
2K   m
2 K cos(ka / 2)
• Combining (1) & (2) & manipulating:
• Cross multiplying & manipulating with (1) & (2):
2 K cos(ka / 2)
2K   2 M


2
2K   m
2 K cos(ka / 2)
ka
4 K cos ( )  4 K 2  2 K 2 ( M  m)   4 Mm
2
2
4 K 2 (1  cos 2 (
2
ka
))  2 K 2 (m  M )   4 Mm  0
2
2
m

M
sin
(ka / 2)
4
2
2
  2K (
)  4 K
0
mM
mM
The 2 roots are:
b  b 2  4ac
x1,2 
2a
2
K
(
m

M
)
m

M
4sin
(ka / 2) 1/ 2
2
2
 
K [(
) 
]
mM
mM
mM
• So, the resulting quadratic equation for ω2 is:
• The two solutions for ω2 are:
• Since the chain contains N unit cells, there will be
2N normal modes of vibration, because there are
2N atoms and 2N equations of motion for masses
M & m.
Solutions to the Normal Mode Eigenvalue Problem
ω(k) for the Diatomic Chain

A
ω+ = “Optic” Modes
B
C
ω- = “Acoustic” Modes
–л/a
0
л/a
2л/a
k
• There are two solutions for ω2 for each wavenumber k. That is, there are
2 branches to the “Phonon Dispersion Relation” for each k.
• From an analysis of the displacements, it can be shown that, at point A,
the two atoms are oscillating 180º out of phase, with their center of
mass at rest. Also, at point B, the lighter mass m is oscillating & M is at
rest, while at point C, M is oscillating & m is at rest.
Diatomic Chain Model: Kittel’s Notation!
The solution is:
K
K
Optic Modes
(Optic Branch)
Near the BZ Center (qa << 1)
The Optic Mode becomes:
(ω+)2  2K(M1 + M2)/(M1M2)
or ω+  constant
Gap
The Acoustic Mode becomes:
qa
Acoustic Modes
(Acoustic Branch)
(ω-)2  (½) Kq2/(M1 + M2)
or ω-  vsq
vs  sound velocity in the crystal.
Just like an acoustic wave in air!
The Diatomic Chain Solution:
K
K
( = 2a)
Optic Modes
(Optic Branch)
Near the BZ edge [q =  (π/a)]
(Assuming M1 > M2)
The Optic Mode becomes:
(ω+)2  2K/M2
Gap
qa
Acoustic Modes
(Acoustic Branch)
The Acoustic Mode becomes:
(ω-)2  2K/M1
So, at the BZ edge, the vibrations of
wavelength  = 2a for the 2 modes
behave as if there were 2 uncoupled
masses M1 & M2, vibrating
independently with identical springs
of constant K.
• Again briefly examine limiting solutions at points 0, A, B & C. In the long
wavelength region near k = 0 (ka«1), sin(ka/2) ≈ ½ka.

A
B
C
–л/a
0
л/a
2л/a
k
A Taylor’s series expansion, using for, small x:
1,2 2
K (m  M )

mM
m  M 2 4sin 2 (ka / 2) 1/ 2
K [(
) 
]
mM
mM

A
–л/a
1,2 2
0
B
C
The root with the minus sign gives the
minimum value of the acoustic branch:
K (m  M )
m  M 2 4sin 2 (ka / 2) 1/ 2

K [(
) 
]
mM
mM
k
л/a mM
2л/a
The root with the positive sign gives the
maximum value of the optic branch:
Substituting these values of ω into the expression for the relative amplitude
α and using cos(ka/2) ≈1 for ka«1gives the corresponding value of α:
OR
2
Substituting min
into the expression for the relative amplitude α:
ac
2
min
2K   2 M

2 K cos(ka / 2)
K(k 2a 2 )

2(m  M)
ac
 1
This solution corresponds to long-wavelength vibrations near the
center of the BZ at k = 0. In that region, M & m oscillate with
same amplitude & phase. Also in that region ω = vsk, where vs is the
velocity of sound & has the form:

Optic
1/ 2


w
K
vs   a 

k
2(
m

M
)


A
B
C
Acoustic
k
–π/ a
0
π/a
2π/a
2

Similarly, substituting max

A
B
C
Acoustic
k
–π/a
0
π/a
 
M
m
This solution corresponds to point A in
the figure. This value of α shows that,
in that mode, the two atoms are
oscillating 180º out of phase with their
center of mass at rest.

Optic
into the relative amplitude gives:
2K   2 M

2 K cos(ka / 2)
2K(m  M)

mM
2
max op
op
2π/a

A
• The other limiting solutions for
ω2 are for ka = π. In this case
sin(ka/2) =1, so

2
max ac

K (m  M )

Mm
1/ 2
 M  m 
4 
K 

 
Mm
Mm




K ( m  M ) K ( M  m)
Mm
B
C
2

2
max ac
–л/a
2K

M
(C)
OR
0
л/a
2л/a
2
min

op
2K
m
(B)
• At point C in the plot, which is the maximum acoustic branch
point, M oscillates & m is at rest.
• By contrast, at point B, which is the minimum optic branch
point, m oscillates & M is at rest.
k
Eigenmodes of chain at q = 0
Optical Mode:
These atoms, if
oppositely
charged, would
form an oscillating
dipole which
would couple to
optical fields with
Center of the unit cell is not moving!
< a
a
 2f
 M
1
2 f (M1  M 2 )

 (q  0) 
, D( q  0) 
 2f
M 1M 2

 M 1M 2
2 fM 1M 2
2f
2f
  u1 

 M (M  M )
 
M
M
1
2
1
2
 1
   0  u   M2 u ,
1
2

2 fM 1M 2   
2f
2f
M1


  
M
(
M

M
)
M 1M 2
2
1
2   u2 

2f


M 1M 2 

2f

M2 
sn1
M
 2
sn 2
M1
Normal modes of chain in 2D space
•Constant force model (analog of TBH) : bond stretching and bond bending
0(  )  0
1(  )
0(  )
1(  )

 r  
r
2
2
1

    r    (s j  si )  rˆij    s j  si 

2 ij 
1 
 (q) 
 r      r2  2  2 r cos(q  a) 


M 
2
Acoustic & Optic Branches
• Despite the fact that diatomic chain model is one-dimensional, it’s
results for the vibrational normal modes ω
contain considerable qualitative physics that
carries over to the observed vibrational
frequencies for many real materials.
• So, much of the physics contained in the diatomic chain results
can teach us something about the physics contained in the normal
modes of many real materials.
• In particular, ALL MATERIALS with 2 atoms per unit cell
are observed to have two very different
kinds of vibrational normal modes.
These are called
The Acoustic Branch
&
The Optic Branch
The Acoustic Branch
• This branch received it’s name because it contains long
wavelength vibrations of the form ω = vsk, where vs is the
velocity of sound. Thus, at long wavelengths, it’s ω vs. k
relationship is identical to that for ordinary acoustic (sound)
waves in a medium like air.
The Optic Branch
• This branch is always at much higher vibrational frequencies than
the acoustic branch. So, in real materials, a probe at optical
frequencies is needed to excite these modes.
• Historically, the term “Optic” came from how these modes were
discovered. Consider an ionic crystal in which atom 1 has a
positive charge & atom 2 has a negative charge. As we’ve seen, in
those modes, these atoms are moving in opposite directions. (So, each unit cell
contains an oscillating dipole.) These modes can be excited with optical
frequency range electromagnetic radiation.
A Longitudinal Optic Mode
The vibrational amplitude is highly exaggerated!
A Transverse Optic Mode
for the Diatomic Chain
The vibrational amplitude is highly exaggerated!
For the case in which
the lattice has some
ionic character, with
+ & - charges alternating:
A Long Wavelength Longitudinal
Acoustic Mode
The vibrational amplitude is highly exaggerated!
A Short Wavelength Longitudinal
Acoustic Mode
The vibrational amplitude is highly exaggerated!
A Transverse Acoustic Mode
for the Diatomic Chain
The vibrational amplitude is highly exaggerated!
For the case in which
the lattice has some
ionic character, with
+ & - charges alternating:
Acoustic vs. Optic Phonons
Which has lower energy? Why?
Acoustic Mode
Lower Energy
Less Compression of Springs
Optic Mode