Transcript Document

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 force-spring
model, with masses m & spring
constants K.
• First, consider just two neighboring atoms. Assume that they
interact with a known potential V(r). Expand that potential in a
Taylor’s series about their equilibrium separation:
V(R)
Repulsive
0
a
Attractive
R
r
min
This potential energy is the same as
that associated with a spring with
spring constant:
 d 2V 
K   2 
 dr r a
Force  K (r  a)
• This one-dimensional chain of identical atoms 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
• The simple case of a
monatomic linear chain with
only nearest neighbor
interactions.
• Expand the energy near the
equilibrium point for the nth
atom. Then, the
Newton’s 2nd Law
equation of motion becomes:
l
a
a
Un-1
Un
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:
a
a
K (u n1  u n )
The force to the left:
K (u n  u n1 )
Un-1
Un
Un+1
• Total force = Force to the right – Force to the left
l
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:
0


un  A exp i  kxn  t 
dun
un 
 i A exp i  kxn0  t  
dt
.
d 2 un
2
u n  2   i   2 A exp i  kxn0  t  
dt
..
xn0  na
Undisplaced Position
..
u n   2un
xn  na  un
Displaced Position
• Put all of this into the equation of motion:
• Mathematical manipulation 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 this result 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.
The “Phonon Dispersion Relations” or Normal Mode Frequencies
or ω versus k relation for the monatomic chain.

max  2
K
m
Vs   / k
C
–л/a
B
0
A
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  uN n
This, in turn requires that there are an integer number of wavelengths in the
chain. So, in the first Brillouin Zone, there are only N allowed values of k.
Na  p
Na  p   
Na 2
2

 Nk 
p
p
k
a
• What is 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
Green:
k = (0.85)π/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
K
m
Vs   / K
2
B
0
π/a
A
2π/a
k
ω(k) (dispersion relation)
Points A & C are 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.
The points k = ±π/a have special
significance
n  n2  n 2n 
k
a
n
x
Bragg reflection occurs at k= ±nπ/a
• 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
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)
(n+1)
K
M
m
(n+2)
K
K
M
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
Un-1
M
Un
m
Un+1
M
Un+2
• As before, the GOAL is to obtain 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
M
m
Un-1
Un
un  A exp i  kxn0  t 
xn0  na / 2
un   A exp i  kx  t 
-1
0
n
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.
u n   2 A exp i  kxn0  t  
..
• Put all of this into the two equations of motion.
• 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.
• More algebra gives:
2 K cos(ka / 2)
2K   2 M


2
2K   m
2 K cos(ka / 2)
• Combining (1) & (2) gives a quadratic equation for ω2:
• 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.
• Briefly, examine the limiting solutions at points 0, A, B & C.
• In long wavelength region near k = 0 (ka«1), sin(ka/2) ≈ ½ka.

A
B
C
–л/a
0
л/a
2л/a
k
Taylor expansion 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 negative root 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 positive root 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 values of α as;
OR
2
Substitute min
ac
into the expression for the relative amplitude α
2K   2 M

2 K cos(ka / 2)
K(k 2a 2 )

2(m  M)

2
min 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
A
1/ 2
B
C


w
K
vs   a 

k
2(
m

M
)


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
• 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



2
K ( m  M ) K ( M  m)

Mm

2
max ac
2K
(C)

M
OR

2
min op
2K

m
(B)
• 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.
Acoustic & Optic Branches
• The Acoustic Branch has this name because it gives rise
to long wavelength vibrations ω = vsk, where vs is the
velocity of sound
• The Optic Branch is at higher energy vibration (higher
frequencies) & an optical frequency (energy) 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 these modes with electromagnetic
radiation.
Transverse optic mode for the
diatomic chain
The amplitude of vibration is strongly exaggerated!
Transverse acoustic mode for the
diatomic chain