Normal Modes of Vibration One dimensional model # 1: The Monatomic Chain Consider a Monatomic Chain...

Normal Modes of VibrationOne 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.

• 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:

This potential is the same as that associated with a spring with spring constant K:

ardr

VdK

2

2

)( arKForce r2 r1

V(r)

0 a

Repulsive

Attractive

• 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 a a a a

Un-2 Un-1 Un Un+1 Un+2

One Dimensional Model # 1:The Monatomic Chain

0

• Consider 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 Lawequation of motion becomes:

a a

Un-1 Un Un+1

l

..

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:

)( 1 nn uuK

The force to the left is:

)( 1 nn uuK

Total Force = Force to the right – Force to the left

a a

Un-1 Un Un+1

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:

.

0expnn n

duu i A i kx t

dt

2.. 2 2 0

2expn

n n

d uu i A i kx t

dt

..2

n nu u

naxn 0

nn unax UndisplacedPosition:

DisplacedPosition:

0expn nu A i kx t

• Put all of this into the equation of motion:

• Now, carry out some simple math manipulation:

..

Equation of Motion for the nth Atom..

1 1( 2 )n nn nmu K u u u

0 0 0 01 12

e e 2 e en n n ni kx t i kx t i kx t i kx t

m K A A AA

kna ( 1)k n a( 1)k n akna

2e e e 2 e e e

ika ikai kna t i kna t i kna t i kna tm K A A AA

Cancel Common Terms & Get: 2 e 2 eika ikam K

2e e 2 e ei kna t i kna ka t i kna t i kna ka t

m K A A AA

• Mathematical Manipulation finally gives:

Solution to the NormalMode Eigenvalue Problemfor the monatomic chain.

• 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.

• After more manipulation, this simplifies to

• The maximum allowed frequency is:

“Phonon Dispersion Relations” or Normal Mode Frequencies or ω versus k relation for the monatomic chain.

max 2

/s

K

mV k

0 л/a 2л/a–л/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.

k

C AB

0

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.

Monatomic Chain Dispersion Relation

4sin

2

K ka

m

pa

Nkkp

NapNa

22

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:

0expn nu A i kx t

pNa

nNn uu must satisfy the periodic boundary condition: 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.

• 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.

x

Black: k = π/aor = 2a

Green:k = (0.85)π/aor = 2.35 a

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.

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

x

2 2nn n nk a

Bragg reflection occurs at k= ± nπ/a

2 24 sin2kam K

u

n

x

u

n

a

-π/a k

KVm

K

s /

2

k

C AB

0

ω(k) (dispersion relation)π/a 2π/a0

k = (/a) = (2/); = 2a

k 0;

The Monatomic Chain

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:

• 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 ei(nπ/a) = Uo(-1)n

4sin

2

K ka

m

Group Velocity, vg in the 1st BZ

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

Standing Wave!1st BZ

Edge

vg (dω/dk) = a(K/m)½cos(½ka)

One Dimensional Model # 2:The Diatomic Chain

Un-2Un-1 Un Un+1 Un+2

K K K K

M Mm Mm a)

b)

(n-2) (n-1) (n) (n+1) (n+2)

a

• This is the simplest possible model of a diatomic crystal.• a is the repeat distance, so, the nearest-neighbor separation is (½)a

• 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.

M m M m M

Un-2Un-1 Un Un+1 Un+2

..

1 1( ) ( )n n n n nM u K u u K u u

1 1( 2 )n n nK u u u

..

1 1 2( ) ( )n n n n nmu K u u K u u ..

1 2( 2 )n n n nmu K u u u -1

-1

• This model is complicated due to the presence of 2 different atom types, which, in general, move in opposite directions.

• 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.

Equation of Motionfor M

Equation of Motionfor m

M m M m M

Un-2Un-1 Un Un+1 Un+2

0 / 2nx na 0expn nu A i kx t

0expn nu A i kx t

..

2 0expn nu A i kx t

• As before, assume harmonic (plane wave) solutions for the atomic displacements Un:

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.

Carry out some simple mathmanipulation as follows:

..

1 1( 2 )n n n nM u K u u u

1 1

2 22 2 22

k n a k n akna knai t i ti t i t

MAe K Ae Ae Ae

2 2 2 2 22 22kna kna kna knaka kai t i t i t i ti i

MAe K Ae e Ae Ae e

Cancel Common Terms

2 2 1 cos2

kaM K

2cosix ixe e x 2 2 22ka kai i

M K e e

Equation of Motion for the nth Atom (M)

22 2 2 2 22 2 22

kna kna kna knaka ka kai t i t i t i ti i imAe e K Ae Ae e Ae e

..

1 1 2( 2 )n n n nmu K u u u

1 1 2

2 2 22 2 2

k n a k n a k n aknai t i t i ti t

A me K Ae Ae Ae

2 2 cos2

kam K

2cosix ixe e x

2 2 21 2ka kai i ikame K e e

2 2 22

ka kai i

m K e e

Cancel Common Terms

Equation of Motion for the (n-1)th Atom (m)

• The Equation for m becomes:

• The Equation for M becomes:

• (1) & (2) are two coupled, homogeneous, linear algebraic equations in the 2 unknowns α & ω as functions of k.

• More algebra gives:

(1)

(2)

2

2

2 cos( / 2) 2

2 2 cos( / 2)

K ka K M

K m K ka

• Combining (1) & (2) & manipulating:

2 2 2 44 (1 cos ( )) 2 ( ) 02

kaK K m M Mm

2 2 2 2 44 cos ( ) 4 2 ( )2

kaK K K M m Mm

24 2 2 sin ( / 2)

2 ( ) 4 0m M ka

K KmM mM

22 2 1/ 2( ) 4sin ( / 2)

[( ) ]K m M m M ka

KmM mM mM

2

2

2 cos( / 2) 2

2 2 cos( / 2)

K ka K M

K m K ka

2

1,2

4

2

b b acx

a

• Cross multiplying & manipulating with (1) & (2):

The 2 roots are:

• 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

• 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.

0 л/a 2л/a–л/a k

A

BC

ω+ = “Optic” Modes

ω- = “Acoustic” Modes

K

The solution is:

K

Near the BZ Center (qa << 1)

The Optic Mode becomes:

(ω+)2 2K(M1 + M2)/(M1M2)

or ω+ constant

The Acoustic Mode becomes:

(ω-)2 (½) Kq2/(M1 + M2)

or ω- vsq

vs sound velocity in the crystal.

Just like an acoustic wave in air!

Diatomic Chain Model: Kittel’s Notation!

Acoustic Modes(Acoustic Branch)

Optic Modes(Optic Branch)

qa

Gap

K

The Diatomic Chain Solution:

K

Near the BZ edge [q = (π/a)](Assuming M1 > M2)

The Optic Mode becomes:

(ω+)2 2K/M2

The Acoustic Mode becomes:

(ω-)2 2K/M1

Acoustic Modes(Acoustic Branch)

Optic Modes(Optic Branch)

qa

So, at the BZ edge, the vibrations ofwavelength = 2a for the 2 modes behave as if there were 2 uncoupledmasses M1 & M2, vibrating

independently with identical springs

of constant K.

( = 2a)

Gap

• 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.

22 2 1/ 2

1,2

( ) 4sin ( / 2)[( ) ]

K m M m M kaK

mM mM mM

A Taylor’s series expansion, using for, small x:

0 л/a 2л/a–л/a k

A

BC

22 2 1/ 2

1,2

( ) 4sin ( / 2)[( ) ]

K m M m M kaK

mM mM mM

The root with the minus sign gives the minimum value of the acoustic branch:

Substituting these values of ω into the expression for the relative amplitudeα and using cos(ka/2) ≈1 for ka«1gives the corresponding value of α:

OR

0 л/a 2л/a–л/a k

A

BC

The root with the positive sign gives the maximum value of the optic branch:

1

Substituting into the expression for the relative amplitude α:

22

2 cos( / 2)

K M

K ka

ac

2 22min

K(k a )

2(m M)

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:

k

A

BC

Optic

Acoustic

1/ 2

2( )s

w Kv a

k m M

ac

2min

0 π/a 2π/a–π/ a

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.

M

m

22

2 cos( / 2)

K M

K ka

op

2max

2K(m M)

mM

Similarly, substituting into the relative amplitude gives:op

2max

0 π/a 2π/a–π/a

k

A

BC

Optic

Acoustic

• The other limiting solutions for ω2 are for ka = π. In this case sin(ka/2) =1, so

1/ 222max

( ) 4ac

K m M M mK

Mm Mm Mm

( ) ( )K m M K M m

Mm

2max

2ac

K

M OR

2min

2op

K

m

(C)(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.

0 л/a 2л/a–л/a k

A

BC

Eigenmodes of chain at q = 0Optical Mode: These atoms, if oppositely charged, would form an oscillating dipole which would couple to optical fields with a

1 1 21 2

1 2

21 2

2 2

2 ( )( 0) , ( 0)

2 2

f f

M M Mf M Mq q

f fM M

MM M

D

1 2 1

1 1 2 1 2 12 21 2

1 2 1 2 1

2 1 21 2 2

22 2

( )0 ,

22 2

( )

n

n

fM Mf f uM M M M M sM M

u ufM Mf f M s M

M M MM M u

Center of the unit cell is not moving!

< a

Normal modes of chain in 2D space

2 2 21( ) 2 cos( )r r r

M q q a

2 21ˆ( )

2 r j i ij j iij

s s r s s

•Constant force model (analog of TBH) : bond stretching and bond bending

( )0 0

( )1

( )0 r

( )1 r

• 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

Acoustic & Optic Branches

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 someionic 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 someionic character, with + & - charges alternating:

Lower EnergyLess Compression of Springs

Optic ModeAcoustic Mode

Acoustic vs. Optic PhononsWhich has lower energy? Why?