Bab 5 Fonon Dan Vibrasi Kristal

Phonon, crystal vibration (atomic vibration in lattice)

Phonon: Quantized energy of lattice vibration (or regarded as heat generation)

Single atomic vibration

Diatomic vibration

Two modes of Vibration

Transverse mode

Longitudinal mode

A row of atoms (only acoustical branch)

Longitudinal acoustical mode

Transverse acoustical mode

In-phase displacement

Out-of-phase displacement

Transverse acoustical standing mode

Diatomic vibration (acoustical and optical branches)

Transverse acoustical mode for diatomic vibration

Transverse optical mode for diatomic vibration

Phonon

Relationship between the vibration frequency () and phonon wave vector (k).

Crystal lattices at zero temperature posses long range order

At T>0 ions vibrate with an amplitude that depends on temperature – because of lattice symmetries, thermal vibrations can be analyzed in terms of collective motion of atom

Collective motion of atoms = “vibrational mode”:

Energy content of a vibrational mode of frequency is an integral number of energy quanta . We call these quanta “phonons”. While a photon is a quantized unit of electromagnetic energy, a phonon is a quantized unit of vibrational (elastic) energy.

How to relate the vibration frequency () and phonon wave vector (k)

Procedure a: consider force between atoms as an elastic spring, so atomic vibration is similar to an elastic wave (phonon wave). So now we have a wave vector (k).

1. One aton per unit cell

x

Spring pulling upF = - kx

Gravity pulling downF =mg

B

we get2 = k/m or = (k/m)1/2 ---------------------(5)

F = ma = mg - kx = md2x/dt2 ---(1) (this is a differential equation)

g: gravity force; x: displacement

The solution to equation (1) isx = A.sin(t - ) ---------------(2)

Substituted (2) into (1) we get

mg - k Asin(t - ) = m[-2 Asin(t - )] ----(3)

if we say atoms are in balance position, i.e. the gravity (pulling down) is equal to spring pulling up, then- kx = md2x/dt2 = m[-2 Asin(t - )] = - m2 x

Equation (5) is a classical simple oscillator, now how do we transfer the classical simple oscillator into quantum simple oscillator?

Why we need to transfer classical oscillator into quantum oscillator?

Remember that phonon is a quantized energy of lattice vibration!

First we consider the form of classical energy conservation, which is the total energy E (for the atom vibration) equal to kinetic energy (mv2/2) + potential energy (kx2/2),

mv2/2 + kx2/2 = E ----(6)

The potential energy means work done

…..(7)

kx2/2 mv2/2

m: atom massv: atom stretching velocity (i.e. displacement / time)k: spring constantx: atom displacement

mv2/2 in (6) is equal to p2/2maccording to equation (5) 2 = k/m the kx2/2 = m2x2/2, so equation (6) can be re-written as p: momentum

E = p2/2m + m2x2/2 ---(8)

According to the uncertainty law ∆p∆x > ћ/2 ---(9) ћ = planck’s constant/2∆x : position uncertainty∆p: momentum uncertaintythen (8) becomes E = ћ2/8m(∆x)2 + m2(∆x)2/2---(10)

Minimizing this energy [eq (10)] by taking the derivative with respect to the position energy and setting it equal to zero gives

- ћ2/4m(∆x)3 + m2(∆x) = 0 ----(11)∆x =(ћ2/2m)1/2 ---(12)

substitute (12) into (10)we get energy E at zero

E0 = ћ22m/8mћ + m2ћ2/2m2 = ћ/4 + ћ/4 = 1/2 ћ ----(13)

The equation (13) means even energy at zero there still a vibration, which called zero-point enery (a very important concept).

If energy is not at zero then En = (n +1/2)ћ ---(14)

Now you have quantized energy of lattice vibration Phonon

N = 1

N = 2

N = 3

N = 4

Faster vibration,temp increasing

Energy is not continuous, but quantized, i.e. jump from one level to another level

One particle

Free to move on highway at any speeds

Why quantization occursV1

V2

V3

V1

V2

V3

Many particlesCannot freely move, but follow lanes

Particles on the same lane moving same speed

300-600 C600-800 C1200-1500 C2000 C

heat-up

A heated metal (Ta) emits different spectra at different temperatures

A discontinuous emission profile

a b c

A chain of atoms is considered as an elastic body and springs link atoms a, b and c, then the force from atom a acting on atom b (or atom c acting on atom b) can be described by Hooke’s law (F = kx; k is an force constantand is replaced by C here.) x = U is atomic displacements

force force

Fb = C[(ua-ub)-(ub-uc)] ----(1)F = ma= md2ub/dt2 = C[(ua-ub)-(ub-uc)] = C(ua + uc – 2ub) ---(2)

Phonon Dispersion Relation

The solution for Eq (2) with time dependent is exp(-it), so Fb = md2ub/dt2 = -m2ub, so Eq (2) can be re-written as -m2ub = C(ua + uc – 2ub) ---(3)

because the phonon is a traveling wave (i.e. traveling from atom a to c)so the form of traveling wave for Eq (3) isua+c = u·exp(isKa)·exp(iKa) ---(4)

combine Eqs (3) and (4), we get -2.m.u.exp(isKa) = C.u{exp[i(s + 1)Ka] + exp[i(s - 1)Ka] – 2.exp(isKa)} ---(5)

cancel u.exp(isKa) from both sides of Eq (5), we get 2m = -C[exp(iKa) + exp(-iKa) –2] ---(6)

because 2cosKa = exp(iKa) + exp(-iKa) ---(7), so we have 2 = (2C/M)(1 - cosKa) ---(8)

M: atomic mass

The equation (8) is so-called phonon dispersion relation, which connects frequency () and wave vector (K).

Dispersion curve

Wave Packet and Group Velocity

What are wave pocket and group velocity ?

Eq (8) is a phonon dispersion relation for one atom in an unit celland frequency is expressed as square (2)

If we re-write Eq (8) as = [2C/m(1-cosKa)]1/2 = 2(C/M)1/2sinKa/2----(9)

Then eq (9) tells us that there is a max value for lattice vibration frequency = 2(C/M)1/2, and lattice vibration frequency cannot exceed this value.

so the group velocity Vg = d/dK = (C/M)1/2 a.cos(1/2Ka) ----(10)

We need to ask what is the physical significance of Vg?

Transmission velocity of wave pocket is called group velocity

The group velocity Vg = 0 at BZ boundary, which means that the phonon traveling wave will become a standing wave at BZ boundary.

k

BZ boundary

Traveling wave

Standing wave

1st BZ

1st BZ

k

Acoustical branch

0 /a

The above discussed the phonon dispersion relation based on one atom per unit cell, now if two atoms per unit cell, what happen?

Unit cell Two atoms M m

unun-1 un+1vnvn-1 vn+1

Each atom can vibrate in both directions

If two atoms have different masses; m and M, and M > m. the phonon dispersion relation will be:

Md2un/dt2 = C(vn-un)-C(un-vn-1) = C(vn+vn-1-2un) ---(11)

Eq (11) means adjacent atoms forcing on atom (un) with mass M.

md2vn/dt2 = C(un+1-vn)-C(vn-un) = C(un+1 +un-1 –2vn) ---(12)

Eq (12) means adjacent atoms forcing on atom (vn) with mass m

if we also treat phonon as a traveling wave in two atoms per unit cell system,then solutions for traveling wave of Eqs (11) and (12) are:

un = A.exp[i(nKa - t)] ---(13)vn = B.exp[i(nKa - t)] ---(14)

A is amplitude for atom un with mass M B is amplitude for atom vn with mass m

Combining Eqs (11) and (13), we get-M2A = C[B(1 + e-iKa) -2A] ----(15)

Combining Eqs(12) and (14), we get-m2B = C[A(eiKa+1) –2B] ----(16)

Combining Eqs (15) and (16), we get mM4 – 2C(m + M)2 + 2C2(1-cosKa) = 0 ----(17)

solutions for Eq (17) is 2 = C(1/m +1/M) C[(1/m +1/M)2 - 4(sinKa/2)2/mM]1/2 ----(18)

Here we have

So we have two situations, 2 = C(1/m +1/M) + C[(1/m +1/M)2- 4(sinKa/2)2/mM]1/2 ----(19)2 = C(1/m +1/M) - C[(1/m +1/M)2- 4(sinKa/2)2/mM]1/2 ----(20)The frequency of Eq (19) is always greater than Eq (20) …..why?

Because m is mass of smaller atom, therefore it vibrates quicker than larger atoms.

High frequency optical branchLow frequency acoustical branch

Eq (19) is called optical branch or optical mode of vibration, which has frequency always higher than (20).

Eq (20) is Acoustical branch or acoustical mode of vibration.

When k 0Eq (19) 2 =2C(1/m +1/M )-----(21) Eq (20) 2 = C/2.[(Ka)2/(m+M )]-----(22)

0 /a

k

1st BZ

boundary

optical branch

acoustical branch

2 =2C(1/m +1/M)

2 = C/2.[(Ka)2/(m+M )]

We know that electrons in a crystal form band structure, Phonons in a crystalalso form a band structure, which is shown above. The gap between optical and acoustical branch is called phonon band gap.

We know electrons, neutrons, photons carry momentum, is it the same that Phonon also carries momentum (or phonon has a momentum)?

Answer: no, phonon does not carry a momentum!

Why phonon does not carry momentum?

Answer: only real atoms can produce momentum, phonon is established on the basis of mathematics, i.e. from reciprocal lattice which is a relative coordinates to real atomic lattice.

In other words, phonon is not a real thing, but only a mathematical term!