bab 5 fonon dan vibrasi kristal
TRANSCRIPT
Phonon, crystal vibration (atomic vibration in lattice)
Phonon: Quantized energy of lattice vibration (or regarded as heat generation)
A row of atoms (only acoustical branch)
Longitudinal acoustical mode
Transverse acoustical mode
In-phase displacement
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
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
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!