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Yu-Jun Zhao
Department of Physics, SCUT
Lecture X
Phonons I. : Crystal Vibrations
www.compphys.cn/~zhaoyj/lectures/
Vibrations of crystals with monatomic basis
Consider the elastic vibrations of a crystal with one atom in
the primitive cell.
Describe with a single coordinate us the displacement of the
plane s from its equilibrium position.
Three mode for each wavevector, one of longitudinal
polarization and two of transverse polarization.
longitudinal mode transverse mode
Note: for N atoms there are 3N mode, N L-modes and 2N T-modes.
The equation of the motion is:
)2( 112
2
ssss uuuC
dt
udM
a traveling wave solution:
us = u exp[i(sKa t)],
Then we have:
)cos1(2]2)exp()[exp(2 KaCiKaiKaCM
the dispersion relation (色散关系):
or
2sin
42/1
Ka
M
C
2sin
4)cos1)(/2( 22 Ka
M
CKaMC
Group velocity
The group velocity is the transmission velocity of a wave
packet, which is also the velocity of energy propagation
in the medium.
dimension onein /
dimension in three )(
dKdv
Kv
g
Kg
With the particular dispersion relation
2sin
42/1
Ka
M
C
2cos
2/12 Ka
M
Cavg
Q: Does the velocity of “sound” depend on its frequency?
Long wavelength limit
When >> a i.e. Ka << 1
22
22
2sin
4K
M
CaKa
M
C
KaM
C2/1
i.e.
aM
CdKdvg
2/1
/
In this limit, the velocity of sound is independent of frequency.
Derivation of force constants from experiment
In metals the effective forces may be of quite long range,
carried from ion to ion through the conduction electron sea.
Considering p nearest planes, the force on s plane
)(20
1
s
p
j
jsj uuCF
The equation of motion is
0
2
2
)(2p
j
sjsjs uuC
dt
udMThe dispersion relation
0
1
2 )cos1(2 p
j
j jKaCM
a
aKp pKadK
MaC
/
/
2 cos2
By multiplying both sides
by cos 𝑝𝐾𝑎 , and
integrating we may get:
Two atoms per primitive basis
Assume that each plane interacts only with the nearest-
neighbor planes. The equations of motion is
)2(
)2(
12
2
2
12
2
1
ssss
ssss
vuuCdt
vdM
uvvCdt
udM
the solution in a form of a traveling wave:
)](exp[
)](exp[
tsKaivv
tsKaiuu
s
s
substitute the solution in the equations
CviKaCuvM
CuiKaCvuM
2)]exp(1[
2)]exp(1[
2
2
1
2
The homogenous linear equations have a solution only if the
determinant of the coefficients of the unknown u, v vanishes.
02 )]exp(1[
)]exp(1[ 2
2
2
2
1
MCiKaC
iKaCMC
or 0)cos1(2)(2 22
21
4
21 KaCMMCMM
the dispersion relation
)cos1()(
211
cos2)(
2
21
21
21
21
21
2
2
2
121
21
2
KaMM
MM
MM
MMC
KaMMMMMMMM
C
at long wavelength limit (Ka << 1)
branch) l(acoustica )(2
branch) (optical 11
2
2
21
22
21
2
KMM
Ca
MMC
1
2
M
M
v
uFor the optical branch
The atoms vibrate against each other, but their center
of mass is fixed.
vu For the acoustical branch
The atoms and their center of mass move together.
at the 1st Brillouin zone boundary (Ka = )
2
2
1
2 /2 ;/2 MCMC
There is a frequency gap at boundary of the 1st Brillouin zone.
With p atoms in the primitive cell and N primitive cells,
there are pN atoms, 3pN degrees of freedom, N LA branches,
2N TA branches, (p1)N LO branches and (2p2)N TO
branches.
If there are p atoms in the primitive cell, there are 3p
branches to the dispersion relation: 3 acoustical branches
(1 LA and 2 TA) and 3p3 optical branches (p1 LO and
2p2 TO ).
The dispersion relation shows new features in crystals
with two or more atoms per primitive basis.
acoustical and optical branches (声学支和光学支)
Q: What is the vibration mode corresponding to𝐾 ≪
𝜋
𝑎, 𝐾 =
𝜋
𝑎?
)](exp[
)](exp[
tsKaivv
tsKaiuu
s
s
The dispersion relation shows new features in crystals
with two or more atoms per primitive basis.
acoustical and optical branches (声学支和光学支)
Q: What is the vibration mode corresponding to𝐾 ≪
𝜋
𝑎, 𝐾 =
𝜋
𝑎?
Quantization of the elastic waves
The energy of a lattice vibration is quantized. The quantum
of the energy is called phonon in analogy with the photon
of the electromagnetic wave.
Thermal vibrations in crystals are thermally excited phonons.
The energy of the elastic mode
)2
1( n
this mode occupied by n phonons with a frequency
the zero point energy 2
1
The amplitude of the elastic vibration
)/()2
1(42
0 Vnu
Consider a standing wave mode
tKxuu coscos0
The time average kinetic energy is
)2
1(
2
1
8
1 2
0
2 nuVE
tKxuMt
uM 222
0
2
2
sincos2
1
2
1
The kinetic energy of an atom is
What is the sign of ?
The energy of a phonon must be positive.
It is conventional and suitable to view as positive.
A mode with imaginary (negative 2) will be unstable.
Phonon momentum
A phonon of wavevector K will interact with particles
such as photons, neutrons, and electrons as if it has a
momentum .K
However a phonon does not carry physical momentum.
For most practical purpose a phonon acts as if its momentum
were , which is called the crystal momentum.K
In crystal there exists wavevector selection rules for allowed
transition between quantum states.
The true momentum of the whole system always is
rigorously conserved.
For the elastic scattering of a photon by a crystal, the
wavevector selection rule is
Gkk
'
GkKk
'
If the scattering of the photon is inelastic, with the creation
of a phonon , the wavevector selection rule becomesK
If a phonon is absorbed in the inelastic scattering, the
wavevector selection rule becomes
K
GKkk
'
Inelastic scattering by phonons
Phonon dispersion relations (K) are most often determined
by inelastic scattering of neutrons with the emission or
absorption of a phonon.
A neutron sees the crystal lattice chiefly by interaction with
the nuclei of the atoms.
The wavelength and the energy of the slow neutron is just
within the order of phonons.
The general wavevector selection rule is:
KkGk
'
The statement of conservation of energy is:
nn M
k
M
k
2
'
2
2222
Where are the wavevector and the energy of a
phonon created (+) or absorbed () in the process.
and K
22
3D Example: Normal modes of Silicon vs. Lead
L — longitudinal
T — transverse
O — optical
A — acoustic
8.828THz
2.245THz
r
Si
Si
M
M
Si
LOA(X) LTO(Gamma)
Pb
Physical processes related to neutron scattering
A new neutron source (CSNC) will be set up at DongGuan in
two years.
• Repetition rate : 25 Hz
•Average proton current: 62.5 A
• Proton kinetic energy: 1.6 GeV
•Average beam power: 100 kW
• Target: Tungsten
• Moderators: H2O,LCH4 ,LH2
• Spectrometers:
HRPD
HIPD
Reflectometer,
SANS
Chopper spectrometer
Homework
2018/4/7
1. Problem 4.1 of textbook.
2. Problem 4.2 of textbook
3. Problem 4.5 of textbook.
http://www.compphys.cn/~zhaoyj/lectures/
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