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Main Topics from Chapters 3-5Due to time, not all topics will be on test.
Some problems ask to discuss the meaning or implication.
Lattice Dynamics (Monatomic, Diatomic, Mass Defect, 2D Lattices)
Strain (compliance, reduced notation, tensors)Harmonic Oscillator (Destruction/Creation, Hamiltonian &
Number Operators, Expectation Values)Energy Density and Heat Capacity (phonons, electrons
and photons)Quasiparticle Interactions (e-e, e-phonon, e-photon,
defect interations)Electrical and Thermal Conductivity
Lattice Vibrations
Longitudinal Waves
Transverse Waves
When a wave propagates along one direction, 1D problem.Use harmonic oscillator approx., meaning amplitude vibration small.
The vibrations take the form of collective modes which propagate. Phonons are quanta of lattice vibrations.
The force on the nth atom;
)( 1 nn uuK
• The force to the right;
• The force to the left;
)( 1 nn uuK
The total force = Force to the right – Force to the left
a a
Un-1 Un Un+1
Eqn’s of motion of all atoms are of this form, only the value of ‘n’ varies
Monatomic Linear Chain
)2( 11
..
nnnn uuuKum
Thus, Newton’s equation for the nth atom is
11 nnnnn uuKuuKum
0expn nu A i kx t ..
2n nu u
Brillouin Zones of the Reciprocal Lattice
1st Brillouin Zone (BZ=WS)
k
a
a
2a
a
2 0
M
K4
a
3a
4a
3
a
4
2nd Brillouin Zone
3rd Brillouin Zone
Each BZ contains identical
information about the lattice
2p/a
Reciprocal Space Lattice:
There is no point in saying that 2 adjacent atoms are out of phase by more than (e.g., 1.2 =-0.8 )
Modes outside first Brillouin zone can be mapped to first BZ 2/sin2 0 kaq m
K0
Fig 4.43From Principles of Electronic Materials and
Devices, Third Edition, S.O. Kasap (© McGraw-Hill, 2005)
Four examples of standing waves in a linear crystal corresponding to q = 1, 2, 4, and N. q is maximum when alternating atoms are vibrating in
opposite directions. A portion from a very long crystal is shown.
Are These Waves Longitudinal or Transverse?
Diatomic Chain(2 atoms in primitive basis)2 different types of atoms of masses m1 and m2 are connected by identical springs
Un-2Un-1 Un Un+1 Un+2
K K K K
m1 m1m2 m2m a)
b)
(n-2) (n-1) (n) (n+1) (n+2)
a
Since a is the repeat distance, the nearest neighbors separations is a/2
Two equations of motion must be written; One for mass m1, and One for mass m2.
2,12111 2 nnnn uuuKum 1,11222 2 nnnn uuuKum
•As there are two values of ω for each value of k, the dispersion relation is said to have two branches
Upper branch is due to thepositive sign of the root.
Negative sign: k for small k. Dispersion-free propagation of sound waves
Optical Branch
Acoustical Branch
• This result remains valid for a chain containing an arbitrary number of atoms per unit cell.
0 л/a 2л/a–л/a k
wA
BC
2/1 22
221
222
21
22
21
2 2/sin4 qa
A when the two atoms oscillate in antiphase
• At C, M oscillates and m is at rest.• At B, m oscillates and M is at rest.
Number and Type of Branches• Every crystal has 3 acoustic
branches, 1 longitudinal and 2 transverse
• Every additional atom in the primitive basis contributes 3 further optical branches (again 2 transverse and 1 longitudinal)
2D Lattice
K
Ulm Ul+1,m
Ul,m-1
Ul,m+1
Ul-1,m
)( ,1 lmml uuK )( ,1 lmml uuK
)( 1, lmml uuK
)( 1, lmml uuK
• Write down the equation(s) of motion
)2()2( 1,1,,1,1
..
lmmlmllmmlmllm uuuKuuuKuM
What if I asked you to include second nearest neighbors with a different spring constant?
)( tamkalkiolm
yxeuu
2D Lattice
C
Ulm Ul+1,m
Ul,m-1
Ul,m+1
Ul-1,m
)2()2( 1,1,,1,1
..
lmmlmllmmlmllm uuuKuuuKuM
)coscos2(22 akakM
Kyx
Similar to the electronic bands on the test, plot w vs k for the [10] and
[11] directions.
Identify the values of at k=0 and at the edges.
Specific Heat or Heat Capacity
The heat energy required to raise the temperature a certain amount
The thermal energy is the dominant contribution to the heat capacity in most solids. In non-magnetic insulators, it is the only contribution.
Classical Picture of Heat CapacityWhen the solid is heated, the atoms vibrate around their sites like a set of harmonic oscillators.
Therefore, the average energy per atom, regarded as a 3D oscillator, is 3kT, and consequently the energy per mole is = 3 3BNk T RT
v
dC
dT
Dulong-Petit law: states that specific heat of any solid is independent of temperature and the same result(3R~6cal/K-mole) for all materials!
nn
nn
n
P
P _
Average energy of a harmonic oscillator and hence of a lattice mode at temperature T
Energy of oscillator
1
2n n
The probability of the oscillator being in this level as given by the Boltzman factor
exp( / )n Bk T
Thermal Energy & Heat Capacity Einstein Model
_0
0
1 1exp /
2 2
1exp /
2
Bn
Bn
n n k T
n k T
_
/
1
2 1Bk Te
Mean energy of a harmonic oscillator
Low Temperature LimitT
2
1
TkB
TkB
12
1_
TBke
2
1_
Zero Point
Energy
exponential term gets
bigger
..........!2
12
x
xex
Tke
B
TBk
1
112
1_
TkB
_ 1
2 Bk T
_
Bk T
High Temperature Limit
Bk T
is independent of frequency of oscillation. This is a classical limit because the energy steps are now small compared with thermal/vibrational energy
<<
Heat Capacity C (Einstein)Heat capacity found by differentiating average phonon energy
2
2
1
k TB
k TB
B
Bv
ke
k TdC
dTe
k
2
2
1
T
T
v B
eC k
T e
where
T(K)
Area =2
Bk
Bk
vC
The difference between classical and Einstein models comes from zero point energy.
Points:ExperimentCurve: EinsteinPrediction
The Einstein model near T= 0 did not agree with experiment, but was better
than classical model.Taking into account the distribution of
vibration frequencies in a solid this discrepancy can be accounted for.
1. Approx. dispersion relation of any branch by a linear extrapolation2. Ensure correct number of modes by imposing a cut-off frequency ,
above which there are no modes. The cut-off freqency is chosen to make the total number of lattice modes correct. Since there are 3N lattice vibration modes in a crystal having N atoms, we choose so that:
Debye approximation to the dispersion
vk
Debye approximation has two main steps
Einstein approximation to the dispersion
D
D
23
9( )
D
Ng
2 3 3 3 3
1 2 3 9( ) 3
2 L T D D
V N N
v v
0
( ) 3D
g d N
2
2 3 30
1 2( ) 3
2
D
L T
Vd N
v v
3
2 3 3
1 2( ) 3
6 DL T
VN
v v
2( ) /g
Density of states (DOS) per unit frequency range g()
• The number of modes/states with frequencies and +d will be g()d.
( ) ( )Sdn k dk g d # modes with wavenumber from k to k+dk=
dk
ddk
d( ) ( )Sg k
d
dk
2cos
2 2
a K ka
m2 sin
2
K ka
m
for 1D monoatomic lattice
1
cos2
K kaa
m
( ) ( )Sg k
( )gL
2 2max
2 1
a
2 2 2sin cos 1 cos 1 sinx x x x
The energy of lattice vibrations will then be found by integrating the energy of single oscillator over the distribution of vibration frequencies. Thus
/0
1
2 1kTg d
e
1/ 22 2max
2N
Mean energy of a harmonic oscillator
for 1D
It would be better to find 3D DOS in order to compare the results with experiment.
Debye Model adjusts Einstein Model
3D Example: The number of allowed states per unit energy range for free electron?
• Each k state represents two possible electron states, one for spin up, the other is spin down.
( ) 2 ( )g E dE g k dk ( ) 2 ( )dk
g E g kdE
2 2
2
kE
m 2dE k
dk m
2
2mEk
( )g E 2 ( )g kdk
dE2
22
V
kk
2
2mE
2
m
k
3/ 2 1/ 22 3
(2 )2
( )V
m Eg E
2
2( ) ,
2
Vkg k dk dk
L
L
L
Octant of the crystal:
kx,ky,kz(all have positive values)
The number of standing waves;
3
3 3 33s
L Vk d k d k d k
/L
k
dk
zk
yk
xk
214
8k dk
3 23
14
8s
Vk d k k dk
2
322s
Vkk d k dk
2
22S
Vkk
The Heat Capacity of a Cold Fermi Gas (Metal)
Close to EF, we can ignore the variation in the density of states: g() g(EF).
TkEgdT
TdUC BFe
22
3
dTfgTU ,0
By heating up a metal (kBT << EF), we take a group of electrons at the energy - (with respect to EF), and “lift them up” to . The number of electrons in this group g(EF)f()d and each electron increased its energy by 2 :
dTfEgUU F ,20
0
F
F E
nEg
2
3
F
BBe E
TkNkC
2
2
The small heat capacity of metals is a direct consequence of the Pauli principle. Most of the electrons cannot change their energy.
kBT
Bam!
Random Collisions
On average, I go about
seconds betweencollisions
with phonons and impurities
electron
phonon
Otherwise metals would have infinite conductivity
Electrons colliding with phonons (T > 0)
Electrons colliding with impurities
0Tphimp is independent of T
The thermal vibration of the lattice (phonons) will prevent the atoms from ever all being on their correct sites at the same time.The presence of impurity atoms and other point defects will upset the lattice periodicity
Fermi’s Golden Rule
fpfi iHfW 22
Transition rate:
i
ff ,
Quantum levels of the non-perturbed
system
Perturbation is applied
Transition is induced
r(E) is the ‘density of states available at energy E’.
• See Fermi‘s Golden Rule paper in Additional Material on the course homepage
Absorption
When the ground state finds itself in the presence of a photon of the appropriate frequency, the perturbing field can induce
the necessary oscillations, causing the mix to occur.
This leads to the promotion of the system to the upper energy state and the annihilation of the photon.
This process is stimulated absorption (or simply absorption).
Einstein pointed out that the Fermi Golden Rule correctly
describes the absorption process.
i
f
fi
iN
fif - degeneracy of state f
Quantum Oscillator
)(2
aam
x
0)(02
22 aa
mx
000000002
22 aaaaaa
m
Atoms still have energy at T=0.
What is <x2> for the ground state of the quantum harmonic oscillator?
mm 2
10002
(1D Case)
For 3D quantum oscillator, the result is multiplied by 3:
mx
2
32
M
GII
2exp
2
0⇒
• These quantized normal modes of vibration are called
PHONONS• PHONONS are massless quantum mechanical particles which
have no classical analogue.– They behave like particles in momentum space or k space.
• Phonons are one example of many like this in many different areas of physics. Such quantum mechanical particles are often called
“Quasiparticles”
Examples of other Quasiparticles:Photons: Quantized Normal Modes of electromagnetic waves.
Magnons: Quantized Normal Modes of magnetic excitations in magnetic solidsExcitons: Quantized Normal Modes of electron-hole pairs
Phonon spectroscopy =
Constraints:Conservation laws of
Momentum Energy
Conditions for: elastic scattering in
In all interactions involving phonons, energy must be conserved and crystal momentum must be conserved to within a reciprocal lattice vector.
Elastic anisotropy
klijklij c
Stress tensor , Compliance C, Stiffness S Message: Wave propagation in anisotropic media is quite
different from isotropic media:• There are in general 21 independent elastic constants (instead of 2 in the isotropic case), which can be reduced still further by considering the symmetry conditions found in different crystal structures. • There is shear wave splitting (analogous to optical birefringence, different polarizations in diff. directions)• Waves travel at different speeds depending in the direction of propagation
birefringenceDeformation tensor
)(2
1
i
j
j
iij x
u
x
u
o
cba
90
x=(a-b)/2 or
The cubic axes are equivalent, so the diagonal components for normal and shear distortions must
be equal.
And cubic is not elastically isotropic because a deformation along a cubic axis differs from the stress
arising from a deformation along the diagonal.e.g., [100] vs. [111]
2CC
C 121144
x
C
CC
CA 44
1211
442
Zener Anisotropy Ratio: