Download Multielectron Atoms * The Independent Particle Approximation

Main Topics from Chapters 3-5
Due 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
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.
Longitudinal Waves
Transverse Waves
Monatomic Linear Chain
The force on the nth atom;
a
a
•The force to the right;
K (u n1  u n )
•The force to the left;
K (u n  u n1 )
Un-1
Un
Un+1
The total force = Force to the right – Force to the left
Thus, Newton’s equation for the nth atom is
mun   K un  un1   K un  un1 
..
m u n  K (un 1  2un  un 1 )
Eqn’s of motion of all atoms are of this form, only.. the value of ‘n’ varies
0

un  A exp i  kxn  t 
u n   2un
Brillouin Zones of the Reciprocal Lattice
Reciprocal
Space Lattice:
2/a

4K
M
k

4
a

3
a

2
a


0
a

a
2
a
3
a
4
a
1st Brillouin Zone (BZ=WS)
Each BZ contains
identical
information
about the lattice
q  20 sin ka / 2
2nd Brillouin Zone
3rd Brillouin Zone
0 
K
m
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
Are These Waves Longitudinal or Transverse?
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.
Fig 4.43
From Principles of Electronic Materials and
Devices, Third Edition, S.O. Kasap (©
McGraw-Hill, 2005)
Diatomic Chain(2 atoms in primitive basis)
2 different types of atoms of masses m1 and m2 are connected by identical springs
(n-2)
(n-1)
K
(n)
K
m1
m2
(n+1)
K
K
m1
(n+2)
m
m2
a)
a
b)
Un-2
Un-1
Un
Un+1
Un+2
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.
m1un1   K 2un1  un 2  un 1, 2 
m2un 2   K 2un 2  un1  un 1,1 

        
2
2
1
2
2
2
1
  4 
2 2
2
2
1
2
2

sin qa / 2
2
1/ 2
•
As there are two values of ω for each value of k, the dispersion
relation is said to have two branches

A
Optical Branch
Upper branch is due to the
positive sign of the root.
B
C
Acoustical Branch
–л/a
0
л/a
2л/a
k
Negative sign:   k for small k. Dispersionfree propagation of sound waves
• At C, M oscillates and m is at rest.
• At B, m oscillates and M is at rest.
• This result remains valid for a chain containing an arbitrary number of
atoms per unit cell.
A when the two atoms
oscillate in antiphase
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)
..
M u lm  K (ul 1,m  ul 1,m  2ulm )  K (ul ,m 1  ul ,m 1  2ulm )
K (ul ,m 1  ulm )
Ul,m+1
K (ul 1,m  ulm )
Ul-1,m
K
Ulm
K (ul ,m 1  ulm )
Ul,m-1
• Write down the
equation(s) of motion
K (ul 1,m  ulm )
Ul+1,m
ulm  uo e
i ( lk x a  mk y a t )
What if I asked you to
include second nearest
neighbors with a different
spring constant? 2D Lattice
..
M u lm  K (ul 1,m  ul 1,m  2ulm )  K (ul ,m 1  ul ,m 1  2ulm )
Ul,m+1
2K
 
(2  cos k x a  cos k y a )
M
2
C
Ul-1,m
Ulm
Ul,m-1
2D Lattice
Ul+1,m
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 Capacity
When 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
 = 3NkBT  3RT
d
Cv 
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!
Thermal Energy & Heat Capacity
Average energy of a harmonic
oscillator and hence of a
lattice mode at temperature T
Einstein Model
n Pn n
_

 Pn
n in
The probability of the oscillator being
this level as given by the Boltzman factor
Energy of oscillator


exp( n / kBT )
1
 
1


n


exp

n


/
k
T




B 
 
_
2
2

n 0 




 
1

exp

n


/
k
T


B 
 
2


n 0



_
1
  
2
e

 / k BT
1
n   n   
2
1

kBT
Mean energy of a
harmonic oscillator

High Temperature Limit
1

2
 << kBT
T
Low Temperature Limit   kBT
exponential
1

    
term gets
kBT
2
e
 1 bigger
_
_
1
  
2
Zero Point
Energy
 is independent of frequency of oscillation.
2
x
e x  1  x   ..........
2!

e
 1
k BT
_
1

   

2
1
1
k BT
_
1
    k BT
2
This is a classical limit because the energy steps are
now small compared with thermal/vibrational energy

k BT
_
  kBT
Heat Capacity C (Einstein)
Heat capacity found by differentiating average phonon energy
d
Cv 

dT
Cv
 
 kB
 k BT 
2

e
 k 1

e
2
k BT
kBT
e
 
Cv  k B   
 T  e T 1
Area =

2
where
T

B

kB

2

2


k
The difference between
classical and Einstein models
comes from zero point energy.
T(K)
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.
Points:Experiment
Curve: Einstein
Prediction
Debye approximation has two main steps
1. Approx. dispersion relation of any branch by a linear extrapolation
2. Ensure correct number of modes by imposing a cut-off frequency
 D, 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
 D so that:
Einstein
approximation to
the dispersion
Debye
approximation
to the
dispersion V
  vk
6 2
D
 g ( )d  3N
0
(
1
2

)D3  3 N
3
3
vL vT

V 1 2 D 2
( 3  3 )   d  3 N
2
2 vL vT 0
V
1
2
3N
9N
(

)

3

2 2 vL3 vT3
D3
D3
g ( ) 
9N
D3
2
g ( ) /  2
Density of states (DOS) per unit frequency range g()
• The number of modes/states with frequencies  and
+d will be g()d.
# modes with wavenumber from k to k+dk=
dn  S (k )dk  g ( )d
dk
for 1D monoatomic lattice
d
d
K
ka
2a
2
sin

dk
m
2
2
1
g ()  S (k )
K
ka
a
cos
m
2
g ( )  S (k )
K
ka
cos
m
2
sin 2 x  cos2 x  1  cos x  1  sin 2 x
1
L 2
g ( ) 
2
 a max
2
Debye Model adjusts Einstein Model
The energy of lattice vibrations will then be
found by integrating the energy of single oscillator
over the distribution of vibration frequencies. Thus

1
    
2
e
0

 / kT

  g   d
1 
2N
Mean energy of a harmonic
oscillator


2
max

 for 1D
2 1/ 2
It would be better to find 3D DOS in order to compare the
results with experiment.
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.
g ( E )dE  2 g (k )dk
2
2
k
E
2m
2
dE
k

dk
m
V
m
2mE dk
kk
g ( E )  2 g ( k2)
2
2
2
dEk
dk
g ( E )  2 g (k )
dE
k
2mE
g (E) 
2
V
2 2
3/ 2 1/ 2
(2
m
)
E
3
Vk 2
g (k )dk  2 dk ,
2
L
Octant of the crystal:
kx,ky,kz(all have positive values)
The number of standing waves;
L
3
V 3
L 3
s  k  d k    d k  3 d k

 
1
 4 k 2 dk
L /
8
V 1
3
 s  k  d k  3  4 k 2 dk
 8
2
Vk
 s  k  d 3k  2 dk
2
ky
Vk 2
S  k   2
2
3
L
kz
dk
k
kx
The Heat Capacity of a Cold Fermi Gas
(Metal)

U T      g   f  , T d
0
Close to EF, we can ignore the variation in
the density of states: g()  g(EF).
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 :
kBT

U  U 0  2 g EF    f  , T d
0
dU T   2
Ce 

g EF k B2T
dT
3
3 n
g E F  
2 EF
Ce 
2
2
Nk B
k BT
EF
The small heat capacity of metals is a direct consequence of the
Pauli principle. Most of the electrons cannot change their energy.
Otherwise metals would have infinite conductivity
On average,
I go about 
seconds between
collisions
electron
Bam!
phonon
Random Collisions
with phonons and impurities
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
Electrons colliding with phonons (T > 0)
Electrons colliding with impurities
 ph T  0  
imp is independent of T
Fermi’s Golden Rule
Transition rate:
2
W fi 

(E) is the ‘density of states
available at energy E’.
f Hp i
2
f
Quantum levels of
the non-perturbed
system
f ,f


Perturbation is
applied
i
Transition is
induced
• 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).
 f - degeneracy of state f
f
 fi
Einstein pointed out that the
Fermi Golden Rule correctly
describes the absorption
process.
 fi
Ni
i
Quantum Oscillator
Atoms still have energy at T=0.
What is <x2> for the ground state of the quantum harmonic oscillator?
x

(a   a )
2m

0 (a   a ) 2 0
2m



0 a 2 0  0 a 2 0  0 a  a  0  0 a  a  0
2m
x2 


0  0  0  1  
2m
2m
⇒
(1D Case)
For 3D quantum oscillator, the result is multiplied by 3:
x
2
3

2m
 G 2 

I  I 0 exp  
 2M 

• 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 solids
Excitons: Quantized Normal Modes of electron-hole pairs
Phonon spectroscopy =
Conditions for: inelastic scattering
Constraints:
Conservation laws of
Momentum
Energy
In all interactions involving phonons, energy must be conserved and crystal
momentum must be conserved to within a reciprocal lattice vector.
Deformation tensor
1 ui u j
 ij  (

)
2 x j xi
Elastic
anisotropy
birefringence
 ij  cijkl kl
Stress tensor 𝜎, 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
abc
      90o
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]
Zener Anisotropy Ratio:
A
2C44
C
 44
C11  C12
x
x=(a-b)/2 or
C44 
C11  C12
2