Summary since last test
Ch.8 & 9 Effect of adding periodic potential
 Ch.10 Tight binding method
 Ch.11 Other methods of calculating bands
 Ch.12 Effective Mass and Band Filling
 Ch.18 Surface Effects (Work Function, quality)
 Ch.19 Types of Bonds and Resulting Properties
 Student Talks (basics only, suggested reading)
 Ch.22 Lattice Dynamics & Phonon Dispersion
 Ch.23 Specific Heat
 Ch.24 Neutron Scattering
 Ch.28 & 29 Semiconductors and Applications

May need the
basics from
old stuff, but
no direction
questions on
old stuff
FCC Primitive and
# of nearest
Conventional
Unit
neighbors
Cells
Nearest-neighbor
distance
# of second
SC
BCC
FCC
6
8
12
a
½ a 3
a/2
12
6
6
Review
Chapter 8 & 9:
 Calculate free electron energy bands and
the number of k states in the bands
 Apply Bloch’s theorem to the KronigPenney model or any other periodic
potential
 Explain the meaning and origin of bands
and “forbidden band gaps”
 Understand
difference between metals,
semiconductors and insulators!
Free Electron Brillouin Zone
Repeated Zone Scheme
2
2
More generically:
h (k  2 / a )
2
2
 (k ) 
h (k  2n / a )
2E m
 nk 
2m
k
–/a
2
 /a
2
hk
 (k ) 
2m
 nk   n ,k  K
 nk (r )   n ,k  K (r )
Bloch Wavefunctions
a

Bloch’s Theorem states that for a particle moving
in the periodic potential, the wavefunctions ψnk(x)
are of the form
 nk ( x)  unk ( x)e , where unk ( x) is a periodic function
ikx
unk ( x)  unk ( x  a )

unk(x) has the periodicity of the atomic potential

The exact form of u(x) depends on the potential
associated with atoms (ions) that form the solid
Using Bloch’s Theorem: The Krönig-Penney Model
Bloch’s theorem allows us to calculate the energy bands of
electrons in a crystal if we know the potential energy function.
V
V0
-b 0
a a+b
x
2a+b 2(a+b)
Solve the SE in each region:
Energy in terms of E0
Each atom is represented
by a finite square well of
width a and depth V0.
2K 2
E
2m
BAND 2
Forbidden band gap
BAND 1
0<x<a
 I ( x)  Ae
 iKx
E0 
-b < x < 0
 II ( x)  Cex  De x
ka/
iKx
 Be
 2 2
2ma 2
 b 2 

 sin Ka   cosKa   cos( ka)
2
K


Figure 1.7: Coupled Well Energies
What happens to these levels as the
atoms get closer (b smaller)?
How would the
energy levels
look for
multiple wells?
Energy Band Overlap
Mixing of bands known as
hybridization (Si=sp3)
Conduction Band
The bottom
unfilled energy
band
Valence Band
The last filled
energy band at
T=0
8
14Si:
3s23p2
Out of 8 possible n=3
electrons (2s and 6p)
Learning Objectives for Today
Chapters 9 - 11:
 Relate
DOS to energy bands
 Be able to use tight binding model to
calculate energy bands
 Discuss the differences, advantages
and disadvantages between various
methods of calculating energy bands
 Predict general features of simple
structures
LCAO: Electron in Hydrogen Atom
(in Ground State)
 1 (r)  Ae
 K x1
Second hydrogen atom
 2 (r)  Ae
 K x2
2.0
1(r)
1.5
1.0
0.5
0.0
-10
-5
0
5
10
r (aB)
Approximation: Only Nearest-Neighbor interactions
Group: For 0, 2 and 4 nodes,
determine wavefunctions
If there are N atoms in the chain c0
there will be N energy levels and
N electronic states (molecular
orbits). The wavefunction for
each electronic state is:
c1
c2
c3
c4
k=/a
Yk = S eiknacn
k=/2a
Where:
a
is the lattice constant,
n
identifies the individual atoms
within the chain,
cn represents the atomic
orbitals

k
is a quantum # that identifies
the wavefunction and tells us the
phase of the orbitals.
a
k=0
The larger the absolute value
of k, the more nodes one has
The Independent Electron (e-)
Approximation & Sch. Equation
2

2
(
  U (r )) k (r )   (k ) k (r )
2m
A SE for each e- is a huge simplification
 The ind. e- approx. however doesn’t ignore all
 U(r)=periodic potential + periodic interactions
 To know U(r), you need 
 To know , you need U(r)
 Guess U(r), use to solve 
 Use to get better guess of U(r), repeat same

The Cellular Method (1934)
Since we have periodicity, it is enough to solve
the S.E. within a single primitive cell Co
 The wavefunction in other cell is then

 k (r  R)  e
 k (r )
 ik  R
 is a sum over spherical harmonics, need BCs
 Computationally challenging to solve B.C.s
 Results in potential with discontinuous
derivative at cell boundary

Muffin-tin potential
Solves both complaints of the last method
(set to 0)
U(r)=V(|r-R|), when |r-R|  ro (the core region)
=V(ro)=0, when |r-R|  ro (interstitial region)
ro is less than half of the nearest neighbor distance
Pseudopotential Method
Began as an extension of OPW
ik r
 If we act H on   e

k

b
c
c
k
(r )
c
In the outer region, this gives ~ free energy
 What goes on in core is largely irrelevant to
the energy, so let’s ignore it

U(r)=0 , when r  Re (the core region)
=-e2/r, when r  Re (interstitial region)
Group: Linear Chain of F
F: 1s22s22px22py22pz1
F
F
F
(a)
F
F
F
(b)
F
F
(c)
F
F
(d)
EF
E(k)
EF
EF
0
k
/a 0
k
/a 0
EF
k
/a 0
k
Which of the following is the correct band structure for a
linear chain of F atoms (atomic #=9)?
/a
Learning Objectives for Today
Chapter 12:
 Calculate effective mass and
cyclotron frequency along certain
direcitons
 Understand the difference between
an electron and an electron
quasiparticle
 Distinguish between two types of
excitons
Two Types of Excitons
Wannier – Matt excitons (free exciton): mainly exist in semiconductors,
have a large radius, are delocalized states that can move freely throughout
the crystal, the binding energy ~ 0.01 eV
Frenkel excitons (tight bound excitons): found in insulator and molecular
crystals, bound to specific atoms or molecules and have to move by
hopping from one atom to another, the binding energy ~ 0.1 -1 eV.
Wannier – Matt
A phonon
is another
excitons:
quasiparticle
stable
at cryogenic
temperature
Frenkel excitons:
stable at room
temperature
Physical Meaning of the Band Effective Mass
2 2
k
E
*
2m
The effective mass
is inversely
proportional to the
curvature of the
energy band.
Near the bottom of a nearly-free electron band m* is approximately constant,
but it increases dramatically near the inflection point and even becomes
negative(!) near the zone edge. What does that mean for electron near BZE?
Deflection of Electrons in a
Uniform Magnetic Field
F  Bev  ma
The centripetal acceleration of the electrons is
Bev
a
m
v 2 Bev
Hence a 

r
m
mv
r
eB
which gives
Cyclotron frequency
Chapter 18
Work function
 How to measure it and other surface
effects
 Atomic force microscopy
 Growth mechanisms (e.g., layer by layer)

Work function with dipoles
Corrections:
For
finite lattices in 3D,
there is a difference
compared to infinite lattices
   F  WS
Also,
non-equivalent crystal
faces will have different
dipole moments  work
function depends on the
orientation
Possibility is that the atoms
near surface move
Atomic Force Microscope
deflection sensor
force
sensor
tip
feedback
tip
sample
scanner
The force between tip and
sample is measured. The
interatomic force between
tip and sample deflect a
cantilever which carries the
tip. In this case the
vibratio
n
dampin
g
Data acquisition
Measure roughness to look at quality of film growth.
Oriented growth is referred to as epitaxy. Thin film growth is
controlled by the interplay of thermodynamics and kinetics.
Three growth modes:
• Layer-by-layer, or Franck van der Merve growth (typically ideal)
• Island, or Vollmer-Weber growth (very rough)
• Layer plus island, or Stranski-Krastanov growth (in between)
The occurrence of the individual growth modes is governed by the bond
strength between the atoms in the layer and the atom-substrate bonds.
Chapter 19
 Define
and compare ionic/covalent
bonds
 Predict properties of materials with
different kinds of bonds
 Guess what kind of bonding should
occur for specific materials and
back it up with things you’ve
learned about bonding
Covalent to Ionic Transition
Purely
Covalen
t
An ion is an atom in which the total number of electrons is not equal to
the total number of protons, giving the atom a net electrical charge.
Cation:
positive ion
Anion:
negative ion
Purely
Ionic
Covalent/Ionic Materials Have Strong Bonds
Property
Melting point
and boiling point
Electrical
conductivity
Hardness
Explanation
Very high melting points because each atom
is bound by strong covalent bonds. Many
covalent bonds must be broken if the solid is
to be melted and a large amount of thermal
energy is required for this.
Poor conductors because electrons are held
either on the atoms or within covalent
bonds. They cannot move through the
lattice.
They are hard because the atoms are
strongly bound in the lattice, and are not
easily displaced.
Brittleness
Covalent network substances are brittle. If
sufficient force is applied to a crystal,
covalent bond are broken as the lattice is
distorted. Shattering occurs rather than
deformation of a shape.
METALLIC PROPERTIES





All conduction e-s in a metal combine to form a sea of
electrons that move freely between cores  high electrical
and thermal conductivity. (but weak bond)
More electrons=stronger attraction. Means melting and
boiling points are higher, and metal is stronger and
harder.
The free electrons act as the bond (or a “glue”) between
the positive ions.
This type of bonding is nondirectional and
+
+
+
rather insensitive to structure.
As a result we have a high ductility of metals: +
+
+
the “bonds” do not “break” when atoms are
rearranged – metals can experience a
+
+
+
significant degree of plastic deformation.
van der Waals Bond





From charge fluctuations in
atoms due to zero-point
motion (due to Heisenberg
uncertainty principle); these
create attractive dipoles
Goes as p2/r6, short ranged
Always present, but
significant only when other
types of bonding not possible
Strength of 0.2 eV/atom
~1% of other bonds
Because force results from
dipole-dipole interactions, it
is short range, varying as r -6
graphite
Usually with oxygen,
fluorine or nitrogen
Hydrogen “Bonding”

Comes from the fact that it’s really hard
(13.6eV) to completely remove an
electron from hydrogen
• In H20, hydrogen covalently
bonds with nearby oxygen,
but it also has a hydrogen
bond with the next nearest
neighbor.
• Since the electron is pulled
toward the closer oxygen,
the positive charge is on the
outside and attracted to
Chapter 22 Objectives
 Model
simple phonon systems in 1, 2 or 3
dimensions
 Analyze the dispersion curves for real crystals
 Explain why neutron scattering is sensitive to
the phonon dispersion curve
 Use to understand experimental specific heat
 A 2 


K
 K

A2   2   2 B cos ka
m
 m

K
B(eika  e ika )  2 A
m

K
 B 
A(eika  e ika )  2 B
M
2

K2
 K
2  K
2
2
2 m    2 M     4 Mm cos ka
2
4
2
K B cos ka
m K
2
2 m   
K
 K

B 2   2   2
A cos ka
M
M


•Click on the picture to start the animation M->m
note wrong axis inKthe movie
1 
1
 2
  2K  

m
m M 
2
K
K
K
K
 2 2  2 2 4  4
cos 2 ka
Mm
M
m
Mm
 K K
   2 
 m M
4
A2
 2
K
M
K2

1  cos 2 ka  0
  4
Mm



2
2
sin2 ka

k
:
2a
 1 1
 1 1
  D    D   
 m M
 m M
2
2
2
1 1 
 1 1  4 sin ka
2
  K    K    
Mm
m M 
m M 
 2
K ,
m
 2
K
M
2
Atomic Displacement
B
m


k 0
A
M
Optic Mode
B
k 0  1
A
Atomic Displacement
Click for animations
Acoustic Mode
Dispersion curves of 3D crystals
•3D crystal: clear separation into longitudinal and transverse mode only possible in
particular symmetry directions
•Every crystal has 3 acoustic branches
1 longitudinal
acoustic
2 transverse
sound waves
of elastic theory
•Every additional atom of the primitive basis
further 3 optical branches
again 2 transverse
1 longitudinal
p atoms/primitive unit cell (
primitive basis of p atoms):
3 acoustic branches + 3(p-1) optical branches = 3p branches
1LA +2TA
(p-1)LO +2(p-1)TO
The energy of other methods
Electron
X-Ray
Neutron
λ = 1A°
λ = 1A°
λ = 2A°
E ~ 104 eV
E ~ 0.08 eV
E ~ 150 eV
interact with electron
Penetrating
interact with nuclei
Highly Penetrating
interact with electron
Less Penetrating
SUMMARY: NEUTRON DIFFRACTION
Major differences as compared to X-rays
– Neutrons are scattered by atomic nuclei
– The scattering intensity does not depend on sin, don’t interact
with electrons
– The energy is ideal for probing phonon dispersion
– Scattering by light atoms might be as intense as scattering by
heavy atoms
– Isotopes of the same element might have different scattering
power
– Scattering depends on magnetic moments of atoms
– Much larger beam and sample are required (due to lower beam
intensity). Neutron sources (like high brightness synchrotrons)
in the world are limited so neutron diffraction is a very special
tool.
Treating Phonons like QM Oscillator
Einstein Model = Heat capacity C can be found by
differentiating the QM average phonon energy
   Pn n
1
 
1


n


exp

n


/
k
T




B 
 
_
2
2

n 0 




 
1

exp

n


/
k
T


B 
 
2
n 0
 


n
1

n   n   
2

k BT
1

2
_

1

2
e

 / k BT
1
T
Einstein Model Fails at Low Temp
 Einstein
model also gave correctly a specific
heat tending to zero at absolute zero, but the
temperature dependence near T= 0 did not
agree with experiment. Why?
 Taking
into account the actual distribution of
vibration frequencies in a solid this discrepancy
can be accounted for
Einstein was aware. He
proposed this theory because it
was a simple demonstration
that QM could solve the
specific heat problem.
C of Diamond
Low Temperature Limit
At low T’s, only lattice modes having low
frequencies can be excited from their ground states,
since the Bose-Einstein function falls to zero for
small values of 

Low frequency
sound waves
at low T
0

k
a
Vk 2 dk
g    2
2 d

k 1
dk
1
vs    

k
 vs
d  vs
  vs k

vs 
k
 2 
V 2 
vs  1

g   
2 2 vs
vs depends on the direction
V2 1
V2  1 2 
g   
 g   
 3
2
3
2  3
2 vs
2  vL vT 
Debye approximation has two main steps
1.
2.
Approx. of any branch by a linear extrapolation
Ensure correct number of modes by imposing a cutoff frequency  D , above which there are no modes.
Since there are 3N lattice vibration modes in a crystal
having N atoms, we choose  D so that: 
D
Einstein
approximation
to the
dispersion
Debye
approximation
to the
dispersion
  vk
g ( ) 
 g ( )d  3N
0

V 1 2 D 2
( 3  3 )   d  3N
2
2 vL vT 0
V
1
2
( 3  3 )D3  3 N
2
6
vL vT
V
1
2
3N
9N
(

)

3

2 2 vL3 vT3
D3
D3
9N
 D3
2
g ( ) /  2
Debye approximation on specific heat at low T

1
    
2
e
0

 / kT

  g   d
1 
 1
d
2
2  k T 
Cv 
 V  2 kB  3  3   B 
dT 15

 vL vT  
3
The lattice heat capacity of solids varies
as T 3at low temperatures; Debye law.
Excellent agreement for
non-magnetic insulators!
Debye motivation: The exact calculation of DOS is hard.
Debye obtained a good approximation to the heat
capacity by neglecting the dispersion of the acoustic
waves, i.e. assuming   s k
for arbitrary wavenumber.
This approximation is still widely used today.
Review
Chapter 28 & 29:
 Distinguish
between direct and indirect
semiconductors and know the difference
between their optical absorption
 Explain photoconductivity
 Calculate carrier density, chemical potential, and
donor binding energies
 Apply doped semiconductors to explain the pn
junction or how to make a transistor 0 or 1
Many solids conduct electricity
There are electrons that are not bound but are able to move
through the crystal.
 = n e2  / m
( = 1/)
Conducting solids fall into two main classes; metals and
semiconductors.
 ( RT )metals ;106  108   m
and increases by
the addition of small amounts of impurities (semiconductors do
opposite).
Impurties have a big
effect on
semiconductors.
Why?
Impurities can add
states within the band
gap that makes
excitation (and thus
conduction) much
more likely (convert)
Electron and Hole Densities

To determine the number of carriers in each
band, we will modify what we derived for free
electrons

n   g ( ) f ( )d


Fermi-Dirac distribution function
f ( )
and v
  (E - ) 
fe (E)  exp

 kB T 
for c

What dependence on energy should the DOS
have?
Density of States
Conduction
Band
Assume: bottom of conduction band
and the top of valence band parabolic
conduction band
valence band
E = Ec + 2k2/2me*
E = Ev - 2k2/2mh*

1
Conduction g e (E) =
2 3 2me
2 
band
Valence
band
g h (E) =
1
2
2

* 3/2

2m 

* 3/2
h
3
Ec = Eg
( E  Eg )
( E )
1
2
1
2
Ev = 0
Valence
Band
D(E)
Conduction
Band
Valence
Band
0
Eg
E
g (E) 
V1
2 2
3/ 2 1/ 2
(2
m
)
E
3
Electron density in conduction band
Density of states
Distribution function
To find the electron
density of occupied
states:
1  2me

g e (E) =
2
2   2
*
3/2
1

 ( E  Eg ) 2


  (E - ) 
fe (E)  exp

 kB T 

Ne 
 g ( E ) f (E)dE
e
e
Eg
3
 me * k BT  2
N e  2
 exp (   E g ) / k BT ) 
2
 2 
As Ne=Nh(intrinsic) we can say: N e  N c N v exp  Eg / 2k BT ) 
This result is important; known as law of mass
action; find carrier density without µ.
True for both intrinsic & extrinsic semiconductors.