Physics 404
Solid State Notes
© J Kiefer 2006
1
I.
Crystallography......................................................................................................... 2
A.
Crystal Lattices ......................................................................................................... 2
B.
Atoms ......................................................................................................................... 8
C.
Diffraction by Crystals ........................................................................................... 15
II.
Lattice Vibrations—Phonons............................................................................. 22
A.
Continuous Media ................................................................................................... 22
B.
Lattice Waves .......................................................................................................... 29
III.
Band Theory ........................................................................................................ 37
A.
Free Electron Model of Metals .............................................................................. 37
B.
Bands ........................................................................................................................ 41
C.
Electrical Conductivity ........................................................................................... 48
1
I.
Crystallography
A.
Crystal Lattices
1.
Inter-atomic Binding
In a solid, atoms/molecules are bound together in a rigid structure. The nature of the
binding virtually determines the structure, as well as the physical properties of the solid.
a.
Binding energy
The binding energy is defined as the negative of the potential energy.
EB   V
For a system to be “bound” its total potential energy must be < 0, so EB  0 . The atoms
will be arranged in geometry and spacing to maximize the EB.
b.
Types of bond
Ionic
Very strong, but not directional. One or more electron is transferred from one atom to
another forming closed electron shells.
Covalent
Strong and directional, according to which orbitals (s, p, d, etc.) are involved in forming
the bond. A hybrid electron orbital is formed so that one or two electrons are shared by
two atoms.
Metallic
Strong, but not directional. The valence electrons are virtually free, forming a “sea” of
electrons in which the + metallic ions are immersed.
Hydrogen bond
Weak and directional. A hydrogen atom forms a kind of bridge between two molecules.
Water ice is the most familiar example. Sketch it. . . .
van der Walls
Weak and not directional. This a kind of average induced dipole-dipole interaction
between atoms having closed electron shells, such as Ar, Kr, Ne, etc.
2
c.
In general
Between any two atoms, the potential energy of interaction looks like this:
At large separation, r, the net force between two atoms is attractive and 
1
. For two
rn
1
1
. For two neutral atoms, it’s more like  6 . However, as atoms are
2
r
r
closer together, the repulsion between the electron clouds increases. That repulsion goes
1
like 12 . The sum of these two terms gives an effective V® like the one depicted above.
r
There is some r  Ro at which V is a minimum.
ions, it’s 
2.
Crystal Lattices
In a crystal, atoms are arranged in a repeating, or periodic, pattern called an array. We
first consider the geometry of a periodic array of points—the crystal lattice. The points
are called lattice points.
a.
Basis vectors
Let Rn be the position vector of a lattice point. For 3-dimensional crystal lattice, the
lattice vectors are
Rn  n1a  n2 b  n3 c
Where n1 , n2 , & n3 are integers and a , b , & c are the lattice basis vectors. The lattice
basis vectors must be non-coplanar, but need not be mutually orthogonal.
b.
Bravais lattice (bra-vay)
3
In a Bravais lattice, all lattice points are equivalent. In a non-Bravais lattice, not all
lattice points can be reached by an Rn , using a specified set of basis vectors.
We refer to such a lattice as a lattice with a basis. Pairs or clusters of points are
uniformly spaced, rather than individual points.
c.
Unit cells
The basis vectors form the edges of a parallelepiped (parallelogram in 2-d), like a block
out which the larger crystal is built.
Volume of the unit cell.
V  a  b  c 
In 2-dimensions, the volume is the area
A  a b
Obviously, a cubic or rectangular unit cell is more
convenient. V  a  b  c .
We can identify any number of repeating blocks of
various sizes, incorporating any number of lattice
points.
c.
Primitive unit cells
A primitive unit cell contains a single lattice point.
So, it is the smallest possible unit cell. The
4
primitive unit cell is still not unique. However, all primitive unit cells of a specified
crystal structure have the same volume (area). E.g., a square lattice
Note: these two primitive unit cells have lattice points at their 4 corners. The corner
1
lattice points are shared with 4 neighboring cells. So, each cell contains 4   1 lattice
4
point.
Note: we are not required, though, to put the origin of our coordinates on a lattice point.
3.
Crystal Systems
The distinguishing feature of a crystal system is that the structure is periodic—the
repeating units repeat in all directions, filling all of space. The key word there perhaps is
filling.
a.
Two dimensional
We might ask, how many distinctly different shapes of tile are there that will completely
cover a flat floor? It turns out, fortunately, that there are only five distinct shapes that
will completely fill a 2-dimensional space.
system
square
geometry
a  b ,  90 o
rectangular
primitive
rectangular
centered
hexagonal
a  b ,  90 o
a  ,  120 o
oblique
a  b ,  90 o
a  b ,  90 o
5
Levy, R. A., Principles of Solid State Physics, Academic Press, 1968.
b.
Three dimensional
Of course, there are a few more possibilities in 3-dimensions, since there is a 3rd basis
vector and 3 angles. Even so, there are only 14 distinct crystal systems in 3-dimensions.
See pages 36 & 37 in Blakemore; pages 8 & 9 in Omar.
c.
Symmetry
What property distinguishes one crystal system from another? It’s their symmetries.
Translation symmetry—slide by integer multiples of a , b , c and the environment is
exactly the same.
Rotation symmetry—Consider the shape of the unit cell. If the cell can be rotated about a
specified axis and appear unchanged, then it has rotation symmetry. Commonly, that
2
angle is
; the axis is called an n-fold axis.
n
Reflection symmetry—The unit cell is unchanged if reflected across a plane. A reflection
symmetry is designated by the letter m.
6
d.
No fives
Here’s something odd. It is impossible to construct an infinite crystal lattice with a unit
cell having 5-fold rotational symmetry. Consider the diagram below.
BC ought to be a lattice vector, but it is shorter than a. So, C cannot be a lattice point,
but it is a lattice point relative to the point A by rotation. So, we have a contradiction.
2
Put it another way, we cannot make an equilateral triangle with side a and angles
.
5
2
3
 .
5
e.
Symmetry of the basis
This is a second use of the word basis in connection with crystal symmetry. It refers to
the physical object associated with each lattice point in the crystal. In the simplest case,
that object is a single spherically shaped atom or ion. However, the object may be a
molecule (as it is in ice, say) or it may be 2 or more ions as in NaCl or it may be
something even more complicated. This has the effect of reducing the overall symmetry
of the crystal.
4.
Point Groups & Space Groups
A more systematic description of crystal symmetries. . .
7
B. Atoms
Now, we imagine an atom occupying each lattice site. In a simple crystal, the atoms are
all the same—argon or iron or carbon, etc. Each substance will form a distinct crystal,
according to the type of binding involved.
1.
Packing
The atoms will assume a structure that minimizes the binding energy.
a.
Cubic
Simple cubic
The side of the cubic unit cell is a  2R , where R is the radius of the atoms. We are
visualizing the atoms as hard spheres.
Now, consider the amount of “empty“ space in the cube: the cell is occupied by one
complete atom (sharing 8 with 8 neighboring cells).
The volume of the cubic unit cell is Vcell  a 3  2R  8R 3 .
3
4

The volume of a sphere is Vatom   R 3  .
3

V

The packing ratio is r  atom   0.5236 for the simple cubic structure.
Vcell
6
Body-centered cubic
The bcc structure has an atom situated at the center of the cube, as well as at its corners.
8
Therefore, the diagonal of the cube is 4R, so a 
16 R 2
 2.309 R . There are 2 atoms
3
4
2  R 3
8.3776 R 3

 0.6802 .
per unit cell. The packing ratio is r  3 3
a
12.317 R 3
Face-centered cubic
The fcc structure has an atom in the center of each of the six faces of the cube, as well as
at the corners. Hence, it’s the cube face diagonal that equals 4R, and the cube edge is
a
16 R 2
 2.8284 R .
2
The atoms in the cube faces are shared between two cells, so that the unit cell contains
4
4  R 3
6
 0.7405 .
1   4 atoms. The packing ratio is r  3 3
2
a
Of the cubic structures, the body-centered-cubic is the most closely packed.
b.
Hexagonal
In the hexagonal structure, the unit cell is a hexagonal right prism. There are two
hexagonal faces joined by six rectangular faces.
Simple hexagonal
The atoms are situated at the corners of the prism and in the centers of the hexagonal
faces.
9
The 12 corner-atoms are shared among 6 cells, while the two in the hexagonal faces are
12 2
  3 atoms per unit cell. [We
shared two cells each. Consequently, there are
6 2
could also visualize the atoms situated at the midline of the hexagonal prism rather than
6
at the corners. Then we’d say there were  1  3 atoms in the unit cell.]
3
The volume of the hexagonal prism is
1
6  b   a  a  cos 30 o  2.598ba 2  2.598  8 R 3  20.785R 3 . The volume of three atoms
2
4
4
 0.605 , which is less
of radius R is 3  R 3 . Therefore, the packing ratio is r 
3
20.785
closely packed than the fcc or bcc structures. However, there is a hexagonal structure
more closely packed than is the simple hexagonal.
Hexagonal closest packed
The hexagonal layers are offset, fitting into the troughs, as it were, of the layer below.
As a result, the inter-atomic spacing is a bit less than it is in the simple hexagonal
structure.
a
8
The volume of the hexagonal cell is 6   a  cos 30 o  a 
 33.941R 3 . There are 6
2
3
atoms contained in the hexagonal prism. Thus, the packing ratio is
4
6  R 3
25.133
3
r

 0.7405 , the same as the fcc structure. By the way, the packing
3
33.941
33.941R
ratio is also called the packing fraction.
2.
Layers
In a simple crystal, each lattice point is occupied by an atom, all of the same kind. The
atoms in the crystal are arranged in parallel planes. Because the crystal can be cut at
different angles, any number of atomic planes can be identified, not just those parallel or
perpendicular to the basis vectors.
We need a scheme for labeling the planes that will tell us the orientation of the specific
plane of atoms. Here is such a scheme:
a.
Direction
Consider the lattice vector joining any 2 lattice points. Let one lattice point be the origin
of coordinates. The lattice vector would be written
10
R  n1a  n2 b  n3 c .
Now imagine factoring out any common factors, if any from the integer triplet:
n , n , n  1  n , n , n  .
1
2
3
m
1
2
3
For example, R  6a  3b  9c
6, 3, 9 1  2,1, 3
3


We say that R is in the [2, 1, 3] direction. For instance, [1, 0, 0] corresponds to a , etc.
b.
Miller indices
Slice the crystal with some arbitrary plane. That plane intersects the a , b , and c
x y
z
directions at x, y, and z. Next take , , and and turn the ratios upside down:
a b
c
a b
c
, , and . Finally multiply by a common factor to obtain three integers.
x y
z
For example,
5
 2 1 1
5
a , y  2b , z  4c   , 2, 4    , ,   20  8,10, 5
2
2
 5 2 4
The numbers 8,10, 5 are the Miller indices of that crystal plane. Now the thing is, all
planes parallel to this plane have the same Miller indices.
x
In practice, we are interested in planes with lower numbers, because they are populated
by more atoms per unit area. E.g., 1, 0, 0 , 1,1,1 , etc.
c.
Plane spacing
Planes having the same Miller indices h, k ,  are parallel to each other and have uniform
spacing between them. They also have the same number of atoms per unit area. We are
interested, or will be interested, in that inter-planar distance, d hk .
11
All we need to figure out is the distance of the reference plane in the unit cell from the
origin.
We envision the vector from the origin perpendicular to the plane. The direction cosines
d d d 
of that vector, d , are  , ,  .
x y z
2
2
d2 d2 d2
1
1
1
1 h k  
 2  2  1  2  2  2  2         
2
x
y
z
d
x
y
z
a b c
1
d
2
2
2
h k  
       
a b c
2
Therefore, the Miller indices, along with the lattice parameters, give us the atomic plane
spacing. Subsequently, we’ll see that the atomic plane spacing determines the diffraction
patterns produced by a crystal.
12
3.
Some Real Crystalline Substances
a.
Sodium chloride
Binding is ionic, between Na+ and Cl-. So, the binding is strong, and the crystal is hard.
Because the ions are spherically symmetric, the crystal structure is cubic, with +/- ions
alternating. In fact, the structure consists of two interpenetrating fcc lattices, offset by
1
a from each other.
2
b.
CsCl
This crystal is also ionic, but in this case the crystal structure is two interpenetrating
simple cubic lattices, forming a bcc configuration. This illustrates how confusing crystal
descriptions can be. In the case of NaCl above, there are two lattices, an fcc lattice of Na
and an fcc lattice of Cl. The Cls, for instance, lie between Na ions along the cube edges.
In the case of CsCl, however, a Cs ion lies at the center of a simple cubic lattice of Cl
ions (or vice versa), hence the term bcc configuration. The crystal might also be
described as being simple cubic with a two-atom basis.
It is evidently the relative sizes of the filled electron shells of the ions that distinguishes
the NaCl from the CsCl structures. Other ionic compounds take on the Sodium chloride
or the Cesium chloride structure. [Table 1.2 in Omar]
c.
Metallic elements
The metal ions in a metallic crystal take on fcc, bcc, or hcp structures, similarly to the
inert elements (Ar, Kr, etc.). The metal crystal may change from one structure to another
with changes in temperature. The metal ions are all positive, yet are held in place by the
surrounding “sea” of almost free electrons.
d.
Diamond structure
The diamond structure is an fcc structure, with a basis of two atoms. In the case of
diamond itself, both those atoms of the basis are Carbon. In other compounds, such as
Zinc sulfide (ZnS), the basis is 1 Zn & 1 Sulfur atom. In relation to the cubic unit cell,
1 1 1
the two atoms of the basis are at (0,0,0) and  , ,  along the cube diagonal. Each
4 4 4
atom is surrounded by four nearest neighbors that form the vertices of a regular
tetrahedron. The binding is covalent—the shapes of the valence orbitals determine the
tetrahedral coordination.
13
Similarly to the NaCl or CsCl structures, water ice forms in what is called the wurtzite
structure which consists of two interpenetrating hcc lattices, rather than cubic. The water
molecules form tetrahedrally coordinated hydrogen bonds with each other. The term
wurtzite is the name of an alternative structure for ZnS.
14
C. Diffraction by Crystals
The atoms in a crystal are arranged in a periodic array. We can identify any number of
identical parallel planes cutting through the crystal at specific angles relative to the basis
vectors. The density of atoms in these planes may vary widely.
1.
Elastic Scattering of X-rays
X-rays scatter from the electrons in the crystal. Let nr  be the electron number density
in the crystal. This will be greatest in the vicinity of an atom.
a.
Fourier analysis
The nr  is a periodic function of r : nr  Rn   nr  , where Rn  n1a  n2 b  n3 c .
Firstly, consider a Fourier expansion in one-dimension: (p is an integer)

px
px 
nx   no    C p cos 2
 S p sin 2

a
a 
p 0 
p
is a point in Fourier space, or in the reciprocal lattice of the crystal. Cp and Sp are
a
real coefficients. Alternatively, we might use the complex exponential form:
2
Where np is complex and n* p


px 
nx    n p exp  i 2
,
a 
p 0

 np .
In three dimensions, there is periodicity in three directions (but not necessarily the
Cartesian directions) along a , b , & c . So the Fourier expansion is in three dimensions:
nr    nG exp iG  r 
G
G is a vector. The question is, what G will keep nr  invariant under a translation of
the crystal by Rn ?
b.
Reciprocal lattice vectors
We will propose some reciprocal lattice vectors, and show that they meet the condition
posed by the question above.
Proposal: reciprocal lattice vectors shall be defined thusly,
b c
a b c
c a
b *  2
a b c
a b
c *  2
a b c
*
*
*
*
*
Notice that a  a  2 , a  b  0, & a  c *  0 , etc.
a *  2
15
The vectors Gn  n1 a *  n2 b *  n3 c * generate the points of the reciprocal lattice. [In
Omar, Rn  R   1a   2 b   3 c at this point.]
The reciprocal lattice is a lattice of points in Fourier space. Here’s why these reciprocal
lattice vectors are the same G that appear in the Fourier expansion:
nr  R    nG exp iG  r exp iG  R 
G
However,



exp iG  R   exp i n1 a *  n 2 b *  n3 c *   1 a   2 b   3 c 
 exp i 2 n1 1  n2  2  n3  3 
1
Therefore, nr  R   nr  if the G in the Fourier expansion are reciprocal lattice
vectors.
c.
Elastic scattering
From a single point scattering center, such as an electron, incident plane waves produce
outgoing spherical waves. In elastic scattering, the scattered waves have the same
energy as the incident waves.
Now, consider scattered spherical waves from two electrons, one at the origin and the
other at a relative position r .
16
The path difference between the two outgoing waves is    2   1  r    r   o . The
phase difference between the two is exp  i    o   r   exp  i  r  . We wish to
determine for which scattering vector,  , will there be constructive interference
between the two scattered x-rays.
The amplitude of the scattered wave is proportional to
F   nr  exp  i   r dxdydz
Substitute for nr 
F 
G

G
n
n
exp i G  r exp  i   r dxdydz
G
G
exp i G     r dxdydz
F is called the scattering amplitude. When is F large, when small? Well, it’s large only
when   G and very small otherwise. That is, constructive interference among the
scattered waves occurs in the directions    o   such that   G , a reciprocal
lattice vector.
d.
Bragg condition
2
2
For elastic scattering,    o .
2
 o     o
o  G
2
 o
2
2
 o2  2 o  G  G 2   o2
2 o  G  G 2  0
2 o  G 2  G 2 , since G  G .
e.
Related to Miller indices
Now we wish to connect the reciprocal lattice vectors to the planes in the crystal, which
we previously labeled with the Miller indices h, k ,  .
17
Recall the definition of the Miller indices. The basis vectors define three coordinate axes
(not necessarily mutually perpendicular). We’ll use x, y, & z nonetheless to be distances
along those three coordinate axes. For example, in the figure above, the plane of interest
intersects the axes at x, y, & z.
x  2a
3
y b
2
zc
x y z  3 
 a , b , c    2, 2 ,1

 

1 2 
 , ,1  6  3,4,6   h, k ,  
2 3 
1 1 1
We can see that h, k ,     , ,  . Now, consider the vectors, u & v , which define
x y z
the plane of interest. u   xa  yb and v  yb  zc . Consider, too, a reciprocal lattice
vector defined as Ghk  ha *  kb *  c * . Our claim is that this Ghk is perpendicular to
the plane of interest, the h, k ,  plane.

u  Ghk   xa  yb   ha *  kb *  c *

 xh2  0  0  0  0  yk 2  0
 2 xh  yk 
 2 1  1  0

v  Ghk  yb  zc   ha *  kb *  c *
Therefore, Ghk
 2  yk  z   0
is perpendicular to the h, k ,  plane.

G
2
Further, the spacing of the h, k ,  planes is d hk  xa  nˆ  xa  Gˆ hk  xa  hk 
.
Ghk
Ghk
18
2.
Laue Equations
Here’s an alternative expression of the diffraction condition.
a.
Scattering vector
The phase difference between waves reflected from two parallel planes is
ˆ  r 
ˆ  r  
ˆ
      r 
2
1
o
ˆ  n . The phase angle is   2 r  ˆ .
Constructive interference occurs if r  

b.
Laue equations
Now suppose r is a lattice (or basis vector), say a . Then the phase angle becomes
2
2
a 
a  ˆ  2  h . By similar tokens, b 
b  ˆ  2  k , and


2
ˆ  2   . The h, k ,  are integers.
c 
c  

Consider the first one.  a 
2

a  ˆ  2  h .
a  ˆ   h


a ˆ cos  a , ˆ   h
a ˆ    h
where  is the direction cosine.
ˆ 
However, 
2 c  sin 
.

2 o sin 
2 a  sin 
2 b  sin 
  . Similarly,
  and
, so
h
k
o
c.
Miller indices
Of course, these h, k ,  are just the Miller Indices of the lattice planes in which the two
scattering centers (atoms) lie. That’s why we used the same symbols in the paragraph
above.
19
This claim can be demonstrated by considering a two dimensional case. N is the normal
to the h, k ,  plane.
By similar triangles,
a
h  Nb
b
Na
k
h
Na
 a
Nb k
b
h

a
Likewise,  
k

and   .
b
c
3.
Brillouin Zones
Here is yet another way of visualizing the diffraction condition.
a.
In the reciprocal lattice
2 o  G  G
2
 0 is the diffraction condition. The reciprocal lattice vector, G , is the
same as    o   . Here is a rectangular reciprocal lattice:
Any x-ray whose  o is such that  o  G 
G2
will be reflected/diffracted by the crystal.
2
20
b.
First Brillouin zone
Brillouin zones are areas/volumes enclosed by perpendicular bisectors of reciprocal
lattice vectors, Gn
The first brillouin zone is defined by the perpendicular bisectors of the reciprocal basis
vectors, a * and b * (for a 2-dimensional lattice).
The first Brillouin zone is the smallest volume enclosed by perpendicular bisectors. It
forms the primitive unit cell of the reciprocal lattice. Any  o that reaches from the
origin of the reciprocal lattice to the zone boundary will diffract. The zone boundaries
are formed by the bisectors.
c.
More zones
Second, third and etc. zones are defined by the bisection of longer reciprocal lattice
vectors, such as a *  b * and so on.
21
II.
Lattice Vibrations—Phonons
A.
Continuous Media
1.
Elastic Waves
If the wavelength of an elastic wave is much longer than the atomic spacing, then we
treat the solid as a continuous medium.
a.
Dispersion relation
Consider a narrow bar of solid material.
The bar has density  and cross sectional area A . Let there be a longitudinal wave
traveling in the bar, so that the elastic displacement at x is u(x) in the  x -directions.
Applying Newton’s 2nd “Law” to the segment of bar dx at x . . .
ma  F
 2u
 F x  dx   F x 
t 2
We can write this in terms of Young’s Modulus and the stress, e, & strain, S.
 Adx
 2u
 S  x  dx)   S x A
t 2
 2 u dS
 A 2 
A
dx
t
 2u
de
 2u
 2 Y
Y 2
dx
t
x
2
2
 u  u

0
t 2 Y x 2
 Adx
e
du
dx
S  Ye 
F
A
This is a wave equation, which has the usual solutions of the form u  Ae i qx t  . The
parameters q and  are evaluated by plugging the solution into the wave equation.
q
Y
q

22
2

The expression for  q  is called the dispersion relation.
The quantity
Y

is the
wave speed, vs, and is the slope of the dispersion relation graph.
The phenomenon called dispersion (as in prisms) occurs when the dispersion relation is
not linear—the slope is not constant with q.
b.
Modes of vibration—density of states
Now, apply periodic boundary conditions to the bar (we are aiming toward crystals).
That is,
u x  0  u x  L 
Ae iq0  Ae iqL
1  e iqL
Therefore, the product qL  n2 , n  0,1,2,. These are plotted as uniformly
spaced points on the q-axis.
Each “allowed” value of q is a mode of vibration corresponding to a particular
2
2 L L


 .
q n n2  n
The density of states is the number of allowed qn that lie in the interval dq on the q-axis.
L
dq , since the modes are uniformly spaced.
That number is
2
In terms of  , rather than  ,
g  d  2
g   
L

L
dq
2
1
 d
23
dq
With   v s q , we get g   
L 1
  constant . At least, g does not depend on  ; it
 vs
does depend on the properties of the medium: L, Y,  .
c.
Extension to three dimensions
Consider elastic waves traveling in 3 dimensions in a cube of side L.
i q x  q y  q z 
u  Ae x y z  Aeiq r
i q L  q L  q L 
With periodic boundary conditions, we require that e x y z  1 . So the allowed
modes are those for which
2
2
2
qx  n
qy  m
qz  
L
L
L
These form a uniformly spaced grid of points in q-space.
How many such points lie in a sphere of radius q? Divide the volume of the sphere by
the cubical volume occupied by each point.
V = L3
24
4
 q3
V 4 3
3

q
3
2 3 3
 2 


 L 
Take the derivative with respect to q.
V
g q dq 
4 q 2 dq
2 3
This is the number of points in the shell of radius q and thickness dq. Again, in terms of
the frequency,   vs q ,
2
   d
V
g  d 
4  
.
3
2   vs  vs
This density of states is proportional to the frequency squared.
Finally, we multiply the density of states by 3, since there are three vibrational modes for
each value of q. (This is only approximately correct, since in fact the
g   are not really identical for all three modes. Particularly, the vs is not
quite the same for transverse and longitudinal waves.)
2.
Specific Heat
According to the First “Law” of Thermodynamics, U  Q  W , where U is the internal
energy of a system (in this case a chunk of crystal), Q is the energy transfer into the
system in the form of heat, and W is the work done on the system.
a.
Classical theory of specific heat
If no work is done, then U  Q , and the constant volume specific heat is obtained from
Cv 
Q  U 


T  T V
It is found experimentally that
25
The classical assumptions are that the atoms of a crystal vibrate on their lattice sites like
harmonic oscillators. Further, at equilibrium, the average energy of each oscillator is the
same, namely kT , where k is the Boltzmann constant and T is absolute temperature. A
mole of 3-dimensional oscillators, then, will have an internal energy of U  N A 3kT .
From this, we get (R is the gas constant.)
U
 3 N A k  3R ,
T
which is independent of temperature. Obviously, this theoretical expression fails to
match experiment at “low” temperature.
Cv 
b.
Einstein solid
Let’s assume once again that the crystal is composed of independent harmonic
oscillators. Let’s assume further that the energies of the harmonic oscillators are
quantized. An atom in the crystal lattice can gain or lose energy only in discrete steps of
 . The average energy of an ensemble of harmonic oscillators in contact with a
thermal bath is

 
 e
n0


n
kT
n
e

n
.
kT
n 0
(The crystal is in equilibrium with the thermal bath. Its temperature is constant the
probability that a specific oscillator has energy  n is given by the Boltzmann
distribution.)
26




 n
ln   e kT 
 1 

   n
 kT 
 





 n
kT
ln   e

 1   n

 
 kT 


1
ln 
 
kT
 1 
   1  e
 kT 




 
 
1

 1  e kT 


  1   1  e   kT
 
 kT 
 
 
 1  e kT  11  e kT 






 e
 
 
1 e

2




 e

kT
kT
kT

e kT  1
The total energy of a mole of 3-dimensional oscillators is, then,
N 3
.
U  A
e kT  1
Finally, the specific heat is

E
e kT
e T
 U 
  
 E 
Cv  

3
R
.
  3N Ak 



2
2
 T V
 kT   e kT  1
 T   E T



 1
e





(The Einstein Temperature is defined as  E 
. The Einstein Temperature depends
k
on the crystal substance, since it has the oscillator frequency in it.)
2
2
While superficially this graph resembles the experimental graph, in fact it has the wrong
shape at very low temperature. Experiment shows that Cv  0 as T 3 . So, we need a
further refinement of the theory of specific heat.
c.
Debye theory of specific heat
The oscillators do not vibrate independently. Rather, there are collective modes of
vibration in the crystal lattice. Eventually, we will solve for the vibrational states of
coupled oscillators. At this point, however, we’ll return to the elastic waves propagating
in the solid. We consider that the energy residing in the elastic wave of frequency  is
27
quantized. Secondly, there is an upper limit of the frequency that can exist in the
crystal—the cut-off frequency.
So, consider sound waves propagating in the crystal, with the dispersion relation   vs q .
The total vibrational energy of the crystal will be
U     g  d .

The average energy of a mode is still    
e
g   

kT
1
and in a continuous medium
3V  2
3V

. Therefore, U 
 2 
d . Now, what is the range of
2
3
2 3 
2 vs
2 vs
e kT  1
frequency? Not 0,   , but 0, D  , where  D is the Debye frequency, or cut-off
frequency. The cut-off frequency arises because the shortest possible wavelength is
determined by the inter-atomic spacing. Put another way, the maximum possible number
of vibrational modes in the crystal is equal to the number of atoms (let’s say a mole) in
the crystal, times 3. I.e.,
D
 g  d  3N
A
0
D

0
3V  2
d  3 N A
2 2 vs3
3V  D3
 3N A
2 2 3vs3
6 N A 2vs3
V
With this as the upper integration limit, he total energy becomes
D3 
3V

3V
U
 2 
d 
2 3 
2 vs
2 2vs3
e kT  1
In turn, the specific heat is
D

D
 3
e

0
3
kT
d .
1

 T  D   
e kT


d


9
R
d .
2
0   kT 2
     kT 




D  0 

kT
 1
 1
e
e




 D
In the last expression, the Debye temperature,  D 
, is introduced.
k
3V  2
 U 
Cv  


2 3
2
 T V 2 vs kT
 4e
kT
N 
      
   D     vs3  6 2 A 
V 
 k  k 
3
3
 k V
vs3  D2 3
6  N A
3
3
3
D
28
4
Let’s look at the high and low temperature limits.
For T   D ,

d  1   D  , so that Cv  3R .
3 T 
3
3
4 4
12 4  T 
  T 3 .
, so that Cv 
R
For T   D , D  , and   d 
15
5

 D
Evidently, the Debye version matches the experimental temperature dependence of the
specific heat at low temperatures, as the classical and Einstein theories do not.
We can relate the Debye temperature to the Young’s modulus of the crystal, through the
wave speed.
D   2 N A 
D 
 vs  6

k
k 
V 
Now, the density is  
d.
1
3
 Y  2 NA 

  6

k  
V 
N AM
, so qualitatively,  D 
V
1
3
Y
. (M is the atomic mass.)
M
Reduced temperature
T
is used when plotting the heat
D
capacity. The Debye temperature contains the information about the particular crystal
substance. When plotted versus reduced temperature, the CV curve is identical for all
substances.
Often, a quantity called the reduced temperature, T * 
B. Lattice Waves
Of course, the crystal is not a continuous medium, but is composed of separated atoms.
Determination of the normal vibrational modes of the system means solving the equations
of motion.
1.
One-Dimensional Monatomic Crystal
We’ll start with a discussion of a one-dimensional crystal consisting of a linear chain of
identical atoms of mass m and lattice spacing a. The displacement of the nth atom from
its equilibrium position is un. The atoms interact with their nearest neighbors only, via a
Hook’s “Law” (linear restoring) force--  u .
29
a.
Dispersion relation
The net force on the nth atom is
Fn   un 1  un    un  un 1 
Fn   un 1  2un  un 1 
In Newton’s 2 “Law”
nd
 un 1  2un  un 1   m
d 2u n
dt 2
We’ll assume a solution of the form un  un,oei  t  qt   un,oei  t  naq . Similarly for un+1,o
and un-1,o. We may as well let un+1,o = un-1,o = un,o = uo. If the assumed solution is
substituted into the equation of motion, we obtain the dispersion relation.
uo ei  t  naq iqa  2ei  t  naq  ei  t  naqiqa  m 2uoei  t  naq




 eiqa  2  eiqa  m 2
qa
 m 2
2
4
qa

sin
m
2
 4 sin 2
Now, in one dimension the density is  


1 iz
e  e  iz
2i
1
sin 2 z   eiz  e  iz
4
sin z 
Using an identity, this equation becomes


2
m
Y
and the speed of sound is vs 
for
a

longitudinal waves. This gives us the expression
2
qa
   v s sin
.
a
2
30
2 qa
vs
 vs q , which is a straight line.
a 2
2 
2
 2
 ,   vs sin  vs , which is a constant.
For q 
2a a
a
2 a
For q  0, sin z  z , so that  
b.
Group velocity
The phase velocity of a traveling wave is v p 

q
. The group velocity is vg 
d
. For
dq
the one-dimensional lattice, we obtain
d 2
qa 
qa
vg 
 vs sin
  vs cos ,
dq  a
2 
2
While on the other hand
2v
qa
.
v p  s sin
qa
2

2v
For q  0, v p  vg  vs . For q  , v p  s , but vg  0 . Now, vg is the speed with
a

which energy is transmitted by the wave. If vg = 0, then the wave is a standing wave, and
no energy is transported. This is the same as saying that a wave of wavelength less than
the lattice spacing, a, cannot propagate in the crystal. Notice, too, that the cut off at

q   corresponds to the boundary of the 1st Brillouin zone in the crystal reciprocal
a
lattice. (by the way, Brillouin’s first name was Le’on.)
2.
Diatomic One-Dimensional Crystal
a.
Dispersion relation
There are 2n atoms, alternating masses m and M. The lattice constant is 2a, with a basis
of two atoms.
The net force on the atom of mass m is
d 2u 2n
F2 n  m
  u 2 n 1  2u 2 n  u 2 n 1 
dt 2
On the atom of mass M,
F2 n 1
d 2 u 2 n 1
M
  u 2 n 2  2u 2 n 1  u 2 n 
dt 2
31
Assume the solutions u2n  u 2n,o e i  t 2nqa and u2n1  u2n1,o e i  t ( 2n1) qa  , and plug into the
equations of motion-  2 mu2 n ,o   u 2 n 1,o e iqa  e  iqa  2 u 2 n ,o
  Mu 2 n 1,o   u 2 n ,o
2
Rearrange
2   mu
2
2 n ,o

e
iqa
e
iqa

  2 u
2 n 1,o
 2 cos qa  u 2 n 1,o  0


 2 cos qa  u 2 n ,o  2   2 M u 2 n 1,o  0
These are two simultaneous linear equations with u2n,o and u2n+1,o as the unknowns. The
determinant of the coefficients must vanish.
2   m
2
 2 cos qa 
 2 cos qa 
2   M   0
2
Expand the determinant and solve for  2 with the quadratic formula.
1 1 
       
m M 
There are two solutions.
2
(+)
2
1 1 
4 sin 2 qa
   
Mm
m M 
1 1 
       
m M 
2
2
1 1 
4 sin 2 qa
   
Mm
m M 
And
1 1 
1
1
(-)          
m M 
m M
These yield two branches on the dispersion curve graph.
2
32
2

4 sin 2 qa
 
Mm

The upper branch is called the optical branch while the lower branch is called the

acoustic branch. The gap at 
represents forbidden frequencies—they do not
2a
propagate along the crystal.
b.
Physical interpretation of the branches
For the acoustical branch, the two atoms of the basis oscillate with the same amplitudes
and in phase. In the optical branch, the two atoms oscillate  out of phase, with
amplitudes inversely proportional to their masses, so that the diatomic center of mass
remains still.
3.
Normal Modes
Those curves in the dispersion curve are not continuous curves, but rather dotted lines.
a.
Boundary conditions
Recall the one-dimensional continuous medium. We required that u n ( x)  u n ( x  L) ,
2 j
, j  1,  2,  If we count
meaning that the allowed q-values are discrete: q j 
L

up how many q-values there are from q = 0 to q 
, we get N, the number of atoms in
2a
the unit of crystal of length L. (N will need to be even, since the basis of the onedimensional diatomic crystal is 2.) There are always N normal modes, then number of
degrees of freedom in the crystal. However, this total of N is distributed in the branches
of the dispersion graph.
33
b.
Density of states
The density of states is the number of vibrational modes in the interval ,   d  .
For the 1-dimensional case, we found g   
dimensional lattice we found  
density of states is g   
Notice that for    n 
L

m 2
4 a
L 1
. For the monatomic 14 d
dq
4
qa
d
, whence
sin

m
2
dq
4 a
qa
cos . Thus the
m 2
2
1
qa
 sec
.
qa
2
cos
2
4
, there are no vibrations.
m
c.
Extension to three dimensions
In three dimensions, the atoms can vibrate in three (3) directions—along the x-, y-, and zcartesian directions, for instance. There are 3N normal modes. The dispersion relation is
different for each kind of motion, so there are three branches in the dispersion relation.
34
The density of states typically looks like the figure below, and possibly different for each
branch of the dispersion relation.
For a diatomic three-dimensional lattice, there are six branches. Imagine even more
branches with more different atoms in the crystal.
4.
Phonons
Regarding each atom as a harmonic oscillator, we said that each oscillator has energy
n . Let us now regard the energy of collective lattice vibrations similarly as being
quantized, and call the quanta phonons.
a.
Quantized lattice waves
At a specified temperature there will be, on average, a certain amount of energy in waves
of frequency  . The average number of phonons of that frequency is
1
,
n  
e kT  1
and the average energy of those phonons is E    n  .
b.
Phonon density
The phonon density is the number of phonons per frequency mode.
kT
kT
V 3
q , whence

. At the same time, n 
 v s q
2 2
dn 3V 2

q .
dq 2 2
So to obtain the total number of phonons of all frequencies or wave numbers, we have to
integrate
qmax
2
3kTVqmax
kT 3V 2
.
N total  
q
dq

2
2

v
q
2

4


v
s
s
0
The total number of vibrational modes is equal to the number of degrees of freedom—
3N. [N is the number of atoms in the crystal.] The number of phonons per unit volume
of the crystal is the following:
N
3kT 3N
3 T
.
N p  total 
 3N
V
2v s qmax
2 D
[Note that E  kT when T   D .]
When E  kT , then n 
35
c.
Phonon scattering
We think of the crystal being filled with phonons. An incident photon (x-ray, for
instance, or infrared) entering the crystal interacts with the phonons. Perhaps the photon
is absorbed, and a phonon of the same energy is created. Perhaps a phonon vanishes and
a scattered photon is created. Perhaps an incident neutron absorbs or emits a phonon by
colliding with an atom of the lattice.
36
III. Band Theory
A. Free Electron Model of Metals
We are seeking to construct a model for the physical properties of metals, particularly
electrical and thermal.
1.
Classical Electron Gas
We assume that when metallic atoms form a crystal lattice, the valence electrons of each
atom are liberated. They become free to move throughout the crystal—they are called
conduction electrons.
a.
Free electron gas
Consider a collection of free particles confined to a box. The particles are charged (-e)
10 29
and there are a lot of them: N  3 . The conduction electrons comprise a dense
m
plasma.
b.
Electrical conductivity

Consider an electron in an electric field, E . The electron will be accelerated. However,
as there are other particles present, the electron will at some point collide with another
particle, and lose its kinetic energy. The average time between such collisions is the
collision time,  .
We represent the effect of the collisions by a frictional-style force, assumed to have the

v
form m . The equation of motion for the electron undergoing a series of collisions


while being accelerated by the electric field is [component in the direction of E .]
dv
v
m
 eE  m .
dt

dv
 e
 0 , whence v 
E  v d , known as the drift velocity.
In the steady state case,
dt
m
The drift velocity is the net movement of conduction electrons through the crystal in the
direction of the externally applied electric field. That will be the electric current. The
current density is J   Nevd in coulombs per cubic meter.
37
  e 
Ne 2
J   Ne
 E 
E  E , which is Ohm’s “Law.” Let us put in some
m
 m 
characteristic numerical values.
m  9.1  10 31 kg
N  10
29
m3
  10 14 sec
(see Table 4.1 Omar)


Ne 2 10 29 m 3 1.6  10 19 C
2.8  10 7
14


10
sec

m
m
9.1  10 31 kg
This is the conductivity,  . So, the simple electron gas model can reproduce the
observed behavior of conductivity in metals. The still open question is, how did we
know what collision time to use?
2
c.
Collision time
We presume that the electron is colliding with the lattice atoms (ions). So that is hardly a
free electron model, really. We do assume that the electron is free between collisions.
Most crudely, we could imagine an electron bouncing from one lattice atom to the next.
But this actually underestimates the mean free path. Instead, we have to picture
deBroglie waves being diffracted by the crystal lattice. At any moment, the deBroglie
h
wavelength of a conduction electron is  
, where vr is the random thermal speed
mvr
of the electron, which v r  v d . If the lattice were perfectly periodic, the electron would
propagate (scatter) through the lattice without colliding, in effect, that is without losing
kinetic energy. The collision time would be infinite. If there are imperfections in the
lattice, the resonance is interrupted; the electron stops. The mean free path is   vr .
We have estimates for the mean free path (typical number of crystal defects observed)
and the velocity (from thermodynamics), from which we derive an estimate for the
collision time.

10 8 m
  6
 10 14 sec
m
vr 10
sec
10
10 m
 10 16 sec , where crystal atomic spacing is typically
Notice that this 10 14 sec  6
10 m
sec
a few Ångstroms.
2.
Quantum Mechanics—Electrons in a Box
Sadly, the simple free-electron model deos not give the correct temperature dependence
of  . Also, the free-electron model does not give the correct contribution of the
3
conduction electrons to the total specific heat of the metal; i.e., not R .
2
38
a.
Quantum mechanics—quick review
We still assume that the conduction electrons are free, but are confined to a line segment
of length, L. The time independent Schrödinger Equation for a free particle is
2 d 2

  E .
2m dx 2
We assume a solution of the form
   o e iqx .
n
The boundary conditions are  (0)   ( L)  0 , so q 
, as usual.
L
2
2q2 2    2
The quantized energy levels are E n 

  n .
2m
2m  L 
If we extend this to a cubical box, then
2 2

   E
2m

   o e iqr o e

q x  nx
En 
L

i qx x q y y  qz z
, qy  ny
2

L

, q z  nz


L

   2
  n x  n y2  n z2 .
2m  L 
2
1
. Finally, the Pauli
2
Exclusion Principle says that no two electrons in a system can have the same set of
quantum numbers.
In addition to nx, ny, and nz, there is a spin quantum number, m s  
b.
Fermi energy
Because of the exclusion principle, all the electrons in the box cannot have the same
energy. The energy levels will be filled from the lowest level upward. The highest level
occupied by an electron is called the Fermi Energy, EF.
The question is, given N and L, what is EF? Put another way, N electrons will fill energy
levels up to what n2?
1 4
NL3    n 3f  2
8 3
 3 NL3 

n f  




39
1
3

2
Plug this into En; E F  E n f
2    2 2

  n f 
3 2 N
2m  L 
2m

2
3
.
c.
Density of states (again)
At T = 0 K, no electrons have energy E > EF. All the states E n  E F are filled, or
occupied by an electron. Therefore,
1  2mE 
0 g ( E )dE  N  3 2   2 
d
1 2m 1
g E  
N 2 2 E 2.
dE
3 
E
3
2
g(E) is the number of energy states per unit energy per unit volume. Since all the states
are occupied (up to EF), it is the same as the number of electrons per unit energy per unit
volume. Consequently, the average energy of the N electrons is
EF
E
 Eg ( E )dE
0
EF
 g ( E )dE
0
EF

E
0
EF
E
3
1
2
dE
2
dE
0

d.
3
EF
5
Zero point energy
Classically, E  0 at T = 0K. But now, E 
3
EF , all because no two electrons can be in
5
exactly the same state.
40
3.
Fermi-Dirac Statistics
a.
Ferni-Dirac distribution
At T = 0, all energy states up to EF are occupied by electrons, one to each energy state. A
graph of the probability that the state of energy E is occupied looks like this:
Now, if T > 0, only those electrons in states of E  EF  kT can jump to an energy state
higher than EF. The distribution would look like this:
The functional form of this distribution is
1
f ( E )  E  EF 
.
kT
e
1
b.
Specific heat of the conduction electrons
kT
Approximately, only
of the conduction electrons are affected by an increase in
EF
temperature. Each of them, on average, absorbs an energy kT. The average thermal
kT
k 2T 2
kT  N A
emergy per mole of electrons is roughly E  N A
. The constant
EF
EF
dE
k2
 2N A
T  T , and << R. For a gas of particles,
dT
EF
even confined to a box, we’d expect 3R.
volume specific heat is CV 
B. Bands
Now we need to incorporate the influence of the lattice atoms on the conduction
electrons. That is, in addition to being confined to a box of volume, V, the conduction
electrons interact with the atoms that form the crystal.
41
1.
Bloch Function
Apart from small vibrations, the atoms are arranged in a periodic lattice.
a.
Schrödinger equation for a crystal
2 2


   V (r )  E
2m


V (r ) is the potential energy function and is periodic. If R is a lattice vector, then
 

V ( r  R)  V ( r ) .
b.
Bloch theorem



 

Claim:  q (r )  e iqr u q (r ) , where u q (r  R)  u q (r ) .

 

Proof: We can always write  (r )  f (r )u (r ) , where u (r ) is a periodic function. What

 2
determines f (r ) ? Well, physically we require that  (r ) be periodic. Consequently,


  2
 2
f (r  R)  f (r ) , which is true only if f (r )  e iqr .



The  q (r )  f (r )u q (r ) is called a Bloch Function. The Bloch function is a traveling

wave, modulated by the u q (r ) . The electron is shared in effect by the whole crystal, not
localized close to any one atom.
2.
Energy Bands
“Solve” the Schrödinger equation
2 2


   V (r )  E
2m



iqr
Plug in  q (r )  e u q (r ) .
a.

 2 2 iqr

 



 e u q (r )  V (r )e iqr u q (r )  Ee iqr u q (r )
2m
Piece by piece. . .

 d
d iqr

e u q (r )  u q iq x e iqr  e iqr
uq
dx
dx
2




d 2 iqr

iq r
iq  r d
iq  r d
iq r d
e
u
(
r
)

u
iq
iq
e

iq
e
u

iq
e
u

e
uq
q
q
x
x
x
q
x
q
dx
dx
dx 2
dc 2

 
d
d2
   u q q x2  2iq x
u q  2 u q  e iqr
dx
dx


e

iq r
2
d

  iq x  u q
 dx

Likewise for qy and qz.
42
 2
2
  






i
q

V
(
r
) u q (r )  Eu q (r )
 2m



For each value of the wave vector q , we would obtain a discrete set of eigen values


E n (q ) . Put another way, for each quantum number, n, the En varies with q . Therefore,
rather than single discrete energy levels, we have energy bands. Qualitatively, an energy
level diagram would look like this:
b.
Number of states ina band

Each energy state is described by a Bloch function,  n (q ) . How many states are there?
Take a one-dimensional case for the sake of argument, in which
2
qn
, n  0,  1,  2, .
L
Now, the q is the wave number of the wave function rather than of a lattice vibration.
Still, the math is the same.
The first Brillioun zone is of length
2
2
. The number of q-states that fit in that reciprocal
a
a  L  N , the number of atoms in the (1 dimensional) crystal. Similarly in
a
L
three dimensions.
length is
2
43
To develop the discussion further, we need to make some assumption about the potential


energy function, V (r ) . The two cases we’ll consider are V (r ) being very weak, or very
strong.
3.
Nearly Free Electron Model
To explicitly solve for the wavefunction and energy levels, we need to put a specific
2 d 2

V (r ) into the Schrödinger equation. In one dimension 
 k  V ( x) k  E k .
2m dx 2
a.
Energy band gaps
We start with no atoms at all, just the free conduction electrons.
2 d 2 o

 k  E ko ko
2
2m dx
The usual plane-wave solution is
1 ikx
 ko 
e
L
E ko 
2k 2
2m
L

2
k
dx  1
0
Plot Eo vs. k.
There is no gap between neighboring bands. Now, we add a weak interaction with the
lattice atoms. Since the interaction is weak, we expect only slight changes in the energy
levels and wavefunctions—perturbations. [Note, in Omar, this is discussed in Section
44
A.6, not A.7]. I.e., E k  E ko  E k1 , where E k1 is small compared to E ko . The perturbation
 2 3
,
, etc.
has the effect of opening gaps at the Brillouin zone boundaries ,
a a a
b.
Effective mass

Next, we consider the effect of an external electric field,  , on a conduction electron.
d
The group velocity of the wavefunction for a conduction electron is v g 
and its
dk
1 dE
energy is E   . Thus, v g 
, whence dE  v g dk . The external electric field
 dk
does work on the electron.
Note:
dW  edx  ev g dt
E is electric
field.
This work changes the energy of the electron
E is energy.
dE  ev g dt  v g dk
dk
.
dt
In Newton’s 2nd “Law”, we’d say that the electric force changes the momentum of the
electron. The acceleration resulting from the external force would be
dv g
d  dE  1  2 E dk
1 2E
a
  


 e .
dt
dt  dk   k 2 dt  2 k 2
e  
1 2E
We can see that the quantity 2
is where the mass-1 should be in the 2nd “Law.”
2
 k
This is the effective mass for a conduction electron experiencing an external electric field.
2
m*  2
 E
k 2
45
For a free electron (not in the crystal), E 
E
2k 2
and m* = mo. But, in a lattice
2mo

2k 2
, especially near the Brillouin zone boundary at k  .
a
2mo
The m * 
1
.
curvature of E(k)
Notice that m* may be greater than or less than mo, and may even be negative!
c.
Meaning of a negative effective mass
1
 d 2E 
m   2 
 dk 
The curvature is the slope of the slope; a negative curvature implies a decreasing slope.
For instance, at the top of the second energy band m* < 0. At the bottom of the third
band, m* > 0.
*
The effective mass is an alternative conceptual device that incorporates the interaction
with the crystal lattice into the conduction electron’s equation of motion.
m 
dv
mo
 Ftotal  Fexternal  FL  Fexternal  o* 
dt
m 
The FL is like a drag force that can be either (+) or (-), not only negative. This FL arises
when Fexternal is non-zero, since an external electric field will polarize the lattice atoms as
well as accelerate the conduction electrons.
4.
Tight Binding Model
At the opposite extreme approximation, the crystal potential is strong—each conduction
electrons is nearly bound to a lattice atom. It moves through the crystal by tunneling
from one lattice atom to another.
a.
Effective crystal potential & the Bloch function
46
We start with an ordinary atomic orbital wave function,  (x) . This might be the 4s
orbital of Cu, for instance. The  k (x) for a conduction electron can be written as a sum
of the  (x) over the N atoms (one atom per unit cell) in the crystal.
 k ( x) 
1
1
N 2
where  j  j  a , and a = lattice spacing.
N
e
j 1
ik j
 j (x   j ) ,
In the tight binding model, there is little overlap of the  (x) s between neighboring
atoms.
N
1 N ik
1
ik ( x   j )
 k ( x)  1  e j  j ( x   j )  1 e ikx  e
 j ( x   j )  e ikx u k ( x) ,
j 1
N 2 j 1
N 2
where u k (x) is periodic.
b.
Energy bands
We construct the Schrödinger equation for a conduction electron thusly:
 2 d 2

 v( x)  V ( x) k  E    2 cos ka k ;

2
 2m dx

E is the energy of the electron in the  th atomic orbital;
v (x ) is the interaction potential with the nearest atom;
V (x ) is the interaction potential with all the other atoms in the crystal;
 is the shift in E due to V (x ) ;
and
 is the so-called overlap integral—the shift in E due to the overlap of  (x) with
 ( x  a) . (This overlap integral is the origin of the exclusion principle.)
So, the single energy level, E , is broadened into a band, depending on k.
47
C.
Electrical Conductivity
48
I.
A.
4.
Point Groups & Space Groups
III.
B.
Band Theory
Bands
3.
Nearly Free Electron Model
a.
b.
c.
Nearly-free electrons
Effective mass
Negative mass?
4.
Tight Binding Model
a.
b.
c.
Effective crystal potential
Energy bands
Effective mass
C.
Electrical Conductivity
1.
Electron Dynamics
a.
b.
Velocity of the Bloch electron
Electric current
2.
Bands and Transitions
a.
b.
c.
d.
e.
Valence & conduction bands
Metals, semiconductors, insulators
Holes
Hole current
Hall effect
49