For review/introduction of Schrodinger equation:
http://web.monroecc.edu/manila/webfiles/spiral/6schrodingereqn.pdf
Section 1.2-3
Homework from this section: 1.5
(We will do a similar problem in class today)
Bloch’s Theorem and
Krönig-Penney Model
Learning Objectives for Today
After today’s class you should be able to:
 Apply Bloch’s theorem to the KronigPenney model or any other periodic
potential
 Explain the meaning and origin of
“forbidden band gaps”
 Begin to understand the Brillouin zone
For another source on today’s topics, see
Ch. 7 of Kittel’s Intro to Solid State Physics.
 Crystal basics to prepare us for next class

“Realistic” Potential in Solids
 Multi-electron
atomic potentials are complex
 Even for hydrogen atom with a “simple”
Coulomb potential solutions are quite
complex
 So we use a model infinite one-dimensional
periodic potential to get insight into the
problem (last time, looked at 1-6 atoms)
Periodic Potential

For one dimensional case where atoms (ions) are
separated by distance a, we can write condition of
periodicity as
V ( x)  V ( x  a  n)
a
Section 1.3: Bloch’s Theorem
This theorem gives the electron wavefunction in the
presence of a periodic potential energy.
We will prove 1-D version, AKA Floquet’s theorem.
(3D proof in the book)
When using this theorem, we still use
the time-indep. Schrodinger equation
for an electron in a periodic potential
 2 2
 



V
(
r
)    E

 2m

where the potential energy
is invariant under a lattice
translation of a
V ( x)  V ( x  a  n)
In 3D (vector): T  ua  vb  wc V (r  T)  V (r)
Bloch Wavefunctions
a

Bloch’s Theorem states that for a particle moving
in the periodic potential, the wavefunctions ψ(x)
are of the form
 ( x )  u k ( x )e
 ikx
, where uk ( x) is a pe riodicfunction
uk ( x)  uk ( x  a )

uk(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
Main points in the proof of
Bloch’s Theorem in 1-D
 ( x )  u k ( x )e ,
ikx
uk ( x)  uk ( x  a)
1. First notice that Bloch’s
theorem
implies: 
 
  







 k (r  T )  uk (r  T )eik r eik T  uk (r )eik r eik T   k (r )eik T
Or
just:
 



 k (r  T )   k (r )eik T
Can show that this formally implies
Bloch’s theorem, so if we can prove it
we will have proven Bloch’s theorem.
2. To prove the statement shown above in 1-D:
Consider N identical lattice points around a
circular ring, each separated by a distance a.
ika
Our task is to prove:

(
x

a
)


(
x
)
e
1
2
N
3
Built into the ring
model is the periodic
boundary condition:
 ( x  Na)   ( x)
Proof of Bloch’s Theorem in 1-D: Conclusion
The symmetry of the ring (and last lecture) implies
that we can find a solution to the wave equation:
If we apply this translation N times we
will return to the initial atom position:
This
requires
1
2
3
C 1
N
N
And has the
most general
solution:
 ( x  a)  C ( x)
 ( x  Na)  C N ( x)   ( x)
C N  e2ni n  0,  1,  2,...
Or: C  e 2ni / N  eika
Where we define the k  2n
Na
Bloch wavevector:
Now that we know C we can rewrite
 ( x  a)  C ( x)  e  ( x) Q.E.D.
ika
Consequence of Bloch’s Theorem
Probability of finding the electron
P( x)  P( x  a )

Each electron in a crystalline solid “belongs”
to each and every atom forming the solid
Very accurate for metals where electrons are
free to move around the crystal!
 Makes sense to talk about a specific x (± n a)

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.
First done for a chain of finite square well potentials model by Krönig
and Penney in 1931 with E<V0
V
Each atom is represented by
a finite square well of width
a and depth V0. The atomic
spacing is a+b.
V0
-b 0
a a+b
2a+b 2(a+b)
x
  2 d 2



V
(
x
)

  E
2
2
m
dx


We can solve the SE in each region of space:
2
2

K
iKx
iKx
E
0 < x < a  I ( x)  Ae  Be
2m
I wish the book had selected
 2 2 different letters than K and
x
x
V0  E 
-b < x < 0  II ( x)  Ce  De
, but staying consistent
2m
 I ( x)  Ae  Be
iKx
iKx
 x
 x
Boundary Conditions and Bloch’s Theorem
The solutions of the SE require that the
wavefunction and its derivative be
continuous across the potential boundaries.
Thus, at the two boundaries (which are
infinitely repeated):
 II ( x)  Ce
x=0
A  B  C  D (1)
x=a
AeiKa  BeiKa   II (a)
Now using Bloch’s theorem for a
periodic potential with period a+b:
 De
iK ( A  B)   (C  D)
 ( x)  uk ( x)e
 II (a)   II (b)eik ( ab)
(2)
ikx
k = Bloch
wavevector
Now we can write the boundary conditions at x = a:
AeiKa  BeiKa  (Ce b  Deb )eik ( ab) (3) The four simultaneous equations
(1-4) can be written compactly in
matrix form 
(iK  ik ) AeiKa  (iK  ik ) BeiKa  ( (  ik )Ceb  (  ik ) Deb )eik ( ab)
(4, deriv.)
Results of the Krönig-Penney Model
1


iK


eiKa

iKa
(
iK

ik
)
e

1
 iK
e iKa
 (iK  ik )e iKa
1

 e b eik ( a b )
 (  ik )e b eik ( a b )
1




b ik ( a  b )

e e

(  ik )eb eik ( a b ) 
 A
B
 0
C 
 
 D
Since the values of a and b are inputs to the model, and 
depends on V0 and the energy E, we can solve this system of
equations to find the energy E at any specified value of the
Bloch wavevector k. What is the easiest way to do this?
Taking the determinant, setting it equal to zero and lots of algebra gives:
2  K2 

 sin Ka sinhb   cosKa  coshb   cosk (a  b)
 2 K 
By reducing the barrier width b (small b), this can be simplified to:
Graphical
Approach
2  K2 

 sin Ka sinhb   cosKa  coshb   cosk (a  b)
 2 K 
 b 2 

 sin Ka   cosKa   cos(ka)
 2K 
small b
Right hand side cannot exceed 1, so values exceeding will
mean that there is no wavelike solutions of the Schrodinger eq.
(forbidden band gap)
Gap occurs
at Ka=N or
K=N/a
2K 2
E
2m
Ka
Plotting left side of equation
Not really much different
Single Atom
Multiple Atoms
Greek Theater Analogy: Energy Gaps
What Else Can We
Learn From This
Model?
Exercise 1.4
 b 2 

 sin Ka   cosKa   cos(ka)
 2K 
Different Ways to Plot It
Extended Zone Scheme
Note that the larger the energy, the larger the band/gap is (until
some limit).
y=cos ka
Different x axis
Ka
 b 2 

 sin Ka   cosKa   cos(ka)
2
K


The range -<ka< is called
the first Brillouin zone.
Atoms
k = Bloch
wavevector
Different Representations of E(k)
Reduced zone scheme
All states with |k| > /a are
translated back into 1st BZ
Frequently only one side is
shown as they are
degenerate.
In 3D, often show one side
along with dispersion
along two other directions
(e.g. 100, 110, 111)
Band diagrams can refer to either E vs.
real space or E vs momentum space k
Momentum space example
Real space examples
Compare to the free-electron model
Free electron dispersion
E
...with first Brillouin zone:
–/a  /a
(a the lattice constant)
E
2
(k x  k y  k z )
2
2m
2
2
–/a
/a
k
Let’s draw it in 3D!
Let’s slowly turn on the periodic potential
21
Electron Wavefunctions in a
Periodic Potential
E
(Another way to understand the energy gap)
Consider the following cases:
  Aei (kxt )
V1  0 Wavefunctions are plane
2 2

–/a
/a k
waves and energy bands E  k
2m
are parabolic:
V1  0 Electrons wavelengths much larger than atomic spacing
k  a a, so wavefunctions and energy bands are nearly the
same as above
V
V1
-b 0
a a+b
2a+b 2(a+b)
x
How do X-rays Work?
The soft tissue in your
body is composed of
smaller atoms, and so
does not absorb X-ray
photons particularly
well. The calcium
atoms that make up
your bones are much
larger, so they are
better at absorbing
X-ray photons.
Consequence of Bloch’s Theorem
Similar to how radio waves pass through us without affecting
E
Electron Wavefunctions in a
Periodic Potential
U=barrier potential
Consider the following cases:
  Aei (kxt )
V1  0 Wavefunctions are plane
2 2

–/a
/a k
waves and energy bands E  k
2m
are parabolic:
V1  0 Electrons wavelengths much larger than a, so
k  a wavefunctions and energy bands are nearly the same
as above
V1  0 Electrons wavelengths approach a, so waves begin to
k  a be strongly back-scattered by the potential:
B A
   Aei ( kxt )  Bei ( kxt )
V1  0 Electrons waves are strongly back-scattered (Bragg
k  a scattering) so standing waves are formed:
   C ei ( kx t )  e i ( kx t )  
1
2


A eikx  e  ikx e  it
The nearly-free-electron model
(Standing Waves)
Due to the ±, there are two such standing waves possible:
 
 
1
2
1
2

Ae

e
A eikx  eikx eit 
ikx
e
ikx
it

1
2
1
2
2 A cos(kx)eit
2iA sin(kx)e
 
it
1
2


A eikx  e  ikx e  it

These two approximate solutions to the S. E. at k  a have
very different potential energies.   has its peaks at x = a,
2a, 3a, …at the positions of the atoms, where V is at its
minimum (low energy wavefunction). The other solution,  
has its peaks at x = a/2, 3a/2, 5a/2,… at positions in between
atoms, where V is at its maximum (high energy wavefunction).
Either:
Nodes at ions
Or:
a
Nodes midway
between ions
The nearly-free-electron model
Strictly speaking we should have looked at the probabilities
before coming to this conclusion:
 
1
2
 
1
2


Ae

e
A eikx  eikx eit 
ikx
 eikx
it

1
2
1
2
~
2 A cos(kx)eit
 *   2 A2 cos2 ( ax )
2iA sin(kx)eit
 *  
2 A2 sin 2 ( ax )
2
2
Different energies for electron standing waves

2
a
Summary: The nearly-free-electron model
BAND GAPS APPEAR AT EACH BRILLOUIN ZONE EDGE
E
The periodic potential
V(x) splits the freeelectron E(k) into “energy
bands” separated by gaps
at each BZ boundary.
In between the two energies
there are no allowed energies;
i.e., wavelike solutions of the
Schrodinger equation do not
exist.
Forbidden energy bands form
called band gaps.
EE+
-2π/a
–π/a
Eg
π/a
2π/a
k
28
Approximating the Band Gap
BAND GAPS APPEAR AT EACH BRILLOUIN ZONE EDGE
a
E g  E  E   V ( x)[
 2

 2
]dx
E
x 0
    2 A cos ( )
*

2
2 x
a
    2 A sin ( )
*

2
2 x
a
E-
a
E  E 
2
a
E+
2x
V
(
x
)
cos
(
a ) dx

x 0
-2π/a
–π/a
Eg
π/a
2π/a
k
For square potential: V(x) =Vo for specific values of x (changes integration limits)
Crystal Directions


Choose one lattice point on the line as an
origin (point O). Choice of origin is
completely arbitrary, since every
lattice point is identical.
Then choose the lattice vector joining O to
any point on the line, say point T. This
vector can be written as;
R = n1 a + n2 b + n 3 c



To distinguish a lattice direction from a
lattice point, the triplet is enclosed in
square brackets [ ...]. Example: [n1n2n3]
[n1n2n3] is the smallest integer of the
same relative ratios. Example: [222]
would not be used instead of [111].
Negative directions can be written as [n n n
Figure shows
[111] direction
]
Also sometimes
[-1-1-1]