Lecture 6
Schrödinger Equation and relationship to electron
motion in crystals
Reading:
Notes and Brennan Chapter 2.0-2.4, 7.4, 8.0-8.2 and
2.5-2.6
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ECE 6451 - Dr. Alan Doolittle
Schrödenger Equation
So how do we account for the wavelike nature of small particles like electrons?
Schrödenger Equation:
•In “Electrical Properties of Materials”, Solymar and Walsh point out that. Like
the 5 Postulates, there are NO physical assumptions available to “derive” the
Schrödenger Equation
•Just like Newton’s law of motion, F=ma, and Maxwell’s equations, the
Schrödenger Equation was proposed to explain several observations in physics
that were previously unexplained. These include the atomic spectrum of
hydrogen, the energy levels of the Planck oscillator, non-radiation of electronic
currents in atoms, and the shift in energy levels in a strong electric field.
In Schrödinger’s fourth paper he ends with:
“ I hope and believe that the above attempt will turn out to be useful for
explaining the magnetic properties of atoms and molecules, and also the electric
current in the solid state”.
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ECE 6451 - Dr. Alan Doolittle
Schrödenger Equation
So how do we account for the wavelike nature of small particles like electrons?
Schrödenger Equation:
•Lets start with the Classical Hamiltonian and substitute in the operators that correspond to the
classical state variables.
KE + PE = ETotal
Kinetic energy
“operator”
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h2 2
∂Ψ
−
∇ Ψ + VΨ = ih
2m
∂t
Potential electron
moves through
Energy
“operator”
ECE 6451 - Dr. Alan Doolittle
Schrödenger Equation
To solve the Schrödinger equation one must make an assumption about the wave function. Lets
assume the wave function has separate spatial and temporal components:
* φ ( x , y , z , t ) = Ψ ( x , y , z ) w( t )
Plugging this (*) into the Schrödinger equation and dividing both sides by (*) we arrive at:

 h 2 ∇ 2 Ψ ( x, y , z )
1 ∂w(t )
 −
+ V  = ih
w(t ) ∂t

 2m Ψ ( x , y , z )
Since the left hand side varies only with position, and the right hand side varies only with time,
the only way these two sides can equate is if they are equal to a constant ( we will call this
constant, total energy, E). Thus, we can break this equation into two equations:
 h 2 ∇2 Ψ

 −
+ V  = E
 2m Ψ

1 ∂w
E = ih
w ∂t
Consider first the time variable version (right side) then later we will examine the spatially
variable portion. This will give us time variable solutions and, later, a separate spatially variable
solution.
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ECE 6451 - Dr. Alan Doolittle
Schrödenger Equation
Consider the time variable solution:
E = ih
1 ∂w
w ∂t
∂w
 E
= − i  w
∂t
 h
w( t ) = e
  E 
 − i  t 
  h 
or w(t ) = e (−iωt )
where E = hω
This equation expresses the periodic time nature of the wave equation.
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ECE 6451 - Dr. Alan Doolittle
Introduction to Quantum Mechanics
Consider the space variable solution:
 h 2 ∇2 Ψ

 −
+ V  = E
 2m Ψ

The combined
“operator” is called
the Hamiltonian
momentum
“operator”
 h2 2

 −
∇ + V  Ψ = EΨ
 2m

Hˆ Ψ = EΨ
 1
(− jh∇ )2 + V Ψ = EΨ

 2m

 pˆ 2 ˆ 

+ V  Ψ = EΨ
 2m

Kinetic
Total
+ Potential =
Energy
Energy
Energy
Classically, momentum, p=mv and kinetic energy is ½ (mv2) = ½ (p2)/m
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ECE 6451 - Dr. Alan Doolittle
Introduction to Quantum Mechanics
Consider a specific solution for the free space (no electrostatic potential, V=0) wave solution is
(electron traveling in the +x direction in 1D only):
 h2 2

 −
∇ + V  Ψ = EΨ
 2m

h2 ∂2Ψ
+ EΨ = 0
2
2m ∂x
Ψ ( x ) = Ae ikx + Be −ikx
where k =
2π
λ
=
2mE
h2k 2
or E =
2
h
2m
Since we have to add our time dependent portion (see (*) previous) our total solution is:
Ψ = Ψ ( x ) w(t ) = Ae − i (ωt −kx ) + Be − i (ωt + kx )
This is a standard wave equation with one wave traveling in the +x direction and one wave traveling in
the –x direction. Since our problem stated that the electron was only traveling in the +x direction, B=0.
Classically, momentum, p=mv and kinetic energy is ½ (mv2) = ½ (p2)/m
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ECE 6451 - Dr. Alan Doolittle
Introduction to Quantum Mechanics
An interesting aside: What is the value of A?
Since Ψ is a probability,
∫ [Ψ Ψdx ] = 1
∞
*
−∞
∫
∞
∫
∞
∫
∞
−∞
−∞
−∞
Aeikx Ae −ikx dx = 1
A2eikx −ikx dx = 1
A2dx = 1
This requires A to be vanishingly small (unless we restrict our universe to finite size) and is the same
probability for all x and t. More importantly it brings out a quantum phenomena: If we know the
electrons momentum, p or k, we can not know it’s position! This is a restatement of the uncertainty
principle:
∆p ∆x ≥ ħ/2
Where ∆p is the uncertainty in momentum and ∆x is the uncertainty in position
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Wave Packet Discussion on Board:
ECE 6451 - Dr. Alan Doolittle
Introduction to Quantum Mechanics
The solution to this free particle example brings out several important observations about the
dual wave-particle nature of our universe:
− i (ωt − kx )
Ψ = Ψ ( x ) w(t ) = Ae
•While particles act as waves, their charge is carried as a particle. I.e. you can only say that there is a
“probability” of finding an electron in a particular region of space, but if you find it there, it will have
all of it’s charge there, not just a fraction.
•Energy of moving particles follows a square law relationship:
Neudeck and Pierret Fig 2.3
Classically, momentum, p=mv and kinetic energy is (mv2)/2 =(p2)/2m
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2
p
h2k 2
E=
=
2m
2m
ECE 6451 - Dr. Alan Doolittle
Introduction to Quantum Mechanics
What effect does this “E-k” square law relationship
have on electron velocity and mass?
The group velocity (rate of energy delivery) of a wave
is:
dE 1 dE
vg ≡
=
dp h dk
So the “speed” of an electron in
the direction defined by p is found
from the slope of the E-k diagram.
Similarly, since
h2k 2
E=
2m
−1
2
*
2 d E 
m = h  2 
 dk 
So the “effective mass” of an
electron is related to the local
inverse curvature of the E-k
diagram
Note: Brennan section 8.1 rigorously derives the equation for vg and m*
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Figure after Mayer and Lau Fig 12.2
ECE 6451 - Dr. Alan Doolittle
What effect does an electrostatic potential have on an electron?
Consider the electron moving in an electrostatic potential, Vo. The wave solution is (electron
traveling in the +x direction in 1D only):
 h2 2

 −
∇ + V  Ψ = EΨ
 2m

h2 ∂2Ψ
+ (E − Vo )Ψ = 0
2
2m ∂x
Ψ ( x ) = Ae ikx + Be −ikx
where k =
2π
λ
=
2m(E − Vo )
h2k 2
− Vo
or E =
2
h
2m
Since we have to add our time dependent portion (see (*) previous) our total solution is:
Ψ = Ψ ( x ) w(t ) = Ae − i (ωt −kx ) + Be − i (ωt + kx )
This is, again, a standard wave equation with one wave traveling in the +x direction and one wave
traveling in the –x direction. Since our problem stated that the electron was only traveling in the +x
direction, B=0.
When the electron moves through an electrostatic potential, for the same energy as in free space, the
only thing that changes is the “wavelength” of the electron.
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ECE 6451 - Dr. Alan Doolittle
Localized Particles Result in Quantized Energy/Momentum:
Infinite Square Well
First a needed tool: Consider an electron trapped in an energy well with infinite potential barriers.
The reflection coefficient for infinite potential was 1 so the electron can not penetrate the barrier.
 h2 2

 −
∇ + V  Ψ = EΨ
 2m

h2 ∂2Ψ
−
− EΨ = 0
2
2m ∂x
∂2Ψ
+ k 2Ψ = 0
2
∂x
General Solution : Ψ ( x ) = A sin(kx ) + B cos(kx )
2π
h2k 2
2mE
=
where k =
or E =
2
h
2m
λ
Boundary Conditions :
Ψ ( 0) = 0 ⇒ B = 0
nπ
Ψ (a) = 0 ⇒ Asin (ka ) = 0 ⇒ k =
for n = ±1, ± 2,±3...
a
n 2π 2 h 2
 nπ x 
Ψn ( x ) = An sin
 and E n =
2ma 2
 a 
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After Neudeck and Pierret Figure 2.4a
ECE 6451 - Dr. Alan Doolittle
Localized Particles Result in Quantized Energy/Momentum:
Infinite Square Well
What does it mean?
n 2π 2 h 2
 nπ x 
Ψn ( x ) = An sin
 and E n =
2ma 2
 a 
A standing wave results from
the requirement that there be a
node at the barrier edges (i.e.
BC’s: Ψ(0)=Ψ(a)=0 ) . The
wavelength determines the
energy. Many different
possible “states” can be
occupied by the electron, each
with different energies and
wavelengths.
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After Neudeck and Pierret Figure 2.4c,d,e
ECE 6451 - Dr. Alan Doolittle
Localized Particles Result in Quantized Energy/Momentum:
Infinite Square Well
What does it mean?
n 2π 2 h 2
 nπ x 
Ψn ( x ) = An sin
 and E n =
2ma 2
 a 
Solution for much larger “a”. Note:
offset vertically for clarity.
Recall, a free particle has E ~k2.
Instead of being continuous in k2, E
is discrete in n2! I.e. the energy
values (and thus, wavelengths/k) of a
confined electron are quantized (take
on only certain values). Note that as
the dimension of the “energy well”
increases, the spacing between
discrete energy levels (and discrete k
values) reduces. In the infinite
crystal, a continuum same as our free
particle solution is obtained.
After Neudeck and Pierret Figure 2.5
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ECE 6451 - Dr. Alan Doolittle
What about an electrostatic potential step?
Consider the electron incident on an electrostatic potential barrier, Vo. The wave solution (1D
only):
V(x<0)=0
V(x>0)=Vo
Vo
Region I
x=0
Region II
x
We have already solved these in regions I and II. The total solution is:
ΨI = ΨI ( x ) wI (t ) = AI e − i (ωt −k I x ) + BI e − i (ωt + k I x )
ΨII = ΨII ( x ) wII (t ) = AII e − i (ωt −k II x ) + BII e − i (ωt + k II x )
where k I =
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2π
λI
=
2mE
2π
and
k
=
=
II
2
h
λ II
2m(E − Vo )
h2
ECE 6451 - Dr. Alan Doolittle
What about an electrostatic potential step?
Vo
cont’d...
Region I
x=0
Region II
x
ΨI = ΨI ( x ) wI (t ) = AI e − i (ωt −k I x ) + BI e − i (ωt + k I x )
ΨII = ΨII ( x ) wII (t ) = AII e − i (ωt −k II x ) + BII e − i (ωt + k II x )
where k I =
2π
λI
=
2mE
2π
=
=
and
k
II
λ II
h2
2m(E − Vo )
h2
When the “wave” is incident on the barrier, some of it is reflected, some of it is
transmitted. However, since there is nothing at x=+∞ to reflect the wave back, BII=0.
Incident Wave
Reflected Wave
ΨI = AI e − i (ωt −k I x ) + BI e − i (ωt + k I x )
Energy is conserved across the boundary so,
Incident Wave
Reflected Wave
hϖ I = Eincident
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Transmitted Wave
ΨII = AII e − i (ωt −k II x )
Transmitted Wave
h 2 k I2
h 2 k II2
=
= E reflected = hϖ I = E transmitted =
+ Vo = hϖ II
2m
2m ECE 6451 - Dr. Alan Doolittle
What about an electrostatic potential step?
Vo
cont’d...
Region I
x=0
Region II
x
We can apply our probability current density concept,
ρ ≡ Ψ * (r )Ψ (r )
j=
(
)
h
Ψ * (r )∇Ψ (r ) − Ψ (r )∇Ψ * (r )
2mi
∇• j+
∂ρ
=0
∂t
to define transmission, T, and reflection coefficients, R, such that the transmission
and reflection probability is
T *T ≡
J Incident
J transmitted
J incident
and
R*R ≡
J reflected
J incident
dΨI* 
h  * dΨI
h
 ΨI
 =
(+ ik I )AI*e +i (ωt −k I x ) AI e −i (ωt −k I x ) − (− ik I )AI e −i (ωt −k I x ) AI*e +i (ωt −k I x )
=
− ΨI
2mi 
dx
dx  2mi
(
)
h
(
i 2k I AI AI* )
2mi
Similarly,
J Incident =
J Transmitted =
h
h
(
(
i 2k II AII AII* ) and J Re flected = −
i 2k I BI BI* )
2mi
2mi
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ECE 6451 - Dr. Alan Doolittle
What about an electrostatic potential step?
Vo
cont’d...
Region I
Thus,
T *T ≡
J transmitted
J incident
k II (AII AII* )
T T=
k I (AI AI* )
*
and
R*R ≡
x=0
Region II
x
J reflected
J incident
k I (BI BI* ) (BI BI* )
and R R =
=
k I (AI AI* ) (AI AI* )
*
This transmission probability is dramatically different from that quoted in MANY quantum
mechanics texts INCLUDING Brennan’s. The above form (based on current flow) is valid for all
cases, but the bottom form is valid for only E≥Vo. Before we consider the case of E<Vo, it is worth
considering where the error in these text (including ours) comes from.
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ECE 6451 - Dr. Alan Doolittle
What about an electrostatic potential step?
Vo
Consider a different approach...
Region I
x=0
Region II
x
ΨI = ΨI ( x ) wI (t ) = AI e − i (ωt −k I x ) + BI e − i (ωt + k I x )
ΨII = ΨII ( x ) wII (t ) = AII e − i (ωt −k II x ) + BII e − i (ωt + k II x )
where k I =
2π
λI
=
2mE
2π
=
=
and
k
II
λ II
h2
2m(E − Vo )
h2
When the “wave” is incident on the barrier, some of it is reflected, some of it is
transmitted. However, since there is nothing at x=+∞ to reflect the wave back, BII=0.
If we use the simpler boundary condition, ψ is a wave, both ψ and it’s first derivative
must be continuous across the boundary at x=0 for all time, t. Thus,
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ΨI ( x = 0) = ΨII ( x = 0)
AI + BI = AII
and
∂ΨI ( x = 0) ∂ΨII ( x = 0)
=
∂x
∂x
and
ik I ( AI − BI ) = ik II AII
ECE 6451 - Dr. Alan Doolittle
What about an electrostatic potential step?
Vo
cont’d...
Region I
AI + BI = AII
x=0
Region II
x
and
ik I ( AI − BI ) = ik II AII
ik I ( AI − BI ) = ik II ( AI + BI )


B 
B 
ik I 1 − I  = ik II 1 + I 
AI 
AI 


BI  ik I − ik II 

= 
AI  ik I + ik II 
We can define a “reflection coefficient” as the amplitude of the reflected wave relative
to the incident wave,
B
k − k II
B
and R * R ≡ I
R≡ I = I
AI k I + k II
AI
2
k − k II
= I
k I + k II
2
Same as previous case.
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ECE 6451 - Dr. Alan Doolittle
What about an electrostatic potential step?
Vo
cont’d...
x
x=0
Region II
Region I
And likewise, we can define a transmission coefficient as the amplitude of the transmitted wave
relative to the incident wave, T=AII/AI
A I + B I = A II
and
ik I (A I − B I ) = ik II A II
ik I (A I − (A II − A I )) = ik II A II
ik I (2A I ) = ik II A II + ik I A II
T=
kI =
2π
λI
=
2mE
2π
and k II =
=
2
λ II
h
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A II
2k I
A
=
and T *T = II
A I k I + k II
AI
2 m (E − V )
h2
2
=
2k I
k I + k II
2
Notice that this approach is missing the
factor (kII/kI). BIG ERROR! For this
erroneous approach, T*T ECE
+ R*R
≠ 1.
6451 - Dr. Alan Doolittle
What about an electrostatic potential step?
Vo
cont’d...
Region I
x=0
Region II
x
Final details:
(
k II (AII AII* )
BI BI* )
*
T T=
and R R =
(AI AI* )
k I (AI AI* )
*
A II
2k I
k II  2k *I  2k I 
*



Since
and T T =
=
A I k I + k II
k I  k *I + k *II  k I + k II 


4k II 
1
T *T =

kI 
k II
1
+

kI

kI =
2π
λI
=
2mE
2π
and k II =
=
2
λ II
h
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


2 



k − k II
R *R = I
k I + k II
2
T *T + R * R = 1
2 m (E − V )
h2
ECE 6451 - Dr. Alan Doolittle
What about an electrostatic potential step?
cont’d...
Consider 2 cases: Case 1: E>V
Both kI and kII are real and thus, the particle travels as a wave of different wavelength in the two
regions.
However, R*R is finite. Thus, even thought the electron has an energy, E, greater than V it will have a
finite probability of being reflected by the potential barrier.
If E>>V, this probability of reflection reduces to ~0 (kI Æ kII)
B
R *R ≡ I
AI
2
k − k II
= I
k I + k II


4k II 
1
T *T =

kI 
k II
1
+

kI

kI =
2π
λI
=
2mE
2π
and k II =
=
2
λ II
h
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2
λI
λII



2 



2m ( E − V )
h2
V
Region I
x=0
Region II
x
ECE 6451 - Dr. Alan Doolittle
What about an electrostatic potential step?
cont’d...
Case 2: E<V
kI is real but kII is imaginary. When an imaginary kII is placed inside our exponential, e(ikIIx), a decaying
function of the form, e(-ax) results in region II.
AII can be complex
Transmitted Wave
ΨII = AII e − i (ωt −k II x ) = AII e
− ( k II x ) − iωt
e
However, T*T is now finite but evanescent. Evanescent waves carry no current (see homework). So
even though the electron has an energy, E, less than V it will have a finite probability of being found
within the potential barrier. The probability of finding the electron deep inside the potential barrier is ~0
due to the rapid decay of ψ.
B
R *R ≡ I
AI
2
=
k I − i k II
2
k I + i k II
z*
=
=1
z
T *T = 0
V
Due to JTransmitted = 0
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kI =
2π
λI
=
2mE
2π
and k II =
=
2
h
λ II
2m (E − V )
h2
Region I
x=0
Region II
x
ECE 6451 - Dr. Alan Doolittle
What about an electrostatic potential step?
Consider the following potential profile with an electron of energy E<Vo.
λII λ III
λI
E>Vo
E<Vo
Vo
x=0
Region I
x
x=a
Region II
Region III
The electron has a finite probability to “tunnel” through the barrier and will do so if the barrier is thin
enough. Once through, it will continue traveling on it’s way.
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ECE 6451 - Dr. Alan Doolittle
What about an electrostatic potential step?
Incident Wave
Reflected Wave from x=-a
ΨI = Ae − i (ωt −k1x ) + Be − i (ωt + k1x )
+x Wave Region II
λII λ III
λI
Consider the following potential profile
with an electron of energy E<Vo.
E>Vo
E<Vo
Vo
Reflected Wave from x=+a
x=0
Region I
ΨII = Ce − i (ωt −k2 x ) + De − i (ωt + k2 x )
x
x=a
Region II
Region III
Transmitted Wave
ΨIII = Fe − i (ωt −k1x )
Energy is conserved across the boundary so,
h 2 k12
=E
2m
k1 =
2π
λ1
=
2mE
2π
and k 2 =
=
2
λ2
h
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h 2 k 22
= E − Vo
2m
h 2 k12
=E
2m
2m( E − V )
h2
ECE 6451 - Dr. Alan Doolittle
What about an electrostatic potential step?
Region I
− i (ωt − k1 x )
As was found before for the currents,
Ae
Vo
Region II
Ce − i (ωt −k2 x )
Be − i (ωt +k1x )
Region III
Fe − i (ωt −k1x )
De − i (ωt +k2 x )
x=a
−i (ωt − k1 x )
x=0
−i (ωt − k1 x )
x
dΨ 
h  * dΨ+ I
h
 Ψ+ I
 =
(+ ik1 )A*e +i (ωt −k1x ) A e
− Ψ+ I
− (− ik1 )A e
A* e +i (ωt −k1x )
2mi 
dx
dx  2mi
h
(
J +I =
i 2k1 AA* )
2mi
Similarly,
h
h
h
h
(
(
(
(
J −I = −
i 2k1 BB * ) and J + II =
i 2k 2 CC * ) and J − II = −
i 2k 2 DD * ) and J + III =
i 2k1 FF * )
2mi
2mi
2mi
2mi
Current continuity at the boundaries x = ± a implies,
*
+I
J +I =
J + I + J − I = J + II + J − II and
(
)
J + II + J − II = J + III
k1 AA* + (− k1 BB * ) = k 2 CC * + (− k 2 DD * ) and k 2 CC * + (− k 2 DD * ) = k1 FF *
Solving for k 2 CC * + (− k 2 DD * ) from the first equation and substituting into the second,
k1 AA* − k1 BB * = k1 FF *
BB * FF *
1=
+
but,
AA* AA*
↑
↑
1 = RR + TT
*
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*
or rearranging and dividing through by k1 AA*
F
∴ TT≡
A
*
2
B
R R≡
A
2
*
ECE 6451 - Dr. Alan Doolittle
What about an electrostatic potential step?
Ae − i (ωt −k1x )
Be − i (ωt +k1x )
Vo
Ce − i (ωt −k2 x )
Fe − i (ωt −k1x )
De − i (ωt +k2 x )
x
x=a
x=0
Region III
Region I
Region II
T T=?
*
R R=?
*
Homework!
Hint: Use a combination of the continuity of the wave function and continuity of
the probability density current (or equivalently in this case, the continuity of the
derivative).
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ECE 6451 - Dr. Alan Doolittle
Multiple/Periodic Potential Barriers:
Kronig-Penney Model
Resonant reflectance/transmission creates “standing waves” in the crystal. Only certain
wavelengths (energies) can pass through the 1D crystal.
By analogy, a multiple layer optical coating has similar reflection/transmission
characteristics. The result is the same, only certain wavelengths (energies) are transmitted
through the optical stack. In a since, we have an “optical bandgap”.
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ECE 6451 - Dr. Alan Doolittle
Now consider an periodic potential in 1D
Kronig-Penney Model: Bloch Functions Explained
Since each unit cell is indistinguishable from the next, the probability of finding an
electron in one unit cell is identical to that of finding it in an adjacent unit cell.
The Bloch theorem states that since the potential repeats every “a” lengths, the
magnitude of the wavefunction (but not necessarily the phase) must also repeat every
“a” lengths. This is true because the probability of finding an electron at a given point
in the crystal must be the same as found in the same location in any other unit cell.
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ECE 6451 - Dr. Alan Doolittle
Now consider an periodic potential in 1D
Kronig-Penney Model: Bloch Functions Explained
To achieve this property, the MAGNITUDE of the wavefunction (but not necessarily the wavefunction) must have
the same periodicity as the lattice. Thus, we choose a wavefunction that is modulated by the periodicity of the
lattice.
Since V(r + a) = V(r)
we choose plane waves modulated by a periodic function,
Ψ (r) = e ikr u nk (r) where u nk (r) = u nk (r + a)
Thus,
The wavefunction in one unit cell is
merely a phase shifted version of the
wavefunction in an adjacent unit cell.
Ψ (r + a) = u nk (r + a)e ik (r +a )
[
]
Ψ (r + a) = e ika u nk (r)e ikr = e ika Ψ (r) or merely a phase shifted version of Ψ (r)
Thus,
Ψ * (r + a) Ψ (r + a) = e -ika Ψ * (r)e ika Ψ (r) = Ψ * (r) Ψ (r)
Thus, the probability of finding an electron in one unit cell is identical to that of finding it in
an adjacent unit cell – as expected.
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ECE 6451 - Dr. Alan Doolittle
An Important Aside: Effect of Bloch Functions
Assuming a large number of unit cells in a material, N, the boundary condition for the system is Na
translations must result in the wavefunction being translated to return to itself? (The probability at the
material edges must be symmetric and equal).
Ψ (x + Na) = e ikNa Ψ (x) = Ψ (x)
Thus,
e ikNa = 1 so taking the Nth root,
(1 )
(1 )
(in2π )
e ika = 1 N = (e in2π ) N = e N
ka = n2π
N
So the allowed states of k are :
2π n
k=
Na
(
)
Thus, if N (the number of unit cells available) is very large, like in a semiconductor, the spacing between the
allowed k-values (kn=2-kn=1 etc... are almost continuous, justifying the treatment of the k-states as a
continuum.
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ECE 6451 - Dr. Alan Doolittle
Now consider an periodic potential in 1D
Kronig-Penney Model
Consider what potentials an electron would see as it moves
through the lattice (limited to 1D for now). The electrostatic
potential, V(x) is periodic such that V(x+L)=V(x).
We MUST have standing waves in the crystal that have a
period equal to a multiple of the period of the crystal’s
electrostatic potential. (Similar to a multilayer antireflection
coating in optics)
It is important to note that since, the wavefunction repeats
each unit cell, we only have to consider what happens in one
unit cell to describe the entire crystal. Thus, we can restrict
ourselves to values of k such that –π/a to +π/a (implying ka
≤1 or (2π/λ)a≤1)
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After Neudeck and Peirret Fig 3.1
ECE 6451 - Dr. Alan Doolittle
Now consider a periodic potential in 1D
Kronig-Penney Model
Assumptions of Kronig-Penney Model:
•Simplifying the potential to that shown
here:
•1D only
•Assume electron is a simple plane wave of
the form,
ikx
e
...modulated by a function with the
same periodicity as the periodic
crystalline potential, U(x)
•The crystalline potential is periodic,
U(x)=U(x+L)
•Thus the wave function is a simple plain
wave modulated by a function with the
same periodicity as the periodic crystalline
potential:
ikx
nk
Ψ (x) = u (x)e
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The question “how does the presence of a periodic
potential effect the free electron” can thus be
converted to the question “what effect on the
electron does the changing of an electron’s free state
(plane wave) to a Bloch state”?
Neudeck and Peirret Fig 3.2
ECE 6451 - Dr. Alan Doolittle
Kronig-Penney Model
For 0<x<a:
For -b<x<0:
0
 h

 −
∇2 + V  Ψ = EΨ
 2m

∂ 2Ψ
2
+
Ψ=0
α
2
∂x
2mE
where α =
h2
2
 h2 2

 −
∇ + V  Ψ = EΨ
 2m

∂2Ψ
2
+
Ψ=0
β
2
∂x
2 m (E − U o )
for E > U o
β =i
h2
β=
Ψa ( x ) = A sin(αx ) + B cos(αx )
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2m(U o − E )
for 0 < E < U o
2
h
Ψb ( x ) = C sin( βx ) + D cos( βx )
ECE 6451 - Dr. Alan Doolittle
Kronig-Penney Model
For 0<x<a:
For -b<x<0:
Ψa ( x ) = A sin(αx ) + B cos(αx )
Ψb ( x ) = C sin( βx ) + D cos( βx )
Applying the following boundary conditions:
Ψa ( x = 0) = Ψb ( x = 0)
dΨa ( x )
dx
=
x =0
dΨb ( x )
dx
BC for continuous
wave function at the
boundary
x =0
Ψa ( x = a ) = e ik ( a +b ) Ψb ( x = −b)
dΨa ( x )
dx
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= e ik ( a +b )
x =a
dΨb ( x )
dx
x =− b
BC for periodic wave
function at the
boundary
ECE 6451 - Dr. Alan Doolittle
Kronig-Penney Model
Applying the boundary conditions, we get:
B=D
αA = βC
A sin(αa ) + B cos(αa ) = eik ( a +b ) [− C sin( βb) + D cos( βb)]
αA cos(αa ) + αB sin(αa ) = eik ( a +b ) [βC cos( βb) + βD sin( βb)]
Eliminating the variables C and D using the above equations, we get:


 α  ik ( a +b )
Asin(αa ) +  e
sin( βb) + B cos(αa ) − eik ( a +b ) cos( βb) = 0
β 


[
[
] [
]
]
A α cos(αa ) − αeik ( a +b ) cos( βb) + B − α sin(αa ) − βeik ( a +b ) sin( βb) = 0
This equation set forms a matrix of the form:
 w x   A 0
 y z   B  =  0

   
A and B are only non-zero (non-trivial solution) when the determinate of the above
set is equal to zero.
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ECE 6451 - Dr. Alan Doolittle
Kronig-Penney Model
Taking the determinate and simplifying we get:
α2 + β 2 
 sin(αa ) sin( βb) + cos(αa ) cos( βb) = cos(k ( a + b) )
− 
 2αβ 
Plugging in the definitions for α and β we get:


E


−
1
2

  2mU
Uo
E   2mU o
o
 sin b

 sin a
2
 
U
h
h2

 
o 
E  E

 2 U  U − 1 
o 
o


 2mU o E   2mU o
 E

 cos b

− 1  + cos a
2
 

U
U
h
h2
o 
 o



 E


− 1  = cos(k ( a + b) ) for E > U o
 Uo
 


E


1− 2
 2mU
  2mU

Uo
E 
o
o
b

sinh
 sin a

2
2


h
h
U

o 
E 
E  

 2 U 1 − U  
o 
o  

 2mU
 2mU o E 

E  
o
b

1 −
 + cos a
cosh
2
2



h
h
U
U
o 
o 





E  
1 −
 = cos(k ( a + b) ) for 0 < E < U o
 U o  
The right hand side is constrained to a range of +/- 1 and is a function of k only. The limits of
the right hand side (+/- 1) occurs at k=0 and +/- π/(a+b) where a+b is the period of the crystal
potential.
The left hand side is NOT constrained to +/- 1 and is a function of energy only.
Georgia Tech
ECE 6451 - Dr. Alan Doolittle
Kronig-Penney Model
The right hand side is constrained to a range of +/- 1 and is a function of k only. The limits of
the right hand side (+/- 1) occurs at k=0 to +/- π/(a+b).
Left or Right hand side of Kronig-Penney Solution
The left hand side is NOT constrained to +/- 1 and is a function of energy only.
Within these “forbidden energy
ranges”, no solution can exist (i.e.
electrons can not propagate.
E/Uo
Various “Bands or allowed energy” exist
where the energy E is a function of the
choice of k (see solution equation)
for the specific case where,
a
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2mU o
2mU o
=b
=π
2
h
h2
ECE 6451 - Dr. Alan Doolittle
Kronig-Penney Model
Replotting the previous result in another form recognizing the lower k limit is shared by + and –
π/(a+b) while the upper limit is for k=0.
There are at most 2 k-values for each allowed energy, E
The slope, dE/dK is zero at the k-zone boundaries at k=0, k= – π/(a+b) and k= + π/(a+b) Thus we
see that the velocity of the electrons approaches zero at the zone boundaries. This means that the
electron trajectory/momentum are confined to stay within the allowable k-zones.
Note: k-value solutions differing by 2π/(a+b) are indistinguishable
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Figures after Neudeck and Peirret
ECE 6451 - Dr. Alan Doolittle
Kronig-Penney Model
Replotting the previous result in another form ...
Free Space E-k
diagram
−3π/(a+b) −2π/(a+b) −π/(a+b)
3rd
2nd
Brillion
Zone
Brillion
Zone
0
1st Brillion
Zone
Note: k-value solutions differing by 2π/(a+b) are indistinguishable. Also
due to animations printed version does not reflect same information.
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π/(a+b) 2π/(a+b) 3π/(a+b)
2nd
3rd
Brillion
Zone
Brillion
Zone
The presence of the
periodic potential
breaks the “free space
solution” up into
“bands” of
allowed/disallowed
energies. The
boundaries of these
bands occurs at
k=±π/(a+b)
ECE 6451 - Dr. Alan Doolittle
Now consider an periodic potential in 1D
Kronig-Penney Model
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After Neudeck and Peirret Fig 4.1
ECE 6451 - Dr. Alan Doolittle
What is the Importance of kSpace Boundaries at k=(+/-)π/a?
Crystal Structures, Brillouin
Zones and Bragg Reflection
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ECE 6451 - Dr. Alan Doolittle
Concept of a Reciprocal Space
(Related to the Fourier Transform)
a
Real
Space
Reciprocal
Space
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2π/a
ECE 6451 - Dr. Alan Doolittle
Importance of k-Space Boundaries at k=(+/-)π/a
Crystal Structures, Brillouin Zones and Bragg Reflection
The crystal reciprocal
lattice consists of a
periodic array of
“inverse-atoms”( more
explanation to come
later).
→
1 →
1 →
k1 • 2 G = 2 G
2
≡ Bragg Reflection Condition for total internal reflection
G
½G
k1
Surface made from
all such planes is
the 1st Brillouin
Zone. The
Brillouin zone
consists of just
those k vectors for
which Bragg
Reflection occurs.
Since |G|=2π/a
then | ½G|=π/a
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ECE 6451 - Dr. Alan Doolittle
Importance of k-Space Boundaries at k=(+/-)π/a
Crystal Structures, Brillouin Zones and Bragg Reflection
Ψ( x )
for k =
≈e
π
iπx
a
+e
−
iπx
a
or e
iπx
a
−e
−
iπx
a
a
Thus,
Ψ( x )
for k =
π
a
 πx 
 πx 
∝ cos  or sin 
 a 
 a 
Thus,
Ψ * ( x )Ψ( x )
for k =
π
a
 πx 
 πx 
∝ cos 2   or sin 2  
 a 
 a 
Electrons are represented by “standing
waves” of wavefunctions localized
near the atom cores (i.e. valence
electrons) and as far away from the
cores as possible (i.e. free electrons).
Given the different potential energies
in both of these regions, the energies
of the electrons away from the atom
cores is higher. This is one
explanation for the origin of the
energy bandgap.
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Real Space Crystal
ECE 6451 - Dr. Alan Doolittle