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Quantum Mechanics for
Electrical Engineers
Dennis M. Sullivan, Ph.D.
Department of Electrical and
Computer Engineering
University of Idaho
1
Why quantum mechanics?
At the beginning of the 20th century, various
phenomena were being observed that could not
be explained by classical mechanics.
1. Energy is quantized
2. Particles have a wave nature
2
Source of
electrons
3
Source of
electrons
4
Conclusion:
Particles have wave
properties.
5
Photoelectric Effect
Incident light
Photoelectrons
Material
The velocity of the escaping particles was dependent on
the wavelength of the light, not the intensity as expected.
6
Kinetic
Energy T
Photoelectric Effect
Planck postulated in 1900
that thermal radiation is
emitted from a heated surface
in discrete packets called
quanta.
f0
Frequency f
7
Einstein postulated that the energy of each photon
was related to the wave frequency.
E    hf
h is Plank’s constant
h  6.625  1034 J  s
 4.135  10
15
eV  s
This is the first major result:
energy is related to frequency
8
In 1924, Louis deBroglie postulated the existence of
matter waves. This lead to the famous
wave-particle duality principle. Specifically that
the momentum of a photon is given by
p
h

Or the more familiar form
p k
Second major result:
Momentum is related to wavelength
9
Bottom line:
Everything is at the same
time a particle and a wave.
E  hf
p
h

10
Physicists formulate everything as an energy problem
Solve using only energy.
1 kg
Problem: Determine the velocity
of the ball at the bottom of the
slope.
1 meter
11
While the ball is on the top of the hill, it has
potential energy.
Since the acceleration of gravity is
g  9.8 m / s 2
1 kg
the potential energy is
PE  g  mass of the ball  height
m

  9.8 2  1 kg 1 m   9.8 J
s 

1 meter
12
When it gets to the bottom of the hill, it no longer has
potential energy, but it has kinetic energy. Since there
have been no other external forces, it must be the same
1 v2
T  9.8 J 
2M
1
T  9.8 J  Mv 2
2

kg  m 2  1
m
v  2   9.8

4.3
 1 kg
2
s
s


1 meter
13
There are several methods of
advanced mechanics that change
everything into energy.
1. Lagrangian mechanics
2. Hamiltonian mechanics
14
Erwin Schrödinger was taking this approach
and developed the following equation to
incorporate these new ideas:
2


1 
2
  2
 V  
2
t
 2m

2
2
This equation is 2nd order in time and 4th order
is space.
15
Schrödinger realized that this was a completely
intractable problem. (There were no computers in
1921.) However, he saw that by considering  to
be a complex function, he could factor the above
equation into two simpler equations, one of which is

2
i
 
   V
t
2m
2
This is the Schrödinger equation.
16
The parameter in the Schrödinger equation, , is a
state variable. It is not directly associated with any
physical quantity itself, but all the information can be
extracted from it. Also, remember that the
Schrödinger equation was only half of the “real”
equation from which it was derived. So to determine
if a particle is located between a and b, calculate
b
P(a  x  b)   
a
*
 x   x  dx
17
This brings us to one of the basics requirements
of the state variable: it must be normalized


  r   r  dr  1
*

Note: The amplitude of the state verctor  does not
represent the strength of the wave in the usual sense.
It is chosen to achieve normalization
18
Computer simulations can show how the
Schrödinger equation can model a particle like an
electron propagating as a wave packet.
The real part is blue, the imaginary part is red.
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In quantum mechanics, physical properties are
related to operators, call observables.
Two of the fundamental operators are:
1. Momentum
2. Kinetic energy
25
The momentum operator (in one dimension) is

p
.
i x
This seems pretty strange until I remember
that momentum is related to wavelength:
p
h

 k.
26
We think of a wavepacket as a
superposition of plane waves
e
i  kx t 
e
p

i  x t 


When I apply the momentum operator
p E 
i x  t 




e
x
p E 
i x  t 


 pe
27
Kinetic energy can be derived from
momentum:
2
p
K .E. 
2m
1   
 


 
2
2m  i x 
2m x
2
2
2
28
If I want the expected value of the
kinetic Energy
 
K . E.    *  x 
  x  dx
2

2m x

2
2
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30
31
Note that smaller values of wavelength lead
to larger values of momentum and kinetic
energy:
p
h

p2
h2
KE 

2
2m 2m
32
Let’s look back at the Schrödinger equation
2

2
i
  x, t   
   x, t   V  x   x, t 
t
2m
V(x) is the potential. It represents the
potential energy that a particle sees. Any
physical barrier must be modeled in terms
of a potential energy as seen by the
particle.
33
Conduction band diagram for n-type
semiconductors
This is modeled in the SE as a change in potential:
34
Similarly, if I want the expected value of the
Potential Energy

P.E.    *  x V  x   x  dx

35
Let’s take a look at the Schrödinger equation and some
of the things we might be able to determine from it.
2

i
 
2  V
t
2m
Total energy
= Kinetic energy + Potential energy
The kinetic energy operator is
The potential energy operator is

2
2m
2
V
36
K .E  
2
2

 
*
  x   
 x dx
2   
 2m x 
40 nm
0
37
38
39
Notice that as the wave interacts with the
potential, some kinetic energy is lost but some
potential energy is gained.
40
41
42
P.E  
 *  x  V  x    x  dx
40 nm
20 nm
The total energy stays the same!
43
44
Now notice that part of the waveform has
continued propagating in the medium and
part of it has been reflected and is
propagating the other way.
45
Does this mean that the particle has split
into two pieces?
46
No! It means there is a probability it is transmitted
and a probablility is was reflected.
reflected  
 *  x    x  dx
20 nm
0 nm
40 nm
transmitted  
20 nm
 *  x    x  dx
47
Another example
Notice that the particle has KE of 0.127 eV,
but the barrier has potential energy of 0.2 eV.
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This is an important quantum mechanical
phenomena referred to as “tunneling.”
55
Fourier Transform Theory
in Quantum Mechanics
Fourier theory is used in quantum
mechanics, but there are a couple
differences.
56
Instead of transforming from the time domain t to
the frequency domain   2 f , physicists prefer
to transform from the space domain x, to the
inverse space domain k.  k  2 /  
57
58
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The other difference stems from the fact that
the state variable  is complex, so the FFT
gives its direction.
60
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64
65
66
Uncertainty
or
Where the @#$% am I?
Part of the mystique of quantum
mechanics is the famous Heisenberg
Uncertainty principle.
67
The uncertainty principle comes in two forms:
1. You can’t simultaneously know the position
and the momentum with unlimited accuracy.
x  p  .
2
2. You can’t simultaneously know the time
and the energy with unlimited accuracy.
t  E  .
2
68
We will start by looking at the first one:
x  p  .
2
Remember that we said that momentum
was related to wavelength through Plank’s
constant:
p
h

.
69
If we substitute this in for p and use the
other definition of Plank’s constant
h

,
2
we get
 h 1 h 
x      
.
   2  2 
70
Finally, we divide h from both sides leaving
1 1
x     
 0.07957.
   4
This certainly took a lot of mystery out of it!
It says that I can only know position and
momentum to within a constant value.
71
The uncertainty of quantum mechanics is like
the standard deviation of statistics.

x 
  x  x    x
2
0
2
dx

72
I can also apply this to the Fourier transformed
Y, which is the (1/) domain:

2
2
1 1 
1 1
      Y  d  
     0 
   
1
73
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75
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All of the above signals and their Fourier transforms
were minimum uncertainty pairs because I started
with a Gaussian waveform in the space domain. The
Gaussian is the only signal that Fourier transforms to
the same shape.
 1
FT 
e
 2
1 x 
  
2  
2

 2 x 2 / 2
e

Note that  is in the denominator in the space domain
and the numerator is the Fourier domain. This is the
mathematical expression of uncertainty.
77
This is the two-sided exponentially decaying function.
It is fairly close to minimum uncertainty.
  x   Ae
 x  x0
2
78
Here is an extreme case of a sinusoid times a
rectangular function. The uncertainty if far from
minimum.
79
Up until now, we have been talking about
propagating waveforms. However, we can
also have stationary solutions.
This is similar to an electromagnetic wave
confined to a cavity.
80
One of the canonical problems in quantum
mechanics is the particle in an infinite well.
V 
V 
0
a
V 0
So we have to solve the Schrödinger equation
in the well subject to the boundary conditions:
  0    a   0
  0   a 

0
x
x
81
We start with the Schrödinger equation
2

i
 
2  V
t
2m
We are looking for a solution at the bottom of the well
where V=0. Furthermore, we are looking for a stationary
solution, so there is no time variation. And since this is
a one-dimensional problem,


  x  E
2
2m x
2
2
Solutions are of the form
  x   Asin Kx
82
n
K
a
2mE
K
2
Lastly, we have to find the magnitude constant A.
This is where the “normalization” comes in.

1    *  r   r  dr


a
0
A2a
A sin Kx  A sin Kx dx 
2
2
A
a
83
Let’s look again at the K parameter
K
2mE
2
n

a
Or solving for E
n 2
E  En 
2ma 2
2 2
This means that the energies are quantized.
The energy of the electron is determined by the
order of the eigenfunction, not by the amplitude.
84
If the base of the well is 100 A or 10 nm
n
2
En 

3.75
n
meV
2
2ma
2 2
2
E1  3.75 meV
E2  3.75  4  15 meV
E3  3.75  9  33.75 meV
85
Only certain functions can exist in the well.
These are the Eigenfunctions.
Corresponding to the
frequency of each function
is an energy given by
En  n
These are the
eigenvalues, or
Eigenenergies.
86
In an infinite well, there will be an infinite
number of these eigenfunctions
n  x   An sin Kn x
Kn 
2mEn
2
87
We start by initializing a particle in the “ground
state,” the lowest energy state, of the 10 nm well.
88
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As time progresses, the waveform oscillates
between the real and imaginary parts, but the
over-all magnitude stays the same.
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Note that the energy, 3.75 meV, is expressed
as kinetic energy, even though the particle is
not moving.
94
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As the simulation progresses, it will
eventually go back to its original state
96
Is there a way we could have predicted this time?
Remember that energy is related to frequency:
1
E  hf  h  
T 
97
h 4.135  1015 eV  s
T 
 1.086 ps
E
0.0038076 eV
98
Here is 4 at 60 meV
99
100
101
102
Notice that the real and imaginary parts
oscillate much faster because it is at a much
higher energy.
103
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One of the most important principles in quantum
mechanics is the following:
If I have a “complete” set of eigenfunctions for a system,
I can write any other function as a superposition of these
Eigenfunctions.
N
  x    cnn  x 
n 1
N
  cn sin  K n x 
n 1
We recognize this as a Fourier series.
105
Here are the first six eigenfunctions of the
10 nm infinite well.
106
We have seen that if we initialize with an
eigenfunction, it will simple stay there.
107
108
109
A propagating pulse will be made up of a
superposition of the eigenfunctions.
110
As the pulse propagates, the magnitude of the
eigenfunction coefficients stays the same, but
the phases changes.
111
N
N
n 1
n 1
  x    cnn  x    cn sin  K n x 
112
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116
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120
121
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This illustrates the importance of finding
the eigenfunctions  and the
corresponding eigenenergies En for a
given system.
N
  x, t    cnn e  E
n
/
t
n 1
Because each eigenfunction evolves
according to its eigenenergy, I can predict
how (x,t) evolves.
123
Banding in
Semiconductors
124
Semiconductors are crystals. The atoms
in a semiconductor, e.g., silicon atoms
are at very specific locations.
125
In many cases, the spacing
of the atoms within the
crystal is more important
that the atoms themselves.
The periodic structure that
indicates the position of
each atom is referred to as a
lattice.
126
Even though a semiconductor crystal is charge
neutral, there will be a variation in the potential
as seen by a free electron in the crystal. For the
sake of discussion, we will simply model this as
a small negative potential.
127
We want to see what effect this lattice has
on an electron attempting to move in the
semiconductor.
128
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133
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135
136
137
138
139
140
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142
143
144
145
146
Notice that some of the positive frequency
has been attenuated, but it appears as
negative frequency. This corresponds to a
wave that is propagating in the negative
direction, i.e., has been reflected.
147
We might be inclined to ask the
following question: If a well is other
than infinite, will it still have preferred
states?
148
The answer to that is yes!
In fact, even if the potential is negative
the spacing will want to retain certain
states: those whose wavelength is a halfinteger multiple of the spacing.
149
The answer to that is yes!
In fact, even negative potentials will attempt
to hold those states corresponding to their
length.
150
Therefore, the forbidden values of 
2a 2a 2a 2a
  2a , a , , , , ,
3 4 5 6
i.e, any integer division of 2a is not allowed
151
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156
157
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Notice that there is virtually no loss to the positive frequency
159
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166
167
There is some loss, but it appears at  = 20.
168
These regions where the particle can’t
propagate leads to the gaps in the E vs. k
diagrams.
169
We would like to use these concepts to
think about transport through a device, like
a FET.
170
Assume that the 3 wells are a very simple model for
a FET. If we can determine currents I1 and I2, we
know the current flow through the devices
171
Start by initializing a particle in the left well
and see if it can flow to the right well.
172
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176
177
178
Obviously, it can flow.
179
Here’s another example.
180
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This one is clearly not flowing. Why?
183
The 0.94 meV is an eigenstate of the 20 nm well.
However, the middle well is 10 nm and it doesn’t have an
eigenstate at 0.94 meV. That is the reason the particle
can’t flow: it doesn’t have an eigenstate in the middle
well!
184
The previous waveform could go through the middle
well because 3.75 meV corresponds to the
ground state eigenfunction of the middle well.
185
We saw earlier that 15 eV was an eigenenergy
of the 10 nm well, so we expect this one to go
through.
186
187
188
Before too long, we can see the 2nd
eigenfunction start to form in the middle well.
189
Eventually, it tunnels through.
190
Suppose I were to go back and store the
values at the middle of the 10 nm well after I
had initialized it in one of the eigenstates.
What would I expect to see when I plotted
the resulting data?
191
It would be an exponential with frequency
determined by
E
0.00375 eV
11
f  

9.07

10
 907 GHz
15
h 4.135  10 eV  s
192
I know that the Fourier transform of an exponential is
a delta function in the frequency domain. In fact, each
one of the eigenfunctions results in a spike at a certain
frequency, which I can convert back to energy.
193
This leads me to believe that I will only get current
flow if the particle is exactly at one of the
eigenenergies of the middle well.
194
Of course, that represents the ideal situation
where we had an infinite 10 nm well. In
actuality, any given eigenfunctions is going to
decay out.
195
196
197
I’d like to know how fast particles decay in
and out of the channel. This will be the key
to determining transport through the
channel.
198
It is easier to measure the decay of the left
well into the middle well.
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205
206
So at one of the eigenenergies, the time
dependence is given by
 t   e
e u t 
 t / 2 n int
This is a causal function, i.e, positive time.
Mathematically, we’d rather deal with the
two-sided function
 t   e
 t / 2 n int
e
207
The Fourier transform of the function
  t   e t / 2 ei t
n
n
is
Y   
1/  n
2
 1 
2
 2     n 
 n
We can convert this to a function of energy
YE 
n
n 
2
    E  n 
2
2
208
209
We say that the eigenenergies have been
“broadened.”
210
The broadening shows that even a waveform
that is not exactly at an eigenenergy of the
middle well has some probability of
transmitting through.
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212
Consider the total current flow through the
channel:
I channel  I1  I 2
213
Of course, if we had initialized the right well, then
I2  q
The total current is
1
I channel  I1  I 2
The current flow depends on the following:
1. Available particles in the left or right well.
2. Corresponding eigenstates in the channel.
3. The escape rate  1 / corresponding to the energy of the
particle crossing the channel.
214
Drain eigenenergies
Source eigenenergies
Channel eigenenergies
We might say if there is an eigenstate occupied by the drain which is not
occupied by the source, and it corresponds to an allowed state in the
channel, we will get current flow.
215
The more realistic situation is determined by
the Fermi-Dirac distribution
1
f E 
1 e
 E  EF 


 kT 
E is the energy of the state
EF is the Fermi energy, a parameter that is
manipulated by the applied voltage
k is Boltzmann’s constant
T is temperature in K
216
Drain eigenenergies
Source eigenenergies
Now we have probabilities of the occupation of states.
This branch of physics is statistical mechanics.
217
The drain and source are usually modeled as
reservoirs. A reservoir may be thought of as
an infinite well. The consequence is, it can
provide or absorb a particle of any energy.
218
Finding the Eigenfunctions
of Arbitrary Structures
We have shown the importance of knowing the
eigenfunctions and corresponding eigenenergies
of a quantum structure. We made an analytic
calculation for an infinite well, but this is one of
the few quantum mechanics. MATLAB can
calculate them for one dimension.
219
220
Let us start by assuming we don’t know the
eigenfunctions of the 10 nm infinite well. I will put in
an impulse and see what happens. (Actually, I use a
fairly narrow Gaussian pulse.)
I will store the time-domain data at the point of origin of
the pulse.
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226
I am also taking the FFT of the time-domain data.
227
228
229
Notice the two peaks in the Fourier domain at
3.75 meV and 15 meV. These correspond to the
first two eigenenergies of the 10 nm well.
230
So now we know the eigenenergies n.
How do we find the eigenfunctions n?
Any function, the input pulse included, must
be a superposition of the n even if we don’t
yet know them:

p  x, t    cnn  x  e
 i  n /
t
.
n 1
231
Let’s look at just one point in the well, x0:

p  x0 , t    cnn  x0  e
 i  n /
t
n 1
I’m going to take the discrete Fourier transform at just
one frequency corresponding to the ground state energy:


0
p  x0 , t  ei1 / t dt  

0
 c11  x0 

 i  n / t  i 1 / t
dt
 cnn  x0  e
e
 n 1

232


0

 i  n / t  i 1 / t
dt
 cnn  x0  e
e
 n 1

 c11  x0 
The reason is the orthogonality of the timedomain exponential functions.


0
e
 i  n /
t
e
 i  m /
t
1 if n  m
dt  
0 if n  m
233
To get the entire eigenfunction 1 we have
to repeat this procedure at every point in
the problem space. It is not practical to
store all the time-domain data off-line and
then perform the DFT. But notice the
following:
DFT  x0   
Tmax
0
N 1
p  x0 , t  ei1 / t dt 
  p  x0 , t  n  e  1
n 0
i  /
tn
 p  x0 , t  n  e  1
i  /
tN
234
PFT  x0   
Tmax
0
N 1
p  x0 , t  ei1 / t dt
  p  x0 , t  n  e  1
n 0
i  /
tn
 p  x0 , t  n  e  1
i  /
tN
In other words, I could have a “running DFT”
that is calculating as the program is running.
235
An arbitrary 3D structure
236
Determination of Eigenenergies
Initialize the FDTD problem
space with a test function.
As the simulation proceeds, record
data at the test point.
When the simulation is finished,
Fourier transform the recorded data.
Identify the eigenenergies from the
peaks in the Fourier transform.
237
238
239
240
241
242
243
Eigenenergies for 3D Structure
Eigenenergies found at: 552, 762,1150,
1530, 1830, 2162, 2560, 3405 meV
244
Determination of the Eigenfunctions
Reinitialize the FDTD problem
space with the same test function.
As the program is running, calculate
a discrete Fourier transform at
every cell in the problem space at
each eigenfrequency.
When the simulation is finished, use
the data to construct the eigenfunctions.
245
First 4 Eigenfunctions
246
Next 4 Eigenfunctions
247
Vertical View of the Eigenfunctions
248
Time Evolution
(2162 meV)
249
Time Evolution (cont’d)
250
Observation
The mathematics presented in
this talk is covered in ECE 350.
251
ECE Semiconductor Classes
ECE 460
Semiconductor Devices
ECE 462/562 Semiconductor Theory
ECE 465/565 Introduction to
Micorelectronics Fabrication
252
ECE 462/562 Semiconductor Theory
Spring 2011: MWF 10:30-11:20 am
This is primarily a crash course in quantum
mechanics as it pertains to semiconductors
specifically aimed at electrical engineers.
HW assignments are often simulation using simple
MATLAB programs.
The final exam is a presentation of a paper from the
literature that uses quantum semicondutor theory.
253
ECE 462/562 Semiconductor Theory
The text book for the course is:
Quantum Mechanics for
Electrical Engineers
by Dennis Sullivan
Available from Wiley, January 2012
254
No one understands
quantum mechanics
Richard Feynman
255
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