EE 362 Electric and Magnetic Properties of Materials

1
EE 362 Electric and Magnetic Properties of
Materials

• Dr. Brian T. Hemmelman
• Chapter 2 Slides

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Introduction to Quantum Mechanics

• Several experimental results around the beginning
  of the 20th century couldnt be explained by
  classical physics, which views energy as a
  continuous parameter.
• One such experiment was/is the photoelectric
  effect.
• Classically, if light of sufficient intensity was
  impinging on a surface then an electron should be
  ejected independent of the incident frequency
  (only total energy matters).

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Introduction to Quantum Mechanics

• This effect is not observed, however.
• You can use some frequencies with great intensity
  and no electrons are emitted.
• The actual experimental observation is that you
  can use a constant intensity to eject electrons
  as long as you use a high enough frequency
  (color).
• In fact, the maximum kinetic energy of the
  electrons is linearly dependent on the incident
  frequency.

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Introduction to Quantum Mechanics

• Light appears to have an energy dependent on
  frequency, not intensity.
• Planck said energy emitted in Blackbody radiation
  (heated surfaces) is given off in discrete
  packets of energy called quanta.
• Einstein later interpreted the photoelectric
  effect similarly saying the energy in a light
  wave also came in discrete packets, which he
  called a photon.

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Introduction to Quantum Mechanics

• Both energy relationships are defined by
• The minimum energy required to remove an electron
  is called the work function of the material.
• Any excess photon energy becomes kinetic energy
  of the electron.

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Wave-Particle Duality

• So waves can act as particles at times.
• In 1924 DeBroglie postulated the reverse, that
  particles can behave like waves at times.
• Momentum of photon
• Wavelength of particle
• Electron microscopes use this property.
• Energy and matter are essentially the same thing.

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Heisenberg Uncertainty Principle

• This famous concept applies to the relationship
  between conjugate variables
• The two most common pairs of conjugate variables
  are
• Position Momentum
• Energy Time

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Heisenberg Uncertainty Principle

• The most basic side effect of the uncertainly
  principle is in the limitation of measurements.
• The more precise you try to measure position of
  electron, the less precise you know its
  momentum.
• Eventually we use probability density functions
  to determine probability of finding electrons at
  specific positions.
• The observer affects the observation!!

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Schrödingers Wave Equation

• Let us define the motion of an electron in a
  crystal (or anything) in terms of wave
  mechanics/motion.
• The 1-D, nonrelativistic Schrödingers wave
  equation is given as

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Schrödingers Wave Equation
Assume the wave function is separable
, then
Now divide through by
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Schrödingers Wave Equation

• The only way the right and left sides of the
  equation can be equal f(x)g(t) is if the two
  sides are equal to a constant.
• Call this constant ? and solve the right side
  of the equation.

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Schrödingers Wave Equation
This solution is then of the form
where
However, we know that energy
So
, but
Thus,
The separation constant is simply total energy, E.
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Schrödingers Wave Equation
Now that we know the separation constant we can
apply this to the lefthand side of Schrödingers
equation.
This is called the Time-Independent Schrödingers
Equation.
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Schrödingers Wave Equation
So where does Schrödingers Equation come
from? We know particles have wave tendencies at
small scales, so lets start with the classical
wave equation. The time-independent classical
wave equation, in terms of voltage (potential),
is written as
where ? ? radian frequency and vp ? phase
velocity. A common form of the equation in terms
of electric field would be
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Schrödingers Wave Equation
Lets take the classical wave equation and simply
rename the variable
But
as
where ? ? frequency and ? ? wavelength.
But wave-particle duality also tells us
so
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Schrödingers Wave Equation
Why rewrite in terms of ?
We also know that E T V, Total Energy
Kinetic Energy Potential Energy So and
thus,
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Schrödingers Wave Equation
Therefore we can substitute this wave-particle
relationship back into the wave equation to get
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The Wave Function
The complete wave function is
But this is a complex value and cannot represent
a physical quantity. Max Born postulated that
??(x,t)?2 is the probability of finding the
particle between x and x dx. That is,
??(x,t)?2 is a probability density function for
the particle.
Thus, under these conditions the probability
density function is independent of time.
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The Wave Function
Since the square of the wave function magnitude
is the probability of finding the particle at
position x, we can state

• That is, the probability of finding the particle
  between -? and ? must be 1.
• Wave function probability leads to the famous
  thought experiment of Schrödingers Cat!
• Other boundary conditions also can be deduced
  assuming the particle energy, E, and the
  potential it is exposed to, V(x), are finite.
• ?(x) must be finite, single-valued, and
  continuous.
• ??(x)/?x must be finite, single-valued, and
  continuous.

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Wave Function Boundary Conditions
Since the wave function, ?(x), is related the
probability density function it must be finite
and single-valued. If ?(x) ?, then there would
be no uncertainty in the particles position
which would violate the Heisenberg Uncertainty
Principle. The first derivative of the wave
function is related to the particle momentum.
Assuming the particle has finite momentum the
first derivative of the wave function must also
be finite (and single-valued) and therefore,
?(x), must be continuous. As we are assuming that
the particle energy, E, and the potential, V(x),
are finite, and we know the wave function, ?(x),
itself must be finite (otherwise no uncertainty),
the second derivative of the wave function must
also be finite. Therefore, ??(x)/?x must be
continuous.
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Electron In Free Space
In free space the potential, V(x), is constant.
Potentials must have a reference point so we can
define the constant to be anything. Let V(x) 0.
The solution for the differential equation is
then
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Electron In Free Space
But remember,
so
This however represents traveling waves. If we
know the electron is travel in the x direction
then B 0 (the coefficient is attached to the
term representing the negative traveling wave).
So we can reduce our expression to
where k is a wave number
Since V(x)0 there is only kinetic energy. Thus
and
Free particles have well defined energy and
momentum.
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Electron In Free Space
The probability density function tells us where
we are likely to find the particle. In this case
the probability density function is
This is just a constant, independent of position
x, which means the particle can be found anywhere
with equal probability. This actually matches
what the Heisenberg uncertainty says. If a
particle has precise momentum, then it has an
undefined position. When we do have localized
particles, we use wave packets (a superposition
of wave functions with different momentum or k
values) to define the location and behavior of
the particle. This class will NOT explore wave
packets.
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Electron In Free Space
Just for fun, lets see what the value of the
constant A would be in this case.
This is somewhat similar to the mathematical
definition of the unit impulse, which we define
to be infinitely tall and infinitely thin
with an area under the curve equal to 1. Here, we
are looking for a single particle over an
infinite space. The probability of finding it at
one specific location is just as good as finding
it at another. Thus the probability for any
specific location must be infinitesimally small,
but that does not preclude the particle from
existing (just as the impulse function still
exists).
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The Infinite Potential Well
If the energy of the particle, E, is finite then
?(x) 0 in Regions I and III since the particle
cannot overcome the infinite potential and
therefore cannot be located in Regions I and III
(remember the wave function is linked to the
probability of finding the particle).
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The Infinite Potential Well
In Region II the potential is 0, so Schrödingers
wave equation becomes
The solution of this 2nd order equation is of the
form
Our boundary conditions tell us ?(x) must be
continuous and we know it is equal to zero at the
edges of the infinite potential well.
These must be the values of the Region II
solution at the edge of the infinite potential
well then too.
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The Infinite Potential Well
Evaluate the solution from Region II at x 0 and
x a.
The second equation has solutions anywhere Ka
n?, with n 1, 2, 3,
A2 can be found be applying another boundary
condition
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The Infinite Potential Well
So
where n 1, 2, 3,
Lets take another look at K which was related to
the energy of the particle.
The particle energy is quantized and cannot take
on any arbitrary value! This result is
completely contrary to what classical physics
would say.
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The Infinite Potential Well
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The Infinite Potential Well
Example 2.3. Consider an electron in an infinite
potential well of width 5 Å. Calculate the first
three possible energy levels for the electron.
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The Electron Volt
Consider a parallel plate capacitor (air gap)
which has an applied voltage (and therefore an
electric field). If an electron is released from
the plate at x 0 at time t 0, the force that
acts on it can be written as
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The Electron Volt
Continuing, we can find position x as a function
of time t (here assuming that v 0 at time t
0).
The electron reaches the other plate at time t
t0, and has position x d. So
The velocity of the electron at this point would
be
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The Electron Volt
Since the electric field is
the kinetic energy can be written as
If an electron is accelerated through a potential
of 1 Volt the energy would be
The electron volt (eV) unit of energy is defined
as
So by definition the energy of an electron
accelerated through a potential of 1 Volt will be
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The Finite Step Potential
We have a finite potential, V0, and we will
assume the particle energy is less than this (E lt
V0).
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The Finite Step Potential
In Region I, Schrödingers equation will be
1st term is a traveling wave in x direction
(Incident Wave). 2nd term is a traveling wave in
x direction (Reflected Wave). So A1A1 is the
probability density of incident particles, and
B1B1 is the probability density of reflected
particles. viA1A1 is then the flux of incident
particles and vrB1B1 is the flux of reflected
particles.
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The Finite Step Potential
In Region II Schrödingers equation will be
(recall V V0 and E lt V0)
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The Finite Step Potential
Now apply our boundary conditions. Since ?2(x)
must be finite, B2 0.
The overall wave function must be continuous at x
0.
The derivative of the wave function must be
continuous at x 0.
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The Finite Step Potential
From the boundary equations we obtain 2 equation
with 3 unknowns (A1, B1, and A2). We cant find
explicit expressions of each variable, but we can
find ratios between the variables. Let us solve
for B1 (the coefficient of the reflected wave)
and A2 (the coefficient of the transmitted wave)
in terms of A1 (the coefficient of the incident
wave). This will let us determine what
percentage of the incident wave is reflected and
what percentage is transmitted.
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The Finite Step Potential
The reflected wave probability density function
is then found to be
Recall that the flux of incident and reflected
particles is viA1A1 and vrB1B1 respectively.
Thus, we can define a reflection coefficient as
the ratio of these two fluxes.
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The Finite Step Potential
But in Region I recall the potential is zero, so
the energy of the particles is purely kinetic
energy.
Since both the incident and reflected waves would
be in Region I they would both have the same
value of K1 and therefore the same velocity
magnitude (i.e. vrvi). The reflection
coefficient then becomes
The particle can penetrate into Region II, but
will ultimately be reflected.
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The Potential Barrier
Assume we have a particle with energy E lt V0
which starts moving in Region I in the x
direction. The solutions in the three regions
are familiar.
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The Potential Barrier
We can conclude B3 0 as this represents a
negative traveling wave in Region III. If the
particle is traveling in the x direction in
Region III there is nothing to cause it to
reflect. Now apply our boundary conditions.
Here we have five unknowns and four equations so
we cant solve directly, but we can solve for A2,
A3, B1, and B2 in terms of A1.
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The Potential Barrier
We can define a transmission coefficient between
Regions I and III like we define the reflection
coefficient for the finite step potential. We get
The final expression for the transmission
coefficient is rather complicated and can be
simplified if E ltlt V0. Under that condition we
have
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The Potential Barrier
The wave function before, through, and after the
potential looks like this
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The One-Electron Atom
We have only considered one dimensional problems
so far, where in reality our semiconductors will
have atoms in three-dimensional
configurations. Let us start by considering a
single atom with one electron (hydrogen). The
potential function due to the coulomb attraction
between the proton in the nucleus and the
electron is
This problem is not completely unlike the
infinite potential well. Note, that as r ? 0,
the potential function becomes infinitely
large. One important difference between this
problem and the infinite potential well is that
here the potential function is spherically
symmetrical and we need to use to spherical
coordinates.
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The One-Electron Atom
The solutions to this three dimensional problem
are not particularly simple to compute. However,
one result that should not surprise us is that
the energy of the electron orbiting the nucleus
must again be quantized
The wave functions can then be used to compute
the most probable location of the electron if it
is in the nth energy level. What is unusual
about this problem is that even the potential
function is spherically symmetrical, the
resulting the wave functions are NOT all
spherically symmetrical!
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The One-Electron Atom
The simplest solutions to the one-electron atom
are spherically symmetrical. The radial
probability density functions for the spherically
symmetrical solutions (with n1 and n2) are
So in the lowest energy state the electron is
most likely to found at a radius of r a0
(regardless of the angles around the nucleus).
For n2, the most likely radius has moved further
away from the nucleus.
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The One-Electron Atom
The represent the 1s and 2s states in the
hydrogen atom (remember your chemistry from way
back?) They are spherically symmetric
solutions. However as we go up into the higher
energy levels (n 2, 3, 4, ) there are
solutions whose wave functions are NOT
spherically symmetrical! These other solutions
are the p, d, f, and higher shells talked about
in your chemistry courses. The shape of these
shells (probability distributions) affects the
atomic bonding process as well as physical and
chemical properties of the various atoms in the
periodic table.
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The p-orbitals
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The d-orbitals
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The f-orbitals