[Slide 1] Welcome everyone to lecture four of
the first week of atoms to materials. In
this lecture we're going to apply what we
learned in the last two lectures about quantum
mechanics to solve a very simple problem called
the particle in a box and we'll do
this in a step-by-step basis, given that it's
our first full example that we'll discuss, we're
going to see out this simple problem allows
us to highlight some key features of quantum
mechanics and even discuss optical properties of molecules
and small crystals. [Slide 2] So let's get
started. The system that we're going to deal
with is a particle that we will consider
to be an electron that's confined inside a
box, that's represented by a potential energy function
that you can see here, represented in black.
So the potential energy is infinitely large outside
of the box. The box goes from X
equals zero to L and the potential is
zero inside of this box. Since this is
a one-dimensional system to solve the quantum mechanical
problem as we all know, we need to
solve the Schrodinger equation, that's the Hamiltonian, applied
to wave function equals energy times the wave
function, the Hamiltonian in this case, is written
here, it's a one-dimensional problem so we only
have derivatives with respect to x, in the
kinetic energy. Remember kinetic energy has to do
with wiggles, with gradients, and the potential energy
is the second term here, V of x.
So to solve it we need to do
as I said, in a step-by-step manner. So
the first thing is to ask ourselves, is
"What is the probability to find the electron
outside of this box?". So because the potential
is infinitely large outside of the box, the
probability of the electron being there is going
to be zero. Now if you remember, the
wave function squared is the probability density, so
if the probability is zero, the wave function
has to be zero. So the wave function
is zero outside of the box, for x
less than zero, x bigger than L and
the remaining question is "What is the wave
function inside of the box?" And to answer
that question we need to solve the Schrodinger
equation inside of the box. Now inside the
box the potential energy is zero, we would
have kinetic energy, so the Schrodinger question reduces
to the equation that you see here at
the bottom. So the solution outside of the
box provides boundary conditions for my wave function.
The wave function has to be continuous so
the wave function inside of the box would
have to tend toward zero as we approach
x equals zero and x equals L to
avoid these continuous jumps that would cause a
gigantic kinetic energy. [Slide 3] So that's our
mathematical problem, we have a differential equation of
boundary conditions. If you're a little bit rusty
with solving differential equations, this is a simple
one, a differential equation is an equation that
allows you to find a function or a
family of functions and the way to think
about this equation is to find a function
such that if you take the derivatives twice
and you change the sign you get same
thing except for a multiplicative constant. So that's
what we're trying to do. And I'm sure
all of you are guessing by now what
type of functions we could try, a good
candidate is the sin function for obvious reasons.
So we are going to try this function,
A times sin of kx and k and
A are constants that we will have to
find out and in order to see whether
this is a solution of the Schrodinger equation
what I need to do is plug in
this function into the equation. So to do
that I first need to calculate the second
derivative of wave function with respect to x
two times, that is negative A k square
sin of kx. And remember the derivative of
sine is cosine, the derivative of cosine is
negative sine, so now what I have to
do is put this solution into the Schrodinger
equation. I have negative H-Bar square over two
m, times the second derivative of the wave
function, that's negative again, Ak squared sin of
kx, and at the right hand I have
energy times the wave function, so EA sin
of kx. The sins go and also for
the A on both sides, so do the
k's on both sides and you can see
here our trial function satisfies the differential equation.
We have a constant on the right-hand side,
E, and also a constant on the left-hand
side; H-Bar Square times k squared over 2m.
So that's the solution but what remains to
be done is to apply the boundary conditions.
[Slide 4] So we need to make our
solutions go to zero at x equals zero
and x equals m. So the x equals
zero condition comes for free, okay, because the
wave function, the sin of zero is zero,
but the wave function on x equals L
imposes constraints on the values of k that
I can allow. So in order to satisfy
the second equation I need that kL- for
kL to be an integer times Pi, so
n times Pi where n is 1, 2,
3, so on and so forth. What happens
is not every value of k is allowed,
k is quantized and n is called the
quantum number and it indexes the solutions that
have. So the values of k that are
allowed going to be the k sub n
and there n times Pi over L. So
those are my solutions, my wave functions, the
allowed wave functions are a family of functions
indexed by n, okay, as we just saw
and the corresponding energies are also quantized E
sub n. So n can be 1, 2,
3, 4... [Slide 5] Let's look at these
solutions, what I'm plotting here on the left
is the wave functions, the first four wave
functions, n equals one in the bottom that's
the lowest energy and we call that the
ground state, n equals two, n equals three,
and n equals four. You can see these
sin functions and these wave functions are shifted
up by an amount equivalent to their energy.
You can see the energy expression here, the
energy goes up, increases with n squared so
that a ground state energy is one times
this constant H-Bar squared times Pi square divide
by 2m squared and then the second solution
is four times that and then nine times
that and 16 times that, so the energies
increase with the quantum number squared. So there
are a couple of interesting observations that we're
going to make from this solution. The first
one is states are quantized okay, not every
energy not every k is allowed in the
system. The second one is that the lowest
possible energy, the ground state energy is not
zero so the kinetic energy of the electron
in the quantum well cannot be zero which
would be the classical solution. If I have
particle in a well, the potential energy in
a box with zero potential energy, the lowest
energy is for the particle to be not
moving, zero velocity, zero kinetic energy and consequently
zero total energy. Now quantum mechanics does not
let us do that and the reason is
very simple, it's the uncertainty principle. The particle
is confined within the box and that means
that I know something about its position. So
given that I have information about its position,
that means that because of the uncertainty principle,
I cannot know the velocity with infinite precision.
If I cannot know the velocity with infinite
precision that means that the velocity cannot be
zero. And that's why there's the ground state
energy is not zero, this is called zero
point energy and that's an amount of energy
that cannot be extracted from the system. You
cannot slow down the electrons further, there's fundamentally
low given by the uncertainty principle. So it's
interesting to note that this energy of the
ground state energy decreases with increasing length, with
the square of the electron so the energies
is inversely proportional to the length Squared so
as I make the particle the box bigger
and bigger and bigger, I know less about
the position of the particle and then I
can know more about the velocity and I
can reduce the velocity and reduced the kinetic
energy. So the larger the particle in a
box the smaller these energy values become. The
other thing, interesting thing to note is to
look at the wiggles in this wave functions
and the nodes in this wave function. So
note that the ground state wave function, that's
the red function there, has no nodes, its
half of the sine wave. Well remember the
no node theorem; the ground state energy has
no nodes. The first excited state has a
single node and the second excited state has
one more node and so on and so
forth so that the higher the number of
nodes, the higher the energy and the more
wiggles the higher the kinetic energy, and so
a lot of very important concepts in quantum
mechanics and the simple system that we're going
to use to understand hydrogen and molecules and
crystals. [Slide 6] So now let's ask ourselves
with this very simple model whether we can
think about optical properties. So let's say I
have this box and I have four electrons
in it. So I'm going to put the
first two electrons in the lowest possible state,
in the ground state and then the next
two electrons are going to go in the
next available state, the next excited state. I'm
going to be ignoring electron-electron interaction and of
course the presence of additional electrons change the
potential energy, change the Hamiltonian but I'm going
to neglect those interactions for now. So that's
my system for electrons and the question is
what happens if I shine light with my
system of these four electrons in the box.
So light is made of photons and these
photons are particles that carry an amount of
energy that's proportional to their frequency. So H-Bar
omega, Planck's constant times the frequency of the
photons, that's the energy, the amount of energy
that they carry and so if a photon
hits the system with enough energy to excite
an electron from its' current state to an
empty state then I can absorb that photon
and conserve energy. And so if the photon
carries an energy that's not able to do
that, if the photon carries, for example, this
amount of energy then it cannot be absorbed
because I cannot promote an electron to a
state such that energy is conserved. So a
photon can only be absorbed if it carries
enough energy to excite an electron to an
empty state that exists in the system. So
that is going to put a constraint on
the amount of energy that I can absorb
and it puts a constraint, it quantizes the
frequencies that the system can absorb of light.
So the energy of the absorbed photon has
to correspond to a difference in energy between
states and we can write that from the
solution that we have of our system. So
if I start in state n, my initial
energy is proportional to n squared, the final
state is n prime and if that is
satisfied then I can absorb the photon, otherwise
I can't. And emission occurs in a similar
way, if I have an excited electron these
electrons can decay to an empty state that's
lower in energy and meet the photon of
that specific frequency. So again photons that don't
satisfy these conditions are not able to promote
an electron from an occupied state to an
empty state, cannot be absorbed and go through
and that's why we can see through a
window. So silicon oxide that makes a glass
that doesn't have energy states that correspond to
the frequencies in the visible range of light
that that's what we can see through them.
They have a band gap that's larger than
the visible range. [Slide 7] Now what we're
going to doing the homework assignment is this
very simple model to explain the colors of
a set of die molecules that you see
here on the left and I'm not going
to pronounce their names but these conjugated systems
have some electrons called Pi electrons that are
essentially free to move in a quasi-one-dimensional box
that's given by the separation between these nitrogen
atoms. So essentially the nitrogen atoms provide the
box, you can see that the size of
the box is slightly different in these three
molecules of quantum confinement, it's different and using
the model that we just developed in the
homework assignment you're going to compute the amount
of energy these molecules can absorb and consequently
the frequency of the phonons and the color
of those photons and you're going to correlate
that with what is known experimentally. and you
would say that you're going to be able
to make semi-quantitative predictions that certainly explain the
trends in the change in colors in these
strains of molecules so even this very simplistic
model that ignores the fact that the potential
energy that the electrons in the real molecules
is not flat and ignores electron-electron interactions it
allows us to understand some interesting phenomena and
explain interesting trends and make predictions are not
too off from what you measure experimentally. [Slide
8] Let me finish today's lecture by pointing
out that in the last few years we've
learned how to build Nano crystals that are
very high quality and very well controlled and
very small size where quantum effects can be
seen. And you can see here an example
of work where the optical properties of these
Nano-crystals depend on their shape and their size.
This quantization, remember the ground state energy that
goes one over nL squared, that number becomes
very, very small, negligible when L becomes macroscopic
so if L is 1 cm, that energy
will be irrelevant but when you get to
the tens of nanometer scales, this the zero
point energy becomes important and you can actually
measure it, measure this spectra that originates from
quantum confinement a. Another interesting example is a
quantum cascade laser, that's created as a hetro-structure
and the size of the hetro-structures control the
wavelength of these lasers so again; this very
simple model allow us to understand some recent
very interesting experiments and then some it interesting
devices. So with that, we've finished lecture four
and I'll see you all in lecture five.
Thank you.
