[Slide 1] Hey everyone, welcome to the course.
This is Nanophotonic Modeling.
I'm Peter Bermel, the instructor, and I teach at Purdue University.
What we're going to be talking about in our first unit, Unit 1, are band structures,
and then of course in the subsequent units, we'll go over different topics.
And so after band structures then we'll go over transfer matrix methods,
and then in the third unit, we'll talk about finite difference time domain methods.
In the final unit, Unit 4, we'll actually talk about some advanced
multi physics problems and finite element methods to solve them.
So, now let's get started with the bandstructure problem.
[Slide 2] So what is the bandstructure problem?
So it's basically an eigenvalue equation, and
the eigenvalue equation may have this form, H psi equals E psi.
So we interpret H is a matrix or an operator, and
then E is the scalar, or a number essentially.
And so what's interesting about the eigenvalue equation in general is that you
end up, when you multiply H times psi you still recover that psi times a constant.
And so of course, there are many different types of bandstructure problems
that we could potentially consider.
One famous one, of course, is Schroedinger's equation.
And so Schroedinger's equation looks like this.
It has both kinetic and potential energy terms equal
to the energy of the photon or electron in the system.
And then you have a photonics bandstructure,
which is basically given by combining Maxwell's Equations.
This is almost like, slightly different than the Helmholtz Equation.
And then you have the so-called phononic bandstructure, which describes
a quanta of heat that would propagate in any kind of solid material, for example.
[Slide 3] So now we're going to talk first about Schrodinger's equation,
which will kind of help us understand the basic premise of the bandstructure problem
before we go in to photonic bandstructures.
So basically Schroedinger's equation describes the wave function
psi to tell you the extent of the particle.
The probability of finding a particle at any given location
it goes like psi squared.
So, of course, it has to be normalized, so integrating over all space would give you
a probability of one, so it has to be somewhere.
So then when you write down Schroedinger's equation,
H psi equals E psi, like I mentioned, has kinetic and potential terms.
So of course you know in classical mechanics that kinetic energy goes like
one half m v squared or also can written as momentum, p
squared divided by 2m, right?
So that's this.
However, in quantum mechanics usually you promote what would normally
be just a simple number like momentum to an operator.
And so when you call that operator minus ih-bar times the gradient function or
Nabla, then you get something like the p
squared becomes like minus h-bar squared times the Laplace function
And then you divide by 2m to get the kinetic energy.
[Slide 4] And so if you didn't have any potential,
then that would be your full Hamiltonian, and so then the Hamiltonian would
look like this kinetic energy operator times psi equal E psi.
And so then if you think about it for a minute, you can imagine based on analogy
with say electrostatics that your solution would just be a plain wave.
So if you actually substitute this expression that side goes e to the plus or
minus ikx, you can see that this laplacian brings down the factor of minus k squared.
And so then you get h squared k squared over 2m, right?
And so it kind of make sense, because we're applying operators that
like the wave functions themselves should be given by compact exponentials,
because we recovered the original value.
So it's very suitable for this sort of eigen problems.
And the other thing that you notice of course is that,
because E goes like h-bar squared k squared over 2 m,
you have kind of this quadratic curve and that's of course for particles.
[Slide 5] If you put the particle in a well, which was infinite say a proton in iron nucleus,
then you can basically say that within the radius of the iron nucleus that there's
no potential basically is stable, but then outside potential is infinite.
Of course this is a little bit of a simplification will just give you an idea
of what's going on, so
there is boundary condition on both sides where psi goes to zero.
And so of course, we can use actually the same kind of solutions that we had before
for the free particle, except that we have to impose this boundary condition,
which means that eigenfunctions are actually a sum of two plain waves that
will cancel out.
But then that means that the k values are actually quantized now.
And so in order to have basically the psi vanish on both sides,
that means we have to have a half integer number of waves.
And so then that tells us that k would be integer n times pi
over a if it has a length of a.
And so then if you plug in the value that we've expressed for the energy
as a function of k, h-bar squared k squared over 2m substitute this k value,
then you get E as h-bar squared n squared pi squared over 2ma squared.
[Slide 6] So then of course what we could also do a finite quantum well.
Like an alpha particle in uranium nucleus.
And so here we would say it's very tall, but not infinite potential.
And so the difference is just that now it can actually
decay exponentially to the outside and maybe even tunnel out.
Which of course would be classically forbidden if U is greater than
kinetic energy, but it's allowed in quantum mechanics.
But it may be very improbable, so that's why uranium-235 nucleus lifetime
would be on the order of hundreds of millions of years.
[Slide 7] A way to actually kind of extend that to a more interesting problem for
bandstructures is to make that whole system I showed you earlier periodic.
So imagine if you had a whole series of periodically repeating wells,
and so you have these little regions where potential is 0, and
it can move freely, but then you have these barriers as well.
So then the question is, what would be the boundary conditions that are appropriate
for this system, and would each electron get stuck in its own little well?
So in the next lecture, we'll actually explore how to solve this problem.