Welcome back, everyone. I'm Charles Clark and this is Exploring
Quantum Physics. In today's lecture, we're going to look
at solving the Schrodinger Equation by a well-known proven technique.
Actually, first used by Schrodinger himself, which is the use of the special
functions of mathematics. I'm using a convenient online reference,
which is freely available around the world for this.
It's a standard repository digital library of information on the so-called
special functions of mathematics, which are the ones that are in most widespread
use in areas of science and engineering. it contains the definitions, properties,
interactive graphics. Something might be of interest to some of
you. TeX encoding of all of all appropriate
material of all the equations and so on. Also, MathML encoding.
And the ability to do Semantic search which recognizes fairly free format
mathematical, mathematical information. Also, I've put some material on the
course page in the additional materials section.
And one of them is a summary of the solutions of frequently encountered
Schrodinger equations in terms of special functions.
and then, something I encourage you to try a demonstration which uses a, a
numerical solution of the Schrodinger equation.
It's coded up very simply on a spreadsheet and basically two lines of
code are all that are required. And this will give you the opportunity to
actually gain some experience with practice in numerical solution which I
think, in my case, has been a really good guide to understanding how things happen
and why and what to expect. I mean, there really is, the only way
that you're going to become comfortable and familiar with solving the Schrodinger
Equations by working on examples. So I do strongly encourage you to take
that opportunity and, you know, perhaps to apply to some of the, the simple
examples we're going to start with now in the lecture.
Well, who was the first to solve the Schrodinger Equation using special
functions? No kidding, it was Schrodinger himself.
What's good enough for Erwin Schrodinger is good enough for us.
in his very first paper on wave mechanics he used special functions to solve the
equations for the hydrogen atom. And I think in the some second or third
paper, he also computed the harmonic oscillator.
And what he did and I think this is one reason why the Schrodinger equation was
so widely adopted quickly is that he found that he could solve the Schrodinger
Equation with standard well-studied equations of mathematical physics.
So, these were functions that were defined in previous contexts and applied
to things like electromagnetism and acoustics.
And thereby, there was, there was a rich technology that had already been
developed and the properties of the solutions of those functions were
mathematically well-defined in them. So that's a pretty good approach to be
following. So, we're going to look at three
one-dimensional examples in this context. the constant potential, the linear
potential, which describes a particle in electric field, and then the quadratic
potential of which you've already seen a good treatment of by Victor in the
harmonic oscillator section. this is going to look at things in
somewhat more general context. So, the solution is here.
A map on to the exponential, the Airy and the Parabolic Cylinder functions
respectively. and I hope to show you from a few simple
examples some of the generic features that you'd expect when you might be
dealing with a more a system that has parts of it that look like this.
But then, they, they might vary in a piecewise manner or something like that.
For example, if you're designing a a multi-layer semiconductor quantum dot or
a quantum wave guide you will, you will encounter segments of the problem where
the potentials might look piecewise, like they do here.
Well, let's proceed. Okay, now, we look at the constant
potential. Something not terribly interesting in its
own right. But we're going to do it in a more, in
the most general way. So, the, the point is we have a, a given
potential. And now, we have the Schrodinger
equation, which we're defining without any boundary conditions.
And we want to find the solutions for all possible energies.
So we want, in other words, we want to be able to solve this equation in the most
flexible way. And the, the standard approach that
applies to all these examples, and will apply to most other practical problems
you work on, is to transfer, transform the Schrodinger Equation into a, a
standard dimensionless form that defines one of the special functions.
So this one is pretty easy. we can see we, you just, you just move
these constants around, and and scale the variable.
And as you [INAUDIBLE], the, there, there is in fact a dividing line between two
classes a solutions associated with sine of this coefficient.
So, here are, here is the transformation I described choose a scaled variable, and
we get these two the two different solutions.
So the point to note here is this is a second order different actual equation.
And so, it always has two independent solutions.
Now, you can choose those solutions in different ways.
Here is a standard choice for the oscillatory solutions.
The sine and the cosine out of phase by half a wavelength.
and any other solution that you get will, is, can be described as some linear
combination of those two there. So, that's that seems fine.
Now, in the in what we call the classically forbidden region, where the
energy is less than the potential energy, then we get the minus sign in the natural
way of decomposing this equation. And now, here's an important point.
Once again, there are two solutions and actually there are, there are any numbers
of ways you can pick the two solutions. So here, I have the exponent of rho and
the exponent of minus rho. Well, sinh rho and cosh rho would be
equally acceptable. They are independent.
but there is uniquely one solution only up to a multiplicative constant.
One solution only, that, that converges in the, in this, this is what we call the
classically forbidden region. I mean the energy is the potential energy
plus the kinetic energy if the energy is less than the, the potential energy, that
corresponds to the kinetic energy being negative.
Which can't be the case because it's well, you can, but, but you do get, you'd
get, you get you'd get suppressed behavior.
So, if you think about it, this, this area is tricky from the standpoint of
implementation, because if you're, if you are, let's say integrating a normal wave
function. That in, into the forbidden region in the
forbidden region, your solution is going to be a linear combination of one
that's divergent and one that's convergent.
And the chances of your hitting the exactly convergent function are very
small. Okay.
I admit, we're not going to win any Nobel prizes for solving the one-dimensional
Schrodinger equation with a constant potential.
But let me say two things about it. First of all, you can well imagine
simulating, let's say, let's say you have a interesting potential that you want to
solve for some problem a waveguide design, or something like that.
Well, you can think about approximating it by by a step, a stepwise
approximation. So here, I've got these.
I, you sort of like approximate an integral by a, a finite sum and you could
then use use this simple mapping onto functions within each region to advance a
wave function efficiently across an arbitrary potential.
The second is sometimes one thinks of these, these step or square well problems
as being highly contrived. But virtually, every day, you encounter
practical example of such a system and that's when you look, you look through a
window. so let's say, here, you look at a window.
So here, we have incoming light e to the ikx is representing the incoming wave.
And then, we have the reflected wave of coefficient R, Re to the minus ikx.
And then, there's a transmitted wave into the glass.
And the relationship between the transmitted wave of normal incidence and
the incident one is that the, the wave vector in the glass is multiplied by the
refractive index, which is about 1.5. So my challenge to you is to determine
what that transmission code, what fraction of the light passes from the air
into the glass and what fraction is reflected?
So, I hope you saw that for glass, about 4% of light is transmitted.
So, that, that's a pretty low loss. but it's, it's enough that you can see
your reflection under certain conditions. Now, we look at the linear potential.
So you might say we started with V naught.
Now, we're adding a term, which is constant.
Now, we're adding a term that's linear and x.
So in some respects, this is sort of the next it, in order of, of mathematical
complexity, this is like the next function in the series away from the
constant, the linear. And so there's Schrodinger's equation,
it's very straightforward. And what we do is here is x in terms of
the, the dimensionless variable row. we, we, we take away the offset and
divide by the slope. here is the here is the equation for the
coefficient alpha. Now, are, the standard solution, which is
the defining solution of the area function is here.
It's just the second really, there's a portion of linear multiple, of the wave
function. And again, we have a pair of independent
solutions. So, in the region rho less than zero, the
wave function is always oscillatory. And we see that in that region, we have
these two independent solutions called Ai and Bi, they're both called Airy
functions. So, so any choice, any choice of wave
function in this region is going to be oscillatory.
You might, you might ask yourself, what does that correspond to physically?
Now, once again, in, there's another region.
And I guess you can see, because if rho greater than zero, this is sort, sort of
corresponds to the negative sine equivalent for the the constant
potential. So now, there is one unique solution.
Again, up to an overhaul multiplicative constant.
But one unique solution that converges as, as rho goes to infinity and another
that diverges. So these two have applications.
We'll talk about this one, the quantum bouncing ball.
and the Bi's is, is arises in field ionization, and scanning tunneling
microscopy. Well, just important, how important is
the quantum mechanical bouncing ball? Well, might you ask.
Well first off, it's, it's a good example, because it, it it's, it's kind
of like the, you might say in some sense the simplest case of a balanced state
problem one has. But in fact, it's been the subject of a
very fascinating recent experiment of trapping neutrons the neutral counterpart
to the proton in the earth's gravitational field.
So, in this experiment very cold neutrons are put into a chamber and they, they
they're reflected by a mirror, and so, they bounce back and forth in the earth,
earth's gravitational field. Now, you remember this map of all the
theories that Victor showed in the introductory lectures, that showed that
gravity was something kind of disconnected from the main themes of
quantum mechanics and, and relativity. While here in this experiment, they're
actually looking at the quantized energy levels of a particle in the earth's,
subject to the earth's gravitational field.
So this is a test of whether a quantum particle performs as expected under the
influence of gravity. In principle, this is a very, in my
opinion, very significant, intellectually interesting experiment.
Now actually, I mentioned a predecessor, predecessor in earlier lecture, which is
neutron interferometery in the gravitational field.
well, no surprises were found here, but you might, you might consult this paper,
it's rather interesting. And we're now going to solve the, solve
the problem associated. Okay, just some numbers for you to keep
in mind. here's the standard value for the earth's
gravity, the mass of a neutron. And now, in the way this problem is set
up, here is the, here is the Schrodinger equation that we're solving.
And so this part includes the the linear potential.
So, x is zero on the surface of the mirror, and as you go up, you're in a
potential which is Mgx. and then, the boundary condition on the
mirror it's a bouncing ball. So we require the wave function to vanish
there. All right, so now I just have a sanity
check for you to answer in terms of an in video quiz.
Okay, I admit, that was another unit check.
But to tell you the truth, I always check units.
I never, I hope never to forget to do that.
You catch so many errors that way. And what I hope many of your saw is that
you didn't really need to do deep analysis to figure out what alpha is.
Because you see in this equation, alpha rho is equal to E, and rho is
dimensionless. So, that means that alpha must be equal
to energy or another way of noting that is when you put Mgx.
That's got to equal to that's got to be equal to alpha rho, and rho again has
been chosen manifestly to be dimensionless.
Okay, now, I have another brief quiz. basically, the same sort of idea of
getting familiar with the details of the problem as it's laid out here.
Okay, so this, the answer was rho is this classical turning point.
It's the highest point reached by the neutron's trajectory.
And I guess you can see that, here, in the sense that for rho equal to zero, we
have E is equal to Mgx. So that means all the energy is in the
potential energy. So that, that rho of zero is the turning
point to the classical motion. So now, you can see the influence of the
turning point here, that's where the wave functions starts to die off.
The sort of the, the lore is there's like a quarter of a wavelength of damped
oscillation there at the just beyond the turning point.
And now, do you see how we're going to solve the the, the problem with the
boundary condition? We, we're going to require that the wave
function vanished at the surface of the mirror.
Well, the surface of the mirror is given by this.
And so, and, but the wave function is, is just is the most general solution here is
Ai of rho minus the sum rho naught. And so, when we choose the values of rho
to correspond to the nodes of the area function here, then you might say, this
here, the profile from here on out. That is the amplitude of the ground
state, because it's the lowest energy solution that vanishes on the mirror.
Now, to find the next one, you're basically moving the mirror down.
You might say you could think of a different experiment where you drop, you
drop the neutron from a certain height, and then you, you keep moving the mirror
down until you get, until you get sort of a perfect reflection.
And so, the sequence of excited states can be, can be determined by picking off
the area functions. And so once again, it's sort of, here's
the, here's the energy. It's, it's the scaling constant times
these rhos, the lowest one, you know, it's, it's 2.34, it's of the order of
[UNKNOWN]. So, this is the case where, just
calculating the scaling constant gives you a good prediction in the energy scale
of the system. Now, here is the the quadratic equation.
now, I'm just going to take the, the the case where it has a positive quadratic
coefficient, which does correspond to the, the harmonic oscillator.
And what you do is you you, you, you, you transform, so this, this potential always
has a minimum somewhere. You just move to that minimum then
there's an energy offset here. And then as we know from the analytic
solution of the harmonic oscillator, the, the spectrum is given by this, but again,
let's look at it from the general, the general perspective.
So, the standard equation for the quadratic potential is what's called a
parabolic cylinder function. And in particular, there is a class of
well-studied solutions traveling under this name, the D, D sub n's.
And the two solutions to the, this equation are Dn of rho and Dn of minus
rho. And it so happens that these are chosen
so that for positive n, Dn of rho always converges.
So this would be something quite useful to know if you were dealing with a
problem, which was sort of piece wise quadratic.
And you wanted to make sure that you were getting a solution to tail off in the in
the class of forbidden region. on the other hand, there in the region
rho less than zero both of these diverge unless n happens to be an integer.
And that of course, is the here's, here's the n, you see.
And that, that, that of course corresponds to the solution from the
harmonic oscillator. So those are just shown here.
Here is n equals zero. It's the ground state of the harmonic
oscillator n equal to three. But now, he sees, n, for n equal 3
halves, you get something that's kind of harmonic oscillator looking, but do you
see it's not crossing? It's not crossing it's not, it's not
intersecting rho equals zero, so it can't be an odd solution.
And it doesn't have a math a local maximum of rho equals zero.
So it can't it doesn't have a zero derivative, rho equal to zero, so it
can't be infinite an even solution. So, these, these all, these all diverge
at large values of negative rho. Okay, I think, now is a good time to
leave off. And just imagine again that I think
you'll, those of you who like to solve problems should find some benefit from
working out the numerical solutions to some of these equations.
You can modify this spreadsheet to treat potentials of your own choosing.
By the way, it's not, it's not a highly competitive method of solving an ordinary
differential equations. It's just something that happens to be
easy to code up, it's [UNKNOWN] when understood.
And you could, you know, you can see the solution happening in real time.
again, there's also a a summary of the cases that are most readily treated by
use of special functions.