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.