Welcome back everyone. 
I'm Charles Clark, and this is Exploring 
Quantum Physics. 
In this part we're going to look at 
solving the Shroedinger Equation for 
Systems that have well defined Angular 
Momentum. 
This means typically systems that have 
some sort of spherical symmetry. 
those include, for example, the Hydrogen 
Atom and the Isotropic Harmonic 
oscillator. 
We'll touch on both of those during this 
part, but we'll focus mostly on the 
hydrogen atom. 
And try to bring matters to closure by 
generating the Eigenfunctions of the 
hydrogen atom, really in a very similar 
to the way that Schrodinger first used. 
Just a refresher, here's what we looked 
at previously concerning the solution of 
a Schrodinger equation for a radial 
function 
And we, we happened to choose a function 
that corresponds to the ground state of 
the hydrogen atom. 
Now we're going to apply the general 
result of the previous part, which is 
that 
We have a universal set of functions that 
characterize the angular momentum of a 
system, of a, a particle moving in three 
dimension potential. 
And so we can look at eigenfunctions that 
separate the radial from the angular 
coordinates. 
And it doesn't matter, because it 
going to, quite a remarkable thing is, in 
quantum mechanics these, these functions 
are, are independent of the particular 
form of the potential. 
So for example, this decomposition holds 
equally well to the isotropic harmonic 
oscillator in the column potential and a 
number of others. 
So let's proceed. 
Here's the Schrodinger equation. 
Time independent Schrodinger equation, 
usual form. 
And now we we have this, this identity 
that you proved a while ago. 
That relates the Laplacian, the kinetic 
energy operator, to some radial 
derivatives in the kinetic energy 
operator. 
And so when you when you plug that in, 
and just take account of the fact that 
the L squared operates only on the Yl,m 
and returns an eigenvalue. 
So l squared ylm equal h bar squared l 
times l plus 1. 
that's, that is the eigenvalue, and so 
that gives what looks like a sort of an, 
it has the appearance of a repulsive 
inverse square potential there. 
So here is the here's, here's where we 
get when we just substitute the the 
separated, it's the product into the, 
into the Schrodinger equation. 
And now, my advice is always to go to 
dimensionless variables at your first 
occasion. 
It's unwise to just start setting all the 
coefficients equal to 1. 
Because, you're bound to make a mistake, 
it then becomes untraceable. 
It's like sort of a one way hash 
function. 
If you set everything to 1, or do it 
different times in the calculation, it's 
very difficult to back out where the 
error is when you make one. 
and, you know, in these, in these sorts 
of problems, errors are frequent. 
So generally speaking this is, this is in 
some sense a generally valid approach. 
And that is you identify a length scale, 
so r has the dimensions of length, you 
know, for a real problem it's in 
nanometers or something. 
And you choose a variable row which is 
dimensionless. 
And then a has the units of length. 
then, then this equation this equation 
takes this form. 
And I've done a few things. 
I've preserved this one half here you 
might say for sentimental reasons. 
It helps me rem, it helps remind me of 
relationships that exist. 
you may prefer it a different way. 
This is really just an arbitrary 
convention but you should find that this 
equation's identical to the original one. 
With the redefinition of the energy and 
the potential, and also something to keep 
carefully in mind Is that the wave 
function has unitive dimension. 
their length of minus three halves for a 
wave function three dimensions. 
And so, although you need not necessarily 
make this this conversion at the start. 
You're going to have to do it at the end. 
So why not, why not do it right off the 
bat? 
So let's just have a quick set check of, 
your graph for the concepts here. 
Okay, I hope that little question was 
very easy for you to solve. 
It's, it's, it just shows you a way, it 
gives you a reading of how quickly you 
can do these sorts of conversions between 
[UNKNOWN]. 
Again the point of being a physicist is 
actually to get beyond that, but its very 
helpful to start, to always have a grasp 
on the units in mind. 
Now, this equation we've gotten, so far, 
in dimensionless form, has singular 
points. 
Some of them are apparent. 
Near row equals zero there's a 
singularity structure and then there's 
also one at large values of row. 
Now its nature depends on the potential. 
We, we're going to address that 
momentarily. 
But there's really a universal problem of 
the singularity in this equation near row 
equal to zero. 
And that has to do, you might think about 
it. 
If you have, here's, let's say, circle 
surrounding the origin. 
So this is the point, r equal to zero. 
And it, for example, that, that the, the 
ground state of the Coulomb wave function 
we looked at actually has its second 
derivatives discontinuous at the origin. 
It's still allowed solution, because that 
discontinuity of the second derivative, 
doesn't involve a discontinuity in the 
energy or anything. 
This occurs as an isolated point. 
Nevertheless it's it's, it's really 
critical to make sure that you're looking 
at the wave function in a way that takes 
account of any possible singular 
structure. 
And I guess the only way to really do 
this is to look at some examples. 
So here, here are two relevant examples 
pictures taken from the digital library 
of mathematical functions chapter on 
Bessel functions. 
The Bessel function, is a, a motion, a 
wave function describes a motion of a 
particle in two dimensions. 
That's just, these are pretty good 
images. 
And so there are, there are, there are 
two types as we saw in examples I covered 
back in the first discussion of solving 
the Schrodinger equation. 
there are functions that are oscillatory, 
and then they're, in this case they're 
called j and y, are the two families. 
And they I want to say j0 and y0, I guess 
are the things that should be compared 
here. 
So if they oscillate together out of 
phase but they're both always 
oscillatory. 
But when one ap, approaches the singular 
point. 
There's one family, the j is regular, 
near r equals 0 and y is irregular. 
Then on the other hand, there are 
functions which travel in the so called 
forbidden regions. 
This is, let's say, what happens when a 
parquil encounters a barrier and can't 
pass over the barrier. 
Well, physically we know the wave 
function has to has to die off as one 
goes into the barrier. 
But, very often, the wave function that 
dies off there, near the singular point, 
will, will always diverge. 
In fact, this is, this is always the case 
for a free particle. 
And then correspondingly, a wave function 
that is regular near the singular point 
will necessarily always diverge. 
So it's, it's building the appropriate 
behavior into the wave function. 
It's very important at the initial stage. 
Okay, now how do we isolate the singular 
points near the, near, near the near the 
cordon origin. 
Well, I'm emphasizing this because it 
turns out that for most potentials, this 
analysis is rather, is independent of the 
details of the potential. 
It's really sort of the very high energy 
limit of the kinetic energy operator 
that's involved in this very close 
approach to the origin. 
And unless, unless, a potential is more 
singular than the inverse square and we 
don't really know of any physical 
potential 
Any potential that can be made in the 
laboratory that has that type of singular 
behavior all the way to the origin. 
unless it would have that, then it will 
be dominated at small rho by the effects 
of kinetic energy. 
And so what one does near the singular 
point is to assume a a a form of this 
type where this, this unknown exponent nu 
will supply the desire. 
By an appropriate choice, will supply the 
desired behavior. 
And so, your row equal to 0, you just, 
you retain the leading order term in this 
expansion. 
And what you find is that nu has got to 
be either, either equal to l, which is a 
positive number, or minus l plus 1. 
And these correspond to the solutions 
that are regular and irregular near the 
origin, respectively, so we choose the 
We choose we choose nu equal to l near 
rho equal to 0. 
And that, that then leads to this, that 
then leads to this equation for phi, 
which is a scaled version of psi. 
Okay, now we face the issue of large rho 
behavior, and this is completely 
determined by the behavior of the 
potential. 
So, a basic message to keep in mind is 
when we're, we're computing a 
normalizable state. 
That is a, a, eigenfunction whose, the 
interval of whose square of all spaces 
equal to one. 
That means that it must converge at large 
distances from the origin. 
And so, in other words, it's gotta be 
confined by something. 
And the confinement typic, condition, is 
typically written this way. 
it's, it's confined in the, in the region 
where the energy Is less than the 
potential energy. 
sorry, the barri, the barrier begins 
where the energy is less than the 
potential energy. 
It's confined to a region where the 
energy's greater than the potent, the 
potential energy. 
And then, that's where we need to make 
sure that we don't pick up diverging 
solutions. 
We actually saw this behavior before and 
it, it was in the context of something 
that was put as a problem in the week six 
homework. 
The quantum mechanical bouncing ball, and 
so in that case this direction here is 
let's say, sorry, the other way. 
That's the direction of gravity pointing 
towards negative rho. 
And really this is just, this just has to 
do with convention for defining the area 
function. 
You could make it the other way if you 
like. 
so basically as you may recall, if we 
have a classical ball and this point 
defines the classical turning point of 
its trajectory. 
It bounces back and forth. 
It goes, it goes, slightly into the 
forbidden region and as described by this 
airy function Ai which decays in the 
forbidden region. 
The, the, the of the complementary 
function, always diverges. 
In fact, the Ai is unique. 
You could add a small amount of Ai to a 
Bi to get another solution, and you'd 
never know the difference, because it's 
so rapidly divergent. 
So we want to end our discussion of 
motion and potential. 
we, we, you know, kind of easily generate 
the motion in the classically allowed 
region. 
But then there's a quantization condition 
that, that only picks up the sorts of 
energies that match onto the converging 
solutions in the forbidden region. 
Okay, so at this stage, we actually have 
to discuss specific potential. 
Because, for example, for the harmon, 
isotropic harmonic oscillator, this new 
rho goes as rho squared. 
At large rho for the coolant potential 
goes as minus 1 over rho, which, which 
dies off. 
So we'll treat the hydrogen atom and then 
I'll make some comments in the end that 
will enable you to think about how you 
would apply it to the isotropic harmonic 
oscillator. 
So we're going to do do the the classical 
problem assuming that the, the nucleus 
has infinite mass. 
That just means that the, all the orbital 
motion is tied up in the electron, so we 
use for the mass of the par, of the 
moving particle, we use the mass of the 
electron. 
And this actually, this is, I showed you 
how to, how to get away from this 
restriction. 
In the material on the Bohr model, but 
its very useful for matching answers to 
precise experiment data. 
And so the interaction is in, this choice 
Gaussian units, this cooling interaction, 
and when we take the mass, the, the mass 
of the electron and the scaling parameter 
A. 
Which originally was just chosen, it 
wasn't identified what it was, we take it 
to be the bohr radius. 
Then as it turns out when you go through 
these these identities, you might try 
that. 
You get this very simple solution for the 
potential in the dimensionless 
coordinates. 
So, here is the equation with what, with 
which we must deal. 
I dunno, maybe it, you know, I guess it's 
a little bit more specific than what 
you've seen previously. 
Nothing here is very complicated, but 
it's solution takes a little bit of 
thought. 
So let's get started. 
Now, as we discussed before, a bound 
state in the hydrogen atom is one for 
which the energy. 
Well, it's for which the energy is less 
than 0, and that means that this 
parameter epsilon is less than 0. 
So I'm writing epsilon as in this form, 
minus kappa squared over 2, where kappa 
is real. 
And again, this is completely this is a 
conventional designation. 
You could you, you could use another 
terminology if you like. 
It doesn't really matter, what you call 
this. 
It's just the, the, the, the point is, 
you want to, 
It is important to retain in the, 
equation. 
A meaningful signature that epsilon is 
negative. 
And maybe, maybe this is the way to see 
why, why this choice is made. 
Because, when, for epsilon, less than 
zero, if we go to large values of row in 
this equation. 
Then everything else dies off with a 
power one over row. 
And basically asymptotically, you, you 
know, there are very large values in a 
row. 
You just have to solve the equation that 
contains these two terms, and of course 
that is one, for which there is also an, 
an exponential plus kappa row solution. 
So what we are we are setting up this 
condition and finding it a new function, 
f, which is associated with the decaying 
exponential. 
And we want f to behave so that, 
basically, it exhausts the function. 
In other words, there can always be a, a 
small amount of the diverging solution 
that gets into the description of our 
wave function. 
And that would, that would poison it. 
So the way that this is done in 
As you'll see the approach, it has some 
similarities with the numerical 
integration method that's discussed in 
the additional materials. 
Okay, so there with this substitution, we 
now, we obtain this equation, this 
differential equation for the the 
function f, this one here. 
Now, you must be wondering what's the 
point? 
It just doesn't look, it doesn't look any 
simpler than, than, than, than the 
original equation, but it's got these 
different parts. 
But here's something to notice. 
The terms underlined in red here, you can 
see it's a second derivative. 
So this is basically like a 1 over rho 
squared acting on f, and this is a first 
derivative divided by 1 over rho. 
So there's a scaling down by a a rho 
squared. 
So if you think, for example, of if you 
have the power series expansion for f and 
you had a term, you know, which was f 
sub, sub j, x to the j. 
Then these terms here would, would, would 
take you down to x to the j minus 2. 
And then the other terms here are, the 
first order in rho, a derivative and, and 
a division, so for the same thing x of j 
x to the j. 
This takes you down to x to the j minus 
one, so this form gives 1 a 2 terms 
recurrence relationship, between the, 
between the coefficients in this, in this 
expa, expansion. 
And that, that recurrence relationship is 
solvable. 
It's basically, you know? 
If you have ai plus 1 is equal f, f sub 
i, ai. 
That's like a geometric series. 
You can very often find an exact 
solution, or, at the least, you can, you 
can numerically solve it quite readily. 
If you have a three term recurrence 
formula, on the other hand. 
It becomes, much more, much more 
demanding. 
Okay, I recognize that there is a lot of 
mathematical detail in this presentation. 
And I can only say this is a sort of 
thing that if you go through it and learn 
it, you will get insight into how to 
solve a larger class of problems. 
But basically, you can see that 
everything that's in the chain of 
reasoning here, they're all these 
transformations. 
and what they lead to is the discovery 
that if this, if this power, if this 
power series terminates at a finite value 
of n. 
Then it turns out that that corresponds 
to energy eigenvalue, which gives back 
the, the result obtained from the 
Well actually we didn't derive the result 
obtained from the symmetry arguments, but 
we outlined t hose symmetry arguments. 
those, those are things you really have 
to do by yourself to see how they, work. 
But we got the same, we got the same 
result from this solution. 
The differential equation as is done by 
the symmetry argument. 
The other thing that is inferred from the 
analysis is that if the, if the power 
series doesn't terminate, then the 
solution will diverge exponentially. 
Because you will have in it, you'll have 
terms in this, the higher order terms in 
the power series here as you, as you 
increase the the value of the index 
indefinitely 
Will turn out to build up into an 
exponentially diverging function. 
Now I haven't made it an assignment. 
But I put on the in the additional 
materials part of the course page, a 
spreadsheet that enables you to solve the 
Schrodinger Equation 
For the even parity states, the harmonic 
oscillator and the isotropics, the s 
states, the l equals zero states of the 
hydrogen atom. 
Solve the Schrodinger equation by 
numerical integration using a boundary 
condition that either has the value of 
the function or its derivative vanishing. 
At a point, and then Well, I maybe need 
to going into the spreadsheet to see it. 
Using the energies and iterative 
parameter, and seeing what happens as you 
vary the, the energy around the exact 
value, and what you find here, here's an 
example. 
Where 1 is solving the Schrodinger 
Equation with the hydrogen, but with a 
slightly wrong energy. 
This isn't a representation of whether 
there's an additional factor of r out of 
a wave functional order to make the 
boundary condition there. 
But basically what you do is you start 
out with uh,with a wave function that 
looks like its going to converge, then 
you, as you get to larger and larger r 
you. 
It, it diverges as you get closer to the 
energy, then you follow the converging 
trajectory for longer time. 
But eventually, eventually you'll take 
off and diverge. 
So finding the exact solution, or finding 
a good approximation of the, a good 
numerical approximation of the exact 
solution of these cases 
Is done by iterating the energy so that 
you basically postpone this divergence, 
which is, just because you're using 
finite precision arithmetic. 
You're always going to get divergence of 
some sort. 
You're sort of postponing it, inevitably. 
And this the, solving this equation is 
sort of like, solving the power series 
termination method that we just 
discussed. 
which was in fact first used by 
Schrodinger. 
So I, I suggest you use other time you 
might spend listening to additional 
lectures to to, acquainting yourself with 
the, with what happens when you actually 
try to solve a Schrodinger equation 
numerically. 
OK that's it, thanks for attending.