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.