Welcome back everybody. I'm Charles Clark.
This is Exploring Quantum Physics. And today we're going to turn to the
problem of angular momentum, one of most fundamental symmetries in quantum
physics. We say everything revolves around angular
momentum and we're going to start looking at some of the explicit solutions.
We'll try to show you different ways of calculating them from first principles,
and then with pointers to the literature so that you can, you can develop better
understanding if you so choose. I hope you enjoy it.
Now, home works of week six had some problems of tensor analysis, a lot of
vector algebra. some of you found that quite frustrating,
I think. others, I believe, found it enjoyable.
in any event, we're going to do a little bit more of that.
And please don't be too alarmed by everything that's up here.
These are really the core, core identities that I'm putting for
completeness. the uh, [INAUDIBLE] tensor.
Just the basic notation about, our change in notation in order to simplify
calculations. the definition and the position vector,
in terms of the old and the new notation, the radius.
And then the definition of the angular momentum operator and the square of the
angular momentum operator. So we're going to now use some of these
to generate Eigenfunctions of the angular momentum operator.
And I think you'll find this approach very useful.
Especially if you work in other fields so electrical engineering, or anything that
studies acoustics or wave motions or o, optics, you'll find a lot of counterparts
in those areas to what we're studying here.
We'll start by showing what I hope you think is quite an amazing result.
If we take any function of the radial coordinate only.
it, it turns out to be an Eigenfunction of any given complement of the angular
momentum operator and of the square itself.
Okay. So let's see how this goes.
we start with, Just applying the l, l sub u upon psi.
And then that is, very straightforward, expansion of the angular momentum
operator. And now what you see is that the, effect
of the derivative on a, a radial function, function of the radius only can
be evaluated by the chain rule. And so you see we get an expression that
boils down to this. So we'll go with a little in-line quiz to
see It get, get, give you a gauge of your
understanding of how to deal with an expression encountered like this.
So I hope that you saw that this vanished, this term vanishes by symmetry
relations because it's just a sum over dummy indices but with a, a sign change
associated with the permutation of. The two indices, so that term must
necessarily be zero. So we've recorded that fact.
And then the next one is is, is easy because if l u of psi of r zero, then
anything else applied to that l u. Of psi of r is also 0, including lu
squared, and then the sum of the squares of the l sub u, which is the total
angular momentum vector is also equal to 0.
So we've shown that any, any function that is a function of the radius alone is
an Eigenfunction. Well, it's an Eigenfunction of any
complement of the angular momentum operator.
It is conventional to take the complement along the third axis.
This traditionally the z axis, as the quantization axis.
But we'll see that that's not always the case.
Anyway, let's proceed and see what the implications of this discovery are.
Several weeks back, we tried a sort of simple constructive technique.
It's not a general purpose technique, but just a way of getting some understanding
of the Schrodinger equation by applying it to functions that we already know, the
ones that we're familiar with. And seeing what, what type of potential
would support those as an Eigenfunction. So what we want is to use a function of
the radius that is, is, is normalized, or normalizable, that is it can be divided
by another constant to make this identity.
Correct. And in the previous case, we chose a a
radial Gaussian wave function, of this, of this form, which is in fact the the
solution for the ground state of a three-dimensional isotropic Harmonic
oscillator. You see that this term here is just equal
to r squared, so this is a radial solution.
And it therefore has the anglar momentum, 0, that we just described.
Well, how many functions do you know that are radially symmetric and die off in
infinity, in a, in a way that's sufficient to,
To allow the normalization integral to be computed.
Well, of course, one can invent a whole number of, contrive functions that do
that. But, to me,
This one here sort of stands out. It's a simple, exponential decay in all
directions. it's easy to, use, and so it's a good
choice for this constructive approach. So here I show a cut through that
function. This is radially symmetric, but I cut
through the origin, r equals 0, along, it doesn't matter which axis, any axis looks
the same. And this shows something that you might
not have appreciated before, which is the wave function has a cusp at the origin.
In fact the second derivative of the wave function, you know, with respect to x, is
actually a singular there. Well that corresponds, in fact, to
something else that we find if you just if you take this function and put it into
this equation you will find that the the, the potential that is implied by the use
of this function is just the Coulomb funct-, Coulomb potential.
Again, Gaussian units, Gaussian units. and this is, in fact, the ground state
wave function for the Hydrogen atom. it's [INAUDIBLE] with energy, given by
this expression. in, whatever the Bohr radius is you've
seen it before, but here it is made more explicit.
Now there was a, one of the problems in, one of the questions in week five
homework had to do with approximating using a Gaussian approximation to this
function in a variational calculation. Well, a Gaussian, you see, is going to
have a rather smooth profile of the origin.
So this type of cusp behavior is not something that's well represented by a
Gaussian wave function. As a matter of fact, there were two
variational calculations in that week's homework.
And in some sense, the. Functions that you were being asked to
approximate were the same function. This is actually also, if you think about
it, the one dimensional cut here that gives functions the same as a function of
a ground state of an attractive delta function potential of one dimension.
So this is a rather interesting correspondence between the delta function
in one dimension and the Hydrogen atom in three dimensions.
So just for convenience of further notation we'll just, we'll write that,
this wave function psi sub 1s. The 1s is the conventional designation
for the quantum numbers that bounced eight of Hydrogen.
Maybe this is also a good time for you to refresh your recollection of the meaning
of the Bohr radius. And so that will be the subject of an
in-video quiz. Okay.
So we're going to break at this point, and you might review this material.
See if you can follow the derivations. I also, recommend that you use your time
just to do a few simple calculations, like verifying that this wave function is
appropriately normalized. You know, I could make a mistake and I'll
give a nice reward to the first person who identifies that in the student form.
to do this, it requires integration over polar coordinate angles.
A theta in phi. Now we're going to discuss those later.
We haven't really dealt with that yet, but you can find pointers on how to do
three dimensional integrals many places in the open literature.
And so this is a sort of exercise for those of you who want some practice.
those of you who had trouble. With the homework problems of week five,
maybe this is a good way a good approach to sharpening your skills.
Okay, thanks for listening and hope to see you again.