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.