Welcome back everyone. 
I'm Charles Clark, and this is exploring 
Quantum Physics. 
Now we're going to continue on from the 
rather si, simple example of zero Angular 
Momentum, that we looked at, in the 
previous part. 
And to go on to see how to systematically 
develop things to look at more and more 
complex systems. 
Just to refresh your memory in the 
previous part we just looked at the 
problem of a purely radial function and 
found that it has zero angular momentum. 
So if you think about angular momentum as 
an operator that induces rotations. 
perhaps that's and intuitive way of 
understanding why a function that just 
is, is, has no variation as one goes out 
in in different directions of the origin. 
It's just, it's just something if you 
draw the, the countours of the function, 
they're circles. 
Pardon my draftmanship. 
maybe this is a good way of thinking 
about why. 
is the case the radial function has no 
angular momentum. 
Well, having settled that, let's look 
among other familiar things, to find 
functions with definite nontrivial 
angular momentum. 
It's always a good idea when starting 
with basic concepts to, look at the 
simplest variables that you have at hand. 
So, in our case certainly the, the 
coordinates of a particle, the three 
spatial coordinates are candidates for 
that sort of investigation. 
So let's, let's see how they what happens 
to them when they're acted upon by the a 
minimum operator. 
Now here, look at this thing here, I just 
want to make it very clear to you. 
This is a case where the index is not 
repeated, we have 1 index u and one index 
a. 
So as a stand alone expression, this is 
really just an indication of what happens 
when you pick a value of u like 1, right? 
And pick a value of a like 2. 
Then you're, you're going to get, you 
should, stick with an expression that 
depends specifically on the variables 1 
and 2, which are taking fixed values. 
Okay, so we just put in the, the fran 
levitivity answer thusly. 
And now the momentum, acting on a 
position operate, on a position 
coordinate, is h bar over i, if the 
momentum has the same coordinate as the 
position, delta w a. 
other words, well it's a delta b a. 
It's one if w is equal to a, it's zero 
otherwise. 
And so when we then now contract, we're 
now so, summing over w here. 
So you see what that means is that we 
replaced w by a. 
e u v w goes to e u v a. 
And then that is well, that, this is, 
this is, this is the answer. 
So it depends upon the two, the two fixed 
index values, u and a, and shows how you 
basically, those, at most one value of v. 
For which this, relativity center doesn't 
vanish. 
Okay, so now, and this is probably 
something, 
It's a good idea for you to try to do 
these manipulations, satisfy yourselves 
that you understand how they work. 
Occasionally, there are things like 
factors of two that show up. 
That aren't necessarily so obvious, but 
again, this is strictly mechanical. 
So to, to, to calculate the L squared 
acting on x of a, we just do this Lu Lu 
xa. 
So now the, the repeated index u is 
summed over, implicitly summed over. 
And so just by expanding, we see this 
expression. 
We have a, a, a, a product of two 
levitivitus symbols. 
And a, a, again, a contraction on 
indices. 
and so this in particular here is an 
example that you might want to 
investigate. 
I think something like this came up in 
the derivation of the relationship 
between the angular momentum And the 
Laplacian operator. 
There's a factor of three there that one 
had to, that, that emerged in the 
calculation that required a little bit of 
thinking about what you're doing. 
Basically taking the trace of the 
identity matrix there. 
but in any event, what we see is that 
indeed if you apply the square of the 
angular momentum operator on the, 
position coordinate alone. 
You get back that coordinate. 
So, each val-, each xa, for a equal one, 
two, or three, is an eigenfunction of the 
square of the angular momentum operator. 
corresponding to an L value of 2. 
Now in fact these functions that we've 
just found, are widely used in quantum 
chemistry. 
they're the so called the, the, p, well 
it's often the px, py the pz orbitals. 
So, if you have a if you have a, if you 
have a molecule with, with several 
several atoms, you often will, you'll put 
a, a line a, a p orbital in between them. 
And so you, you want a, a directed 
orbital. 
that has well, let's, let's see, let's 
see what it looks like. 
So just what I've done here is multiplied 
these functions by the common gals in 
here, just to get a compact, get a 
compact orbital. 
That has this form, so a blue means a 
positive sign, red a negative. 
And so the P1 here is oriented along the 
x1 axis. 
P2 along the x2 axis, P3 along the x3, 
and so on. 
Now I'm going to suggest that you answer 
a question about the parity of these free 
orbitals. 
Do they have, do they have a definite 
parity, that is, do they have a symmetry 
under the inversion of all the coordinate 
systems. 
And are their parities the same or 
different? 
Okay, so I think that the answer to that 
question about the parity should be, 
should have been the way to think about 
it, is. 
These are the functions of the coordinate 
themselves. 
So of course, they just change sine when 
you change the sin of all the 
coordinates. 
Now, we made an easy choice, and found 
three eigenfunctions that have l equal to 
one. 
And if you remember the summary from a 
past lecture, and the bonus homework 
The num, the number of independent number 
of independent states with a given 
angular momentum Is 2L plus 1. 
So they're, there are just three 4L equal 
1 and these are, these are perfectly good 
independent choices. 
But there's another choice of 
eigenfunctions which is more often used 
by physicists than by the chemists. 
they're complex value. 
So I've, I've just made up a notation 
here, chi 1, 1. 
Is minus the quantity x1 plus ix2 divided 
by the root 2. 
Chi 1,0 is x 3, as before. 
And the chi 1 minus 1 is x1 minus ix2, 
divided by root 2. 
And know there's a sign change, sign 
change between here and here, and that 
is, introduced, so that the well, let's 
just see. 
You can verify, I suggest that you do 
this, that when you apply L3 to any one 
of these, chi 1,m. 
You get an eigenfunction, which is mh bar 
times chi 1m. 
And then, these, the raising and lowering 
operators, which are useful for actually 
generating the entire set of angular 
momentum I, eigenfunctions for a given L. 
If you have one of them alone with this, 
this choice of phases, we get the 
correct, the conventional, factor 
associated with the raising and lowering 
operator. 
In principle, this, this factor can have 
random phase, random phase. 
It's magnitude is it's magnitude is 
determined. 
But by this choice of convention here, we 
make sure that the phase is uniform. 
Now there's a very important property of 
the chi 1. 
And this again is something that is 
fairly easy to work out for yourself. 
That is, if you take this to the Lth 
power, then, then that it is, turns out 
to be an eigenfunction of with of, of L3. 
With the projections L, which is the 
maximum projection allowed. 
And then it also is an eigenfunction of 
the total angular momentum. 
And this is, this is worth thinking about 
for a while. 
and you can use see how this, is affected 
by L minus and L plus. 
And construct L squared from L plus L 
minus, or, the other way around, and Lz 
squared. 
And I think you'll, you'll be able to 
show this relationship with a little, 
with a little bit of practice. 
So here's a summary sheet if you want to 
work on that. 
And this enables you I think you, you may 
be able to show with some thought that 
these two identities are true. 
And so, you see, we haven't actually, 
we've only specified a rather primitive 
object here. 
It's, it's actually just a, a Linear 
function of X1 and X2. 
When we find it, when we ta, when we 
raise it to a certain power. 
We get an eigenfunction of the angular 
momentum of the operator of the direction 
three, and of the square. 
And then, from that we can construct the 
general, an arbitrary state of angular 
momentum. 
Like arbitrary, eigenfunction angular 
momentum as follows. 
we just, we just we have the LL state 
which is given by this, this primitive 
here. 
The L kind one, one. 
And then we just apply L minus to it, and 
applying L minus to any object is quite 
straight forward. 
And we know that the action must be of 
this form. 
So from L, L we can generate. 
The next date down by application of 
lowering operator, and keep on going. 
I'd like to point out that, by the way, 
that we have built these up starting with 
something which is just, you know, a 
power of the sum of two coordinates. 
If you think about it, this means that 
the the LL item function. 
Is a polynomial of homogeneous degree in 
the spatial coordinates. 
But let me say it another way. 
It's a multinomial of homogeneous degree, 
meaning, it's, consists of a, of a sum of 
our terms, 
Actually there is I, there is no X3 here. 
I mean I've just, I've just well what I 
guess I, let me generalize this. 
But when, when we when we the expression 
is valued is, is, the expression here is 
valid LM. 
So but basically, the way that you can 
think of an eigenfunction of angular 
momentum is in terms of polynomial form. 
That sometimes leads to great advantages 
in solving certain types of problems. 
now the actual, the actual set of, range 
of coefficients is fairly complicated. 
And these things have been extensively 
studied mathematically. 
And so the, the best way to get a better 
impression of what's going on is to 
consult reference works. 
Now, here's another remark that you can 
infer from our discussion. 
So, we were, we were, constructing 
eigenfunctions of the angular momentum 
operator from various sums and products 
of these coordinates. 
Well every one of these contains an r. 
So, we know that the, the radial 
coordinate itself has no angular momentum 
connotations. 
It can just basically be removed from the 
from the definition of an eigenfunction. 
And so, the, the most useful set of 
eigenfunctions, for many applications of 
angular momentum operation or functions 
of the angles. 
Only in the conventional, folder 
coordinate system. 
So, I think you can see that if you, you 
just take the previous argument about 
having products of these various things. 
We would get an overall r to the k comma 
to everything. 
And then, then there would be products 
of, you know, sin to the a theta cosin to 
the b of theta. 
sin to the c of v, cosin to the d of v. 
so this means that in fact, there's a 
universal description that's valid for 
all central potentials. 
All potentials that depend on the radial 
coordinate only. 
That is, where angular momentum is a good 
quantum number. 
We can define it just in terms of the 
angles only is independent of the, of the 
distance from the origin. 
Now to check your understanding, I'm 
going to pose to you a question that has 
to do with identifying the angular 
momentum. 
By inspection of the equation of mo, of 
the equation of, of a system, in terms of 
these Cartesian coordinates. 
Okay, I hope that was comprehensible, 
And you could see that the, the one of 
those functions was radially symmetric, 
that what has an increment of zero, the 
other was 
The the, the product of the the two the 
X1 plus iX2 function. 
Then the third was actually just a 
weighted difference of those two. 
So the third one couldn't possibly have 
be an eigenfunction of angular momentum. 
Because it consisted of linear 
combination of eigenfunctions is two 
different values of angular momentum. 
Now, I mentioned that I've repeatedly 
said, you know, how, how, important this 
subject is, and it does occur in the most 
diverse type of systems. 
For example, here's a recent publication 
from April 2013, about the use of 
spherical harmonics in programming of 
video games. 
now, you might ask why, and here's what 
the, is stated, 
The use of Spherical Harmonics as a basis 
allows for very fast rendering and it 
sort of provides some optimal accuracy 
for a given speed of processing. 
So you might think of a spherical 
harmonic as being the equivalent over the 
surface of a sphere. 
To what the Fourier Transform is in 
space. 
And as you saw in Victor's lectures 
Fourier Transform is an incredibly 
powerful tool in quantum physics. 
Also widely used in engineering signal 
processing and to be like. 
And so again, there are many applications 
of spherical harmonics in, in fields such 
as imaging, mapping the Earth's 
gravitational potential. 
And wide variety of others, and so even 
if you're not going to be a professional 
physicist. 
You might benefit quite a bit from 
applying the knowledge you've got from 
the subject here to problems of 
characterization of motions on a sphere. 
Okay, hope to see you again. 
Be back soon.