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.