>> Welcome back to
Chemistry 131A.
Last time we talked about
more realistic oscillators.
We talked about the
6-12 potential.
We talked about Morse potential.
And then we got away from
one-dimensional systems,
systems with just one variable
which we were calling
either X or R, whatever.
And we did a particle in
a box in two-dimensions.
And we discovered that under
certain conditions there was
degeneracy if the lengths of the
box were the same and so forth.
And three-dimensions
is the same.
And the main weapon we had was
that we took a two-dimensional
equation that had two variables
in it and we separated it into
two one-dimensional equations,
both of which we already
knew the solution to,
and then we just
substituted in the solution.
Today, rather than talking
about a particle in a box,
I want to talk about a
particle confined to a ring,
first of all, which will be
an interesting thing to solve.
And a particle confined to
the surface of a sphere.
And both these things will be
important as a prelude to atoms.
But both of them are
completely artificial,
because if you say well,
I've got a particle confined
to a ring, you have to
ask yourself what kind
of potential does that have
that the particle stays
just on a circular ring?
And that's not a very
realistic looking potential
that if you just get off
-- if you're on the ring,
you're just there
with kinetic energy.
And then if you're off the
ring, by epsilon, it's infinite
or something like that.
And so we could get some kind
of anomalous looking behavior.
But really what it means
is we're going to set
up the equation in
two dimensions.
And then we're going
to freeze one dimension
and solve the other, which
will be an angular variable.
And we'll leave the
radius of the ring
as something we fix
at the outset.
And then we go from there.
And we'll leave the radius of
the sphere as the same thing.
And luckily, when we do real
atoms, at least for simple ones
where we don't have
too many electrons,
just one being the right amount,
we can actually factorize
the thing.
And so we can use these
solutions for the ring
and the sphere and then just
paste them together exactly
like an onion, like
growing an onion in shells.
We argued early on that a
particle confined to a ring,
which we could think of as
sort of like an electron
in a classical orbit, would have
to have a wavelength that fit.
And the idea is that
every time it goes around,
it has to match up perfectly.
Because if it doesn't
it cancels out.
And in fact, there's nothing
left for the probability
because it's basically
minus itself half the time.
And so, that was an argument
that the de Broglie
wavelength had to match.
And our qualitative
condition, then, is that we have
to have an integral number
of wavelengths and lambda has
to equal the circumference
of the ring, which is 2 pi r.
if that condition is met, then
we can go around like that
and we can come back and
we'll be at the same place.
And that means that we've
got a stable standing wave,
a probability pattern that's
not changing in space or time.
Now, the de Broglie wavelength
is related to h divided
by the momentum, therefore
we can just substitute
for that n times h upon
p is equal to 2 pi r.
and then we can write that
in a very suggestive way.
We can divide both sides by 2
pi, turn the h into an h-bar.
And we can multiply
both sides by p.
and then we have nh-bar is equal
to r times p. That's interesting
because r times p is going
to have something to do
with angular momentum.
And if we have a
particle on a ring,
we automatically think
of angular momentum.
Whenever we've got anything
going in a circle or confined
like that, we think
of the angular momentum
of such a particle.
We learned about that
in classical physics.
Let's have the ring
oriented in the x, y plane.
Then that means that
the particle has an
angular momentum.
Vector J is equal to r times
the vector cross product of p.
In this case, r points
to the ring
and p is the way the
particle is going.
So r and p are always
at right angles.
And in fact, J is equal
to r cross p which points
in the z direction, if we've got
the particle in the x, y plane.
And so we can take
the z component
of the angular momentum and
relate it to the particle.
And that means that
we've got Jz,
which turns out to be r
times p times sine theta.
Sine theta sine 90 degrees,
or sine pi over 2, that's 1.
So Jz is equal to rp, and
that's equal to nh-bar,
by the condition that
the integral number
of wave lengths fit in.
And now sort of by this backdoor
route we have the angular
momentum seems to be quantized.
Photons came in units h-nu,
angular momentum comes
in units of h-bar.
Now our time dependent
Schrodinger equation
for the particle on a
ring is the following.
It's the kinetic
energy with respect
to x plus the kinetic
energy with respect to y,
second partial derivative.
Same thing as the particle in
a two-dimensional box is equal
to E times psi of x, y.
And the problem is x and y
in this problem are very,
very, very awkward variables.
And that's because
r is the square root
of x squared plus y squared
and so on and so forth.
And if you just try
to just bully your way
through this equation using
Cartesian coordinates,
which are set up
for square problems,
you'll never get anywhere.
What you first have to
do is you first have
to change your variables so
that you're in the same kind
of funhouse mirror
as the problem is.
And then the problem
will seem very easy.
And that's what we're
going to do.
We've got a circular problem,
we're going to use
polar coordinates.
So we're going to set
up two new variables.
Rather than x and y, we're
going to set up r. Why?
Because r is constant.
That's the perfect
thing to have.
And when I get a derivative
with respect to r, it's zero.
So r is constant.
And then the other variable is
just where I am on the ring.
Let's call that phi.
Then I have a relationship
that x is equal
to r times cosine phi.
That's this leg of the triangle.
And y is equal to r sine phi,
that's that leg of the triangle.
And of course, r
squared is equal
to x squared plus y squared,
which is always constant.
Now, we have to make,
unfortunately, a transformation,
not only of the variables x
and y, but of the derivatives.
And that's trickier.
So we need some actual
multi-variable calculus
to do that transformation.
And I'm not going to go through
the transformation over and over
and over because it gets very
tedious to go through it.
But I want you to see one
time exactly how you do it
so that you'll understand
where these terms come from.
To do this, then here's
what we have to do.
We have to use the
chained rule from calculus.
And we have to understand that
if we have a function of more
than one variable and I change
something, I want to figure
out the change, I should figure
out the slope that it does
with respect to the x
direction if I change x.
And then the slope it does
with respect to the y direction
if I change y. And that'll
give me the total change
in the function.
I need to take two
things and add them up.
And for each one of them, if
I use a different variable,
I can use the chain rule.
So here we go.
The derivative of psi, the wave
function with respect to r has
to be d psi dx times dx dr. You
can think of these things just
like fractions where you
just cancel out the dx.
That's how you can
remember the chain rule.
d psi dr is d psi dx dx dr.
That's the one direction.
But I have to -- it's a
function of two variables.
So I have to add d psi dy, dy
dr. and now I have a formula
for y with respect to r.
and so what I get is I get d
psi dx cosine phi plus d psi dy
sine phi.
The second derivative is the
derivative of this thing.
And it gets messy fast, but
it's a really good exercise
in thinking clearly, because
what you do is you just take
each of these terms and call
d psi dx some other thing.
And then put that into the
formula and then expand it back
out very carefully
and you'll see.
So don't skip any steps.
So let's take another
derivative.
The second derivative
of psi with respect
to r squared is just d
by dr of what we got,
d psi dx cos phi,
d psi dy sine phi.
And I can put that in by
factoring out a cosine phi
and get the second
derivative of psi dx squared,
dx dr plus the second derivative
of psi with respect to x
and then with respect to y,
dy dr. And you might say well,
how -- what about the
order of the x and y?
And the answer is for wave
functions and nice things
that we deal with, any kind
of function we have we don't
really care about the order
if we take the derivative,
partial derivative with respect
to y first or the
partial derivative
with respect to x first.
Those are equal.
So we don't worry
too much about that.
And then we have sine phi times,
again, two terms, both of them
with a second derivative, dx
dr, dy dr. And if we add all
that up, we get cos squared
phi times the second derivative
of psi with respect to x squared
plus sine squared phi times the
second derivative
of psi with respect
to y squared plus 2 sine
phi cos phi times the mixed
partial derivative.
The derivative with
respect to phi,
I'm going to leave as a problem.
We do it the same way.
We start with d psi d phi, and
we write it as the chain rule.
And then whenever we have x
over r or something like that,
we express it in terms of
cosine phi and sine phi.
And what you'll find is that
you get a little bit longer
expression here -- or
actually about the same length.
But you get minus r times d psi
dr plus r squared times three
terms, which look very
similar to the other three,
except with a slightly
different ordering.
And now we can get the
relationship we need
between the Cartesian and
the polar second derivatives
because now we have the second
derivatives with respect to r
and with respect to phi.
And we know what they are in
terms of with respect to x
and y. It's still
quite a bit of algebra.
And again, I'll leave
that for you.
I'll quote the result so
you can see what it is.
But there's still quite
a bit of algebra to do.
And what you find out, then,
is that the second derivative
with respect to psi with
respect to x squared,
plus the second derivative
of psi with respect
to y squared is equal to
the second derivative of psi
with respect to r squared plus
1 over r times d psi dr plus 1
over r squared times the second
derivative of psi with respect
to phi, with respect
to the angle.
And we can check that we
haven't made any mistakes
because we can put in units.
These all have units of
length squared on the bottom.
Forget about the wave function
units for the time being.
We have second derivative
dr squared on the bottom,
1 over r dr, 1 over
r squared, d phi.
Phi doesn't have any length.
The angle is just a ratio of
things, because it's in radians.
And now at this point, we
make the totally artificial
assumption, we just say
hey, particles on a ring,
let's just freeze r. And so
wherever we see a derivative
with respect to r,
let's just say it's 0.
Because r is not changing.
And that's great,
because now we just have 1
over r squared times the second
derivative with respect to phi.
And that's looking, except
for phi instead of x,
that's looking awfully similar
to things we've already done.
Therefore, the Schrodinger
equation simplifies to this.
Instead of having the
Cartesian coordinates,
we have minus h-bar
squared over 2m times 1
over r squared times the second
derivative with respect to phi
of psi, which is now
only a function of phi,
because r is fixed
as a constant.
And that's E equal to
E times psi of phi.
This is beginning
to look really good.
And here's why.
We know that m times r squared
is the moment of inertia
of a classical particle orbiting
around an orbit of radius r.
And we're doing a
rotational problem
by keeping the particle
on a ring.
And we found that we just
got the rotational constant,
the moment of inertia coming
into the problem just naturally,
just by the way it worked out.
The differential equation,
then, if we write it in terms
of the moment of inertia times
the energy is just the second
derivative with respect
to phi is equal
to minus IE upon h-bar
squared times psi.
And the solution of
an equation like that,
not surprisingly,
is an exponential.
And so we can write
down the solution.
And we can write down, look
it's AE to the i mu phi plus Be
to the minus i mu phi.
mu is the square root
of 2IE upon h-bar.
And we can verify
if we shove that in
that we solve the Schrodinger
equation, we get the answer.
We need to have i
because we have a minus.
After we take the derivative
twice, we get minus.
That could be i squared is
minus 1 or minus i squared,
you could go around the
other side to minus 1.
But we can't have any
real exponentials.
And so these functions are
things that are corkscrewing
around one way or the other.
And when they corkscrew
around, they come back and meet.
And they could go the
other way, but they meet.
The quantization arises
because the wave function has
to meet itself on
the way around.
And that means that if we add
2 pi to the wave function,
it has to be the same thing.
If that weren't true, the
wave function wouldn't be
single-valued in space.
We could, depending what
variable we pick to call it,
it would have a different value.
But it's the same
point on the ring.
So it has to have
the same value.
So that means it has
to exactly come around.
And that means the mu times
2 pi is equal to 2 pi n,
so that n is an integer.
So if we add 2 pi to the angle,
it has to be 2 pi
times an integer.
And conventionally, rather than
using n, because n gets used
for other things to do
with energy, we use m,
where m is an integer.
And m is called the
magnetic quantum number.
Why is it magnetic?
Because if we've got
a charge on a ring
and we imagine it's moving
around, a charge on a ring,
then, is a current loop.
And a current loop
makes a magnetic field.
That's exactly how you
make an electromagnet.
You take a ton of wire and
you wind it around a core.
And then you put some current
through it and you can pick
up all kinds of stuff.
And that's quite
a fun thing to do
when you're in elementary
school.
And I remember spending
considerable time doing exactly
that and seeing what I
could and couldn't pick up.
In fact, we can make
another connection
with classical mechanics.
So the angular momentum,
Lz is just r cross pz.
And we can put in our
quantum operators.
We can put in x-hat,
py-hat minus y-hat px-hat.
That's what this e component of
r cross p is, if we set it up.
And that's equal to minus
ih-bar times x, the derivative
with respect to y, minus y, the
derivative with respect to x.
So that's our operator.
And we can take that
and we can convert
that to polar coordinates
by exactly the same tricks
as what we did before.
And if you take that
particular combination
and you're very careful and you
convert to polar coordinates,
you find out it comes out to
be this real simple thing:
minus ih-bar d by d phi.
Just the first derivative
with respect to the angle.
That's what we get.
Well, that's really interesting
because what that means is
that when we did the energy
we said well we could have e
to the im phi or e
to the minus im phi.
And we could have A of
that plus B of that.
We could have any
amount of each.
But if we want the particle
to be in an eigenstate,
not only of energy, but
of angular momentum,
then that means it's
one corkscrew.
It's either e to the
plus im phi going one way
or e to the minus im
phi going the other way.
And so what we do when
we set up these problems
for neatness is we
either set A equal to 0
and say it's negative m,
or we set B equal to 0
and we say it's positive
m. and so m can vary
from minus some value to
plus some value and anywhere
in between, including
apparently 0.
And the interpretation, then,
is that the energy is equal
because a particle going
this way at some rate
and a particle going the
other way at some rate, well,
they have the same
kinetic energy.
This, then, is our final
solution: e to the im phi,
m as any positive
or negative integer.
And apparently 0.
It could be 0.
Why not? That solves it as well.
And that certainly
doesn't' have a problem
if there's no twist
at all, just is flat.
Of course, it meets up.
If we normalize the wave
function over the ring,
that means that the probability
that the particle is
somewhere on the ring is 1.
If we do the integral, since
e to the im phi times e
to the minus im phi is
1, the integral is 2 pi.
We don't integrate
over r because r is not
in the problem any
longer, it's fixed.
Just integrate over phi.
Then we can get our normalized
wave function for the particle
on the ring, 1 over
the square root
of 2 pi times e to the im phi.
The lowest energy here is m
is equal to 0, which is 0.
And this seems to run counter
to what I'd been saying,
which is that whenever you
have a confined particle,
you should have some
0 point energy.
Why? Because you want to satisfy
the uncertainty principle.
And the reason why this seems
to violate the uncertainty
principle is just kind
of glossed over in the book.
But the reason why it seems
to violate the uncertainty
principle is first
of all, we just threw r out.
We really had a two-dimensional
problem,
but we froze one
of the variables.
Who says you can freeze one
of those variables
exactly like that?
That's point one.
And point two is that phi seems
to have an artificial range.
We say phi goes from 0 to 2
pi, but it would be the same
if phi went from 0 to infinity
because it keeps wrapping
around over and over and over.
And so if you argue,
well, you don't know phi.
It could be anywhere between
minus infinity and infinity,
and it would still be
somewhere on the ring,
then you could have
the momentum be 0.
Of course, you have to
have the momentum be 0
if you have the energy 0.
And you could still have
the position in terms
of the actual value of
phi be indeterminate.
And it's kind of a
mathematical dodge.
But you have to be careful
if you set up a problem
and then impose a constraint
that might not be physical.
Say, it has to be exactly on the
ring, and then get in a tizzy
that something doesn't
seem to be quite right.
Because the arguments may be
quite subtle at that point.
The probability density
for an energy
and an angular momentum
eigenstate is independent
of the angular variable phi.
The probability density is flat.
And of course, it
would have to be,
because there's no
reason why I should expect
to find the particle more
on one side of the ring
than the other side of the
ring when there's no difference
between the sides of the ring.
So it's got to be flat.
And that makes perfect sense.
If there's nothing to
distinguish the two sides,
how are we going to tell?
And quantization, again,
arises from the fact
that the wave function
has to match up,
has to be single value.
Now, the next step is
to expand to a sphere.
And a sphere, you could argue
a sphere is a bunch of rings.
And that might be a
good way to look at it,
in fact, is to pile up rings.
It's artificial, again,
to assume that the particle
can go anywhere on the sphere
but can't move radially at all.
But nevertheless, it's a really
good stepping stone to getting
to the point where we can
write down atomic orbitals
and figure out what's going on.
We've got a sphere.
We know if we write
down the kinetic energy
and because we've got a
sphere and we're keeping it
on the sphere, we've got no
potential energy except this
completely arbitrary potential
energy that's keeping the
particle right on the sphere.
We have to convert to
spherical polar coordinates.
And a coordinate system here is
x is -- we have two variables,
we have phi, which
goes around the z-axis.
It changes x to y and so forth.
And we have theta, which
starts here and goes
from the north pole, where
theta is 0, to the south pole,
where theta is pi,
or 180 degrees.
But we don't go around again
because every time we go here,
we make a ring, like
a tree ring.
And then we go here and we make
another ring like a tree ring.
And we go here to the
equator, we do that one.
And we go down here and
then we finish down here.
And if we went around again,
we'd be counting all
the rings again twice.
So, phi varies from 0 to 2
pi or 360 degrees around.
And the other one, theta
varies from just 0 to 180.
We adopt a right handed
coordinate system.
And that means that x
cross y, your thumb points
in the plus z direction.
And that's by far the
easiest way to do it,
because if somebody
draws a figure
where x is shooting this
way and y is that way,
and you try to reorient it in
your mind so x is to the right
and y is into the paper, it
takes a long time mentally.
It's quite a gymnastic.
But if you just grab
your hand and do it,
you can figure it out.
Be careful, though,
because what you tend to do
when you're doing problems
is you have the pen,
if you're right-handed, you
have the pen in your right hand.
And you want to figure out
which way something's, going,
you use your free hand.
And that's the wrong hand.
And if you do that on an
exam in physics, especially,
you just get marked wrong
because you didn't use a
right-handed coordinate system.
And I've watched
that happen to people
at various stages of my career.
So here's a right-handed
coordinate system.
Theta and phi are set out.
And our position is given
by three numbers: r,
the distance out;
theta, the distance
down from the north pole;
and phi, the distance
around from the x-axis.
What's very confusing is that
if you take a course in math,
for some reason that
I cannot fathom,
the variables theta
and phi are swapped.
And so theta is the one this
way and phi is the one this way.
And I think that's
because in math,
when you do two-dimensional
problems,
you always call it theta.
And they just like to keep
theta for the same variable
and use phi for the other one.
But in physics, it's
phi around the z-axis
and theta away from the z-axis.
And you have to keep
them straight
because if you open
the wrong book,
you'll get things backwards.
And you'll get in a
terrible mess then.
Let's try a practice problem.
Let's consider a volume element,
dv is equal to dx dy dz.
In Cartesian coordinates
it's a little cube.
What's the volume element for
spherical polar coordinates?
So, let's take a radius, r,
angle theta from the z-axis.
What's the formula
for the volume?
Well, we don't care what phi is
because it's always
the same here.
But we do care what theta is.
And here's why.
The size of the onion ring here
at the top near the north pole
is tiny whet theta is small.
As theta gets smaller,
the total size
of the ring gets
smaller and smaller.
As we go toward the equator,
the same ring around,
360 degrees in phi, is much
bigger in terms of the volume
that it'll hold if we
take a little slice in r.
And so we have to
weight the volume element
by how close we are
to the north pole.
It's like imagining you can
walk around the North Pole
in a little circle
and you've walked
around all possible
longitudes there.
But if you try to do it at the
equator, it's a very long walk.
Same thing.
So here's the beauty
of calculus is
that we actually draw
this thing as a wedge
because r is changing.
And so the inner part is
smaller than the outer part.
And it's like a little cone.
But when we have just
very tiny differences,
then it's like a cube.
And all we need to do is
take the sides of the cube.
And that's the beauty of
taking very small things dx is
that no matter how
curvy something is,
if you take it small
enough it's a straight line.
That's why calculus is so great.
And so we can figure
out the distance here
out is r sine theta.
And so the distance
along if I move
by d phi is this distance
here is r sine theta d phi.
And the distance the
other way is just r,
because that's a full
distance, times d theta.
And the distance in the
third direction is just dr,
neither theta nor phi changes.
And so we can take those
and multiply them together.
And we get the volume is r
squared sine theta dr d theta
d phi.
That's important to know because
you're going to have integrals
to do with psi and
so forth in them.
And they're going to
have dv because you have
to integrate over all space.
And you need to know what dv
is in terms of these variables.
And now you know what it is.
In exactly the same way that
we did before, but I don't want
to take an hour to go
through it carefully,
but I'll just leave it to you
if you're interested
to work it out once.
We can take the Cartesian second
derivatives, second derivative
with respect to x, second
derivative with respect
to y plus the second
derivative with respect to z,
and we can cast it in
the following form.
The second derivative
with respect to r plus 2
over r times the first
derivative with respect
to r plus 1 over r squared
times this operator I've called
lambda squared.
And lambda squared has nothing
in it except theta and phi.
It has some sine squared theta,
second derivative
with respect to phi.
and the second term, which is
written in a very funny way,
1 over sine theta d by d theta
of sine theta, d by d theta.
Unless you're used to dealing
with operators, though,
this is written in a very
compact, very nice way
so you don't have
a lot of terms.
But you have to be quite
careful when you actually put it
on a wave function because
you only put the wave function
on the right of the operator.
You don't start inserting
it in between things.
You just put it on the right
and then you go sequentially.
Right, take the derivative
with respect to theta,
multiply by sine theta, take the
derivative again and so forth.
But if you don't understand
the operator notation,
then you're very likely
to get things wrong
because you may stick a psi
in wherever you think
there's a blank.
And that's not correct.
This thing, lambda squared,
this is called the Legendrian.
It is very famous.
The Legendre polynomials
and so forth.
And as I said, the operator
takes a bit of getting used to.
Only put the wave
function on the right.
Be careful about that.
Don't put psi in front of
both those derivatives.
The operator lambda
squared, the Legendrian,
has all the angular
energy in the Hamiltonian,
because the other operators had
derivatives with respect to r.
If we freeze r, we don't
allow any change in r,
there's no energy that way.
That means that this
thing, lambda squared,
is what we want to focus on.
And it's just the energy to do
with all the possible angular
motions of things on a sphere.
So let's fix r, throw that
out again, same was as we did
with the particle on a ring.
And now we've got this
new equation to solve.
Minus h-bar squared over 2mr
squared times lambda squared
on the wave function, which is
a function of theta and phi,
is equal to some energy times
a function of theta and phi.
And if we can find the wave
functions that solve that,
eigenvalue equation,
and the eigenvalues,
then we know the energy
and we know the possible
wave functions on a sphere.
And of course, we expect that
it's going to be quantized
and so on because it's
a trapped thing, again.
And as we go around
in theta and phi,
it's much more complicated
this time because there's two
of them, so it's harder to see.
But it's got to be similar
as what we had before.
There is no potential
energy here.
The only requirement on
the quantization is just
that the wave function
fit into the space.
Okay. Let's do a
practice problem here.
Then practice problem 14.
Let's show that the
angular Schrodinger equation
is separable.
And what does that mean?
That means that whatever
this wave function in theta
and phi is, we can
write it as a product
of something that's
only a function of theta
and something else that's
only a function of phi.
And if we can do
that, then we'd guess
that the something else
that's only a function
of phi is what we had before.
Because last time when we
did two-dimensional particle
in a box, then okay,
it was a product.
And the x was the same
one as what we had before.
And so, theta is different
from phi because phi goes
around in 2 pi and
theta only stops here.
So we wouldn't expect it's
going to be so easy as that,
because theta could
be different than phi.
But still, phi should be the
same as what it was before.
And so that, we got that
e to the im phi stuff.
That saves us a lot of work.
Okay. Here's what
we've got to show.
It's separable if we can write
the solution as a product.
And if we want to prove that,
what we have to do is we have
to rearrange the equation
so that we have two terms,
one of which only depends on
theta plus or minus another term
that only depends on phi
is equal to a constant.
And then we make
the same argument.
If we fix theta and change
phi, if it's a constant,
that means that both
of them are constant.
Otherwise that wouldn't work.
And that means that we can
do a one-dimensional equation
for each one.
So, let's try the
trial product solution.
Again, we used capital X
of x. Let's use capital
theta of theta.
It's just some function.
We don't know what it is, but
we don't care at this point.
Capital phi of phi.
And substitute it in.
So we've got the
Legendrian on this thing.
And now we get a number.
And here, epsilon, this number.
Minus epsilon is just
2IE over h-bar squared.
And I is the moment of inertia.
So that just cleans it
up so that we don't have
to write a lot of extra terms.
Now, if we substitute this
in, here's what we find.
We have 1 over sine squared,
second derivative with respect
to phi of this product.
Plus 1 over sine theta, first
derivative with respect to theta
of sine theta , first
derivative with respect to theta
of the product is equal to
minus epsilon times the product.
I do the same thing.
I divide both sides
by the product.
And first, I say aha!
I've got the product
of two things.
Second derivative, with
respect to phi, of some function
of theta and a function of phi.
The function of theta is a
constant, so I can pull it
out because it doesn't
matter where it is.
That's what the partial
derivative means.
In the second term, where the
derivative is with respect
to theta, I can pull the
function big phi out in front
as a constant because that's
just a bunch of derivatives
with respect to theta.
And we treat phi as a
constant, so let's pull it
out just the same way we'd pull
out a constant in a derivative.
And if we pull those
out in front,
that makes it much
easier to see.
Now we've got big theta out
in front, sine theta squared,
the second derivative with
respect to phi squared
and then the rest that
you can see is equal
to minus epsilon
times the product.
And now if I divide by
the product on both sides,
the theta goes away
in the term with phi.
And the phi goes away
in the term with theta.
But we still have
this sine squared.
So we have to multiply
through the whole equation
by sine squared.
And then, if we do that, we
finally find the following.
We have 1 over phi times
the second derivative
with respect to phi.
That's one term.
Plus a bunch of gobbledy-gook.
But it doesn't matter what it
is because it doesn't have phi.
It's all theta.
And it has an epsilon
sine square theta.
That's equal to 0.
That's good enough,
because I have this thing
over here, which is just phi.
And this thing over here,
which is just theta.
And so I've got two equations.
One of them's just phi.
The other one's just theta.
And I'm in.
And we can switch, once we've
got it down to one variable,
we can switch to the regular
d instead of the funny d
because it's the same thing.
And we can solve the
differential equation.
So the first term, as I said,
is going to give us
exactly what we had before
in the particle on a ring.
Because it's the same
equation, basically,
as the particle on
a ring equation.
What it's actually
going to be will depend
on what theta happens to be.
But that doesn't' bother us.
We have E to the im phi.
And then whatever theta
happens to be is going
to be some other function
that's going to be our business
to solve in the second
part of the equation,
which is a different
equation to do.
And so we can break this
very complex problem
up into this series of
problems and solve them.
And it turns out for a
real problem with an atom,
what we're going to do is
we're going to first break
up the problem into the
particle on a sphere.
And then we're just going to
let r be the other variable.
And not surprisingly, we're
going to try a product.
We're going to say gee, I think
that the wave function is
a product of some function
of r times some function of
theta times some function
of phi, and see if
that doesn't work.
And then that gives
us a clue, then,
of how to factorize the
thing and see that it works.
When will that fail?
Well, it'll fail right
away, unfortunately,
if we have two electrons because
if we have two electrons,
then where each of
them happens to be,
and they're repelling each
other and they're attracting
to the nucleus, it
gets too difficult.
We can't separate the equation.
And so we run into problems.
In that case what we do is we
treat each electron separately,
solve it and then take the
electron-electron repulsion part
that we couldn't handle
and we couldn't separate.
And we treat that
as a perturbation.
How good that will
be will depend a lot
on how close the
electrons are getting.
You could imagine that
if the electrons happen
to get pretty close in space
that the potential
could get quite high.
And so the electrons will
tend to avoid each other.
And that means that their
motion is correlated.
They're sort of like a
cat chasing its tail.
When one starts going
this way, the other may go
that way and so forth.
And so the electrons may not
be independently moving around.
And that kind of electron
correlation is a very important
aspect of multi-electron atoms
and higher dimensional systems.
But we'll touch on that
later on in the course.
And for now, we'll
leave it there.
Please take time and look
through these transformations.
And spend a little time with,
in a quiet room, with a pencil
and a piece of paper
and just methodically go
through and take each step.
And at each step say
what does it mean?
What am I doing?
Why can I do that?
And go through it and
see if you can't figure
out why these things have
the structure they do.
The d by d theta sine theta
stuff or the 1 over sine theta
out in front, that's going to
be a little bit tricky to get.
But you can get that,
too if you work on it.
And next time what we'll do
is we'll pick up our solution.
We can guess the solution
in phi, but we can't figure
out the solution in theta
yet because that's a
different differential equation
with sine theta in it.
And we haven't figured
out anything to do
with that equation yet.
So, that's a separate
equation for us to solve.
And we'll have to figure out
what kind of techniques we need
to solve that equation
and then figure
out what these functions are.
And then hopefully
we can get some idea
of what these functions on
a sphere actually look like.
So we'll leave it there and then
pick it up next time to figure
out the actual wave functions
for a particle on a sphere.
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