>> Welcome back to Chem 131A
today we're going to pick
up where we left off last time.
We're going to continue our
exposition on a particle
on a sphere which is going to
be our entree into the theory
of angular momentum that
later on we're going to use
when we talk about real atoms.
Today then particle on a
sphere and angular momentum.
Remember that we had
our separated equations.
We had by dividing by the
product of theta, of theta
and phi of phi we added up
with a sum of several terms.
One of the terms only has
phi and the function of phi
and the other has only theta.
And therefore by the same
arguments that we used
in the 2-dimensional
particle in a box,
we can separate the solutions
and deal with them separately.
And that's of course a very
important simplification
because that lets us deal with
just a function of 1 variable
and regular derivatives.
So let's look at
the first term then.
The first term has 1 over phi of
phi times the second derivative.
And that term is exactly
like the solution we had
for the particle on a ring.
So this shows the wisdom
of doing the particle
on a ring first because
now we simply recognize
that we've basically got the
same situation all over again
and we don't actually
have to solve it.
We anticipate that we're going
to have a quantization condition
because we know that we have
some sort of a bound system.
And we saw before that
we had a quantum number M
that I called the
magnetic quantum number,
we'll see why in a minute.
And now I'm going to call this
new number M sub L to keep track
of the fact that M may
depend on something else.
And of course since I already
know the solution I can seem
to be rather prescient
about what notation to pick.
But anyway, the differential
equation,
then the eigenvalue equation
is that the second derivative
of phi of phi is equal to some
constant minus M sub L squared
times phi of phi.
And here M sub L is a
positive or negative integer.
And it could even be
zero just like it was
for the particle on a ring.
So our solution then is
going to be E to the IM phi
and I won't bother
normalizing the Eigen function
because we'll normalize
the whole Eigen function
over the entire sphere
at the end.
But we know that
this makes sense.
And this is basically what we
saw for the particle on a ring.
The other part then,
if this first part sums
to minus ML squared, then
the other part must sum
to plus ML squared
where ML is an integer.
So we have this rather
intimidating, I guess,
differential equation in theta,
with sine theta and derivative
of the function theta plus
epsilon sine squared theta is
equal to M sub L squared.
And at this point, the
text, in my opinion kind
of dodges everything and just
says well mathematicians seem
to know how to solve
this kind of equation.
And while that's true, I'd like
to do a little bit of guess work
as to how we might try to solve
this kind of equation and see
if we can get a little
bit more insight
into what the solutions are
looking like by trying to do it.
And one way to try to do it is
to say well, maybe it's sort
of like a particle
on a ring again,
and to try similar
kinds of things.
And that would make sense
because we have trigonometric
functions
in this differential equation
and so it probably is true
that the solution is also going
to have trigonometric
functions in it.
And because we have this
sphere, this round thing,
we expect that the wave function
will have some periodicity
with respect to theta like
it does with respect to phi.
And that's another argument
as to why we can do that.
So rather than simply quoting
all the results and giving them
in a table and leaving you
none the wiser as they say,
I'd like to go through
just a little bit
of the thinking behind it.
And so let's go back then
to the original equation,
which had both theta and phi in.
And we have an idea
of what we're going
to have for the phi part.
And we want to see
what we're going
to have for the theta part.
To simplify the notation, I'm
going to kind of go backwards
and not write it as explicit
product anymore, but I'm going
to adopt the notation that's
actually used for the solution
which is YLM of theta phi.
And that makes the
equations easier to fit
on the slides as well.
Therefore what we've got
now is minus H bar squared
over 2MR squared, 1 over sine
squared, the second derivative
of YLM with respect to phi.
Plus 1 over sine theta,
DD theta, sine theta,
the first derivative to
YLM with respect to theta.
That should equal some energy
as yet to be determined,
which we expect is going
to depend on L and M
and we'll see what
they are in a second.
M we know, L we don't.
Again times YLM.
And the key is we have to get
an energy that is independent
of the variables theta and phi.
If the energy is some variable
function then we haven't solved
the eigenvalue equation.
We just have to get some
constant, some number
and then look at it and
see if we've got it right.
M was zero was one of the
solutions for the particle
on a ring, just flat
probability all around.
And so I could try just using a
constant for YLM and have a look
at that and see if that might
actually solve this particular
differential equation.
And if we assume that the M part
has to do with the Z component
of the angular momentum,
that's true because the angular
momentum follows the R cross P
rule and so we know that if
it's going around in a ring
in the XY plane, which is what
we had for the variable phi,
that it's really the Z
component of angular momentum.
Then it makes sense that the
rest of the stuff has to do
with the X and the Y components
of the angular momentum.
Those must be in there.
And the question is could
those be zero as well.
If we pick a constant and
there's no curvature anywhere,
then the energy is
going to be zero.
So we try it and we
call it Y zero, zero.
Because whatever powers
we have in L potentially,
we're setting them
constant which is the same
as raising them to
the zero power.
And M we know is zero.
That was the constant solution
for the particle on the ring.
So we just put Y zero zero
theta phi is just equal
to some constant.
We don't know what
the constant is
because we haven't
normalized the wave function.
If we normalize the
wave function
to get the constant we just put
in Y zero zero square modulus
and then we have to integrate
over phi from zero to 2 pi,
that's around the ring.
And then we have to integrate
theta, but we have to remember
when we integrate theta
to remember to look back
to the volume element for
spherical polar coordinates
and include sine theta.
So we integrate sine theta
D theta from zero to pi.
And if we do that we find we get
4 pi times Y zero zero squared
is equal to 1.
And neglecting the phase
which we usually choose
to be just real, Y zero zero
theta phi should just be the
constant 1 over the
square root of 4 pi.
And as I remarked don't
forget the sine theta
when you do these integrals.
It's easy to do so if you
aren't actually looking
at the volume element and
just deciding to integrate
to normalize the function.
If you do that you get
another factor of pi
and everything gets
all crossed up.
In this case, for a constant,
the angular energy is zero
because if we put a constant
in, we take the derivative
with respect to phi we get zero
and if we take the
derivative again we get zero.
And same with theta we get zero.
So we just get zero
for everything.
For an atom in this state which
we'll see later is called S,
which has to do with some
historical observations
in atomic spectroscopy
that I'll explain.
The energy, the energy of the
atom must be in the radial part,
because remember when we
converted the kinetic energy
for the particle on a sphere
we threw away the radial part.
We just said, oh well we're
dealing with a particle
on a sphere here and we're going
to throw away the radial
part because R is fixed.
And of course, that, the same
reason why that's an artificial
thing to do with the
particle on the ring,
although it simplifies
the problem,
is an artificial
thing to do there.
So we don't worry too much
about, that the energy is zero
and that means that the
uncertainty principle would be
violated if we believe
that because we couldn't have
any fluctuation in the square
of the kinetic energy then or we
would have some sort of energy
because that's a
positive number.
Now we have no idea
other than this argument
that a constant seems
to work, what might work
for the other solutions
in theta.
But what we can do is we
can just try the same thing
that we tried with phi.
We could try E to I theta.
And see what happens
if we try that.
And I will let you do that
as an interesting problem.
Try putting in E to the I
theta and see what comes out.
And usually what
you can find out is
that you get something
close to what you want.
But it's not quite
what you want.
It has some extra parts.
And then usually by inspecting
those you can get an idea
of what to throw away.
That's in fact what I did on
the side on a piece of paper.
And then I decided I should
try just cosine theta.
Let's stick with
phi as a constant.
Let's try cosine theta
for the theta dependence.
That's a function
that has a node
and therefore we would expect
that that would be sort
of giving us some
angular momentum in the X
and Y components because now we
have some curvature with respect
to theta along that direction.
If we put that in
and that's going
to actually be called Y 1
zero when we finally get done
with figuring out with
the proper notation is
for these functions, for now
we just put it in and we set
up our item value equation.
Same thing, second
derivative with respect to phi,
1 over sine squared theta and so
forth and the funny derivative
with respect to theta that
we're going to work with.
Now we just put through
cosine theta
since there's no phi
dependence at all
that partial derivative
vanishes, we just can throw
that part out and we have minus
H bar squared over 2MR squared,
1 over sine theta, D
by D theta, sine theta,
D cosine theta, D theta.
And that should be equal
to E, what we're going
to call E 1 zero cosine theta.
Because we have to replicate
the function that we put in.
The derivative with respect
of the cosine with respect
to theta is minus the sine.
If we then in the next
line take the derivative
of minus sine squared theta.
That's minus U squared and
the rule is minus 2 U D U D X,
and so therefore I get minus
2 sine theta cosine theta.
And now magically the 1 over
sine theta that was out in front
on this term cancels the
sine theta that I got
from taking the derivative
of sine squared theta.
And I just end up
with cosine theta.
And the minus sine cancels the
minus sign on the H bar squared,
which is good because I
want a positive energy.
And I end up with 2H bar squared
over 2MR squared cosine theta.
Which is 2H bar squared
over 2Y cosine theta.
And that's equal to E
1 zero cosine theta.
And therefore I figured
out what the energy
of this state is that's cosine
theta and nothing in phi.
It's an Eigen function and the
Eigen energy is 2H bar squared
over 2 times the moment
of inertia, capital I.
I should be careful
to say capital I
so that you don't think it's
the square root of minus 1.
Good. But that's just
again 1 solution we're sort
of in the situation we were
with the harmonic oscillator,
that we could get the
ground state by arguing,
well it should be something
related to an exponential.
We could try E to the X, no good
and then we do [inaudible]
and bingo it works.
Here we tried cosine
theta and it worked.
And what we can do
then is try sine theta.
Because cosine theta and sine
theta are closely related,
both have similar curvature.
And let's look at this
then and try sine theta.
But first let's normalize this
wave function, cosine theta,
that's kind of a good
exercise because we have
to do those integrals.
And if we do the integrals,
we have to integrate
over D phi, cosine squared.
Because remember we have
to take the wave function
and square it before
we integrate it.
And we do the integral
with respect to phi
and we just get 2 pi because
there's no phi function
of phi in there.
And then the sine theta for the
volume element is out in front.
So now we have an integral to
do, the integral from zero pi
of co squared theta,
sine theta, D theta.
And we can look up that
integral or we can work it
out by the substitution U
is equal to cosine theta
and if we do that we find that
that integral is equal to 2/3
and so we have 2 pi
from the phi part
and 2/3 from the theta part.
That should equal to 1
and therefore the constant
out front should be the
square root of 3 upon 4 pi.
That's our final solution then
for another energy Eigen
function that we hadn't seen
and we call it Y1 zero
because L is 1 for this.
What if we try sine theta?
Well it's kind of a good thing
to just try it and
see what happens.
And here's what happens.
If we try sine theta
we then have
for the chain term the
derivative of sine theta
with respect to theta.
That's cosine theta.
So then we have the
derivative with respect to theta
of sine theta, times
cosine theta.
We do, that's a product
so that's the derivative
of the first times the second
which is cosine squared theta.
Plus the second times the
derivative of the first
which is minus sine
squared theta.
And we end up with these 2 terms
and now there's no cancellation.
We have minus H bar squared over
2 MR squared, 1 over sine theta,
cosine squared theta,
minus sine squared theta.
And that should be equal to E,
some constant, times sine theta.
That cannot possibly be true
because we have this cos squared
and sine squared and they
aren't cancelling out and so
that cannot possibly be true
for a constant E
when theta's varying.
Therefore we have to
think well how can we sort
of make up for this?
And the only way to
make up for this is
to have some other part come
in from the derivative
with respect to phi.
But that means that we
have to put in something
for the phi part that gives
a non-zero derivative.
Well we know what we're going
to put in for the phi part,
we're going to put in E to
the IN phi and the easiest one
to try is M is equal to 1 since
M is equal to zero was this one
and this one didn't work.
So we put in as our
second attempt,
E to the I phi times sine
theta and we put that in.
Now the second term is
exactly the same thing
as before except
there's another term E
to the I phi riding along.
So I don't have to do all
those derivatives over.
I can just write down 1 over
sine theta cos squared theta,
minus sine squared theta,
all times E to the I phi
and then there's some
constants out in front.
Now however, for the phi
part, the second derivative,
I get something because
I have a phi in there.
And that actually
comes to the rescue.
So recall that part
is the same constant 1
over sine squared theta
times the second derivative
with respect to phi of
sine theta E to the I phi.
The second derivative of that is
I squared, times E to the I phi
because I just bring down the
constant whether it's a real
number or an imaginary number.
It does not matter
in the slightest.
I bring down the same
constant, multiply it
and I follow the rules.
I squared is equal
to minus 1, great.
Perfect. That gets rid of the
minus sign on the H bar squared.
So that's out of my hair.
And what I end up with
then because of the 1
over sine squared
theta out in front,
is I end up with H bar squared
over 2 MR squared times 1
over sine theta,
times E to the I phi.
And now there's a trick.
And the trick often times in
these eigenvalue equations.
There is a trick.
And until you see the trick,
you don't think of it.
Because I've got 1,
1 is a simple thing,
but 1 could be written in
many, many different ways.
And the trick is to
figure out how to write 1
in a way that's useful
to add to the other term.
Since the other term
has cos squared theta
and sine squared
theta and I know
that cos squared theta plus sine
squared theta is equal to 1,
that's what I'm going to use.
So in the last line
on this slide 293,
I've written H bar squared
over 2 MR squared times
cos squared theta,
plus sine squared theta divided
by sine theta, times
E to the I phi.
If we add the 2 terms up we had
cos squared theta minus sine
squared theta over sine theta
and then cos squared theta
plus sine squared theta
over sine theta.
And so we end up, the
cos squared goes away
and the sine squared adds
up and therefore we end
up with 2 sine squared
theta divided by sine theta.
And then bingo, we end up with
the same function that we had,
namely 2H bar squared over
2 MR squared sine theta,
E to the I phi is
equal to the energy,
times sine theta,
times E to the I phi.
Well, that's perfect because
that means that the energy has
to be that constant 2H bar
squared over 2 MR squared.
And interestingly enough that's
the same energy as we got
when we used to cosine theta.
So it seems like at first blush,
the power in theta has something
to do with the eigenvalue
for the energy that we get.
And that's in fact
going to be true.
We have the exact same
energy and so that means
that this system has degeneracy.
And recall degeneracy is
often related to symmetry
when we had the particle
in a box we had degeneracy
when the 2 dimensions
had the same size
and in this case
we could imagine
that all we're doing is swapping
around whether we
have the Z component
of the angular momentum
or the X or Y component
of the angular momentum and
they would have the same energy,
but they would just be rotating
about a different
axis classically
and they would be
rotating at the same rate.
So that seems okay.
And that's not troublesome.
Another equally valid choice,
which won't surprise you
if you can pick E to I
phi is that you can pick E
to the minus I phi,
again, times sine theta.
And that also has
the same energy.
And we'll call that
one E1 minus 1.
So we have E1 1, E1
zero, and E1 minus 1.
And there are no other ones.
If we try sine theta or cos
theta with E to the 2 I phi,
we end up with a mess.
So therefore those are the 3.
We know it has to be
integers so those are the 3.
And we start to see a pattern.
For the bottom one where
the theta part's a constant,
we end up with 1 function in
phi which is also a constant.
M is equal to zero.
For the next one where we have
some trig functions in theta
that are raised to the
first power, sine theta
and cosine theta, then
we have 3 values of phi.
We have 1, zero, and minus 1.
And so it looks like the values
of M are bounded by the value
of L and that would make sense
because L is going to turn
out to be the square of the
total angular momentum and M has
to do with just the Z component.
And the Z component of angular
momentum whatever it is can't
exceed the total of all of them.
And so it would have
to be bounded by them.
And of course they come in
units of H bar each of them
so L squared comes in
units of H bar squared.
LZ squared comes in units of H
bar squared and LX and LY come
in units of just H bar.
The, so the solutions have 2
times H bar squared over 2Y.
And the second powers in
theta, if we pick cos squared
and cosine sine,
and sine squared,
not surprisingly there are 5
values of M that go with that 2,
1, zero, minus 1, and minus 2.
And those are the 5 and so
we see a pattern 1, 3, 5,
and that pattern continues.
And in general there are
always 2 L plus values of M
that are possible to choose
where L is the highest power
in the trig functions
that you pick for theta.
Now without some more powerful
methods to analyze the equation,
it's going to take us
forever to play around
and guess different kinds of
functions in theta that happen
to work and it could be an
extremely frustrating thing
like doing a crossword puzzle,
rather than doing mathematics
where you just knock it out.
But for us at this point
we aren't going to go
into the theory of differential
equations to such a depth
because it's going to
pull us too far afield
and that would be a proper
course to take in math.
But I think you can see why such
a course is incredibly valuable
because if you come
up against equations
like this you're either
forced to sort of type,
henpeck your way through the
equations which can be good
if you only want a few
solutions, but it's very slow.
Or you can sort of become
the master of how to figure
out these equations and
move a lot more quickly.
Of course it takes time to learn
how to type and it takes time
to learn how to approach
these kinds of problems
in a more sophisticated way
so you actually can get all
the solutions and not have
to spend a tone of
time writing them down.
Therefore, we can generalize
this in the light of the pattern
and we can say look the
energy, ELM is equal
to L times L plus 1, times H
bar squared over 2 capital I,
the moment of inertia.
And classically the energy of a
rotating body is just L squared
over 2 I where L is again the
square, L squared is the square
of the angular momentum.
And therefore what we
can associate then is
that the square of the
angular momentum in the case
of a quantum object, particle
on a sphere is now
quantized in units of H bar.
And that's perfect because
that's exactly what we expect.
The funny thing about
it is that rather
than just L squared times H
bar squared, little L squared,
we get L times L plus 1.
We couldn't tell any
different is L is zero
between L squared
and L times L plus 1.
But when L is 1, when we
have some angular momentum,
there is a difference and
there's this extra plus 1
and as we shall see that
actually has something to do
with the uncertainty principle.
It just came in out
of the equation here,
but it also is very
closely related to the fact
that there has to be some wobble
that we couldn't actually have
all the angular momentum be
along Z.
If we could have
it all be along Z
and there were nothing
left over,
then there wouldn't be any
uncertainty in any of the 3.
Z would be known and LX
and LY would also be zero
because that square
would have to add up.
And to make sure that doesn't
happen we have an extra plus 1
that comes in conveniently,
another unit of H bar that even
when Z is as big as it can
possibly be, the top value of M,
there's still a little extra.
So there's still wobble left and
that is exactly in accordance
with the uncertainty principle.
These functions YLM are well
known, as our textbook says.
And they're called the
spherical harmonics.
And you can look them up and
you can read all about them.
They come in for all kinds
of differential equations
when you take the
divergence of the gradient
of a scalar field you end up
with these kinds of functions.
But as far as we're concerned,
just like sine and cosine,
or E to the IX, are fundamental
things for linear motions,
or for waves moving on a line.
These functions YLM
are the same kind
of fundamental bedrock starting
point for anything that has
to do with waves on a sphere.
As opposed to waves just
moving in a 1-dimensional line.
And the Eigen functions of
the operators L hat squared.
The eigenvalue is L times L
plus 1 times H bar squared.
And simultaneous
Eigen functions of LZ
and the eigenvalue
there is M H bar.
There are tables of these
functions but please, please,
please do not ever try
to memorize all the
spherical harmonics.
If we need them on an
exam or you need them
to do a problem, look them up.
Don't try to memorize
what they are.
Understand where they came from.
But don't try to commit to
memory all the constants
and whether it's minus
plus or plus minus
or what the exact
trigonometric dependence is.
If you understand
where they came from,
how they are normalized and what
they mean and you can use them
to actually do something,
that's plenty.
You don't need to
commit them to memory.
And again if you take
a more advanced course,
there are in fact neat
ways to generate them.
So you can generate these
functions by taking derivatives
of things and you
just start with one
and then you turn a handle
and you get the next one
and then you turn a handle
and you get the next
one and so forth.
And of course knowing
the generating function
as it's called would obviate
the need to memorize anything,
except the generating function.
And for that you
would need to remember
which generating function goes
with which set of polynomials,
because not surprisingly
there are generating functions
for all kinds of polynomials.
And that's of course
how they're made.
As we're going to see, the fact
that this angular momentum
is quantized that it comes
in steps, is very, very
important when it comes
to understanding
atomic spectroscopy.
And atomic spectroscopy
is very important
because the observation that
atoms gave discrete energy lines
of light and not just
a continuum was one
of the crucial things that
led physicists to believe
that they didn't understand
what was going on with atoms.
They had no idea.
They had the classical theory
the atom will spiral away.
It could have any old energy.
And then when they got
these lines on film,
they were discrete lines and
they were always the same
and they had to do with what
kind of element you had.
And that was ever so
useful because that meant
that you could find out what
kind of element you had.
And if you get a telescope
and you look out as far
as you can look what
you find out is
that the whole visible universe
seems to be made out of stuff
on the periodic table.
That may not seem
surprising now.
But that didn't have to be.
It could be that if you
look far enough away
that there's something
that you don't understand.
There's some chemistry
occurring and some blast
from some star left over
that you don't understand.
But when we look out we see
molecules that we see on earth,
we see hydrogen everywhere
in the universe.
We see all these beautiful lines
and red glows from galaxies
and it all makes perfect sense
because we can do
the same experiments
in the laboratory
and see these lines.
And so it seems like
there's nothing mysterious
about the stuff we can see.
Now stuff we can't see,
so-called dark matter could be
some different story altogether
because if you have something
that you really can't see at all
or not very well then it's very
hard to figure out what it is
because it's more or
less invisible to you.
The photon has 1 unit
of angular momentum.
And it has an intrinsic twist
and that's really important
because when the photon
comes into the atom,
the photon is annihilated.
The photon's here,
the photon is gone.
But the angular momentum we
believed to be conserved just
like the energy and therefore it
must be that the twist that was
in the photon is
converted into a twist
in the wave function
for the atom.
And vice versa.
So when the atom emits a
photon a twist has to happen
and the photon comes out.
And that requirement that the
angular momentum be conserved
along with energy really
simplifies the atomic spectrum
otherwise we would
have jillions of lines
and it would be virtually
impossible for us
to figure out what is what.
But because we only
have a few and most
of them are missing it
was possible for Rigburgh
to see s pattern in
the energy levels.
Therefore if we have atomic
transition as we're going to see
when we talk about atomic
spectroscopy in more detail,
there is a selection rule.
The selection rule is delta L,
little l from the
spherical harmonic is equal
to plus or minus 1.
In other words the photon has
to remove or absorb a twist
and the atom has to
compensate accordingly.
There is another player
in the game, called spin.
And it doesn't surprise us that
if we've got a charged particle
like an electron and it's in
something that we understand
to be on a ring or on a sphere,
it's moving around circulating,
that it looks like
a current loop
and it has a magnetic field.
But experiments that were done,
more detailed experiments,
in cases where the atom should
not have electrons moving
around because they
should basically be
in this constant state, this
M equals zero kind of state
that doesn't have any twist and
can't have any angular momentum,
they still saw that there
was some magnetic phenomenon.
And the pivotal experiment,
much like the photoelectric
effect was the quantization
of light.
And the Davisson
Germer experiment showed
that electrons could
be have like waves,
the Stern Gerlach
experiments showed
that electrons themselves have
an intrinsic magnetic field
and in addition to charge and
mass, electrons also behave
on their own like a
little bar magnet.
And because we know that
particles on a sphere and atoms
as we'll see are quantized,
according to these equations
with units of H bar,
would make sense
that the electron would
somehow be quantized.
Now where would this
magnetic field be coming from?
Well if we blow up the
electron and we have some kind
of a planetary model for it
as a uniform sphere of charge,
then if the electron
were spinning,
then this uniform sphere of
charge, if it were just spinning
and maybe it couldn't stop,
would create a magnetic field
because that would be a
bunch of current loops.
And unfortunately that kind
of really concrete picture
of the electron actually
spinning can't possibly be true.
And so rather than saying
the electron is spinning,
the dodge is to say the electron
has spin, rather than thinking
of it as a mechanical
thing actually occurring,
it's now just a property
like charge.
It's something we give
it a name and it's there.
And we can't quite explain it
in terms of a planetary model
of a big blow up
of the electron.
But as I mentioned before
the electron appears
to have null size so it
would be difficult to figure
out what size to assign for
this planetary model anyway.
Stern and Gerlach used
a beam of silver atoms
to do their experiment.
And the reason why they did
that is that silver atoms have,
basically a closed
shell of electrons.
And closed shell of electrons
when they're all filled up,
the angular momentum is zero.
That's one reason why closed
cells are especially stable.
And then there is one
electron in an S orbital.
And an S orbital has no
orbital angular momentum.
And the spin angular
momentum if there is any
for all the closed
shell is also zero.
That can be worked out.
And therefore we've got 1
electron in an outer shell.
And we should just be able
to use the silver atoms then
to figure out what value
of the angular momentum
this 1 electron has.
That's why you pick silver.
The other reason why you pick
silver is that you can heat
up silver in an oven,
in a vacuum
and you can get a
beam of silver atoms.
You can make silver vapor.
You have to heat
it up pretty hot,
but you don't need
a lot of atoms.
And then you can send
them through a detector,
which I'll talk about
in a second.
And you can get sliver stripes.
Just like making a mirror
and you can see whether you
get the classical prediction,
or whether you get
something else.
Now how do you actually
interrogate on something as tiny
as an atom, with the
magnetic dipole moment,
because remember this is
like a little bar magnet.
How are we going to do an
experiment that gets us
to get this thing to move?
Well if we stick these magnets
through a uniform magnetic
field, nothing happens,
they tend to reorient, but
nothing happens to them.
And so Stern and Gerlach used
a magnetic field gradient.
Not a uniform magnetic field.
They had a beam of
silver atoms and then up
and down they had a
magnetic field gradient.
It's not easy to think of what
the magnetic field gradient does
to a little bar magnet because
magnetism is a little bit more
mysterious than just
electrostatics.
But it's basically the same
as if I have an electric
dipole in an electric field.
And it turns out that a
magnetic field gradient is going
to put force on the
little bar magnet and tend
to accelerate it and deflect it.
And which way it deflects can
tell us whether the North Pole
is up, or whether the North Pole
was down when it went through.
As I said it's much easier
to see this in the case
of an electric dipole
and an electric field.
And so I've drawn by
analogy the situation
where we have a capacitor
let's say
and it has uniform
positive charge on one plate
and uniform negative
charge on another plate.
And so we have the electric
field lines going through,
at the edge they have to
bend, but let's say we're
in the middle and they're
going through nice and straight
and we have a uniform
electric field in space.
If we have a dipole which is
always a negative charge hooked
to a positive charge, and
the dipole is like this,
horizontal in this
electric field.
It will feel a torque.
Because the positive
charge will tend
to get pulled toward
the negative plate.
And the negative charge
will get turned toward the
positive plate.
And therefore it
will tend to align,
but it's not going
to move anywhere.
Because the total force on the
thing from the positive charge
and the negative charge and
they're hooked together,
otherwise it wouldn't
be a dipole, is zero.
So it will tend to twist, but
it won't tend to go anywhere.
If it's going through
it will twist
and then stay going through.
But now suppose I make
the positive charge more
concentrated on the top
say by shrinking the plates
so that the field lines
have to converge together,
then I get this picture
shown on slide 302.
Where now because there's
more positive charge,
even if it's aligned, let's
say it's a line with a negative
on top and positive on bottom,
but there's more
positive charge here,
nearer to the negative charge
then there is negative charge
on this more spread out
plate to the positive charge.
And therefore these 2 will be
pulled up along the field lines
where the gradient is
getting closer together.
And that will put a
force then on this thing.
And if it were moving through
it would tend to deflect.
Great. Now let's do the same
thing with the magnetic case.
So let's take silver, let's
make a beam out of an oven,
individual silver atoms in
a vacuum of course, no air,
or we'll just get silver oxide
and the experiment will be off.
Put them through some choppers.
Why? Because we want to
let only silver atoms
with a certain velocity
go through.
Put them through a small pipe.
Align them up so that we know
that they're going this way
and they're all going at a
certain speed, give or take.
And then put them through
a magnetic field gradient
and the silver has
this single electron
in the 5S1 configuration.
We'll talk about that
later when we talk
about the electronic
configuration
of atoms in more detail.
Now if the intrinsic bar magnet
of the electron were a
classical thing it could just be
pointing anywhere.
I mean why not?
Then we would expect to get just
a spray of possibilities if,
depending how it was pointed
in the magnetic field it
would get deflected up or down
to a certain maximum among
and then anywhere in between.
But in fact, when they
did the experiment
which I've sketched
here out on slide 304,
the classical prediction
would be a smear
and when they did the
experiment carefully.
And when I say when
they did it carefully.
I believe historically
when they first did it,
they may not have had
the beam narrow enough
and going enough
at the same speed.
And of course if you
have things going
through at different
speeds so they're in there
for different times and you
don't have them aligned very
much, then it's sort of
like having a camera,
where you've deliberately
blurred the image,
then you can't read anything, it
just looks like it's all grey,
you can't read the
newsprint then.
And I think the first
time, or the initially,
they got what seemed to
be something that agreed
with a classical result.
But then when they did the
experiment more carefully,
they got 2 spots.
If they get 2 spots that means
that M could be some positive
thing or some negative thing,
but the theory of
angular momentum says
that the difference between
the levels should be integer.
And the only way you can
have 2 levels and not have 3,
not have 1, zero, and minus 1.
Only have 2 levels, would mean
that it would have to be 1/2
which was really strange
because that's hard
to see how the intrinsic
angular momentum could be 1/2,
but nevertheless, that's
exactly what was seen.
We got 2 spots and
the conclusion then is
that the electron magnetic
moment is quantized.
It corresponds to some kind
of angular momentum that's
intrinsic to the particle.
We don't have a detailed picture
for it, but we know it's there.
And the 2 allowed magnetic
states, which we tend to call up
and down are just Ms of S
rather than Ms of L. Ms of S
for spin plus or minus
1/2 and it follows then,
that the spin angular
momentum which is just called S
like L, is just 1/2 H bar.
That's the allowed value of
the spin angular momentum.
And there is no other solution
this is just an intrinsic
property of electron itself.
Has nothing to do with anything
except the intrinsic property
of the electron.
And it's very interesting
then, this was puzzling
as to how this could
be like this.
And next time I'll talk
a little bit about spin
and how important it has been,
whether particles have a 1/2
integer spin or an integer spin
because other particles
like protons, and deuterons,
and other things when
they look carefully,
and there were Nobel
Prizes in those fields
as well, also had spin.
They also had little
magnetic moments.
And they were 1/2 or
sometimes they were 1.
The electron's always 1/2.
And they separate into 2 groups.
The halves and the non
halves, the integers.
And they have completely
different properties
in many kinds of experiments
in physics and chemistry.
So next time we'll pick it
up from there and continue
on our exposition of atoms.
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