PROFESSOR: We're talking
about angular momentum.
We've motivated angular
momentum as a set of operators
that provided observables,
things we can measure.
Therefore, they are important.
But they're particularly
important for systems in which
you have central potentials.
Potentials that depend
just on the magnitude
of the radial variable.
A v of r that depends just on
the magnitude of the vector
r relevant to cases where you
have two bodies interacting
through a potential
that just depends
on the distance
between the particles.
So what did we develop?
Well, we discussed the
definition of the angular
momentum operator.
You saw they were permission.
We found that they satisfy
a series of commutators
in which lx with
ly gave ih bar lz,
and cyclical versions
of that equation, which
ensure that actually you
can measure simultaneously
the three components
of angular momentum.
You can measure,
in fact, just one.
Happily we found there
was another object
we could measure, which was
the square of the total angular
momentum.
Now, you should
understand this symbol.
It's not a vector.
It is just a single
operator. l squared
is, by definition, lx
times lx, plus ly times ly,
plus lz times lz.
This is this operator.
And we showed that any
component of angular momentum,
be it lx, ly, or lz,
commutes with l squared.
Given that they commute,
it's a general theorem
that two permission
operators that commute,
you can find simultaneous
eigenstates of those two
operators.
And therefore, we set up for the
search of those wave functions
that are simultaneous
eigenstates of one of the three
components of angular momentum.
Everybody chooses
lz and l squared.
lz being proportional to
angular momentum has an h bar m.
We figured out by looking at
this differential equation
that if we wanted single
valued wave functions--
wave functions would
be the same at phi,
and at phi plus 2 pi,
which is the same point.
You must choose m
to be an integer.
For the l squared
operator we also
explained that the
eigenvalue of this operator
should be positive.
That is achieved when
l, whatever it is,
is greater than 0,
greater or equal than 0.
And the discussion that led
to the quantization of l
was a little longer,
took a bit more work.
Happily we have this
operator, and operator
we can diagonalize, or we
can find eigenstates for it.
Because the Laplacian,
as was written
in the previous
lecture, Laplacian
entering in the
Schrodinger equation
has a radial part
and an angular part,
where you have dd
thetas, and sine thetas,
and the second defies square.
All these things were
taken care of by l squared.
And that's very useful.
Well, the differential
equation for l
squared-- this can be though as
a differential equation-- ended
up being of this form,
which is of an equation
for the so-called Associate
Legendre functions.
For the case of m equals 0
it simplifies very much so
that it becomes an equation
for what were eventually
called Legenre polynomials.
We looked at that differential
equation with m equals 0.
We called it pl 0.
So we don't write the zeros.
Everybody writes pl
for those polynomials.
And looking at the
differential equation one
finds that they have
divergences at theta
equals 0, and a theta
equal pi, north and south
pole of this spherical
coordinate system.
There aren't divergences unless
these differential equations
has a polynomial solution that
this is serious the recursion
relations terminate.
And that gave for us
the quantization of l.
And that's where we stopped.
These are the
Legendre polynomials.
Solve this equation
for m equals 0.
Are there any questions?
Anything about the
definitions or?
Yes?
AUDIENCE: Why do we care about
simultaneous eigenstates?
PROFESSOR: Well, the
question is why do we care
about simultaneous eigenstates.
The answer is that
if you have a system
you want to figure out what are
the properties of the states.
And you could begin by saying
the only thing I can know
about this state is its energy.
OK, well, I know
the energy at least.
But maybe thinking harder
you can figure out, oh, you
can also know the momentum.
That's progress.
If you can also know
the angular momentum
you learn more about the
physics of this state.
So in general, you will be
led in any physical problem
to look for the maximal
set of commuting operators.
The most number of operators
that you could possibly
measure.
You know you have success
at the very least,
if you can uniquely characterize
that states of the system
by observables.
Let's assume you have
a particle in a circle.
Remember that the free
particle in a circle
has degenerate
energy eigenstates.
So you have two
energy eigenstates
for every allowed energy,
except for 0 energy,
but two energy eigenstates.
And you would be baffled.
You'd say, why do I have two?
There must be some difference
between these two states.
If there are two states,
there must be some property
that distinguishes them.
If there is no property
that distinguishes them,
they should be the same state.
So you're left to search
for another thing.
And in that case the
answer was simple.
It was the momentum.
You have a particle with some
momentum in one direction,
or in the reverse direction.
So in general, it's a
most important question
to try to enlarge the set
of commuting observables.
Leading finally to
what is initially
called a complete set of
commuting observables.
So what do we have to do today?
We want to complete
this analysis.
We'll work back
to this equation.
And then work back to
the Schrodinger equation
to finally obtain the
relevant differential
equation we have to
solve if you have
a spherical symmetric potential.
So the equation will be
there in a little while.
Then we'll look at
the hydrogen atom.
We'll begin the hydrogen
atom and this task why?
Having a proton
and an electron we
can reduce this system to
as if we had one particle
in a central potential.
So that will be also very
important physically.
So let's move ahead.
And here there is a simple
observation that one can make.
Is that the differential
equation for p l m depends on m
squared.
We expect to need values of m
that are positive and negative.
You have wave functions
here, of this form.
The complex conjugate
ones should be
thought as having m negative.
So we expect positive and
negative m's to be allowed.
So how did people
figure this out?
They, in fact, figured out that
if you have these polynomials
you can create automatically
the solutions for this equation.
There's a rule, a simple
rule that leads to solutions.
You put p l m of x is
equal to 1 minus x squared,
to the absolute
value of m over 2.
So there are square
roots here, possibly.
An absolute value of m
means that this is always
in the numerator, whether
m is positive or negative.
And d, dx acting exactly
absolute value of m times on p
l x.
The fact is that this definition
solves the differential
equation star.
This takes a little
work to check.
I will not check it, nor
the notes will check it.
It's probably
something you can find
the calculation in some books.
But it's not all that important.
The important thing to
note here is the following.
That this provides solutions.
Since this polynomial is like
x to the l plus x to the l
minus 2 plus
coefficients like this.
You can think that most
m equal l derivatives--
if you take more than l
derivatives you get 0.
And there's no great
honor in finding
zero solution of this equation.
These are no solutions.
So this produces solutions
for an, absolute value of m,
less than l.
So produces solutions
for absolute value
of m less or equal to l.
And therefore m in
between l and minus l.
But that's not all that happens.
There's a little more that
takes mathematicians some skill
to do.
It's to show that there
are no more solutions.
You might seem that
you were very clever
and you found some
solutions, but it's a theorem
that there are no
more solutions.
No additional regular solutions.
I mean solutions
that don't diverge.
So this is very important.
It shows that there is one
more constraint on your quantum
numbers.
This formula you may
forget, but you should never
forget this one.
This one says that if you
choose some l which corresponds
to choosing the magnitude
of the angular momentum,
l is the eigenvalue
that tells you
about the magnitude of
the angular momentum.
You will have several
possibilities for m.
There will be several
states that have the same l,
but m different.
So for example you'll
have l equal 0, in which
case m must be equal to 0.
But if you choose
state with l equals 1,
or eigenfunctions
with l equal 1,
there is the possibility of
having m equals minus 1, 0,
or 1.
So are three waves
functions in that case.
Psi 1, minus 1, psi
1, 0, and psi 1, 1.
So in general when we
choose a general l,
if you choose an
arbitrary l, then
m goes from minus l, minus l
plus 1 all the way up to l.
These are all the values
which are 2l plus 1 values.
2l and the 0 value in between.
So it's 2l plus 1 values.
The quantization in
some sense is done now.
And let me recap about
these functions now.
We mentioned up there that
the y l m's are the objects.
The spherical
harmonicas are going
to be those wave functions.
And they have a normalization,
n l m, an exponention,
and all that.
So let me write,
just for the record,
what a y l m looks like
with all the constants.
Well, the normalization
constant is complicated.
And it's kind of a thing you
can never remember by heart.
It would be pointless.
OK.
All of that.
Then a minus 1 to
the m seems useful.
e to the i m phi, p
l m of cosine theta.
And this is all valid for
0 less 0 m positive m.
When you have negative m you
must do a little variation
for m less than 0 y l m of
theta and phi is minus 1
to the m y l minus m of theta
and phi complex conjugated.
Well, if m is negative,
minus m is positive.
So you know what that is.
So you could plug
this whole mess here.
I don't advise it.
It's just for the record.
These polynomials
are complicated,
but they are normalized nicely.
And we just need to
understand what it
means to be normalized nicely.
That is important for us.
The specific forms of these
polynomials we can find them.
The only one I really remember
is that y 0 0 is a constant.
It's 1 over 4 pi.
That's simple enough.
No dependents. l
equals 0, m equals 0.
Here is another one.
y1 plus minus 1 is minus plus
square root of 3 over 8 pi
e to the plus minus
i phi sine theta.
And the last one,
so we're giving
all the spherical
harmonics with l equals 1.
So with l equals 1
remember we mentioned
that you would have
three values of m.
Here they are.
Plus or minus 1 and 0.
