The following
content is provided
under a Creative
Commons license.
Your support will help MIT
OpenCourseWare continue
to offer high quality
educational resources for free.
To make a donation or to
view additional materials
from hundreds of MIT courses,
visit MIT OpenCourseWare
at ocw.mit.edu.
PROFESSOR: All right.
Today we'll be talking a
little about angular momentum.
Continuing the discussion
of those vector operators
and their identities
that we had last time.
So it will allow us to make
quite a bit of progress
with those operators, and
understand them better.
Then we'll go through
the algebraic analysis
of the spectrum.
This is something
that probably you've
seen in some way or another,
perhaps in not so much detail.
But you're probably
somewhat familiar,
but it's good to see it again.
And finally at the end
we'll discuss an application
that is related to your last
problem in the homework.
And it's a rather
mysterious thing
that I think one
should appreciate
how unusual the
result is, related
to the two dimensional
harmonic oscillator.
So I'll begin by reminding
you of a few things.
We have L, which is r cross p.
And we managed to
prove last time
that that was equal to p
cross r, with a minus sign.
And then part of the
problem's that you're
solving with
angular momentum use
the concept of a vector
and the rotations.
So if u is a vector
under rotations--
to say that something is a
vector under rotations means
the following, means that if you
compute Li commutator with uj,
you can put a hat.
All these things are
operators, all these vectors.
So maybe I won't put a hat
here on the blackboard.
Then you're supposed to get
i, epsilon, ijk, ih bar.
Epsilon, ijk, uk.
So that's a definition
if you wish.
Any object that does that
is a vector under rotations.
And something that in the
homework you can verify
is that r and p are
vectors under rotation.
That is, if you put here xj,
you get this thing with xk.
If you put here pj, you
get this thing with pk.
If you compute the commutator.
So r and p are vectors
under rotation.
Then comes that little theorem,
that is awfully important,
that shows that if u and v are
vectors under rotations-- u
and v vectors under rotations--
then u dot v is a scalar.
And u cross v is a vector.
And in both cases,
under rotations.
So this is something
you must prove,
because if you know how u and
v commute with the angular
momentum, you know how u
times v, in either the dot
combination or the cross
combination, commute with j,
with L. So to say that
something is a scalar,
the translation is that
Li with u dot v will be 0.
You don't have to
calculate it again.
If you've shown that
u and v are vectors,
that they transform like
that, this commutes with this.
So r-- so what do you
conclude from this?
That Li commutes with r squared,
commutes-- it's equal to p
squared.
And it's equal to Li, r dot p.
They all are 0.
Because r and p are
vectors under rotation,
so you don't have to
compute those ones anymore.
Li will commute with r squared,
with p squared r cross p.
And also, the fact
that u cross v
is a vector means that Li
commutated with u cross v,
j-- the j component of u cross
v is ih bar, u cross v. I'm
sorry-- epsilon,
ijk, u cross v, k.
Which is to say that u cross
v is a vector under rotations.
This has a lot of
important corollaries.
The most important
perhaps is the commutation
of angular momentum with itself.
That is since you've shown
that r and p satisfy this,
r cross p, which is
angular momentum,
is also a vector under rotation.
So here choosing u
equal r, and v equal p,
you get that Li, Lj is equal
to ih bar, epsilon, ijk, Lk.
And it's the end of the story.
You got this commutation.
The commutation you wanted.
In earlier courses,
you probably found
that this was a fairly
complicated calculation.
Which you had to put the x's and
the p's, the x's and the p's,
and start moving them.
And it takes quite
a while to do it.
So, that's important.
Another property
that follows from all
of this, which is sort of
interesting, that since L
is now also a vector under
rotations, Li commutes with L
squared.
Because l squared is L dot
L, therefore it's a scalar.
So Li commutes with L squared.
And that property is
absolutely crucial.
It's important that it's worth
checking that in fact, it
follows just from this algebra.
You see, the only
thing you need to know
to compute the commutator
of Li with L squared
is how L's commute.
Therefore it should be
possible to calculate this
based on this algebra.
So this property is true just
because of this algebra, not
because of anything
we've said before.
And that's important
to realize it.
Because you have
algebra like si,
sj, ih bar, epsilon,
ijk, sk, which
was the algebra of
spin angular momentum.
And we claim that
for that same reason
that this algebra
leads to this result,
that si should commute
with s squared.
And you may remember that
in the particular case
we examined in this
course, s squared--
that would be sx squared plus
sy squared plus sz squared-- was
in fact h bar over 2 squared.
And each matrix was
proportional to the identity.
So there's a 3 in
the identity matrix.
And s squared is really in the
way we represent that spin,
by 2 by 2 matrices,
commutes with si.
Because it is the identity.
So it's no accident
that this thing is 0.
Because this algebra,
whatever l is,
implies that this with the
thing squared is equal to zero.
So whenever we'll be
talking about spin angular
momentum, orbital angular
momentum, total angular
momentum, when we
add them, there's
all kinds of angular momentum.
And our another generic name
for angular momentum will be j.
And we'll say that ji, jj,
equal ih bar, epsilon, ijk,
jk is the algebra
of angular momentum.
And by using j, you're
sending the signal
that you may be talking about l.
Or may be talking about
s, but it's not obvious
which you're talking about.
And you're focusing on those
properties of angular momentum
that hold just because this
algebra is supposed to be true.
So in this algebra, you will
have that ji commutes with j
squared.
And what is j squared?
Of course, j squared
is j1 squared
plus j2 squared plus j3 squared.
Now this is so important,
and this derivation
is a little bit indirect,
that I encourage you all
to just do it.
Without using any
formula, put the jx here,
and compute this commutator.
And it takes a couple of lines,
but just convince yourself
that this is true.
OK, now we did have a
little more discussion.
And these are all things
that are basically related
to what you've been
doing in the homework.
Another fact is
that this algebra
is translated into j
cross j equal ih bar, j.
Another result in
transcription of equations
is that the statement that u
is a vector under rotations
corresponds to a
vector identity.
Just the fact that
the algebra here
is this, the fact
that l with u is this,
implies the following algebra.
j cross u plus u cross
j equal 2i h bar.
So this is for a
vector under rotations.
Under rotations.
So this I think is in the notes.
It's basically
saying that if you
want to translate this equation
into vector form, which
is a nice thing to have,
it reads like this.
And the way to do that is to
just calculate the left hand
side.
Put and index, i.
And just try to get
the right hand side.
It will work out.
OK.
Any questions so far
with these identities?
OK.
So we move on to
another identity
that you've been
working on, based
on the calculation of what
is a cross b dot a cross b.
If these things are
operators, there's
corrections to the
classical formula
for the answer of of what this
product is supposed to be.
Actually, the classical
formula, so it's not
equal to a squared, b squared,
minus a dot b squared.
But it's actually equal
to this, plus dot dot dot.
A few more things.
Classically it's just that.
You put 2 epsilons.
Calculate the left hand side.
And it's just these 2 terms.
Since there are
more terms, let's
look what they are for a
particular case of interest.
So our case of interest is L
squared, that corresponds to r
cross b, times r cross b.
And indeed, it's
not just r squared,
p squared, minus
r dot p squared.
But there's a little extra.
And perhaps you have computed
that little extra by now.
It's ih bar r dot p.
So that's a pretty
useful result.
And from here, we typically
look for what is p squared.
So for p squared-- so what we
do is pass these other terms
to the other side.
And therefore we have
1 over r squared,
r dot p squared,
minus ih bar, r dot p.
Yes.
Plus 1 over r
squared, l squared.
And, we've done this
with some prudence.
The r squared is here in
front of the p squared.
It may be fairly different from
having it to the other side.
And therefore, when I apply
the inverse 1 over r squared,
I apply it from the left.
So I write it like that.
And that's very
different for having
the r squared on the other side.
Could be completely different.
Now, what is this?
Well this is a
simple computation,
when you remember that p vector
is h bar over i gradient.
And r dot p, therefore
is h bar over i, r, dvr.
Because r vector is
r magnitude times
the unit vector in
the radial direction.
And the radial direction
of gradient is dvr.
So this can be simplified.
I will not do it because
it's in the notes.
And you get minus h squared, 1
over r, d second, d r squared,
r.
In a funny notation,
the r is on the right.
And the 1 over r is on the left.
And you would say, this
doesn't sound right.
You have here all
this derivatives
and what is an r doing to
the right of the derivatives.
I see no r.
But this is a kind of a
trick to rewrite everything
in a short way.
So if you want,
think of this being
acting on some function of r.
And see what it is.
And then you put a function
of r here, and calculate it.
And you will see,
you get the same.
So it's a good
thing to try that.
So p squared is given by this.
There's another
formula for p squared.
p squared is, of
course, the Laplacian.
So p squared is also
equal to minus h squared
times the Laplacian operator.
And that's equal to minus
h squared times-- in fact,
the Laplacian operator is 1
over r, d second, dr squared, r,
plus 1 over r squared,
1 over sine theta, dd
theta, sine theta, dd theta.
It's a little bit messy.
Plus 1 over sine squared
theta, d second, d phi squared,
times closing this.
So a few things
are there to learn.
And the first thing is if you
compare these 2 expressions,
you have a formula
for l squared.
You have l squared is 1 over
r squared on the upper right.
And here you have minus h
squared times this thing.
So l squared, that
scalar operator
is minus h squared,
1 over sine theta,
dd theta, sine theta, dd
theta, plus 1 over sine
squared theta, d
second, d phi squared.
So in terms of functions of
3 variables, x, y, and z,
L squared, which is a
very complicated object,
has become just a function
of the angular variables.
And that this a very
important intuitive fact.
L squared.
L is operator.
That's rotation.
So it shouldn't really affect
the r, shouldn't change r,
modify r in any way.
So it's a nice thing
to confirm here
that this operator can be
thought as an operator acting
on the angular variables.
Or you could say, on functions,
on the units here for example.
It's a good thing.
The other thing that
you've learned here--
so this is a very nice result.
It's not all that easy to
get by direct computation.
If you had to do Lx squared
plus Ly squared plus Lz
squared, first all this
possible order-- well,
there's no ordering
problems here.
But you would have to write this
in terms of x, and py, and pz,
and xy, and z, then pass
to angular variables.
Simplify all that.
It's a very bad way to do it.
And it's painful.
So the fact that we got
this like that is very nice.
The other thing that we've
got is some understanding
of the Hamiltonian for a
central potential, what
we call a central
potential problem.
v of r.
Now, I will write
a v of r like this.
But then we'll simplify it.
In fact, let me just go to a
central potential case, which
means that the potential just
depends on the magnitude of r.
So r is the magnitude
of the vector r.
So at this moment, you
have p squared over there.
So this whole
Hamiltonian is minus h
squared over 2m, 1 over r, d
second, dr squared, r, plus p
squared over 2m.
So 1 over 2m, r squared,
l squared plus v of r.
So our Hamiltonian has
also been simplified.
So this will be
the starting point
for writing the
Schrodinger equation
for central potentials.
And you have the
operator l squared.
And as far as we can, we'll
try to avoid computations
in theta and phi
very explicitly,
but try to do things
algebraically.
So at this moment,
the last comment
I want to make on this
subject is the issue
of set of commuting observables.
So if you have a
Hamiltonian like that,
you can try to form a set of
commuting observables that
are going to help you
understand the physics
of your particular problem.
So the first thing
that you would
want to put in the list of
complete set of observables
is the Hamiltonian.
We really want to know the
energies of this thing.
So what other
operators do I have?
Well I have x1, x2, and x3.
And well, can I add
them to the Hamiltonian
to have a complete set
of commuting observables?
Well, the x's commute
among themselves.
So can I add them?
Yes or no?
No.
No you can't add
them, because the x's
don't commute with
the Hamiltonian.
There's a p here.
p doesn't commute with x's.
So that's out of the question.
They cannot be
added to our list.
How about the p's?
p1, p2, and p3.
Not good either,
because they don't
commune with the potential term.
The potential has x dependents,
and will take a miracle for it
to commute.
In general, it won't commute.
So no reason for it to commute,
unless the potential is 0.
So this is not good.
Nor is good to have r squared,
or p squared, or r dot p.
r squared, p squared, r dot p.
No good either.
On the other hand, r
cross p is interesting.
You have the angular
momentum, L1, L2, and L3.
Well, the angular momentum
will commute, I think,
with the Hamiltonian.
You can see it here.
You have p squared,
and Li's commute with p
squared because p is a
vector under rotations.
p doesn't communicate with
Li, but p squared does.
Because that was a scalar.
So this term commutes with
any angular momentum operator.
Moreover, v or r, r is this.
So a v of r is a
function of r squared.
And r squared is the
vector r squared.
So ultimately, anything that is
a function of r is a function
of r squared that involves
the operator r squared,
that also commutes
with all the Li's.
So h commutes with all the Li's.
And that's a great thing.
So this is absolutely important.
h commutes with all the Li's.
That's angular
momentum conservation.
As we've seen, the rate
of change of any operator
is equal to expectation
value of the commutator
of the operator with
the Hamiltonian.
So if you put any Li,
this commutator is 0.
And the operator is
conserved in the sense
of expectation values.
Now this conservation
law is great.
You could add this operators
to the commuting set
of observables.
But this time, you have
a different problem.
Yes, this commutes with h.
This commutes with h.
And this commutes with h.
But these one's don't
commute with each other.
So not quite good enough.
You cannot add them all.
So let's see how
many can we add.
We can only add 1.
Because once you have 2 of
them, they don't commute.
So you're going to add 1, and
everybody has agreed to add L3.
So we have H, L3.
And happily we have
1 more is L squared.
Remember, L squared commutes
with all the Li's, so that's
another operator.
And for a central
potential problem,
this will be sufficient to
label all of our states some.
AUDIENCE: So how do we know
that we need the L squared?
How do we know
that we can't get--
how do we know that just
H and L3 isn't already
a complete set?
PROFESSOR: I probably wouldn't
know now, but in a little bit,
as we calculate
the kind of states
that we get with
angular momentum,
I will see that there are many
states with the same value
of L3 that don't correspond
to the same value of the total
or length of the
angular momentum.
So it's almost like saying
that there are angular
momenta-- here is--
let me draw a plane.
Here is z component of
angular momentum, Lz.
And here you got it.
You can have an angular
momentum that is like that,
and has this Lz.
Or you can have an
angular momentum
that is like this, L prime,
that has the same Lz.
And then it will be difficult
to tell these 2 states apart.
And they will correspond
to states of this angular
momentum, or this angular
momentum, have the same Lz.
Now drawing these arrows is
extraordinarily misleading.
Hope you don't get
upset that I did it.
It's misleading because this
vector you cannot measure
simultaneously the 3 components.
Because they don't commute.
So what do I mean
by drawing an arrow?
Nevertheless, the
intuition is sort of there.
And it's not wrong,
the intuition.
It will happen to be
the case that states
that have same amount of Lz
will not be distinguished.
But by the time we have this,
we will distinguish them.
And that's also a peculiarity
of a result but we'll use.
Even though we're talking
about 3 dimensions,
the fact that the 1 dimensional
Schrodinger equation
has non degenerate bound states.
You say, what does that have
to do with 3 dimensions?
What will happen is that the
3 dimensional Schrodinger
equation will reduce to a 1
dimensional radial equation.
And the fact that that
doesn't have degeneracies
tells you that for
bound state problems,
this will be enough to do it.
So you will have
to wait a little
to be sure that this will do it.
But this is pretty much
the best we can do now.
And I don't think you will
be able to add anything
else to this at this stage.
Now there's of
course funny things
that you could add
like-- if there's spin,
the particles have spin, well
we can add spin and things
like that.
But let's leave
it at that and now
begin really our calculation,
algebraic calculation,
of the angular momentum
representations.
So at this moment,
we really want
to make sure we work with this.
Only this formula over here.
And learn things about
the kind of states
that can exist in a
system in which there
are operators like that.
So it's a funny thing.
You're talking about
a vector space.
And in fact, you don't
know almost anything
about this vector space so far.
But there is an action
of those operators.
From that fact alone, and
one more important fact--
the j's are Hermitian.
From these 2 facts,
we're going to derive
incredibly powerful results,
extremely powerful things.
And as we'll see, they
have applications even
in cases that you would
imagine they have nothing
to do with angular momentum,
which is really surprising.
So how do we proceed
with this stuff?
Well, there's a hermeticity.
And you immediately
introduce things
called J plus minus, which
are J1 plus minus i J 2.
Or Jx plus minus y Jc.
Then you calculate
what is J plus J minus.
Well J plus J minus will be
a J1 squared plus J2 squared.
And then you have
the cross product
that this doesn't cancel.
So J plus times J minus would
be J1 plus i J2, J1 minus i J2.
So the next term would
be minus i, J1, J2.
And that's i h bar, J3.
So this is J1 squared plus
J2 squared plus h bar J3.
So that's a nice formula
for J plus, J minus.
J minus, J plus would
be J1 squared plus J2
squared minus h bar J3.
These 2 formulas are summarized
by J plus, J minus-- minus,
plus-- is equal to
J1 squared plus J2
squared plus minus h bar J3.
OK.
Things to learn from this.
Maybe I'll continue
here for a little while
to use the blackboards,
up to here only.
The commutator of
J plus and J minus
can be obtained
from this equation.
You just subtract them.
And that's 2h bar, J3.
And finally, one last
thing that we like to know
is how to write J squared.
So J squared is J1 squared
plus J2 squared plus J3
squared, which
then show up here.
So we might as well
add it and subtract it.
So I add a J3 squared, and I
add it on the left hand side.
And pass this term
to the other side.
So J squared would be J plus,
J minus, plus J3 squared,
minus h bar, J3.
Or J minus, J plus, plus
J3 squared, plus h bar, J3.
OK.
So that's J squared.
OK.
So we're doing sort
of simple things.
Basically at this
moment, we decided
that we like better
J plus and J minus.
And we tried to
figure out everything
that we should know
about J plus, J minus.
If we substitute Lx, and Jx,
and Jy for J plus and J minus,
you better know what
is the commutator
of J plus and J minus.
And how to write J squared in
terms of J plus and J minus.
And this is what
we've done here.
And in particular, we have a
whole lot of nice formulas.
So one more formula
is probably useful.
And it's the formula for the
commutator of J plus and J
minus with Jz.
Because after all, the J
plus, J minus commutator,
you've got it.
So if you're systematic
about these things
you should figure
out that at this I
would like to know what is
the commutator of J plus and J
minus with Jz.
So I can do Jz, J plus.
It's not hard.
It's Jz.
I'm sorry.
I'm calling it 3.
So, I think in the notes
I call them x, y, and z.
But never mind.
J1 plus i, J2.
The plus is really
with a plus i.
So J3 with J1 by the cyclic
ordering is ih bar, J2.
And here you have plus i, and
J3 with J2 is minus ih bar, J1.
So this is h bar, J1, plus i,
J2, which is h bar, J plus.
So what you've learned
is that J3 with J plus
is equal to h bar, J plus.
And if you did it
with J minus, you'll
find a minus, and
a plus minus here.
So that is the complete result.
And that should remind you
of the analogous relation
in which you have in the
harmonic oscillator, N
commutator, with a dagger.
With a dagger.
And N commutator
with a was minus a.
Because of the fact that I
maybe didn't say it here,
and I should have, that the
dagger of J plus is J minus.
Because the operators
are Hermitians.
So J plus and J minus are
daggers of each other,
are adjoins of each other.
And here you see a very
analogous situation.
a and a dagger were
adjoins of each other.
And with respect to N, a
counting number operator.
One increased it.
One decreased it.
a dagger increased the
number eigenvalue of N. a
decreased it, the same way
it's going to happen here.
J plus is going
to increase the C
component of angular momentum.
And J minus is going
to decrease it.
OK.
So we've done most of the
calculations that we need.
The rest is pretty easy work.
Not that it was
difficult so far.
But it took a little time.
So what happens next
is the following.
You must make a declaration.
There should exist
states, basically.
We have a vector space.
It's very large.
It's actually
infinite dimensional.
Because they will be related
to all kinds of functions
on the unit sphere.
All these angular variables.
So it's infinite dimensional.
So it's a little scary.
But let's not worry about that.
Something very nice happens
with angular momentum.
Something so nice that
it didn't happen actually
with a and a dagger.
With a and a dagger,
you build states
in the harmonic oscillator.
And you build
infinitely many ones.
The operators x and p, you've
learned you cannot represent
them by finite
dimensional matrices.
So this is a lot more
complicated, you would say.
And you would say, well,
this is just much harder.
This algebra is so much
harder than this algebra.
Nevertheless, this algebra
is the difficult one.
Gives you infinite
dimensional representations.
You can keep piling
the a daggers.
Here, this is a
very dense algebra.
Mathematicians would say this
is much simpler than this one.
And we'll see the simplicity
of this one, in that you
will manage to get
representations
and matrices that are
finite dimensional
to work these things out.
So it's going to be
nicer in that sense.
So what do we have?
We have to think of our
commuting observables
and the set of Hermitian
operators that commute.
So we have J squared, and J3--
I call it Jz now, apologies.
And we'll declare
that there are states.
These are Hermitian,
and they commute.
So they must be
diagonalized simultaneously.
And there should
exist states that
represent the diagonalization.
In fact, since they commute,
and can be diagonalized
simultaneously, the
vector space must
break into a list of vectors.
All of them eigenstates
of these 2 operators.
And all of them
orthogonal to each other.
Matthew, you had a question?
AUDIENCE: I was
just wondering when
we showed that Jz is Hermitian?
PROFESSOR: We didn't show it.
We postulated that J's
are Hermitian operators.
So you know that when J
is L, yes it's Hermitian.
You know when J is spin,
yes it's Hermitian.
Whatever you're doing we'll
use Hermitian operators.
So not only they can
diagonalize simultaneously,
by our main theorem about
Hermitian operators,
this should provide
an orthonormal basis
for the full vector space.
So the whole answer is
supposed to be here.
Let's see.
So I'll define
states, Jm, that are
eigenstates of both
of these things.
And I have 2 numbers to
declare those eigenvalues.
You would say J squared.
Now, any normal person
would put here maybe h
squared, for units,
time J squared.
And then Jm.
Don't copy it yet.
And Jz for Jm.
It has units of
angular momentum.
So an h, times m, times Jm.
But that turns
out not to be very
convenient to put
the J squared there.
It ruins the algebra later.
So we'll put something
different that we
hope has the same effect.
And I will discuss that.
I'll put h squared,
J times J plus 1.
It's a funny way
of declaring how
you're going to
build the states.
But it's a possible thing to do.
So here are the states, J and m.
And the only thing I
know at this moment
is that since these are
Hermitian operators,
their eigenvalues must be real.
So J times J plus 1 is real.
And m is real.
So J and m belong to the reals.
And they are
orthogonal to-- we can
say they're orthonormal states.
We will see very soon that
these things get quantized.
But basically, the overlap of
a Jm with a J prime, m prime
would be 0 whenever the J's
and the m's are different.
As you know from our
theory, any 2 eigenstates
with different eigenvalues
are orthonormal.
And in fact, you
can choose a basis
so that in fact,
everything is orthonormal.
So there's no
question like that.
So let's explain a little what's
happening with this thing.
Why do we put this like that?
Or why can we get
away with this?
And the reason is the following.
Let's consider
Jm, J squared, Jm.
If I use this, J squared
on this is this number.
And Jm with itself will be 1.
And therefore I'll put
here h-- I'm sorry.
This should be an h squared.
J has units of angular momentum.
h squared, J times J plus 1.
And I'm assuming that
this will be discretized
so I don't have to put the
delta function normalization.
At any rate, this
thing is equal to this.
And moreover, it's
equal to the following.
Jm sum over i, Jm, Ji, Ji, Jm.
But since J is Hermitian, this
is nothing but the sum over i
of the norm squared of
Ji with J acting on Jm.
The norm squared of this state.
Because this times the
bra with Ji Hermitian
is the norm squared.
So this is greater
or equal than 0.
Perhaps no surprise, this
is a vector operator,
which is the sum of squares
of Hermitian operators.
And therefore it
should be like that.
Now, given that, we have
the following-- oops--
the following fact that
L times L plus-- no.
J times J plus 1 must be
greater or equal than 0.
J times J plus 1 must be
greater or equal than 0.
Well, plot it as a function
of J. It vanishes at 0.
J times J plus 1 vanishes at
0, and vanishes at minus 1.
It's a function like this.
The function J times J plus 1.
And this shows that all you need
is this thing to be positive.
So to represent all the states
that have J times J plus 1
positive, I could label them
with J's that are positive.
Or J's that are
smaller than minus 1.
So each way, I can label
uniquely those states.
So if I get J times
J plus 1 equals 3,
it may correspond
to a J of something
and a J of some other thing.
I will have just 1 state,
so I will choose J positive.
So given that J times
J plus 1 is positive,
I can label states
with J positive, or 0.
So it allows you to do this.
Whatever value of this
quantity that is positive
corresponds to some J positive
that you can put in here.
A unique J positive.
So this is a fine
parametrization of the problem.
OK.
Now what's next?
Next, we have to understand
what the J plus operators and J
minus operators
do to the states.
So, first thing is that J plus
and J minus commute with J
squared.
That should not be a surprise.
J1 and J2 commute.
Every J commutes with J squared.
So J plus and J minus
commute with J squared.
What this means in words
is that J plus and J minus
do not change the eigenvalue
of J squared on a state.
That is, if I would have J
squared on J plus or minus
on Jm-- since I can move the J
squared up across the J plus,
minus-- it hits here.
Then I have J plus
minus, J squared, Jm.
And that's there for h
squared, J times J plus 1,
times J plus minus on Jm.
So this state is also a
state with the same value
of J squared.
Therefore, it must have the
same value of J. In other words,
this state J plus minus
of Jm must be proportional
to a state with J and maybe
some different value of m,
but the same value of J.
J cannot have changed.
J must be the same.
Then we have to
see who changes m,
or how does J plus
minus changes m.
So here comes a little
bit of a same calculation.
You want to see what is
the m value of this thing.
So you have J plus minus on Jm.
And you act with it with
a Jz, to see what it is.
And then, you put, well,
the commutator first.
Jz, J plus minus, plus J
plus minus, Jz on the state.
The commutator,
you've calculated it
before, was Jz with J
plus minus is there,
is plus minus h
bar, J plus minus.
And this Jz already act.
So this is plus h bar
m, J plus minus on Jm.
So we can get the
J plus minus out.
And this h bar m plus
minus 1, j plus minus, Jm.
So look what you got.
Jz acting on this state is
h bar, m plus minus 1, Jm.
So this state has m equal to
either m plus 1, or m minus 1.
Something that we can write.
Clearly-- oops--
in this way, we'll
say that J plus minus,
Jm-- we know already
it's a state with J
and m plus minus 1.
So it raises m.
Just like what we said
that the a's and a daggers
raise or lower the number.
J plus and J minus
raise and lower Jz.
Therefore, it's this is
proportional to this state.
But there's a constant
of proportionality
that we have to figure out.
And we'll call it
the constant C, Jm.
To be calculated.
So the way to calculate
this constant-- and that
will bring us almost pretty
close to what we need--
is to take inner products.
So we must take the
dagger of this equation.
So take the dagger, and you get
Jm, the adjoin, J minus plus.
And hit it with this equation.
So you'll have here--
well maybe I'll write it.
The dagger of this
equation would
be C plus minus star of Jm.
Jm plus minus 1.
And now, sandwich
this with that.
So you have Jm, J minus
plus, J plus minus,
Jm equals to norm
of C plus minus Jm.
And then you have this state
times this state, but that's 1.
Because it's J, J,
m plus 1, m plus 1.
So this is an orthonormal basis.
So we have just 1.
And I don't have to write more.
Well the left hand
side can be calculated.
We have still that formula here.
So let's calculate it.
The left hand side,
I'll write it like this.
I will have C plus
minus, Jm squared,
which is equal to the norm
squared of J plus minus, Jm.
It's equal to what?
Whatever this is,
where you substitute
that for this formula.
So you'll put here Jm.
And you'll have--
well, I want actually
the formula I just erased.
Because I actually would
prefer to have J squared.
So I would have this is equal
to J squared, minus J3 squared,
plus minus h, J3.
So let's see.
I have the sign minus
plus, plus minus.
So I should change
the signs there.
So it should be J squared,
minus J3 squared, minus plus J3,
and Jm, minus plus
h bar, J3, Jm.
So this is equal to h bar
squared, J times J plus 1,
minus an m squared,
and a minus plus.
So minus, plus, minus here.
I think I have it here correct.
Plus minus 1.
And that's it.
J squared is h squared this.
J3 squared would give that.
And the minus plus here is
correctly with this one.
So m should be here.
Plus minus m.
So this is h squared,
J times J plus 1,
minus m, times m plus minus 1.
OK.
So the C's have
been already found.
And you can take
their square roots.
In fact, we can ideally
just take the square roots,
because these things
better be positive numbers
because they're norms squared.
So whenever we'll
be able to do this,
these things better
be positive, being
the square of some states.
And therefore the C plus
minus is-- C plus minus of Jm
can be simply taken to be
h bar, square root of J
times J plus 1, minus
m, times m plus 1.
And it's because of
this thing, this m times
m plus 1, that it was convenient
to have J times J plus 1.
So that we can compare
J's and m's better.
Otherwise it would have
been pretty disastrous.
So, OK, we're almost done
now with the calculation
of the spectrum.
You will say, well, we seem
to be getting no where.
Learned all these
properties, these states,
and now you're just
manipulating the states.
But the main thing
is that we need
these things to be positive.
And that will give us
the whole condition.
So, for example, we need
1, that the states J
plus, Jm, their norm
squareds be positive.
So for the plus sign-- so you
should have J times J plus 1,
minus m, times m
plus 1 be positive.
Or m times m plus 1 be
smaller then J times J plus 1.
The best way for my mind to
solve these kind of things
is to just plot them.
So here is m.
And here is m times m plus 1.
So you plot this function.
And you want it to be less than
some value of J times J plus 1.
So here's J times J
plus 1, some value.
So this is 0 here.
This function is 0 at minus 1.
So it will be
something like this.
And there's 2 values at which
m becomes equal to this thing.
And one is clearly J.
When m is equal to J,
it's saturates an inequality.
And the other one
is minus J, minus 1.
If m is minus J,
minus 1, you will
have minus J, minus
1 here, and minus J
here, which would
be equal to this.
So, in order for these states
to be good, the value of m
must be in between J
and minus J, minus 1.
Then the other case is that
J plus on-- J minus on Jm.
If you produce those
states, they also
must have positive norms.
So J times J plus 1, minus
m, times m minus 1 this time,
must be greater than 0.
So m times m minus
1 must be less than
or equal then J times J plus 1.
And again, we try to
do it geometrically.
So here it is.
Here is m.
And what values do you have?
Well, if you plot here
m times m minus 1.
And that should be
equal to some value
that you get fixed, which is
the value J times J plus 1.
So you think in terms of
m's, how far can they go?
So if you take m equals
J plus 1 that hits it.
So this is 0 here, at 1, at 0.
So it's some function like this.
And here you have J plus 1.
And here you have
minus J. Both are
the places for m equal
J plus 1, and minus J
that you get the states.
You get the saturation.
So you can run m
from this range.
Now, m can go less than
or equal to J plus 1,
and greater than or
equal to minus J.
But these 2 inequalities
must hold at the same time.
You cannot allow either one to
go wrong for any set of states.
So if both must hold at the
same time for any state,
because both things have to
happen, you get constrained.
This time for the upper range,
this is the stronger value.
For the lower range, this
is the stronger value.
So m must go between J and
minus J for both to hold.
Oops.
To hold.
Now look what happens.
Funny things happen
if-- this is reasonable
that the strongest value
comes from this equation.
Because J plus increases m.
So at some point
you run into trouble
if you increase m too much.
How much can you increase it?
You cannot go beyond J,
and that makes sense.
In some sense, your
intuition should
be that J is the
length of J squared.
And m is mz.
So m should not go beyond J.
And that's reasonable here.
And in fact, when m is equal to
J, this whole thing vanishes.
So if you reach that state
when m is equal to J,
only then for m equal to J,
or for this state, you get 0.
So you cannot raise
the state anymore.
So actually, you see if you
choose some J over here,
we need a few things to happen.
You choose some J, and some m.
Well you're going to
be shifting the m's.
And if you keep adding
J pluses, eventually you
will go beyond this point.
The only way not to
go beyond this point
is if m reaches the value J.
Because if m reaches the value
J, the state is killed.
So m should reach the value
J over here at some stage.
So you fix J, and you try
to think what m can be.
And m has to reach
the value J. So
m at some point,
whatever m is, you add 1.
You add 1.
You add 1.
And eventually you must
reach the value J. Reach
with some m prime.
m here.
You should reach the
value J, so that you
don't produce another
state that is higher.
If you reach something before
that, that state is not killed.
This number is not equal to 0.
You produce a state and it's
a bad state of bad norm.
So you must reach this one.
On the other hand,
you can lower things.
And if you go below minus
J, you produce bad states.
So you must also,
when you decrease m,
you must reach this point.
Because if you didn't, and you
stop half a unit away from it,
the next state that
you produce is bad.
And that can't be.
So you must reach this one too.
And that's the key logical
part of the argument
in which this distance
2J plus 1-- no.
I'm sorry.
This 2J must be equal
to some integer.
And that's the key
thing that must happen,
because you must reach this
and you must reach here.
And m just varies by integers.
So the distance between
this J and minus J
must be twice an integer.
And you've discovered
something remarkable by getting
to that point,
because now you see
that if this has to
be an integer, well
it may be 0, 1, 2, 3.
And when J-- then J-- this
integer is equal to 0,
then J is equal to 0.
1/2, 1, 3/2.
And you get all these spins
with-- consider particles
without spin having spin 0.
Particles with spin 1/2.
Particles of spin 1,
or angular momentum 1,
orbital angular momentum 1.
And both these things
have a reason for you.
Now if you have 2J being
an integer, the values of m
go from J to J minus
1, up to minus J.
And there are two
J plus 1 values.
And in fact, that
is the main result
of the theory of
angular momentum.
The values of the angular
momentum are 0, 1, 1/2, 3/2.
So for J equals 0,
there's just one state.
m is equal to 0.
For J equals to 1,
there's two states.
I'm sorry for 1/2, two states.
One with m equals 1/2.
And m equals minus 1/2.
J equals 1, there's
three states.
M equals 1, 0, and minus 1.
And so on.
OK.
This is a great result.
Let me give you an application
in the last 10 minutes.
It's a remarkable application.
Now actually, you
would say, so what
do you get-- what vector
space were we talking about?
And what's sort of
the punchline here
is that the vector space
was infinite dimensional
and it breaks down into
states with J equals 0.
States was J equal 1/2.
States with J equal 1.
States with J equal 3/2.
All these things
are possibilities.
They can all be present
in your vector space.
Maybe some are present.
Some are not.
That is part of figuring
out what's going on.
When we do central
potentials, 0, 1, 2, 4
will be present for the
angular momentum theory.
When we do spins, we have 1/2.
And when we do other things,
we can get some funny things
as well.
So let's do a case where
you get something funny.
So the 2D, SHO.
You have ax's, and ay's, and
a daggers, and ay daggers.
And this should
seem very strange.
What are we talking about
2 dimensional oscillators
after talking about
3 dimensional angular
momentum and all that?
Doesn't make any sense.
Well, what's going to happen
now is something more magical
than when a magician takes
a bunny out of a hat.
Out of this problem, an angular
momentum, a 3 dimensional
angular momentum,
is going to pop out.
No reason whatsoever there
should be there at first sight.
But it's there.
And it's an abstract
angular momentum,
but it's a full
angular momentum.
Let's see.
Let's look at the spectrum.
Ground state.
First excited
state is isotropic.
So 2 states
degenerate in energy.
Next state.
ax dagger, ax dagger.
ax, ay.
ay, ay.
3 states, degenerate.
Go up to ax dagger
to the n, up to ax--
no ax, or ax dagger to the 0.
And ay dagger to the n.
And that's n a daggers up to 0
a daggers, so n plus 1 states.
3 states, 2 states, 1 state.
And you'll come here
and say, that's strange.
1 state, 2 states,
3 states, 4 states.
Does that have
anything to do with it?
Well, the surprise is it
has something to do with it.
Let's think about it.
Well, first thing is to put
these aR's and aL oscillators--
these were 1/2, 1 over square
root of 2, ax plus iay.
And a left was 1 over square
root of 2, ax minus iay.
I may have-- no,
the signs are wrong.
Plus and minus.
And we had number operators.
n right, which were a
right dagger, a right.
And n left, which was
a left dagger, a left.
And they don't mix a
lefts and a rights.
And now, we could build a
state the following way.
0.
a right dagger on 0.
a left dagger on 0.
A right dagger squared on 0.
a right, a left on 0.
and a left dagger,
a left dagger on 0.
Up to a right dagger
to the n on 0.
Up to a left dagger
to the n on 0.
And this is completely
analogous to what we had.
Now here comes the real thing.
You did compute the angular
momentum in the z direction.
And the angular momentum
in the z direction was Lz.
And you could compute
this. xpy minus ypx.
And this was all legal.
And the answer was h
bar, N right, minus NL.
That was the Lz component
of angular momentum.
So, let's see what
Lz's those states have.
This one has no n rights, or
n lefts, so has Lz equals 0.
This state has Nz equal h bar.
And this has minus h bar.
OK.
h bar and minus h bar.
That doesn't quite
seem to fit here,
because the z component
of angular momentum
is 1/2 of h bar, and
minus 1/2 of h bar.
That's-- something went wrong.
OK.
You go here.
You say, well, what is Lz?
Lz here was h bar, minus h bar.
Here is 2h bar, 0,
and minus 2h bar.
And you look there, and say, no,
that's not quite right either.
This-- if you would
say these 3 states
should correspond
to angular momentum,
they should have m equal plus 1,
plus h bar, 0, and minus h bar.
So it's not right.
OK.
Well one other thing maybe
we can make sense of this.
If we had L plus, should
be the kind of thing
that you can't annihilate.
That you annihilate
the top state.
Remember L plus, or J
plus, kept increasing
so it should annihilate
the top state.
And I could try to
devise something
that annihilates the top state.
And it would be something
like aR dagger, a left.
Why?
Because if aR dagger, a
left, goes to the top state,
the top state has
no a left daggers,
so the a left just zooms in,
and hits the 0 and kills it.
Kills it here.
So actually I do have
something like an L plus.
And I would have the
dagger-- would be something
like an L minus-- would
be aL dagger, a right.
And this one should
annihilate the bottom one.
And it does.
Because the bottom
state has no aR's,
and therefore has no aR daggers.
And therefore, the aR comes
there, and hits the state,
and kills it.
So we seem to have more
or less everything,
but nothing is working.
So we have to do a
last conceptual step.
And say-- you see, this
is moving in a plane.
There's no 3 dimensional
angular momentum.
You are fooling
yourself with this.
But what could exist is an
abstract angular momentum.
And for that, in
order to-- it's time
to change the
letter from L to J.
That means some kind of
abstract angular momentum.
And I'll put a 1/2
here, now a definition.
If this is what I called Jz,
oh well, then thing's may
look good.
Because this one for Jz has now
angular momentum 1/2 of h bar,
and minus a half of h bar.
And that fits with
this, these 2 states.
And with the 1/2, the
other ones, the Jz's, also
have something here.
So Jz here now becomes
h bar, minus h bar,
and it looks right.
And now you put the
1/2 here, and in fact,
if you tried to
make these things
J-- call it J plus and J minus.
Now you put a number
here, and a number here.
If you would have
put a number here,
if you try to enforce
that the algebra be
the algebra of angular
momentum, the number
would have come out to be 1/2.
But now we claim that in this
2 dimensional oscillator,
there is-- because
there's a number here
that works with this 1/2.
Something you have to calculate.
And with this number, you
have some sort of Jx, Jy, Jz,
where this is like 1/2 of Lz.
And those have come
out of thin air.
But they form an algebra
of angular momentum.
And what have we
learned today, if you
have an algebra of
angular momentum,
the states must
organize themselves
into representations
of angular momentum.
So the whole spectrum of the 2
dimensional harmonic oscillator
has in fact all spin
representations.
J equals 0.
J equals 1/2.
J equals 1.
J equals 2.
J equals n, and all of them.
So the best example of
all the representations
of angular momentum
are in the states
of the 2 dimensional
simple harmonic oscillator.
It's an abstract angular
momentum, but it's very useful.
The one step I didn't do
here for you is to check.
Although you check that
all of these Ji commute
with the Hamiltonian.
Simple calculation to do it.
In fact, the Hamiltonian
is NL plus N right,
and you can check it.
Since they commute with them,
these operators act in states
and don't change the energy.
And they're a symmetry
of the problem.
So that's why they fell
into representations.
So this is our first example
of a hidden symmetry.
A problem that there
was no reason a priori
to expect an angular
momentum to exist,
but it's there, and helps
explain the degeneracies.
These degeneracies you could
have said they're accidental.
But by the time
you know they have
to fall into angular
momentum representations,
you have great
control over them.
You couldn't have
found different number
of degenerate states
at any level here.
This was in fact discovered
by Julian Schwinger
in a very famous paper.
And is a classic example
of angular momentum.
All right.
That's it for today.
See you on Wednesday if
you come I'll be here.
