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ROBERT FIELD: Now the
main topic of this lecture
is so important and
so beautiful that I
don't want to spend any time
reviewing what I did last time.
At the beginning when we
talked about the rigid rotor,
I said that this is not just a
simple, exactly solved problem
but it tells you
about the angular part
of every central force problem.
And it's even more than that.
It enables you to do a
certain kind of algebra
with operators which
enables you to minimize
the effort of calculating
matrix elements
and predicting
selection rules simply
by the basis of commutation
rules of the operators
without ever looking
at wave functions,
without ever looking at
differential operators.
This is a really beautiful
thing about angular momentum
that if we define the angular
momentum in this abstract way--
and I'll describe what I
mean by this epsilon ijk.
If we say we have an operator
which obeys this commutation
rule, we will call it
an angular momentum.
And we go through
some arguments and we
discover the properties of all
"angular momentum," in quotes.
Now we define an angular
momentum classically
as r cross p.
It's a vector.
Now there are things, operators,
where there is no r or p
but we would like to describe
them as angular momentum.
One of them is electron spin,
something that we sort of take
for granted, and nuclear spin.
NMR is based on these things
we call angular momentum
because they obey some rules.
And so what I'm going
to do is show the rules
and show where all
of this comes from
and that this is an abstract
and so kind of dry derivation,
but it has astonishing
consequences.
And basically it
means that if you've
got angular momenta, if
you know these rules,
you're never going to
evaluate another matrix
element in your life.
Now, it has another
level of complexity.
Sometimes you have
operators that
are made out of combinations
of angular momenta,
and you can use these
sorts of arguments
to derive the matrix
elements of them.
That's called the
Wigner-Eckart theorem,
and it means that the angular
part of every operator
is in your hands without ever
looking at a wave function
or a differential operator.
Now we're not going
to go there, but this
is a very important area
of quantum mechanics.
And you've heard of three j
symbols and Racah coefficients.
Maybe you haven't.
But there is just
a rich literature
of this sort of stuff.
So today I'm going
to talk briefly
about rotational spectra
because I'm a spectroscopist.
And from rotational spectra we
learn about molecular geometry.
Now it's really strange
because why don't we just look
at a molecule and measure it?
Well, we can't
because it's smaller
than the wavelength
of light that we
would use to
illuminate our ruler,
and so we get the
structure of a molecule,
the geometric structure,
at least part of it,
from the rotational spectrum.
Now I'm only going to talk
about the rotational spectrum
of a diatomic molecule.
You don't want me to go further
because polyatomic molecules
have extra complexity
which you could understand,
but I don't want to go
there because we have
a lot of other stuff to do.
I also had promised to
talk about visualization
of wave functions,
and I'll leave that
to your previous experience.
But I do want to comment.
We often take a sum
of wave functions
for positive and negative
projection quantum numbers
to make them real
or pure imaginary.
When we do that, they are still
eigenfunctions of the angular
momentum of squared,
but they're not
eigenfunctions of a projection.
And you could-- in fact,
maybe you will on Thursday--
actually evaluate Lz times
a symmetrized function.
So by symmetrizing them, you
get to see the nodal structure,
which is nice, but
you lose the fact
that you have eigenfunctions
of a projection of angular
momentum, and
that's kind of sad.
OK, so spectra-- so we have
a diatomic molecule, mA,
mB, and we have a center mass.
And so we're going to be
interested in the energy
levels for a rotation of the
molecule around that axis which
is perpendicular
to the bond axis.
When we do that, we discover
that the energy levels
are given by the
rotational Hamiltonian.
And for a rotation--
it's free rotation, so
there's no potential.
And the operator is L
squared or J squared.
That's another thing.
You already have experienced
my use of L and J and maybe
some other things.
They're all angular momentum.
They're all the
same sort of thing,
although L is usually referring
to electronic coordinates,
and J is usually referring
to nuclear coordinates.
Big deal.
But you get a sense
that we're talking
about a very rich idea
where it doesn't matter
what you name the things.
They follow the same rules.
So we have a single term in
the Hamiltonian, mu r0 squared,
or this might be the
equilibrium instead of
just the fixed
internuclear distance.
But we're talking
about a rigid rotor,
so r0 is the
internuclear distance.
And now I want to
be able to write--
OK, so I want to
have this quantity.
I want to have this quantity
in reciprocal centimeter units
because that's what
all spectroscopists do,
or sometimes they use megahertz.
In that case, the
speed of light is gone.
And when I evaluate the effect
of this operator on a wave
function, we get an h bar
squared, which cancels that.
We would like to have an
energy level expression
EJM is equal to hcBL L plus 1.
So the units of B just
accommodate the fact
that we want it in wave numbers.
But this is energy,
So we need the hc.
And when the operator operates,
we get an h bar squared,
and that's canceled
by this factor here.
And so the handy
dandy expression
for the rotational
constant is 16.85673 times
the reduced mass in AMU
units times the internuclear
distance in angstrom
units squared reciprocal.
So if you want to
know the energy--
if you want to know
the rotational constant
in wave number units,
this is the conversion.
Big deal.
So the energy
levels are simply--
I'm going to stick with LM even
though I'm hardwired to call it
J. Now if I go back and
forth between J and L,
you'll have to forgive
me because I just can't--
yes.
All right, so we have the energy
levels, hcB times L L plus 1.
Now L is an integer,
and for the simple
diatomics that you're going
to deal with, it's an integer.
You can start at zero.
And so the energy levels, the L
L plus 1, the L L plus 1 is 2.
I want to make this look right.
B-- this is 1 times 2.
This is 2 times 3.
And the 3 times 4 is 12.
And the important thing is that
this energy differences is 2B.
This energy difference is 4B.
This energy difference is 6B.
And so what happens
in the spectrum--
here's energy.
Here's zero.
We have a line here at 2B,
a line here at 4B, 6B, 8B.
So if you were able to look
at the rotational spectrum,
the lines in the spectrum
would be evenly spaced.
The levels are not.
That's very
important, especially
when you start doing
perturbation theory
because you're going to have
energy denominators which
are multiples of
a common factor,
but they're not
equal to each other.
But we have a spectrum, and it
looks really, really trivial.
And textbooks don't
talk about this,
but if you have a relatively
light diatomic molecule
and you have a laboratory which
is equipped with a microwave
spectrometer which is able
to generate data that got you
tenure and whatever,
it's probably
a spectrometer where the
tuning range of the microwave
oscillator is about 30%.
That's a lot.
If you think about NMR,
the tuning range is--
30% is huge.
So, you see this and
you say, oh yeah.
I could assign that spectrum
because an obvious pattern.
But what happens in the
spectrum is you get one line.
And so you say, well, I need
to know the internuclear
distance of this molecule
to 6 or 8 or 10 digits,
but I get one line.
There's no pattern.
The textbooks are
so full of formulas
that they don't indicate
that, in reality, you've
got a problem.
And, in fact, in
reality you've got
something that's also a gift.
So there's two things that
happen, isotopes and vibration.
So we have this one line.
We have a very, very
strong, very narrow--
you can measure the daylights
out of it if you wanted to.
And then down here,
there's going to be--
well, actually, sometimes
like in chlorine and bromine,
there's a heavy isotope
and a light isotope,
and they have
similar abundances.
And so you get
isotope splittings,
and that's expressed
in the reduced mass mA
mB over mA plus mB.
Now the isotope splittings
can be really, really small,
but these lines have a width
of a part in a million,
maybe even narrower.
And so you can
see isotope stuff.
That doesn't tell you
anything at all that
you didn't know except
maybe that you were confused
about what molecule it
was because if you have
a particular atom,
it's always born
with the normal isotope ratios.
Except here we have
a little problem
where, in sulfur, if
you look in minerals,
the isotope ratios are
not the naturally abundant
of sulfur isotope.
And this has to do with
something really important that
happened 2 and 1/2
billion years ago.
Oxygen happened.
And so isotope ratios are
of some geological chemical
significance, but here, if
you know what the molecule is,
there will be isotope lines.
And they can be pretty
strong depending
on the relative abundance
of the different isotopes,
or they can be extremely weak.
So there's stuff, so some grass
to be mowed on the baseline.
In addition-- and
this is something
that really surprises people.
So here is v equals 0, and
way up high is v equals 1.
Typically, the
vibrational intervals
are on the order of a
thousand times bigger
than the rotational intervals.
And typically, the
rotational constant
decreases in steps
of about a tenth
of a percent per vibration.
Now we do care about how much
it decreases because that allows
us to know a whole
bunch of stuff
about how rotation and
vibration interact.
And I'm not probably going to
do the lecture on the rotation
and vibration
interaction unless I
have to give a lecture on
something that I can't do
and I'll slip in that one.
So what happens is there
are vibrational satellites.
So here's v equals 0.
It has rotational structure.
And here is v equals 1.
It has rotational structure.
The v equals 1 stuff
is typically a hundred
to a thousand times weaker
than the v equals 0 stuff.
And that's basically
telling you,
how does a molecule
changes its average 1
over r squared as it vibrates,
and that's a useful thing.
It may even be
useful on Thursday.
So in addition to hyperfine,
there's other small stuff
having to do with vibrations.
And in some
experiments that I do,
we use UV light to
break a molecule.
And the fragments that we make
are born vibrationally excited.
And so by looking at the stuff
near the v equals 0 frequency,
you see a whole
bunch of stuff which
tells you the populations of the
different vibrational levels.
And that's strange because
vibration is not part
of the rotational spectrum.
Vibration is big, but we
get vibrational information
from the rotational spectrum.
And because the
rotational spectrum
is at such high resolution,
it's trivial to resolve
and to detect these
weak other features.
So as much as I'm going to
talk about spectroscopy,
it's a little bit more than
I had originally planned.
And now we're going to move
to this topic which is dear
to my heart, and it's an
example of an abstract algebra
that you use in
quantum mechanics.
And there are people who
only do this kind of thing
as opposed to solving
Schrodinger equation
or even just doing perturbation
theory on matrices.
So the rest of
today's lecture is
going to be an excursion through
here as much as I can do.
It's all clear in the
notes, but I think
it's a little bit strange.
Oh, I want to say
one more thing.
How do we make assignments?
You all took 5.111
or 5.112 or 3.091,
and there are things that you
learn about how big atoms are.
And so you can sort
of estimate what
the internuclear distance is--
maybe to 10% or 20%.
That's not of any
chemical use, but it's
enough to assign the spectrum.
So what you do is you say OK, I
guess the internuclear distance
is this.
That determines what rotational
transition you were observing.
And that has
consequences of suppose
you're observing L to L plus 1.
Well, what about L plus to
L plus 2 or L minus 1 to L?
So if you make an
assignment, you
can predict where
the other guys are.
And that would require
going to one of your friends
who has a different
spectrometer and getting
him to record a
spectrum for you,
and that's good for
human relations.
And that then enables you
to make assignments and know
the rotational constants
to as many digits
as you possibly could want,
includ-- all the way up to 10.
It's just crazy.
You really don't care about
internuclear distances
beyond about a thousandth
of an angstrom,
but you can have them.
So first of all, you know
we can define an angular
momentum as r cross p, and we
can write that as a matrix.
Now I suspect you've
all seen this.
These are unit vectors along
the x, y, and z directions.
And this is a vector, so
there are three components,
and we get three
components here.
Now you do want to make
sure you know this notation
and know how to use it.
So here is the magic equation.
Li, Lj is equal to ih bar
sum over k epsilon ijk Lk.
Well, what is epsilon ijk?
Well, it's got many
names, but it's
a really neat tool which is
very wonderful in enabling
you to derive new equations.
So if i, j, and k correspond
to xyz in cyclic order--
in other words, xyz,
yzx, et cetera--
then this is plus 1.
If it's in anticyclic
order, it's minus 1.
And if any index is
repeated, it's 0.
So it packs a real
punch, but it enables
you to do fantastic things.
So if we have Lx, Ly, it's
equal to ih bar plus 1 times Lz.
And the point of this
lecture is with this,
you can derive all of the
matrix elements of an angular
momentum--
L squared, Lz, L plus
minus, and anything else.
But these are the
important ones,
and this is what
we want to derive
from our excursion in
matrix element land.
So the first thing
we do is we extract
some fundamental equations
from this commutator.
So the first equation
is that L squared
Lz is equal to Lx squared Lz
plus Ly squared Lz plus Lz
squared Lz.
And we know this
one is 0, right?
This one, you have to
do a little practice,
but you can write this
commutation rule as Lx times Lx
comma Lz plus Lx comma Lz Lx.
So if you have a square, you
take it out the front side
then the back side.
And now we know what this.
This is minus ih bar Ly.
And this is minus ih bar Ly.
And we do this one, and we
discover we have the same thing
except with the opposite sign.
And so what we end up
getting is that this is 0.
Now I skipped some steps.
I said them, but I want you to
just go through that and see.
So you know what this is.
It's going to be
Ly, and it's going
to be minus Ly times ih bar.
And you get the same thing here.
But then you have an LxLy,
and you have an LxLy.
And when you do the
same trick with this,
you're going to get
an Ly and an Lx again,
and they'll be
the opposite sign.
So this one is really
important because what it says
is that you can take any
projection quantum number
and it will commute with
the magnitude squared.
The same argument works
for Ly and Lz and Lx.
So we have one really
powerful commutator
which is that L squared Li
equals 0 for x, y, and z,
which means since we
like L squared and Lz--
we could add like
Lx instead of Lz,
but we tend to favor these--
that L squared and
Lz are operators
that can have a common
set of eigenfunctions.
If we have two
operators that commute,
the eigenfunctions of one
can be the eigenfunctions
of the other.
Very convenient.
Then there's another
operator that we can derive,
and that is--
let's define this
thing, a step up
or step down or our raising
or lowering operator--
we don't know that yet--
Lx plus or minus iLy.
So we might want to know
the commutation rule of Lz
with L plus minus.
We know how to write this out
because we have Lz with Lx,
and we know that's going
to be a minus ih bar Ly.
And we have Lz with iLy, and
that's going to be a minus Lx.
Anyway, I'm going to just
write down the final result,
that this is equal to plus or
minus h bar times L plus minus.
The algebra of this
operator enables
you to slice through
any derivation
as fast as you can write
once you've loaded this
into your head.
Yes?
AUDIENCE: So for the epsilon,
how do you [INAUDIBLE]??
Is it like xy becomes--
if it's cyclical it's positive?
ROBERT FIELD: I'm sorry?
AUDIENCE: When you say the
epsilon thing, epsilon ijk,
so you're saying that if
it's in order, it's 1?
ROBERT FIELD: Let's just
do this a little bit.
Let's say we have Lx and Ly.
Well, we know that
that's going to give Lz.
And xyz, ijk,
that's cyclic order.
We say that's the home base.
And if we have yxz, that
would be anticyclic,
and so that would
be a minus sign.
You know that just
by looking at this,
and you say if we switch this,
the sign of the commutator
has to switch.
There's a lot of
stuff loaded in there.
And once you've sort
of processed it,
it becomes automatic.
You forget the beauty of it.
So are you satisfied?
Everybody else?
All right, so now
let's do another one.
Let's look at L
squared L plus minus.
Well, this one is super
easy because we already
know that L squared commutes
with Lx, Ly, and Lz.
So I just need to
just write 0 here
because this is Lx
plus or minus iLy,
and we know L squared
commutes with both of them.
Now comes the abstract
and weird stuff.
We're starting to use
the commutators to derive
the matrix elements
and selection rules.
So let us say that we
have some function which
is an eigenfunction
of L squared and Lz.
And so we're entitled to
say that L squared operating
on this function gives an
eigenvalue we call lambda.
And we can also say that Lz
operating on the function
gives a different concept mu.
Now this lambda and mu
have no significance.
They're just numbers.
There's not something that's
going to pop up here that says,
oh yeah, this means something.
So now we're going to use the
fact that this function, which
we're allowed to have as a
simultaneous eigenfunction of L
squared and Lz with its own set
of eigenvalues, this function,
we are going to operate on it
and derive some useful results
that all are based on
the commutation rules.
So let us take L squared
operating on L plus minus
times f.
And we know that L plus minus
commutes with L squared.
So we can write L plus
minus times L squared f.
But L squared operating
on f gives lambda.
We have L plus minus lambda f.
Oh, isn't that interesting?
We have-- I'll just write it--
lambda times L plus minus f--
L plus minus f.
So it's saying that this
thing is an eigenfunction of L
squared with eigenvalue lambda.
Well, we knew that.
So L plus minus
operating on f does not
change lambda, the
eigenvalue of L squared.
Now let's use another one.
Let's use Lz L plus minus.
Well, I derived it.
It's plus or minus h
bar times L plus minus.
And if I didn't derive
it, I should have,
but I'm pretty sure I did.
And so now what
we can do is write
Lz L plus minus
minus L plus minus Lz
is equal to h plus or
minus h bar L plus minus.
Let's stick in a function
on the right, f, f, f.
So now we have these operators
operating on the same function.
Well, we don't yet know
what L plus minus does to f,
but we know what Lz does to it.
And so what we can
write immediately
is that Lz operating
on L plus minus
f is equal to plus
or minus h bar
L plus minus f plus
mu L plus minus f.
Well, that's interesting.
So we see that we
can rearrange this,
and we could write plus
minus h bar L plus minus f is
equal to mu L plus minus
f plus Lz L plus minus f--
that's h bar, OK.
Oh, I'm sorry, L
plus minus f there.
So what's this telling us?
So we can simply
combine these terms.
We have the L plus minus f here.
And so we can write mu plus
h bar times L plus minus f.
That's the point.
So we have this
operator operating--
AUDIENCE: I don't think
you want the whole second--
ROBERT FIELD: I'm sorry?
AUDIENCE: The first line
goes straight to there.
I think your second
line's [INAUDIBLE]..
ROBERT FIELD: I took
this thing over to here.
So let's just
rewrite that again.
We have Lz L plus minus
f is equal to this.
And so here we have
mu, the eigenvalue,
and it's been
increased by h bar.
And so what that
tells us is that we
have a manifold of levels--
mu, et cetera.
So we get a manifold of levels
that are equally spaced,
spaced by h bar.
AUDIENCE: I think it also should
be plus or minus h bar, right?
ROBERT FIELD: Plus minus h bar--
yeah.
So we have this
manifold of levels,
and so what we can say is, well,
this isn't going to go forever.
This is a ladder of
equally spaced levels,
and it will have a highest
and a lowest member.
And so we can say, all right,
well, suppose we have f max mu,
and we have L plus
operating on it.
That's going to give 0.
And at the same time we can
say we have L minus min mu
and that's going to give 0.
We're going to
use both of these.
Now I'm just going
to leave that there.
Oh, I'm not.
I'm going to say, all
right, so since we
have this arrangement--
all right, I am
skipping something,
and I don't want to skip it.
So if we have L plus operating
on the maximum value of mu,
we get 0.
And the next one down is down
by an integer number of L,
and so we can say that
Lz operating on f max mu
is equal to h bar
L, some integer.
Now this L is chosen
with some prejudice.
Yes?
AUDIENCE: Why is
there an f of x?
ROBERT FIELD: Now
I have to cheat.
I'm going to apply an
argument which is not
based on just abstract vectors.
We have an angular momentum.
It has a certain length.
We know the projection of that
angular momentum on some axis
cannot be longer
than its length.
I mean, I'm uncomfortable
making that argument
because I should be able to
say it in a more abstract way,
but this is, in fact--
we know there cannot be an
infinite number of projection
quantum numbers, values of the
projection quantum number that
aren't reached by applying
L plus and L minus.
It must be limited.
And so we're going to call the
maximum value of mu h bar L
or L.
Now I have to derive
a new commutation
rule based on the original one.
No, let's not erase this.
We might want to see it again.
So let's ask, well, what does
this combination of operators
do?
Well, this is surely
equal to Lx squared,
and we get a plus
i and a minus i,
and so it's going to
be plus Ly squared.
And then we get i times
LyLx, and we get a minus i
times LxLy.
This is L squared
minus Lx squared.
We have the square root of
2 components of L squared,
and so this is equal
to the difference.
And now we express
this as i times LyLx.
And what is this?
This is plus ih bar Lx.
AUDIENCE: I think you
wrote an x [INAUDIBLE]..
ROBERT FIELD: OK, this
is, yes, x, and that's Lz.
I didn't like what
I wrote because I
want to have everything
but the z and the L squared
disappearing, and so
we get that we have
L squared minus Lz squared.
And then we have plus ih bar Lz.
I lost the plus and minus.
No I didn't.
OK, that's it.
And so we can rearrange
this and say L squared
is equal to Lz squared minus
or plus h bar Lz plus L
plus minus L minus plus.
So we can use this equation--
OK, I'd better not--
to derive some good stuff.
I better erase some
stuff or access a board.
We're actually pretty
close to the end,
so I might actually finish this.
So we're going to use
this equation to find--
so we want lambda, the value
of lambda for the top rung
of the manifold over here.
So we apply L
squared to f max mu.
And we know we have
an equation here
which enables us to evaluate
what the consequences of that
will be, and it
will be Lz squared
f max mu minus and plus
h bar Lz f max mu plus L
plus L minus f max mu.
So if we take the bottom
sign, that L plus on f max
is going to give 0.
So we're looking at the bottom
sign, and we have a 0 here,
and so we have L squared
f max mu is equal to Lz
squared f max mu minus or
plus h bar Lz f max mu plus 0.
Isn't that interesting?
So we know that Lz
is going to give--
so we're going to get an h bar
squared and mu max squared.
We're going to get a
minus h bar h bar mu max.
So what this is telling us is
that L squared operating on f
max mu is given by l
because we said that we're
going to take the maximum
value of mu to be h bar L.
So I shouldn't have had
an extra h bar here.
So we get this
result. So lambda--
so L squared operating
on this gives
the-- oh yeah, maximum mu.
So it's telling
us that L squared
f max mu is equal to h
bar squared l l plus 1.
Now that is why we chose
that constant to be l.
And we do a similar argument for
the lowest rung of the ladder.
And for the lowest
rung of the ladder,
we know there must
be a lowest rung,
and so we will simply
say, OK, for the lowest
rung of the ladder we're going
to get that mu is equal to h
bar lambda bar mu min.
And we do some stuff,
and we discover
that lambda has to be equal
to h bar squared l l plus 1.
And using this other
relationship and the top sign,
we get h bar squared
l bar l bar minus 1.
And there's two
ways to solve this.
One is that l is
equal to minus l bar,
and the other is that l
bar is equal to l plus 1.
Well, this is the lowest
rung of the ladder.
Wait a minute, let
me just make sure I'm
doing the logic correctly.
It's this one, OK, here.
So this is the lowest
rung of the ladder,
and l bar is supposedly
larger than l.
Can't be, so this is impossible.
This is correct.
And what we end up getting
is this relationship,
and so mu can be equal to--
and this is l l minus
1 minus l stepped to 1.
This seems very weird
and not very interesting
until you say, well,
how do I satisfy this?
Well, if l is an
integer, it's obvious.
If l is a half integer, it
shouldn't be quite so obvious,
but it's true.
So we can have integer
l and half-integer l.
That's weird.
We can show no connection
between the integer
l's and the half-integer l's.
They belong to completely
different problems,
but this abstract
argument says, yeah,
we can have integer l's
and half-integer l's.
And if we have electron--
well, we call it electron
spin because we want
it to be an angular momentum.
Spin is sort of an angular
momentum or nuclear spin.
And we discover that there
are patterns of energy levels
which enable us to count
the number of projection
components.
And if you have an
integer l, you'll
get 2l plus 1 components,
which is an odd number.
And if it's a half
integer you get
2l plus 1 components,
which is an even number.
And so it turns out that
our definition of an angular
momentum is more
general than we thought.
It allows there
to be both integer
and half-integer
angular momentum.
And this means we can
have angular momenta where
we can't define it in
terms of r cross b.
It's defined by the
commutation rule.
It's more general.
It's more abstract.
It's beautiful.
And I don't have time to finish
the job, but in the notes
you can see that we
can derive the matrix
elements for the raising
and lowering operators too.
And the angular
momentum matrix elements
are that L plus minus
operating on a function
gives this combination
square root,
and it raises or lowers m.
So it's sort of like what we
have for the a's and a daggers
for the harmonic oscillator,
but it's not as good because you
can't generate all the L's.
You can generate the m sub L's.
And that's great, but there's
still something that remains
to be done to generate
the different L's.
That's not a problem.
It's just there's not
a simple way to do it,
at least not simple to me.
And so now anytime we're
faced with a problem involving
angular momenta,
we have a prayer
of writing down the matrix
elements without ever looking
at the wave function, without
ever looking at a differential
operator.
And we can also say,
well, let's suppose
we had some operator
that involves L and S,
now that we know that
we have these things,
and L plus S can be
called J. So now we
have two different operators,
two different angular momenta.
We have S and we have
the total of J. Well,
they're all angular momenta.
They're going to satisfy their
selection rules and the matrix
elements, and we
can calculate all
of these matrix elements,
including things like L dot S
and whatever.
So it just opens up
a huge area where
before you would
say, well, I got
to look at the wave function.
I've got to look
at this integral.
No more.
But there is one thing,
and that is these arguments
do not determine--
I mean, when you take the
square root of something
you can have a positive
value and a negative value.
That corresponds to
a phase ambiguity,
and these arguments
don't resolve that.
At some point you have to decide
on the phase and be consistent.
And since you're never
looking at wave functions,
that actually is a
frequent source of error.
But that's the only defect
in this whole thing.
So that's it for the exam.
I will talk about something
that will make you a little bit
more comfortable
about some of the exam
questions on Wednesday, but
it's not going to be tested.
