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>> Good morning.
Today, we have various things that we need to take care of,
finishing up our discussion of symmetry and bonding.
We're mostly done talking about it.
But we do need to talk about how we figure
out if various integrals go to zero in a particular space
or in a particular point group.
And then, we're going to talk about some terminology
that we need to get into rotational spectroscopy.
And maybe we'll get to talking
about quantization of angular momentum.
So, I hope to be done with rotational spectroscopy
by the end of Friday's lecture
so that we can start vibrational spectroscopy next week.
Let's see how that goes.
OK. So, just a follow-up on the discussion last time,
we were looking at these bonding examples and seeing
which orbitals can form sigma bonds and pi bonds.
And I asked you to, you know, go through
and reduce our reducible representations that we came
up with in class and make sure that you give the right answer.
And I just want to follow-up on that
to make sure everyone gets it.
OK. So here was what we got.
So, we had our representation of the pi bonds.
We know that they have to be perpendicular to the sigma bonds
which was the only condition that we started with.
And so, we had to consider the in plane set
and the out of plane set.
And by making our reducible representations based
on the symmetry of those objects and reducing them,
we came up with the following combinations
of irreducible representations.
So, for the out of plane set,
we came up with E double prime and E double prime.
And for the in plane set,
we came up with E2 prime and E prime.
And if we go and look at the character table and look
at what objects belonged
to these particular irreducible representations,
we see that we've got a PZ orbital belonging
to the out of plane set.
The in plane set has one symmetry species
that doesn't correspond to any orbital.
And then, the other one contains the X and Y orbitals,
and then also some D orbitals.
So, what we know from this is that PX
and PY wouldn't make any sense
because they're already involved in the sigma bond.
We know that the pi bond has to be perpendicular.
And we also know that D orbitals are not realistically going
to be involved in a nitrogen compound.
That's what we are talking about.
So they're energetically not available.
And so, we know that the PZ orbital on that central nitrogen
and also the oxygens is what's involved in the pi bond
which again is something that we already knew.
So, it's useful to do these boding examples
when we're learning these things for a couple of reasons.
One is because you don't need me to make up practice examples.
You can go through sort
of all the random Lewis structure examples
from general chemistry and do problems like this.
There's really a limited number of point groups
that have few enough symmetry species
that you could realistically be expected to do this in class.
So, if you do a few of those and a few other ones
that are harder, you can really get a good handle on this
and you can check your answer yourself because you know
in the case of these bonding examples what you should get
and it should be consistent with your chemical knowledge.
OK. So we also had a bonding example in the homework
where we wanted to do the same kind of analysis for oxylate.
And I'm not going to go through the whole thing.
I just wanted to give you a couple of hints
in case you're having trouble getting started.
So, first of all, here's your basis.
You can draw little arrows representing the carbon
oxygen bonds.
And why did I not represent any single and double bonds here?
Anybody know?
I know you do.
These people talked about it in office hours.
Are they all the same?
Yes, right, it has resonant structures.
And so, we need to make sure and take that into account.
So, we assigned it to a point group.
We know our basis.
There are four things in our basis.
And so, we know that the character
of the identity matrix is going to be zero.
And then, if we do the--
let's just do one example of setting up a matrix.
So, if we do a C2 rotation in the Y direction, so,
remember the way this is set up, the molecules in the XY plane
and Z is coming out at you.
So, if we rotate about the Y axis, here is what we get.
We're just flipping it over this way.
And we need to come up with the matrix that gives us
that when we multiply it by the original vector.
So, A1 switches places with A2, and A3 switches places with A4.
So, there's the matrix that we get for that.
And the character for that operation is zero.
So, this is the kind of stuff that you should be able to do.
And it's useful to practice.
You know, when you come to office hours,
we can do more examples like this.
Again, you can make them up yourself and do them
and make sure that you get it.
OK. So that concludes our discussion of bonding.
I'm not going to go to the rest of the way through this.
I just wanted to give you a hint.
Let's talk about it in a little bit more general way
because we're going to need this sort of symmetry arguments
when we start talking about spectroscopy.
So, spectroscopy is all
about the interaction of light and matter.
And we're going to have different kinds of states
that the molecule can be in.
First, we're going to talk about rotational states.
But then, by rational states,
electronic states, nuclear spin states.
And we're going to have selection rules where we'll see
that only certain transitions are allowed.
And the reason that happens is
because of this symmetry argument.
So, we need to be able to do one more thing
with these symmetry sorts of things.
So, one thing that I'm sure you're familiar with
or at least have seen before in calculus
and some other math classes is just the even, odd rule.
So if we have even functions, they're symmetric
over a symmetric interval, odd functions, or anti symmetric.
So if you integrate an even function
over a symmetric interval, you get some number.
If you integrate an odd function
over a symmetric interval, you always get zero.
And this is a nice thing to remember because it means
that there integrals that could look pretty nasty
and you can just say they go to zero by inspection.
Of course, if they don't, then you have to work it out.
But, that's a trick that mean professors like to pull to seer
that into your frontal lobes
to input some really horrible looking thing on an exam.
And, you know then, if you remember this rule,
you could see that it just goes to zero.
We're going to look at how to use this
in a little bit more general way with symmetry group.
So just here are some more examples.
If we want to multiply these functions together,
if you multiply two even or two odd functions,
you get an even function.
If you multiply an even times an odd, you get an odd function.
We can also look at how their derivatives work.
If you take the first derivative of an even function,
you get an odd function and vice versa.
And again, if we go back to our general chemistry intuition,
this is something that we all understand
that are really intuitive level.
So, if we talk about molecular orbitals,
let's just do a really simple case, H2.
If we have our 2S orbitals in phase, they add constructively.
And we get a bonding molecular orbital.
And if they're out of phase, they interfere destructively.
There's a node in the middle.
And we get an antibonding molecular orbital.
And again, this is something that we all remember.
So we can look at this in a little bit more systematic way.
We can talk about, you know, if we have an S orbital
and a PX orbital, you know,
assuming that we get it oriented the right way,
there's non-zero overlap.
But if we look at something like an S orbital with the PZ orbital
like in this coordinate system, in this orientation,
we have a situation where lobes have opposite signs cancel.
And if we have an integral like this where we have the product
of two functions and we're integrating it
over some symmetric interval here all space,
we can find the symmetry species of each function
in whatever point group we're in.
Then, we want to multiply them together
and that will give us a reducible representation.
And when we reduce it, we have to look at it
and see if it contains an A1.
And if it doesn't, then there's no overlap.
So let's think about what that means.
A1 remember is the symmetry species that is in variant
with respect to all transformations.
It has a character of one under every operation.
And what that means is that if I have say a chemical bond
and my orbitals overlap, that has to be
in variant to all operations.
If that wasn't true, my chemical bond would appear and disappear
when I rotate the molecule for example
and that wouldn't make any sense.
So, that's how you can understand how this works.
So let's look at some pretty simple examples.
So, if we look at a molecule like ammonia, and if we want
to know whether there's overlap of the S orbital on the nitrogen
with this particular linear combination of S orbitals
on the three hydrogens, we can do that the same way.
So again, this doesn't tell you
that whether that's the only thing going on,
obviously, we know it's not.
There are P orbitals involved also.
We just want a yes or no answer.
Do these things overlap?
Another thing I want to point out is that just
because you can make particular linear combinations of orbitals,
it doesn't mean that that's the ground state
or that that's necessarily what's going
on in a given system.
This is important because particularly when we look
at electronic spectroscopy, we are going to see excited states
and things that looked pretty weird.
But we have to worry about them because we're putting in energy
and kicking the electrons up there.
So again, this is just telling us it's powerful
but it's limited in what it can give us.
We get a yes or no answer
as to whether these things have any overlap and that's about it.
But that is useful.
OK. So, we just go through and do this by inspection.
So F1 is our S orbital in the nitrogen and it's a sphere.
So it's going to have--
it's going to be on variant under all these transformations.
Then if we look at the linear combination that we have
of the three hydrogen orbitals all in phase with each other,
we have to actually look at that and see what it does.
Of course we know it's in variant under the identity
because everything is.
For4C3, we do get something that looks the same.
It doesn't change sign.
And for sigma V, we get something that looks the same.
And then, if we multiply all of these characters together,
we do get something that looks like A1.
And so, it has overlap.
So, we know that there can be some interaction
between this S orbital and the nitrogen,
and this particular linear combination.
So, now let's look at another one
that has a different sort of symmetry.
So, we've got this S orbital on the nitrogen again.
And now, we have a linear combination that consists of two
of the three S orbitals of the hydrogens
out of phase with each other.
So, you know, again, does this look like a realistic orbital
that we usually talk about in terms of bonding?
Not really, but we can make linear combinations like this.
And when we get into excited states, we will see some things
that looked kind of weird.
OK. So again, we know what happens with the S orbitals
and variant under all transformations.
The second linear combination, it has a character
of 2 for the identity.
And then for C3, we get an overall character of minus 1.
And for sigma V, we get an overall character of 0.
And then, if we multiply these together and reduce it,
we get the symmetry species of E. And there's no A1 in that.
So there's no overlap.
And I think we're going to come back and talk about this later.
I just want to introduce it
so that we get a feel for how it works.
And when we start talking about selection rules,
we're going to talk about it a little bit more.
So for now, the most important thing
to remember is the even odd rule and also, you know,
how you go through this general procedure.
Question?
[ Inaudible Remark ]
Yeah, I took a shortcut and ended up reducing this.
Why don't I write up a little description of this
with more steps in it and post it for you guys?
That might be-- that might be a useful thing to do.
You're right.
I did skip some steps.
OK. So, now we're moving on.
And we're going to get to direct notation.
Have you seen direct notation before?
This is something that came up last quarter.
OK. It's something that's really important to be able
to read the literature in quantum mechanics.
A lot of things are written down this way.
And it takes a little bit of getting used to it first.
But it's just a short hand notation for writing
down wave functions and writing down integrals.
And it just saves a lot of time.
So, we're going to see a lot of cases
where we have a lot of complex integral.
We're integrating wave functions together.
And everybody knows what the function is.
We don't need to keep writing it down over and over again.
This is just an easy way to write it.
It is a very compact notation.
And it contains a lot of information.
And you need to know what's going
on in order to be able to use it.
So, we have to be careful not to make mistakes and make sure
that everybody knows what's in there.
But once you get the hang of it, it's really useful
and it saves a lot of time.
OK. So, if we have a normalization and I know
that you have seen this and you know what that's about,
that's just an integral of the complex conjugate
of one wave functioning
with some other wave function overall space.
And we know what you get here.
So, that equals 1 if N prime equals N.
So if your two wave functions are the same.
And N equals 0 otherwise.
And that's just a consequence of the fact
that these things are orthonormal sets.
OK. So, in direct notation,
this is how we write down that integral.
It's just a shorter way to do it.
So, this funny little front half
of the bracket thing is called a bra
and the other one is called the ket.
And so, direct notation is also called bra-ket notation.
And if you just have a ket--
yes, let's all have a middle school moment
and giggle at the bra.
That's fine.
[ Laughter ]
I can see people trying to hold it back.
There's no point.
We might as well just give in.
All right.
The-- So when you see a bra by itself,
that's just the complex conjugate of a wave function.
That's all it means.
The ket by itself is a wave function.
When you see them together like this,
that means take the integral of that overall space.
So, there is an integration implied in that operation.
And then, we can also write down this condition for what you get
in a different way as the Kronecker delta function.
Did you see this last quarter?
>> Not really.
No.
>> Not really?
OK. You did because everybody looked happy and looked
like it was familiar when I talked
about the normalization condition where it equals 1
if the wave functions are the same and 0 if they're different.
That's all this is.
The Kronecker delta function is just a compact way
of writing that down.
So when you see delta sub N end prime, that's a function.
I mean, it's a function kind of.
It's just-- You know, it's just telling you that that equals 1
if they're the same and 0 if they're not.
OK. So now, we can go through and talk about how
to do some other things.
So here is how we write down our normalization.
And, you know, I'm saying it's a shorthand notation
and it saves space.
It doesn't look like it saves that much space right now.
But imagine that we have to put
in what the actual wave functions are
and it's a big mess.
Whereas, if we can just-- if we use the bra-ket notation as long
as long as we're in a specified system,
we know what the [inaudible] functions are.
We can just specify them
with their quantum numbers for example.
OK. So, let's look at the matrix element.
So, I want you to write down a matrix element of some operator
which I'm calling omega.
So we know that quantum mechanical operators are--
they are linear operators we can represent them with matrices.
And the matrix element, NM, for omega, is just this integral.
[ Pause ]
And here's how we go about writing that down.
And so, if we wanted to make a whole matrix
for our representation of omega, we would just have to go through
and set up all of our matrix elements.
OK. Does anybody have any questions about this?
It's really important and it's going to keep coming
up over and over again.
And I'm not going to write all the wave functions every time.
So you're definitely going to see it.
Yes?
[ Inaudible Remark ]
OK. So, if you just have like--
so, N prime in the bracket that it's in, that's a ket.
>> OK.
>> That's just the wave function.
The other one that's the bra, that's complex conjugate
of N. So, there's a lot in here.
So the N and M prime are the quantum numbers that represent
that particular wave function.
So, you have to know what system you're in.
So if you say it's a particle in the box, then we're talking
about the particle in the box wave functions.
And you have to know what the one for N equals 0
or N equals 1, or N equals 2 is.
And you have to already know that the system
that you're working in and what the wave functions are have
to be specified.
But once you know that, this is a shorthand way
of writing them down.
Then, when you get the bra and the ket together like that,
that's when it implies an integration overall space.
[ Inaudible Remark ]
Display in the middle mean?
There's-- So there--
[ Inaudible Remark ]
It-- OK. So, if you look at the picture in the bottom right
of the matrix element, so the M in that bracket is a ket.
And then, the other thing is a bra.
And so imagine, you know, take the omega out of the middle.
You're putting them together.
The, you know, line that's on one part of the bra
and the one that's on the other part
of the ket are superimposed on top of each other.
That's all it is.
You're just sandwiching them next to each other.
And then, when we put the operator in between,
that means that we operate omega on M first.
Remember the order of operations for these things.
We operate the omega on M, and then, we take the integral
of whatever the result of that is
with the complex conjugate of N overall space.
Does that help?
>> Yes.
>> Anymore questions?
If you don't understand, please do ask now
because it's going to come up a lot.
Yes?
[ Inaudible Remark ]
Well, it depends on the context of what you're doing, right?
So again, we have to know-- we have to know what the system is.
So, if you have a harmonic isolator that's
in a particular potential, it's over the space of that system.
And, you know, this is really powerful
because it's very general.
When we start to talk about NMR spectroscopy
which we're going to, then we'll be talking
about things in spin space.
It's not even in space.
It's not even in real physical space.
It'll be in spin space which we'll learn about later.
So, it's extremely general and this is used for writing
down all kinds of things in physics and chemistry.
And, you know again, it's not-- it's just notation.
But, you will see in all over the place if you want
to go read the literature in quantum chemistry or physics.
And it is going to come up a lot.
So, your book mostly doesn't use it.
There's a little section somewhere in the early chapters
on how to do it as kind of an aside.
It's probably useful to go look it up and read it
if you want some extra clarification on it.
The Wikipedia page on this is really good.
I also recommend looking at that.
But, for the most part your book doesn't use it.
They just write that on the integrals.
So there will have to be some translating back and forth
because I'm going to use it in class.
You'll see it in the literature.
And your book mostly does not do it.
OK. So let's move on.
Oh, sorry, anymore question?
[ Inaudible Remark ]
So, in that case, there's-- there was no omega.
That was just-- I integrated the complex conjugate of N
with N prime overall space.
And the answer that I've got is the Kronecker delta
which just means 1 if they're the same,
0 if they're different.
I did two different things with the notation.
The first thing I did was I just set
up a normalization condition.
Then, the second thing was an example
of a matrix element for omega.
They're separate issues.
Makes sense?
OK. We'll see lots more of this.
I just want to introduce it at the beginning.
OK. So, now we're ready
to really start talking about spectroscopy.
And what we're going to do here is we're just going to go
through the different types of spectroscopy that there are
that we can use to solve chemical and physical problems.
And I wanted to put up this kind of big picture view
of what spectroscopy is all about.
So, in the most general sense, it's about the interaction
of electromagnetic radiation with atoms and molecules.
And we can use this interaction to probe all sorts of properties
that we want to learn about the molecules.
And we're going to go roughly in general--
in order of things that go from lowest energy to highest energy.
And so, this is an energy level diagram showing, you know,
not really to scale but hopefully it gives you the idea
of how much energy it takes
to do different things with the molecule.
So, we have the electronic transitions.
So, we have these two potentials here
for the electronic transitions.
And of course the ground state is the bottom one.
And this blue arrow is showing the system absorbing a photon
and jumping up to the next excited state.
So, that's what we're talking about when we talk
about fluorescence or, you know, absorption spectroscopy
when you do that in the lab
and measure the optical density of something.
You know, we're talking about, you know, just absorption here.
It takes a lot of energy
to perform these electronic transitions.
So, this usually happens
in either the visible region or in the UV.
And if we don't have enough energy to excite that,
we can excite vibrational or rotational transitions.
And all of these things tell us different things
about the molecule.
So, I should back up a little bit and point out that I'm going
to tell you the quantum numbers that belong
to these different things.
So, for electronic states,
the quantum number we're usually going to use is epsilon.
And then, if we look at vibrational transitions, so now,
we're confined to the ground state
of the electronic transition
because now we're just putting it in for end radiation.
We don't have enough energy
to excite those electronic transitions.
And that little red arrow from the ground state
to the next excited state
within that well is a vibrational transition.
And that happens in the infrared.
And the quantum number that we use for that is mu.
So, don't get confused.
It's not a V. It's not the frequency.
You know, mu gets used for a lot of things.
But, we have to pay attention to context.
For rotational transitions which are these little tiny ones
on between the vibrational transitions, that's--
those happen in the microwave.
And we use the quantum number J. So again, this is a mixture
of just terminology, you know,
what are we going to call these things?
And also, looking at the big picture
on a single energy scale to--
and this will give us a feel hopefully
for why we see certain things and certain kinds of spectra.
So for instance when we do vibrational spectroscopy,
we're going to see a whole bunch of fine structure that comes
about from the rotations because vibrations take a lot more
energy to excite than rotations.
When we excite something to an excited vibrational state,
we get a bunch of rotational excitation for free.
The same thing when we excite electronic transitions,
we're going to see fine structure due
to the vibrational transitions.
But, it doesn't go the other way.
If you're just putting in microwaves,
you don't have enough energy to excite a vibrational transition.
And so, you don't see it.
And so, one aside here is, you know, your friends and relatives
who tell you that a microwave oven works
by exciting the vibrational frequencies of water,
what do you think about that at this point?
Not enough energy, right?
So, microwaves just excite rotational transitions
and you need IR to excite those vibrational transitions.
So, your microwave works
by having an electric field that's oscillating
and moving the dipoles around.
OK. So as a general matter when we go to record spectrum,
it's going to tell us something about a molecule.
We're going to sweep through a range of frequencies
and measure the signal.
So, this is I of mu.
And this mu really is the frequency.
And we're going to plot it as a function frequency.
So, who here is actually taking the spectrum
of a molecule yourself of any kind?
It should be almost everybody, right?
Don't you do this in general chemistry?
Have you-- So have you take in an absorption spectrum?
Raise your hand if you have Beer's Law.
OK, good. How about IR, do you do that?
Yeah. NMR, have you done that yourself?
OK, good. So, you have some experience
with all of these things.
So, one of the things that you probably know is that at least
in NMR and fancy IR spectroscopy, this description
of you vary the frequency and you sweep through
and see the response isn't 100 percent correct.
There is other-- you can do it that way
but it's not the only way to do it.
And we will talk about what happens
in these modern instruments.
But, OK. So here's another view of the same kind of thing.
What energy range does stuff happen in?
So here's the electro magnetic spectrum.
And it turns out people are pretty clever at making use
of electromagnetic radiation and its interaction with molecules.
We can use just about every part of the spectrum
to learn something about the things that we're interested in.
So, if we talk about radio frequency
that is the resonant frequency of nuclear spins.
So, NMR spectroscopy is down here.
We will talk about it later.
Then once we get into the microwave,
that's where we excite rotational transitions.
IR is where we look at vibrations.
UV-Vis spectroscopy is where we can excite transitions
of the valence electrons.
So, this is what we're usually looking
at in the little spectrophotometers
and molecular biology labs, and when we investigate Beer's Law.
That's what's going on here.
We're looking at transitions of valence electrons.
If you want to look at core electrons and for instance find
out what atoms are present on a surface, then you need X-rays.
So, X-ray photo electron spectroscopy can be used
for that.
So we need higher energy
to excite those more tightly bound core electrons.
This starts to get a little bit exotic.
You need a big X-ray source.
This is something that you would do at maybe a synchrotron,
you know, a source that's at a big national lab.
It's not an instrument
that people would typically have sitting around in a lab.
These other things are.
And then, the last one, the gamma rays,
this also sounds pretty exotic, right?
This is-- We can actually look
at excitations of nuclear states.
It's called Mossbauer spectroscopy
and that's done with gamma rays.
So, we will go through many of these types of spectroscopy.
Not all, we're not going to talk about XPS
or Mossbauer spectroscopy so much.
But the others, we will go through
and give you a feel for how it works.
Some of these things are things you you're very likely
to use in your research.
In order to talk about the mechanics
of how spectroscopy works and start really getting into it,
we need to talk about the Born-Oppenheimer approximation.
And I know that you've seen this last quarter.
But, let's just review it quickly because, you know,
maybe it'll be put in a little bit more practical context.
So the Born-Oppenheimer approximation just says
that the electrons move around a lot faster than the nuclei.
And that means that we can separate them
which is really good because we would have very ugly problems
if that didn't work.
And of course, it doesn't--
it's not a good approximation in all cases.
But for many other things that we want to do as chemists,
it is a good approximation.
And so, what that tells us is that the--
you know, we have some overall wave function for the molecule.
And it involves the motions of the electrons and the nuclei.
And we can separate variables and treat them separately.
Just because the electrons are tiny,
they're moving around really fast.
The nuclei are big and heavy.
It takes them longer to catch up.
So, the electronic wave function does depend on the positions
of the nuclei which we know.
We've been talking about molecular orbitals and things
like that, you know, where the electrons are around in bond.
So, it's not that it has nothing to do with the nuclei at all.
The positions are important.
But their motion isn't really important on the time scale
of the electrons under this approximation
which is usually pretty good.
So, we can consider that the nuclei are just sitting still
on the time scale where we're worried about the electrons.
And so, here's our Schrodinger equation for the electrons.
So, we've got our Hamiltonian and our wave function.
And now, notice this have a subscript of epsilon to indicate
that we're talking about the electronic states.
That's its quantum number.
And this is a function of the electron coordinates
and the nuclear coordinates.
But the nuclear coordinates,
we're going to be treating as fixed.
And then, when we go to talk about the nuclear motion,
and this is rotation and vibration,
that just sees an overall smeared out potential
from the motion of the electrons.
It sees the average of what the electrons are doing.
[ Pause ]
And so, here we have our Hamiltonian
for the nuclear transitions.
And it's got subscripts of mu and J. Remember,
those are quantum numbers for vibrations and rotations.
And those just depend on the nuclear coordinates
because it's just seeing some overall smeared
out potential from the electrons.
[ Pause ]
So again, this is a really useful approximation
because it means that for the most part,
we can treat our electronic spectroscopy as being separate
from rotations and vibrations under many conditions.
OK. So that is it for the kind of, you know,
basics and housekeeping kind of stuff and review.
Let's move on to actually talking about quantization
of rotation, angular momentum, things like that that we need
to know for rotational spectroscopy.
So, I guess it's not entirely true
that we're done with review.
We are going to talk about some things
that you learned last quarter.
But give it a little bit of a different spin if you like.
So, please review chapter three
if you don't remember it really well.
And I really recommend reviewing all of the stuff
from last quarter as we talk about it just
because quantum is one of these things where at least I found
when I was learning it as a student.
It's not very intuitive.
And you really have to, you know,
when you get more information about how it works,
you really have to go back and review the basics.
And make sure that you understand it.
And it makes more sense every time you go back and do that.
So, please do review it.
So, one of the things that came up last quarter is the case
of a particle on a ring.
And you might wonder why you're interested
in a particle on a ring.
And that's a fascinating question.
One thing about basic quantum mechanics is that you can end
up with a lot of these examples that don't sound very practical.
So you look at a particle in a box and a particle on a ring,
and a particle on a sphere in a harmonic oscillator.
Is that right?
Is that what you did?
>> Yes.
>> So, why do you think that we pick these particular things?
Is it because they're extremely realistic
and they describe everything in chemistry and physics?
Yeah, it's because those are the only things
that you can solve analytically
like anything more complicated than that.
You need computational methods.
There are lots of computational methods.
There's a huge field of computational chemistry
where people do electronic structure calculations.
There's a big center for that at UCI.
But these simple cases where you do get things
that you can solve analytically do give us some intuition
about processes that we care about.
So, if we think about a particle on a ring, in and of itself,
that's not necessarily the most exciting thing.
You know, you have some particle going around.
But you can also think about it as a rigid rotor.
So if you picture a diatomic molecule that's just
in the gas phase, it's off by itself.
It's not interacting with anything.
And so, we have the diatomic molecule
and it's rotating around.
If you pick a point on that molecule
and follow its rotation,
it looks like a particle on a ring, right?
We can use the same mathematical treatment to talk
about our rigid rotor.
So, what do I mean by a rigid rotor?
That means that the bond between those two nuclei isn't flexing
at all.
It's not changing.
It's not bending.
It's just rotating around.
As we'll see, that's an approximation
that works pretty well at low temperature.
So, if we're in low energy states and, you know,
also it depends on the molecule.
If we have a very stiff bond that works better
than if we have a very floppy bond.
But it is in approximation that we can start out with.
OK. So we can go through angular momentum.
And if we're talking about a particle on a ring,
we have something that's in cylindrical coordinates.
That's sort of the natural coordinate system to use here
because we have our diatomic molecule
and we can say it's rotating about the Z axis in this plane.
So, it makes sense to do this in cylindrical coordinates.
And another thing to go review and look up the Wikipedia page
if you're not really up on it is how to transform back and forth
between Cartesian coordinates and cylindrical
and spherical coordinates because that's something
that we're going to need to know how to do.
OK. So here's our angular momentum in the classical case.
So, we have our angular momentum, about Z, is plus
or minus the momentum times the radius.
And we can get the moment of inertia here.
And we can-- you know, we can calculate the moment of inertia
for diatomic molecule.
In the book, it goes into all of the different kinds
of rigid rotors that you can have
that have different moments of inertia.
And that's a useful thing to look at.
It's good to go through it and understand how it works.
It's not something we're going to spend a huge amount of time
on in class because it boils down to a lot
of a crap to memorize.
There are a lot of formulas.
And that's not really what we're about.
We want to learn how to solve the problems.
So in class, we're mostly going to focus on the diatomic case
with the understanding that there's all
of these other stuff that you can do.
We're going to look at the case
where we can solve things analytically and while you can,
for some of these other things too.
But anyway, it's worth going and looking at it just
to make sure you understand how it works
but we're not really going to focus on in class.
All right.
So if we take, you know, I'm not going to go through
and solve this Schrodinger equation for this for you.
I assume you did that last quarter.
But we'll just go through the argument for how you get this.
So in the quantum version of this, we can, you know,
use an analogy to the De Broglie wavelength
for the momentum here.
And the fact that it has to be quantized comes from the fact
that that it has periodic boundary conditions.
So, it's a little bit strange but it makes sense
when you think about it.
So our wave function has to be single valued.
So that means that, you know, if I have a wave function
for my rotational state, if my molecule was rotating around,
when it comes all the way around the circle,
it makes a complete circuit.
So it goes 2 pi rotation.
It has to come back to the same place.
And that just intuitively makes sense, right?
You can't have something discontinuous happen to it
as it's going around in the circle.
And the quantization comes out of that condition
because an integral number of wavelengths has
to fit in that circle.
So, here's how we would write that down.
So, some integral number of wavelengths has to go
within that circumference for the--
our little point on the molecule rotating around the ring.
And so, that's where we get the quantization from.
OK. So here's the quantum version
of our angular momentum in the Z direction.
And what we get out of it is that it comes in increments
of M sub L which is its quantum number times H bar.
And M sub L can be 0 plus or minus 1,
plus or minus 2, et cetera.
[ Pause ]
OK. So now we've described the Z component of angular momentum.
Let's look at this a little bit more.
So, the natural place to put this is
in cylindrical coordinates.
And of course, R is fixed because we're talking
about a diatomic molecule rotating.
And we said it's a rigid rotor.
So its radius can't change.
It can't vibrate.
It's stuck at that particular radius.
So R just becomes a constant.
That part integrates out.
And so, our Hamiltonian is simplified.
And so now, here's how I write
down that equation in direct notation.
So again, if you don't remember what the wave functions
for this look like, go look them up in chapter three.
But instead of calling them psi, I'm just going to put
that M sub L in the ket and that indicates that indicates
that I'm talking about the wave function
for that particular M sub L value.
And so, if I write out the Schrodinger equation, you know,
substituting in for what the Hamiltonian actually is,
here's what that looks like.
And I also want to point out that angular momentum
and the angle are complimentary values.
What do I mean by complimentary values in this context?
Do you know?
Does that ring a bell from last quarter?
Yeah?
[ Pause ]
Exactly, right.
Yeah, they're complimentary in the sense
of the uncertainty principle.
So it's like it's an analog to position and momentum.
You can't know the angular momentum and the angle.
So they're complimentary observables.
And I bring this up because I wanted to point out that the--
there is a zero angular momentum state and that's legal.
And that would seem to violate the uncertainty principle,
right, because we know the angular momentum
with absolute precision there.
It's zero.
And this is a little bit counterintuitive when we have--
you know, for people who are used to thinking
about quantum mechanics a little bit.
Because if you think about vibrational energy levels or,
you know, your harmonic oscillator potential,
there's a zero point energy, right?
Like you never have zero energy for that.
For rotational state, you can.
You can have zero angular momentum.
The reason for that is that in that case where you know the--
where you know that with infinite precision,
you don't anything about the angle.
It's just out in space.
You know nothing about it all.
So, that's why you're able to have a zero state for that.
OK. So now, we've talked
about the Z component of angular momentum.
We have something rotating on a ring.
Now, what about if we have the particle-free
to rotate over a whole sphere?
Did you do this last quarter too?
So you talked about the hydrogen out of wave functions.
Again, everybody has seen those from general chemistry as well.
So, now we need to talk about the general angular momentum
for a particle on a sphere.
And I think we're not going to finish that at this time.
It'll be too rush.
So, I'm just going to pick it up next time.
Does anybody have any questions about what we did today?
It's kind of jumping around between different topics.
But there were some terminology things that we needed
to clear up before moving on.
All right.
Have a good day and I will see you Friday.
[ Inaudible Discussions ]
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