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>> Good morning everybody, so we've been--
we've been starting our discussion of spectroscopy
and how lights interacts with the matter and.
So far we've talked about rotational spectroscopy.
We're going to continue this.
We're going to talk about rotations and vibrations,
and we're also going to go
into a little bit more conceptual ideas
about the different ways that light can interact with matter.
So, before we do that,
does anybody have any questions leftover
from the last time or about anything?
Yes?
>> There's like equations that are in the test
from the sections, you know, reading, are we responsible
for those although they are now posted on [inaudible] sites.
>> Well-- so, the question is if there are equations in the book
and they're not in the lecture, do you need to know them?
The answer is maybe.
So things that are going to be covered
on the test will be stuff that we do in class,
things that your TAs cover in discussion,
those are also important concepts,
and things that you need to be able to do
to do the practice problems.
So a lot of times in lecture I'm going to focus on just trying
to make sure that everybody understands concepts.
That's what we're going to do today.
And we might not necessarily get a chance to do examples
of every particular kind of problem
that you might need to do.
Hopefully you get a chance to do that in discussion,
the practice problems give you an idea of what you need to do.
There are other things that are in the book and, you know,
that's just there for your information.
Does that help?
OK, anything else?
All right, let's start talking
about how lights interacts with matter.
OK, so what I want to go over today is the difference
between absorption and scattering.
So this is something that people get confused about.
It's one of the main concepts that we need to learn
in [inaudible], to understand the different kinds
of spectroscopy that we use.
And I want to put it more in terms
of things that we see everyday.
So, you know, again it's all
about photons interacting with matter.
And there are lots of different complicated formulas
when she described, the things
that can happen in specific cases.
But what it boils down to conceptually is that,
if a photon interacts with matter,
it can do one of two things.
It can get absorbed or it can be scattered.
So, you know, it either gets absorbed and it causes some kind
of a transition in the outer more molecule
or it can bounce off.
And it can scatter elastically or inelastically,
we'll get into that toward the end of the lecture,
but for right now, I want to use some sort of familiar concepts
to introduce just the idea of what's the difference
between absorption and scattering.
And there are some things that that are obvious here
and there are some things that are not obvious.
So, all right, we're going to start with the basics.
Why is the sky blue?
>> Rayleigh scattering.
>> Rayleigh scattering.
>> Rayleigh scattering, that's right.
What does that mean?
[ Pause ]
So-- so light scatters in all directions.
OK, so if we got-- if we got that far,
why isn't the sky white?
So if we have all the light just scattering off particles
in all directions, why does it have a color?
>> 'Cause then it will going to scatter in a directions.
>> Yeah, it has a really strong wavelength dependence.
What are we scattering off by the way in Rayleigh scattering.
What are the particles involved?
>> Gas molecules.
>> Yeah, gas molecules so mostly nitrogen, things like oxygen,
whatever is in the atmosphere.
All right, so what's happening here is the light is interacting
with molecules or particles that are much smaller
than the wavelength of light.
So remember we know that, you know,
we know the wavelength range of visible light is, you know,
hundreds of manometers,
molecules are a lot smaller than that.
And so Rayleigh scattering is the solution
to Maxwell's equations that describes what happens
in that limit when the scatterers are much smaller
than the wavelength of light.
And so, it has a really strong wavelength dependence,
and what that means essentially is that smaller--
the small particles look a lot bigger to blue light
than they do the red light, and so we got everything scattered
and we see this nice blue color.
So here's the functional form just to remind you
if you haven't seen it recently.
It has an inverse dependence on lambda to the forth,
so that why we have this extremely strong wavelength
dependence, and it also has an orientational factor.
So if we look at our little molecule we see
that there is more light scattered forward
and backward than down.
And so here's how we see this, the blue sky.
OK, so related issue, why your sunset is red?
[ Inaudible Remark ]
Exactly, so, we still have Rayleigh scattering but now,
you know, instead of having the sun overhead,
it's on the horizon and so it has to travel through more
of the atmosphere to get to us.
So we still have the same wavelength dependence
but all the blue light has already been scattered out
and what we see left are the reds and oranges.
So, that's part of the story, there's also Mie scattering
which is what happens when you get particles that are larger
than the wavelength of the light.
And in that case, much more of the light is scattered forward,
and Mie scattering does not have a strong wavelength dependence.
So, this is really responsible for just scattering
of white light in effects like when it's foggy you see halos
around street lights and things like that.
That's because, you know, in order to get to your eye
that photon is, you know,
bouncing off these little particles and so you see the--
you perceive the street light as being, you know,
larger than it actually is because some
of these photons are getting bounce off little water droplets
that are in the air.
So you often hear it said
that pollution makes the sunsets more red and more beautiful
because you have more scatterers.
So, is that true or not?
The answer is that it depends on how big the particles are.
So if they're small then they're contributing
to the Rayleigh scattering
and you see the more brilliant red color whereas
if they're larger, so if we have larger aerosols then that's just
adding Mie scattering.
So like this-- the picture on the right,
it just makes everything kind of duller
and in some cases it can make the colors softer.
You can get more pink and tones
like that rather than bright red.
There was a question in the back.
>> Yeah. I was wondering, what do you mean
by scatters are smaller wavelength
than the light [inaudible]?
>> No, I mean that the physical size of the particles is smaller
than the dimension of the wavelength
of the light, that's all.
So, it matters whether we're talking about molecules
so those are obviously much smaller,
these are Rayleigh scattering, or if we're talking
about larger particles, so like gas-- sorry, vapor droplets.
Yeah, liquid droplets that are suspended in gas like a cloud
or fog, those are larger scatterers,
and there are some pollution particles, aerosols that fit
into that category as well.
OK, so it's also important to remember that both
of these things are not different phenomena.
It's the same thing.
It's just, there are different--
there are different assumptions being made,
there are different regimes.
We're just solving Maxwell's equations
and we're making the distinction between like, OK,
if we are in the regime where the molecules are much smaller
than the wavelength of the light then we can make a certain set
of assumptions.
And if they are larger then we can make another set
of assumptions.
If they are about the same size, then weird things happen.
It's pretty interesting.
But, remember that these people didn't have computers
when these things were discovered.
So it was really, you know, getting any sort
of mathematical form to anything was a really hard one,
and so that's why we have all these things that are treated
as another different phenomena when, you know,
actually at this point we can just stick
into mathematic and solve it.
So, it does-- a lot of times this is taught, you know,
as the same way that it came out historically
where we have these different regimes and that's fine.
Those are convenient approximation that you can use.
But do remember that it's all the same phenomena.
And I'm not giving you a nice pretty functional form
for Mie scattering and that's because there really isn't one.
If you want to do that there are programs that you can use
to do it but computational methods are the way to go there.
Yes?
>> Sorry, this is off topic, how does the northern lights work?
>> How does the northern lights work?
So that's-- there are ionized particles
that are emitting radiation.
And, you know, what we'll talk about that when we get
into electronics spectroscopy, I'll save it for later.
OK, so we have covered the somewhat obvious,
let's get in to less obvious.
Why is the sky blue at twilight?
So remember we have our nice Rayleigh scattering explanation
for why the sky is blue overhead.
So our sun is up here, the light is scattering down to us,
we see the blue light then it's on the horizon,
it goes through a longer path length.
There are more particles in the way and now it's red.
But the sky is still blue even though the sun is not
up there anymore.
What's that about?
[ Pause ]
So, the answer is a lot of it is due to the color of ozone.
So here's the absorption spectrum of ozone
and you can see this, you know,
again these things have the historical names.
The Hartley bands are in the UV and those are really big.
But then there's this little smaller band the Chappuis bands
that are kind of more in the visible region.
So ozone is actually absorbing
and it has a color and it's blue.
So, you know, again remember
when we say something absorbs and has a color.
It's absorbing everything but blue so that's what we see.
That's something that we're going to talk about again
when we get into electronic spectroscopy.
The details of how we perceive color and how we talk
about it are natural to our perceptual system.
But when we go to think about spectroscopy
and what we can measure with the instruments,
it's not necessary the best way to describe things.
OK, so that's why the twilight sky is blue.
And we saw our first example of the difference
between scattering and absorption.
And notice that we get similar effects,
what we perceive can be the same whether it's based
on scattering or absorption.
The sky looks blue to us either way.
And, you know, a lot of us have never thought
about why it's blue at twilight when, you know,
the mechanism looks like it might be different but,
these are different phenomena.
And again, these are the only--
these are basically the only things that photons are going
to do when they interact with matter.
OK, so why is water blue?
And particularly, if I have my bottle of water,
it looks transparent here except for where it's going
through you the-- where light is going through the blue label.
But if we look at the ocean or a large amount of water,
it looks blue, why is that?
[ Inaudible] matter.
So, Rayleigh scattering again that's--
it's part of the answer.
So, it turns out that water itself is actually blue very
weakly, it's weakly absorbing.
And I will show its spectrum.
So here is just the, you know, the overall absorption spectrum
of water with the visible region drawn in there.
And, you know, notice it has giant absorption bands
at both longer and shorter wavelengths.
There's just this really narrow region
where it doesn't absorb very much.
And if you look at where the water starts, it does start
to absorb in the visible region.
It's absorbing outmost to the red so it is very weakly blue.
And the reason that it looks transparent
when you have a small amount of it is
that you just don't have very much.
Whereas when you get in a more bulk quantity you can actually
starts to see the color.
Now, we need to get back to the issue of scattering,
because if that was the whole story,
if we just had the weak chromophore then water wouldn't
look blue when we look down on it, right?
When we went scuba diving and look up,
we would see blue water.
But if it was just absorbing when we look down on it,
it would always look black.
When we look down and see the blue water, a photon has to go
in there bounce around and then come back out to our eyes.
So scattering definitely does have to be part of the answer.
So the answer there is both.
Another thing that's pretty interesting to think
about is a couple of issues involving the spectrum.
So one is if you look at this absorption spectrum of water,
you know, we've got strong absorptions, you know,
both above and below the visible region.
It's a, you know, it's kind on an interesting coincidence
that there is any region where water is transparent and,
you know, so visual systems in humans and all kinds
of other animals we are able to evolve.
If the spectrum of water didn't look like that we won't be able
to see, because, you know, it would be opaque.
So that's one interesting thing.
Another thing that I'm not going to get into right now but I want
to point it out is why is water transparent?
It, you know, or glass or other media like that.
It's not that the photons are just warming
through without interacting with anything.
And if that doesn't confuse you, it may be should.
So this is-- this is something to think about.
If you want to know the more detailed version of the answer,
there's a little book by Richard Feynman called QED:
Quantum Electrodynamics that explains all of these
in a pretty intuitive way.
That's a good thing to read.
OK, so we've seen how combinations of absorption
and scattering can produce these color effects
that are pretty familiar to us.
I want to talk about one more phenomena that sort
of illustrates the difference which is,
why is it that when you put water on media it looks darker?
So as we talk about the water is transparent,
it has a really weak blue color but we don't see
that in the thin layer that would be on some object.
And, when it evaporates, you know, it doesn't leave a stain,
the pigmentation or whatever we're getting goes away.
Let's think about why that is?
So, demonstration, can everybody see that?
>> Yeah.
>> So, it's just water, it's not going to hurt anything.
As we discussed it's going to dry and go away.
OK, so why does it look like that?
Why is it darker when it's wet?
And it gets light again when it's dry?
[ Inaudible Remark ]
>> It's absorbing more that's definitely true,
but let's talks about why, OK.
So, let's go back to sort of the, you know,
more physics interpretation of light, you know?
We've been thinking about photons.
Let's think about light rays for a minute.
It's another perfectly valid interpretation
that explains some of these phenomena.
So if we look at this, what's going on there is the difference
in refractive index between the air in the carpet is large.
And so when light is coming, you know, when light rays are coming
from the air and they're going to the carpet,
they get scattered a lot.
These light rays get bent over a wide range of angles
because there's more bending
when you have a larger change in refractive index.
When we saturate it with water, now the difference
in refractive index between the water and the carpet is lower,
so the light rays get bent a smaller amount.
And so they bounce around more before they make it back
out to our eyes.
So basically what's happening is instead
of just scattering right of, some of the photons are going
into the-- another parts of the carpet, never to reemerge
and that's why it looks dark when it's wet.
OK, so that is the difference
between absorption and scattering.
And now let's apply that to actual spectroscopy.
Question?
>> Sorry, can you say that last part one more time?
>> Sure, so when you--
>> Did you say scattered more?
>> Well so it get scattered at a wider range of angles
when it's dry because the difference
in refractive index is greater between the air and the carpet.
Then when we pour water on it,
that has a higher refractive index
and so the difference is less.
If I poured benzene on it, it would be even less
but that would be dangerous and I'd probably get
in trouble since it's on video.
[ Laughter ]
So, you know, we're making the refractive index difference--
smaller between the, you know, the carpet
and the surrounding medium and so that means
that the light gets bent under smaller angles and so it has
to bounce around more and more before it gets back
out to our eyes.
And some of it gets absorbed before that happens.
So that's why it looks like that.
OK, so that's fun but the real reason why I want
to make sure everybody understands the differences
between absorption and scattering
and how they can give us some of the same phenomena is
that we need to talk about the difference
between absorption types of spectroscopy
and Raman spectroscopy.
So Raman spectroscopy is what we get
when there's an inelastic scattering of light.
So, when we shine light at something,
if it doesn't get absorb it has to scatter we know that.
We already know that most of it just bounces right off.
It scatters elastically.
That's really scattering which we understand.
But what can happen sometimes is it can either--
the light could lose a quantum of energy to the molecule
or it could take a quantum of energy from the molecule.
And when that happens we get a Raman spectrum.
And so I'm going to put up this picture quite a few times
as we go through our discussion
of different types of spectroscopy.
So we've talked up direct rotational spectrum.
We have said, like OK, we need to have a dipole moment
to see a pure microwaves spectrum from the rotation
of a molecule because we're, you know, we're watching some part
of the molecule go around and we have to be able
to tell the difference with our electric field.
Raman spectroscopy is a different effect.
So instead of absorbing that photon and bumping the molecule
up to an excited rotational state,
we're having a photon bounce off
but it doesn't scatter elastically.
It either leaves behind a quantum of energy
or it takes a quantum of energy from that molecule and in
that way, it bumps it up and down in the rotational state.
So this is rotational Raman.
We can also have vibrational Raman
which we'll get to later on.
And I just want to give sort of the big picture version of this.
We're going to have, you know, rotational spectroscopy
and vibrational spectroscopy.
And there are two types of each.
We can either have direct absorption
or we can have the Raman spectrum which is based
on scattering and they have different selection rules
and they give you complementary information.
They're not identical.
OK, so some terminology.
The Rayleigh line-- this picture is actually not draw
in the scale.
The Rayleigh line is way, way, way bigger than any
of these Raman spectra.
And it also makes sense that when we're talking
about vibrational Raman those are off
to the sides quite a bit more.
And you see these breaks on the horizontal axis.
It takes a lot more energy to excite vibrational states
and it does rotational states.
So for now, we're just going to be talking
about the rotational Raman spectroscopy.
But this picture is going to come back again when we talk
about vibrational states.
So some terminology we have the stokes line
and the anti-stokes line depending
on whether the photon is losing energy to the molecule
or gaining energy from the molecule.
And you have to be careful
because you will see the spectrum flooded in terms
of wavelength like this one is and so remember
that a longer wavelength equals lower energy.
You will also see it flooded sometimes in terms of frequency.
So a higher frequency is a higher energy.
And so, which one is which depends on how it's flooded
and we need to pay attention.
These names are just historical, but, you'll see them.
All right, so the spacing
between these lines is four times B the rotational constant.
So remember, the spacing between the lines
in the direct absorption spectrum was 2B, and we're going
to talk about why that is hopefully
by the end of this lecture.
OK, so I want to again put up this description
of energies and frequencies.
So B is our rotational constant.
If it has a tilde over it then it's in wave numbers.
And we can also define it hertz.
And somebody ask this about this before.
I think I had them, you know,
inappropriately said equal or something.
They're different things.
What mattes if it has a tilde over it or not.
And so, you know, we can use the fact that E equals HC
over lambda to express this in hertz.
And then we've also got these energy levels that we talked
about before and we don't need to go through it again.
OK, so how do we know when something is going
to have a rotational Raman spectrum?
So we said that for a direct rotational spectrum the molecule
needs to have a dipole moment.
So, Raman is based on scattering, not absorption,
it's different effect.
So it makes sense that we might have different selection rules
and we do.
The gross selection rule for rotational Raman is that we have
to have an anisotropic polarizability.
So let's talk about what that means.
The polarizability is the ability of the electron clad
of a molecule do distort when you put it in electric field.
So we're talking about an induced dipole moment.
It doesn't have dipole moment necessarily on its own
but when we put it in electron field, the electron cloud is,
you know, big and can move around a little bit
and can get distorted.
So, polarizability is a measure of how much that can happen.
And it is a tensor it's not just a number so it's represented
by a three by three matrix and three dimensional space.
Here's a picture kind of illustrating what I mean,
so if we put this molecule in electric field,
that electron cloud is just kind of distort it a little bit.
And some molecules might distort a little,
some might distort a lot.
That's the magnitude of the polarizability.
But we also have to describe it spatially.
So we can also define the mean polarizability
which is just a number and that is a third times the trace
of the matrix.
So, of course from our group three discussions do you
remember how to get the trace of the matrix?
Just add the elements on the diagonal.
And then if you divide it
by three you get the mean polarizability.
And that's mainly useful for comparing,
you know, among the molecules.
If we want to talk about in detail what this--
what these things do, we need to use the full tensor.
So, just in case it's hard
to visualize what the tensor looks like.
Here's another unrelated example but that
at least the engineers have definitely seen before
and hopefully it's a little bit easier to visualize.
So, we're just, you know, we're just looking at, you know,
how this property varies in space.
So stress is another example of a tensor.
There are lots of them in physics.
This is a really important concept and I recommend that,
you know, looking up more about if it's not clear.
OK, so we said that our gross selection rule
for Raman spectroscopy is that our molecule has
to have an isotropic polarizability.
So we stick the molecule in electric field and it
that has some frequency, we can call it omega, so omega I,
and it induces a dipole.
So, here's an expression for our induced dipole.
So we have an alpha in front of it which makes sense.
That polarizability factor tells us something about the magnitude
of the dipole that we can induce with that electric field.
And then of course it also depends on the amplitude
of the field which is varying in time
so it has this time dependence.
So, omega i is the frequency of the induced field.
The molecule is also rotating at some frequency.
We're going to call that omega r. And so,
its polarizability is time dependent.
Because remember the molecule was an isotropic so here I drew.
It shaped kind of like a football.
So if the molecule is, you know, has the football standing
up with respect to the electric field,
its polarizability is different and if it's lying down.
And it ranges from these between these values.
So delta alpha is the change in polarizability
from the parallel orientation to the perpendicular one.
And, so we can define its time dependence in terms
of the rotational frequency of the molecule.
And one thing that's important to this is that, it comes back
to the same value, two times per rotation.
Why? Because we're assuming that we have small electric fields
and it doesn't matter whether I hold the football
like this or like that.
And that's true, even
if the molecule isn't exactly symmetric, at least for the case
of small electric fields.
And so the specific selection rule for linear rotor is
that we can have either no change in rotational state
or we can have things jumping up and down by two,
and that's because we get the same effect two times per
rotation rather than one time.
So that's why the specific selection rules are different
for rotational Raman than they are for rotations spectroscopy.
OK, so I also want to point out that
if we have this delta j equals zero condition.
That contributes to the Rayleigh line.
So what we're saying there is the photon hit the molecule
and it just bounce off in it
and it didn't change its rotational state.
That's really scattering.
>> I have a question.
>> Yes?
[ Inaudible Remark ]
So when you actually-- so that's a good question.
So the question is, do you have to do this for every wavelength
of light that you want to measure.
What we would do in a Rayleigh experiment is just sweep
through a whole bunch of wavelengths
and see total response which--
and we'll see some examples of that spectra later.
So, essentially yes, but we can--
we can learn some things
about what the overall spectrum is going to look
like without doing that.
So if we want to predict what the spectrum looks
like in a qualitative sense we don't need
to do it every wavelength.
If we want to know specifically what it looks like that we do.
OK, so here is just a picture illustrating what I just said,
just in case.
So we have our electric field, we start ours with this thing
in the perpendicular orientation.
And then if we rotate it 90 degrees now it's parallel
and the polarizability is different.
And then as it keeps going around, you know,
now the blue atom and the red one are flipped they might
be different.
I'm saying I don't care about that it's a good approximation
that we're just looking
at the polarizability this way versus this way.
And so that's why we get the same thing two times per
rotation and that leads to that selection rule.
And we're not going to talk about this right this minute
but I'm just throwing it in here for future reference.
For a symmetric rotor we have the same conditions for delta j
and delta k equals zero.
OK, so I am going to--
>> I have a question.
>> Yes?
>> So this-- is this like the quality [inaudible] based
looking bold in one like in opposite directions
or it's just a bond that's actually stretching or is it--
>> The bond doesn't really stretching.
That would be vibrational spectroscopy.
It's just that the polarizability is changing a
little bit.
It's just like the electron cloud is deforming.
>> Oh so the entire electron is kind of shifting--
>> Yeah, it's just kind of moving around.
It's making an induced dipole
but it's not changing the bond length.
That would be vibrational Raman and we'll see that later on.
OK so, if we substitute in what are--
if we substitute in this expressions into our expression
for the induced dipole what we get is
that it has a component oscillasting
at the incident frequencies.
So that's the electric field that we're putting in.
And that's why we care about that frequency.
That's the Rayleigh line.
And then we also see two components that have to do with,
you know, that incident frequency plus
or minus two times the rotation frequency.
So that's how we get that pattern
where we have the Rayleigh line in the middle and then all
of those little spikes for the individual rotational
transitions on either side of it.
And those only occur if delta alpha or the difference
between polarizability when the molecule is parallel
versus perpendicular doesn't equal to zero.
And so here's that picture again showing
where these frequencies fall out.
So that's how we know where they are.
We know what they look like,
we know why the selection rules the way they are.
We're going to talk about this selection rules
and how this relates to symmetry in more detail later after we go
through vibrational spectroscopy.
But I think for now we are going to move
on to vibrational spectroscopy and talk
about the basics of that.
Does anybody have any questions about rotational Raman
and the difference between that
and absorption rotational spectroscopy?
OK, let's talk about the next level up in this diagram.
OK so, so far we've been looking
at these tiny little rotational transitions
that don't take very much energy and that are, you know,
honestly not all that useful for telling us about properties
of the molecule that we care about as chemist.
Let's move to saying, OK, now we have more energy to put in.
We're looking at infrared
and we can excite vibrational transitions.
This type of spectroscopy is fantastically useful
and people use it all the time in chemistry
to characterize all kinds of things and we'll talk
about some specific applications.
So now, we're in the IR region of the spectrum
or exciting vibrational transitions,
and I should mention that, you know, we will talk
about absorption spectroscopy and Raman.
When people just say, Raman spectroscopy
and they don't specify what kind it is, they're talking
about vibrational Raman.
As we just saw, rotational Raman exist too
but vibrational Raman is much more common to use.
OK, so let's talk about the selection rules first
because I think it's pretty intuitive and then we're going
to move on to going through a lot of the details
about what you actually do with this.
So a vibrational mode of the molecule is IR active,
that means we can see it.
It's actually changing the dipole moment of molecule.
So water obviously has a dipole moment.
But what's important for being able
to see these modes is whether the motion
of the molecule actually changes the dipole moment.
So if we have a water molecule, you know, we have the oxygen
and the two hydrogens, those bonds can stretch
and they can do that in a symmetric way
like they can move together and face with each other,
and that obviously changes the dipole moment, right?
Like those bonds are getting longer,
things are spreading out.
They can also do it in an asymmetric way, so like this.
And that changes the dipole moment too.
And you can also have the bending.
And so it's-- we can see pretty intuitively that all
of these motions are going to be IR active and change the dipole.
Incidentally here's where they show up in wave numbers.
What we're going to learn how
to do is use our group theory operations to figure
out which vibrational modes are infrared and Raman active.
Because in the case of water,
you can do the little vibrating water molecule dance and figure
out which ones are going to change the dipole moment.
It's not always very obvious particularly
if the molecules that all complex.
[ Pause ]
So, how do we know if something produces a vibrational
Raman spectrum?
There the vibrational mode is going to be Raman active
if it changes the electric polarizability,
and we just learned what that is.
So, CO2 doesn't have a dipole moment but if we think
about its polarizability the size of it's electron cloud,
if we have the mode where the two oxygens are stretching
that's going to lengthen the overall molecule
so now we are getting to the case
where the bond length is changing, right,
because we are talking about vibrations.
The overall molecule is lengthening
and shortening as it vibrates.
And that is going to change the shape of the electron cloud so,
the polarizability changes in molecules as Raman active.
So, we're going to see that mode.
You can also imagine modes that aren't really going
to change the polarizability, so for example if this thing goes
in an asymmetric fashion, that's going to look the same, right?
So, not-- not active.
So, now let's-- we're not going to analyze this
in any great detail but let's just look
at the vibrational modes of SF6.
So here are a bunch of them.
And my objective here is to convince you that for molecules
that are simpler than something like water in CO2,
you don't actually want to try
to visualize all the possible vibrational modes
and decide whether they change the dipole moment and/or
the polarizability.
So we need a better way to do this,
and we're going to use group theory.
So, remember I said when we talk about bonding and we're doing,
you know, something that looks like the hardest thing
in the world to get answer that we already know.
That was like martial arts practice.
So, you want to be able to do this over and over again
in the situation where it doesn't matter, so that,
when you get to the problems where it's not obvious,
you can do it automatically.
So here's what we're actually going to use this for,
where we need the answer.
All right, so when we are talking about molecular motion
and we want to use group theory to describe it.
Our basis is now a little x, y,
and z coordinate system on each atom, why?
So when we are talking about bonding, we set up our basis
so that the objects we are looking
at had the symmetry of the bonds.
Well, now what we care about is how those atoms
and the molecule are moving relative to each other.
And so, we're interested in looking at the, you know,
the spatial coordinates of each atom.
So that's how we set up this basis.
[ Pause ]
So we have our little x, y, and z unit vectors on each atom.
And so, you can see what's coming.
If I set up the transformation matrices for all
of these things, I'm going to get huge matrices
because we have nine elements in our basis for water.
So I'm not actually going to do that.
You can do it if you want, it works fine, but what I'm going
to do instead is use the shortcut where we look
at what changes and what doesn't change.
So remember if we have something
when we do a particular operation,
if we have something it doesn't change, it contributes plus one
to the character of the operation.
If it changes sign, it contributes minus one.
And if it switches places with another element
in the basis, we get zero.
So that's what we're going to do.
[ Pause ]
And again, you can right down the 9 by 9 matrixes
if you want to but enjoy that.
OK, so down here on the bottom we have our representation
for C2V a point group of water.
And we're going to look at what character we get
under all of these operations.
So if we do the identity, obviously nothing changes
so we get nine for the character.
Now, let's do a C2 rotation.
So remember the way I define this,
the screen is the xz plane so,
x and y are changing sign z stays the same.
So what do I mean by z2y2 and z2?
Those are for the oxygen atom.
So the red hydrogen is one, the oxygen is 2
and the blue hydrogen is 3.
So, the little unit vectors in the middle change sign.
And all of the arrows on one and three,
the hydrogen atoms get swapped so they contribute zero.
And so I'm going to add up all of those little vectors.
So we get two of them contributing minus 1
because they change sign.
One of them contributing one because it stays the same
and then a bunch of zeros and that all adds up to minus one.
What do you think?
Anybody have any questions about that?
Yes?
>> Can you explain how they're all zeroes through the C2?
>> Sure, because remember we said that if something switches
into another element of the set, it contribute zero.
So when I do a C2 rotation my oxygen is in the middle
and the hydrogens are on the sides, they swap around.
So the oxygen that was here is--
or sorry, the hydrogen that was here is now over there.
And so--
>> Orientation for like the x is like going to be
on that direction or does the orientation
for the x does not matter?
>> Well, but it's not-- remember we have a different access
system define on each atom because that's our basis.
So the x that's over here was the x for hydrogen one.
And the x that was over here is the x for hydrogen two
and they change places so they get zero, whereas the one
in the middle just flip around and so it change sign.
It's stayed in the same place and change sign
as supposed to swapping places.
So, we are going to spend sometime on this.
We're going to go over it in more detail next time.
I think I'm going to stop here just because I want
to give you a little more time to think about this.
But we still have 4 minutes
and I have some other stuff I need to say.
So, you don't have to run away immediately.
I want to point out that I have posted a bunch
of practice problems so I posted some problems having to do
with direct notation because some people ask
for some extra practice with that.
There is also a tutorial that I found, you know,
sort of explaining a little bit more about how
to deal with direct notation.
I am going to have Thursday office hours.
They're going to be 2:30 to 3:30 so it's not possible
to make everybody happy as far as the timing but I'm
at least trying to sandwich it between two classes
so maybe you can just come for the first half or the last half.
Office hours have been really great.
So, you know, those who haven't made, I encourage you to show
up we have a good discussions.
A lot of people come I think, you know, it's fun for me.
I think people are getting something out of it.
All right, that's about it and I will see you next time.
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