Tip:
Highlight text to annotate it
X
[ Silence ]
>> Good morning.
Let's get started.
It's time.
>> It's time.
>> Yes, it does.
>> It is. It's time for P chem.
So, it's really cool that everyone comes to office hours.
I like office hours.
It's fun to get to interact with everyone
on a more personal basis.
I'm thinking about having more of it maybe on Thursday.
So, if you have an opinion about when it should be on Thursday,
please post that on the Facebook page and I'll look at it
and see what-- and see if there's any sort of consensus.
My schedule isn't completely flexible
but I will take people's preferences into account.
So, you know, go ahead and if you-- if you have an opinion,
please put it on the Facebook page.
So we're-- At this point, we're talking
about rotational spectroscopy.
We're going through the different kinds
of excited states that molecules can be put into as a result
of interacting with electromagnetic radiation.
And last time, we looked at this picture that's--
that's the big picture of spectroscopy.
And let me see if I can fix the screen 'cause that's going
to be annoying.
All right, much better.
OK. So right now, we are way in the bottom
of the ground electronic state and we're all the way
in the bottom of that well in the excited--
in the ground vibrational state.
And we're just talking
about exciting rotational transitions on their own.
And last time, we got around to talking about a rigid rotor
in a plane, so like a linear diatomic molecule that's
in a plane and it's rotating about the Z axis.
Now, we need to talk about the more general case of something
that is rotating on a sphere and it has more degrees of freedom.
And again, you've seen this last quarter.
This was described as a particle
on a sphere and/or the hydrogen atom wave functions
which of course are familiar from general chemistry too also.
So we can write down our Schrodinger equation for this.
And now in our kets, we have two quantum numbers
to keep track of, L and M sub L 'cause those are the quantum
numbers for the spherical harmonics.
And we remember what their values need to be.
And if we write down the Hamiltonian
in spherical coordinates with R fixed
because of course we're talking about a molecule
that we are assuming to be a rigid rotor,
so the atoms aren't vibrating around.
Here's what we get for the Hamiltonian.
And that should look really familiar from last quarter.
And I definitely recommend if it's a little rusty, go back
and remind yourself how to convert things into spherical
and cylindrical coordinates, and check out the Hamiltonian score
for these things 'cause I'm not going to go through
and solve the-- shorting your equations for these systems.
I'm assuming that you've already done it.
We can just use the results.
I am however writing them down in direct notation just
so we can get use to making that transition between these things.
OK. So here's the results.
We can write our Hamiltonian in terms of the angle
in the random operator, L squared.
And here's what we get.
And so, from that, we can pull out our energy eigenvalues.
[ Pause ]
So, things that should look familiar, this Hamiltonian,
the solution to Schrodinger equation in this kind
of a system, and this energy eigenvalues.
I also want to briefly talk about commutators
because the commutation relationships of the angle
and the random operators are going to be important
for things that we're doing.
And so, you know, again this is totally review
from last quarter.
If two operators commute,
then it doesn't matter what order you do them in.
And an example I know you have all seen is position
and momentum.
And we talked about this last time that in terms
of angular momentum, the equivalent pair
of complementary observables are angle and angular momentum.
We can't know those two things with infinite precision.
And so, I'm sure this is familiar enough.
But now, let's look at it in terms
of the angular momentum operators.
So, angular momentum is going to come up over
and over again in P chem.
So this is kind of the most literal version of it
when we're actually talking about a molecule rotating around
and we're looking at its rotational states.
But it's worth spending a lit bit of extra time thinking
about angular momentum because we're also going to need to deal
with it in terms of spin.
And things like electrons, protons,
C-13 nuclei have this intrinsic property called spin
that is kind of mysterious actually.
But it behaves in the same way as angular momentum.
Mathematically, we can treat it using the same formalism
as something moving around.
And so, this is going to come up over and over again
and it's worth looking at these things.
OK. So if we look at our angular momentum operators in the X, Y,
and Z directions, here's how they are defined.
And, you know, again this should be familiar from last quarter
but maybe we're looking at in a little bit different context.
The main thing that I want to point out here is well first
of all just to remind you what they are.
And also I want to point
out that they don't commute with each other.
And in fact, they have a special commutation relation.
You can prove this to yourself.
It's kind of tedious but if you look at these at the commutators
of this angular momentum and work them out,
here's what you get for the commutators.
The commutator of LX and LY is ILZ.
The one for LZ and LX is ILY, and et cetera.
And we call that a cyclic commutation relationship.
So, we have the set of three operators
and their commutators are related
to each other in a cyclic way.
Question in the back?
[ Inaudible Remark ]
The capital D is the-- it's the partial derivative with respect
to whatever the subscript is so it's defined
down here on the right.
Just it's a useful short hand for later on.
We're not going to have so much space.
OK. So that's just some-- those are some useful properties
of the angular momentum operators.
I'm going to have you prove
in the homework another one of their properties.
But-- So now, let's look at the actual spherical harmonics.
So we've been talking about a particle on a sphere
or the general case of a molecule that is free to move
around in any way in space and we have to deal
with its angular momentum about each of the three axes.
So we have, you know, we can have rotation around Z or Y,
or X. And we have to be able to deal with that.
So, OK, they don't commute with each other.
Let's look at the eigenfunction functions of L squared.
So L squared is the total angular momentum operators.
This is a pretty fundamental property in quantum mechanics.
And just to remind you,
here's what the spherical harmonics look like.
And I see people writing stuff down.
Don't. You don't want to sit here and try
to draw all these things.
You know, these will be posted on line.
You can also Google spherical harmonics
or hydrogen atom wave functions.
And you'll see there are lots of, you know,
neat 3D representations of these that you can play with.
But I want to remind you what they are and draw a connection
to what the functions looked like mathematically because one
of the things that you are going to have to do
in the practice problems is we're going
to look at selection rules.
And we're going to say, OK, can you have a transition
from a particular state,
a particular rotational state to another one.
And you're going to have to do that based on symmetry
which means that you're going to have to take intervals
with respect to these functions and say, all right,
do these things overlap by symmetry?
And there are a few ways to do that.
One is if you're really, really good at visualizing stuff
in your head you can look at these things
and imagine whether they overlap.
Unless you are really great at drawing stuff really fast,
that's not going to work in the context in the exam
or something like that.
So, you need to, you know, remind yourself
about the symmetry properties of these things.
And so, it's important to know what the functions actually
looked like mathematically.
So, again, we're going to represent our states.
You know, you have seen them as Y sub L,
M sub L. That just means they are described
by these two quantum numbers.
We're going to look at that in direct notation
by just sticking those two quantum numbers that we have
to keep track of in the ket.
So, the state 0,0, it looks like this.
And then, as we go through, we can put in the rest
of the general polynomials.
Again, if you're practically writing this down, don't.
You can-- You can look it up.
I just want to remind you that these familiar shapes
from the hydrogen atom wave functions have mathematical
forms that are easy to write down and we know what they are.
And we can take intervals with respect to them and do things
like figure out whether they overlap and get--
derive the selection rules for different rotational states.
So, another thing that I want to mention here is
that in quantum mechanics, we call--
you know, all of these states
that we're talking about, wave functions.
And you get really used to thinking about that in terms
of an electronic wave function.
So, don't get confused about that.
Here, we're talking about different rotational states.
Later, we'll be talking about vibrational states.
There are all kinds of different things
that we have wave functions for.
OK. So that brings us to practice problems
which these are going to be posted.
So what-- So things that I would like you to be able to do,
I want you to show that L squared commutes with LZ.
So we already said that LX, LY,
and LZ don't commute with each other.
You can prove that to yourself if you want.
It takes a long time.
But I do want you to show that L squared and LZ commute.
And I would also like you to take this LZ operator
that we have in Cartesian coordinates and convert it
to spherical coordinates.
Have you done this before?
Is that something that came up last quarter?
OK. So you've seen it but haven't necessarily done it?
Is that-- yeah, it's good practice.
So go ahead and do it.
And there are also a bunch of extra practice problems.
Many of them are from the books.
Some of them are not.
Some of them I made up.
They are not posted yet but I'll do it as soon
as I get back from lunch.
So, the practice problems are going
to be posted on the website.
Two things that you need to know about them.
One is there's a lot of them.
The other one is you don't have enough information
to them all right now.
So they are practice problems for rotational spectroscopy
and vibrational spectroscopy.
So, if you don't know how to do all of them yet,
don't worry, you will.
You'll see it as we go along in class or you can read ahead
in the book if you want to.
Somebody had a question in the middle.
Yes?
[ Inaudible Remark ]
I'm not going to post the answers.
However, you know, if you ask your TAs nicely,
they'll probably help you with it in discussion.
I will definitely help you with it in office hours if you want
to come, anything like that.
But I'm not going to disclose the key.
OK. So, we talked about where rotational spectroscopy fits in,
in the grand scheme of spectroscopy.
You know, it's a very humble, modest, little spectroscopy.
It doesn't take very much energy to do it.
We talked about some properties of angular momentum
which were going to be important for a lot of different things.
Let's get into the details of rotational spectroscopy.
OK. So, one of the things that we really need to know
to get started learning about this is the rotational constant.
So, it's called B. It has a tilde over it which indicates
that it's in strange units.
And here's how it's defined.
It's just H bar squared over 2 times moment of inertia,
for the particular molecule.
Now, as we see when we look at the pictures
of different molecules that have different shapes in the book,
some things have more than one moment of inertia.
And that has some implications
for what their spectral look like.
But this rotational constant tells you something fundamental
about the molecule because the moment of inertia is in there.
And it comes up in the spectra.
So let's talk about what that tilde means.
That means that its units are in wave numbers.
And in the context of rotational and vibrational spectroscopy,
whenever you see something that has a tilde
over it, that's what it means.
It means it's in wave numbers.
And part of the-- I think one of the hardest things
about learning spectroscopy in general is that it is the land
of messed up units and sloppy notation.
And we just have to deal with it if we want
to read the literature.
It's an old field.
A lot of this stuff is historical.
It's not necessarily consistent among different parts of it.
And there are sort of different units.
So, all right.
So the wave number unit in and of itself isn't sloppy.
That's just-- It's defined as reciprocal centimeters.
And why do we use it?
Historically, it's because, you know, when we're talking
about rotational and vibrational spectra, this is a unit
that gives us reasonable values.
You know, we don't have values of, you know, gigantic numbers.
So, a typical rotational constant for little molecules
of that type that were talking about is something like of 10th
of a wave number to 10 wave numbers.
And for vibrational ones, it'll be-- they'll be larger.
So now, why do I say it's sloppy?
Well, when you get into the sloppy notation is
when people start expressing energies in wave numbers.
So that doesn't make sense, right?
We have a reciprocal wavelength
and people are referring to it as an energy.
And you'll hear this like if you go to seminars, people say it.
They're skipping some steps.
So, if we have something that's
in wave numbers we can get the frequency
of that electromagnetic radiation.
And we know that the frequency is related to the energy,
you know, if you multiply it by points constant.
So, there is a really straightforward relationship
between this and an energy,
and you'll see people use that as a short hand.
OK, somebody had a question over here.
Yeah?
>> What is the C in that equation?
>> Speed of light.
[ Inaudible Remarks ]
Yeah, that shouldn't be there.
OK. That's-- Thanks for pointing that up.
[ Inaudible Remarks ]
So, yeah. That's relevant when we start to--
talking about the energy.
But it shouldn't be in the rotational constant.
All right.
So, if we're talking about a molecule that's free to rotate
about three different axes,
now we need to consider different moments of inertia.
So, if we look at our classical rotational and kinetic energy,
we've got these three moments of inertia.
They are labeled A, B, and C just to emphasize
that it's out in free space.
You know, we could call them X, Y, Z. But, however,
we wanted to find the coordinates.
This is the general case.
So, this is a molecule-- this is the case of a molecule
that doesn't have any symmetry.
It has three separate moments of inertia.
And so, its classical angular momentum around any one
of these axes is related to its frequency.
And so, here's its overall energy.
And what we're going to be dealing
with is the quantum analog of the situation.
And we're going to look at what that looks like for the cases
of different shaped molecules.
Again, we're not going to get into huge levels of detail
about how you calculate the different moments of inertia
for molecules of different shapes.
That's a good thing to look up.
OK. So here's the general case
where we have three different moments of inertia.
And we're going to spend a little bit of time talking
about simplified cases.
So, in the rigid rotor approximation,
we're making the approximation that the bonds are rigid
and they're not moving around.
So, the inner nuclear distance stays the same.
And we also have to worry about the selection rules.
So, selection rules are just telling us, you know,
by symmetry, which transitions can we observe in the spectrum?
And the gross selection rule is sort of you can think of it as,
you know, it's sort of the large scale course rule,
is that a molecule can only have a pure rotational spectrum
if it has a dipole moment.
So, let's think about why that is.
So if we're talking about doing rotational spectroscopy,
we have some electromagnetic radiation.
It's exciting rotational transitions.
And, you know, it's interacting with the E field
of that electromagnetic radiation.
And so, you are only going to see anything if it has a dipole.
So the molecules are rotating around.
And if there's no change in the electron density of the molecule
as that happens like say you have N2 as it's rotating around,
there's nothing for you to observe.
It's like a tree falling in the forest.
You can't see it.
It doesn't interact with that radiation.
So, yes?
>> Does N equal MA plus ME?
>> Yes, it does.
So, you know, again, I'm not going to go into, you know,
too much how you calculate these things.
There's a really nice table in the book that shows you,
you know, how to get the moments of inertia.
We're mostly not going to focus on it.
OK. So we are only able to observe these transitions
if the molecule actually has a permanent dipole moment.
And of course its chemistry so you can't have a rule
without exceptions to the rule.
We'll talk about what they are as we get further on.
But the gross selection rule is you have
to have a dipole moment.
You have to have some change in electron density
as the molecule is rotating around in order to observe it.
And we get the transitions when the molecule absorbs a photon.
And it's at resonance.
It's at the right energy to excite it
to a higher rotational state.
And then, it changes from J initial to J final.
And so, here's how we write that down in a more formal way.
So, we have our transition dipole, you know,
for that rotational transition.
And we can write down its matrix element in this form
where J sub I and J sub F are the initial and final states.
And a formal way of expressing the gross selection rule is
that that transition dipole has to be nonzero.
And it turns out that the answer you get for what it has to be is
that you can have delta J being zero.
The molecule can just stay in the same state
or it can be plus or minus 1.
And I'm not going to prove that you at this point.
I will for other types of spectroscopy later.
For this one, we're just going to leave it at that.
The derivation is in your book if you want to check it out.
But for now, let's just use the result
and if everybody understands how we're writing this down,
I'm happy with that.
OK. So let's look at what the energy levels look like.
So this is for a diatomic molecule.
So, it's really simple.
So, it's a diatomic molecule.
We know that it has a dipole
or we wouldn't be able to see anything.
And here's what the spectrum looks like.
So, the notation is, you know, you'll see J plus 1 and J
or J prime and J. And the arrow is going
in the direction of the transition.
So you'll see these things written down.
All right.
So our rotational constant again is H bar squared over 2I.
And the energy for a particular level J can also be expressed
in terms of the rotational constant.
So it's just that rotational constant times J times J plus 1.
You know again, just from the eigenvalues
of the LZ operator which, you know again,
we have the result here of what it--
what it is in polar coordinates which you're going
to show in your homework.
OK. So as a result of this,
we see we have these equally spaced levels
and the rotational spectrum has these lines
that come in increments of 2B.
And remember that whenever you see a line in the spectrum,
that represents a transition.
So we have the levels.
And then, it's tempting to look at all those lines
in the spectrum and think that those correspond to the levels.
But remember that a spectral line is
where you have a transition from one state to another.
OK. So if we look at the separation
between adjacent lines and this F with a tilde is your,
you know, energy of a particular state in wave numbers.
And we can write down our separation
between adjacent lines and get relationship
between that and the frequency.
So, this is a fancy way of saying
that we can look at the spectrum.
And we know that the lines are spaced in increments of 2B.
And from that, we can calculate the rotational constant
and we can get the moment of inertia of the molecule.
And so, we can figure out, you know, something fundamental
about the molecule from this kind of spectroscopy.
OK. So, as spectroscopics method go, this one is a little lame.
It doesn't actually contain that much information.
I mean, so a lot of times you are going to know the moment
of inertia, that molecule anyway
or there are better ways to get it.
This is not, you know, the most useful method in a lab setting.
There are some situations where this is useful which we're going
to talk about a little later.
The main thing is in space.
It's really cold out there and you don't have the luxury
of aiming a giant laser at some galaxy
and seeing what molecules are there.
You have to deal with the ambient radiation.
OK. So before we talk about applications,
let's just go through, you know again,
the notation is a little bit confusing.
Let's just go through and recap what everything is.
OK. So E sub J is the energy of some rotational level J.
And that's in normal girdled energy units.
F of J with the tilde or root is the energy
of the level J in wave numbers.
So again, we can convert readily between real energy units
and wave numbers because we know the relationship there.
And if we have the-- if we have mu of J of the transition J
to J plus 1, so again, mu with a tilde
over it is your spectral frequency
but it's in wave numbers.
And that is for a particular transition J to J plus 1.
And that corresponds to the position of the line
that you see on the spectrum
when something changes from J to J plus 1.
And that sounds a little bit convoluted in terms of,
you know, thinking about the energy level diagrams.
But it's important because that's what we actually measure.
If we take a rotational spectrum,
that's what we're going to see.
And so, we have to know how to look at that, and then back
out all of these other stuff that tells us about the states.
And then, I have one more confusing notational issue
to remind everyone about which is that mu is the reduced mass.
And that's a constant.
And there's also an operator called mu
which is the dipole moment.
And of course than an operator.
How do you know the difference?
Context. And if you get confused, please ask.
So, I know there are a lot of notational things
that are confusing and hard to get used to.
We just have to deal with them.
It's an old field.
It's been around for along time.
It's something that we just have to learn to read.
OK. So let's talk about our rotational energy levels
in a little bit more detail.
So, we're back to talking about the diatomic molecule.
And these are things that we've already seen.
OK, we won't talk about that.
So we don't need to go into more detail about it.
OK. So what I want to point out now is
that real molecules might not always follow the rigid
rotor approximation.
And that's something that we should be aware of.
OK. So, I'm going to make this point by showing real data.
So, it's just a table of data.
Don't write down all these numbers.
But I think it really makes the point
if you see what's going on.
OK. So HCl, we can measure real numbers
for these rotational transitions.
And so, I have some
of the actual measured numbers for these states.
So, the radius that we measure for HCL if we look at, you know,
again, we can just look at the spacing between the states
and get the rotational constant and measure these things.
And if we do that for different transitions in the spectrum,
here's what we get for radius, in nanometers.
So if we take the one from-- going from three to four,
that's the frequency it has.
And here is the bond length that we get if we calculate it.
And now, if we take the higher energy states,
the bond lengths starts
to increase a little bit that we measure.
And again, if we keep going, it increases a little bit more,
and a little bit more.
And as we go up to higher energy states,
our bond is actually starting to stretch.
So what's happening is we have our HCl molecule.
We're putting some energy in and it's rotating it around.
And at low energies, the bond does stay rigid.
But at higher energies, it's rotating faster and faster
and there's some centripetal distortion there.
And we can compensate for that.
There is a correction term for diatomic molecules.
I'm not going to make you use it for anything
in particular right now.
I just want you to know that it exists.
So, it's important to be aware that a lot
of times we are using approximations
because that makes things easy to treat.
And, you know, we can understand the basics
of how something works.
But we should always know
about the assumptions behind the approximations
that we're making.
And understand when they're appropriate
and when they're not.
So, if we're looking at really high energy states,
this rigid rotor approximation might not be the best.
It also depends on the particular chemical bond
that you're looking at.
So, if it's a really rigid bond, if the--
you know, if it's a very stiff kind of bond,
then this isn't going to happen until much higher energy
than it will for fluffier bond of it that can move around more.
OK. So let's talk about the types of rigid rotors.
So again, there's a nice table on your book
of what all the moments of inertia are.
But I just want to talk conceptually
about what they look like.
OK. So we have diatomic and other linear molecules.
And the moments of inertia are defined differently here.
And in the case of diatomic molecules,
we really only have one axis of rotation
that we're worried about.
So, if we have a diatomic or a linear molecule, you know,
we're looking at, you know, the Z axis is here and we're talking
about rotation in this plane.
We don't have to worry about the larger picture of, you know,
what's happening if it's rotating
on a sphere in that case.
And so, the degeneracy of those states, G is the degeneracy
of state J is just 2J plus 1.
And so, all that means is that as we go to higher
and higher energy, the states get more degenerative.
There are more ways to generate that state
than if you're in the low energy.
And so, if you have, you know, zero angular momentum
about a particular axis, there's only one way to do that.
But then as we add more energy,
the higher states become more populated.
And part of the reason for that is
that they have higher degeneracy.
OK. So we can also look at an asymmetric molecule
that has three different moments of inertia.
And what that means is-- so, say for a water molecule
if I rotated around the Z axis or if I rotated around X or Y,
each of those things is different.
It doesn't have any symmetry in that sense.
So, you know, in this case, we are not talking about, you know,
we have spent all this time talking about rotations
as symmetry operations.
Here we are not talking about it in that sense.
We're just talking like, all right, there's a water molecule
in the gas phase minding its own business.
And it can rotate about the X, Y and Z axes.
You know, in the X and Y cases, that's not a symmetry operation.
But it's still doing that, and in terms
of rotational spectroscopy we have to worry about it.
And remember, you know,
on the picture here the second molecule here is CO2.
One of these things is not like the others.
Remember that, you know, the gross selection rule is
that you have to have a dipole moment to see
that pure rotational spectrum.
So, I put it up here because it's an example
of a linear molecule for this particular type of spectroscopy.
We're not going to see a spectrum for it.
OK. So the other types of rigid rotors
that we have are symmetric rotors and spherical rotors.
And again, these names are a little confusing.
So the symmetric rotor is something like ammonia
where we have two different types of rotation
that we have to worry about.
So, we can rotate it around the Z axis.
So that's around its principal axis of rotation.
And then, it has two other equal moments of inertia and that's
because it's symmetric in the sense that if we rotate around X
or we rotate around Y, those look the same.
So, that's the sense in which it's a symmetric rotor.
It's not the same as the symmetry operations we talked
about in the point group.
And so, one consequence of that is it has two different
rotational concepts and they are called A and B.
And they are defined as parallel and perpendicular.
Again, parallel and perpendicular to what?
The principal axis of rotation.
And on the bottom is an example of part
of the rotational spectrum of CF3I
which of course is a symmetric rotor.
So, there are lots of transitions going on there.
OK. So for the spherical rotor, all the moments
of inertia are the same.
And that's all it-- that's all of it is meant by spherical.
So something like methane and I posted [inaudible] molecule
like SF6, a buckyball, anything like that is going
to be a spherical rotor.
And we can simplify things by noting that all of these moments
of inertia are the same.
OK. So let's look at this further
in the case of a symmetric rotor.
And I just want to draw the parallel
between the classical and quantum cases.
So for the classical case, here's the angular momentum.
We've got two moments of inertia that are equal.
We are calling them B and C. And, you know,
so that has to do with I perpendicular.
And we've also got I parallel which is the unique one.
That's about the principal axis.
And so, we can write down the total angular momentum.
We can write down the energy in terms of that.
And then, we can also look at this in the quantum case
by just making the analogy that we know what the item values
of the total angular momentum are.
And we can relate it to the expression for the position
of the lines in the spectrum.
So, here's what you're going to get in terms
of where the lines show up in the spectrum with respect
to the two rotational constants, A and B. All right.
So, other things that we need to think about.
We have different rotational quantum numbers here
because rotation is quantized around each axis
that we're worried about.
So we've got a rotation about the principal axis,
and then we've also got these other two rotational--
these other two sets of rotational motion.
And we have quantum numbers for all of those things.
And so, what that means is that, you know, if we have--
if K equals zero, that means there's no rotation
about the principal axis.
So, the molecule is in space and it's rotating, you know,
purely around X or Y or somewhere in between there.
And if K equals plus or minus J, that means all the rotation is
about the principal axis.
So, it's just rotating like this.
So that's how you can think about the relationship
between those quantum numbers.
It's just-- We're just talking
about what direction is it quantized?
And again, those are always quantized
in increments of H bar.
So, it's a-- it's written as H there.
It should be H bar.
And so, for symmetric rotors, the specific selection rule is
that we can have rotational transitions where K can--
the change in K is zero.
And we can have delta J being plus or minus 1.
And then, K also has to take these values up to
and including plus and minus J. Question?
[ Inaudible Remark ]
It is H bar.
All right.
So what that means is that for spherical rotors,
we have a lot more degeneracy because there are more axes
that are-- where things are quantized to be worried about.
So in general, this rotor has a 2J plus 1 fold degeneracy
because of its orientation in space.
And it has another one with respect to its orientation
in the molecular frame.
So we've got an axis in the molecule.
We've got an axis because of its orientation in space.
And the degeneracy for this thing gets larger
really quickly.
So if we have J equals 10, so we're only in the, you know,
10th state, there are 441 ways to get that.
And this has some important consequences
for what the spectra looked like.
So, this is a simulated spectrum for FCl03.
And it's at 1 Kelvin.
So its really, really cold.
This molecule is not rotating very much.
So when it's really cold, you know, we're used to think
about if we don't have much energy, everything must be piled
in the ground state, right?
Well, in these kinds of experiments, that's not true.
And the reason for is that the ground state is the lowest
energy, sure, but there's only one way to get it.
That state is non-degenerate.
Everything-- You know, there'd be--
there are entropy considerations,
if you will, to getting that.
There's only way to do it.
So, it's rare.
Whereas, the little bit higher energy states just have more
ways to get to that value.
And so, we see that the maximum population is not piled
into the lowest energy state even
at pretty cold temperatures.
If we look at something at more like room temperature,
we see a couple of things.
One is that the distribution is shifted
and also it's broadened out a lot.
So, some really high energy states can be populated
because there's a lot of degeneracy.
There's a lot of different ways to get that.
And this is the introduction to kind of the first part
of [inaudible] which we'll see at the very end of the class.
But we'll try to bring in at least conceptual representations
of this all along because it's good
to have a feel for how it works.
So, to use another analogy, it's like saying, you know,
the most likely state for the first midterm is
that everyone gets a 100 because everybody is really smart.
And that's true.
That's the lowest energy state, right?
But it's really, really unlikely because there's only one way
to get everything right.
And there are lots of ways to make little mistakes.
So those states are populated.
All right.
So the last thing I want to mention is an actual application
of rotational spectroscopy.
So, I mentioned that this is the main one.
It's really useful for looking interstellar molecules.
So here's a picture of this cloud of gas that has a bunch
of molecules at it that's in it that's out in the space.
And many of molecules that are known to be
out there were discovered near this feature.
And, you know, how do you know what molecules are in space?
So again, you can't shine a giant laser out there
and do laser spectroscopy.
You have to deal with the ambient radiation that's there
and it's really low energy.
Space is cold.
And so, the way people do that is by, you know,
measuring the spectra using a radio telescope.
And then, they make mixtures of molecules in their labs.
So you get these spectra that are a big mess.
There's a whole bunch of different rotational states.
And then, they can kind of guess based on pattern recognition
and knowing what the spectra of different molecules look like
and make up mixtures of molecules in the lab
that can match the spectra.
So, this is from-- this data is from the lab
of Professor Lucy Ziurys who is at University of Arizona.
I visited her lab a couple of years ago.
It's pretty interesting.
So, she has two things.
She has these giant telescopes like she's in charge of one
of these facilities in Hawaii where you can--
you know, she can login from her computer in Arizona
and run these giant telescopes.
And she gets spectra from space that have a bunch
of rotational features of different molecules.
And then, in order to figure out what's there, she goes
and makes mixtures of molecules that she thinks might match
in vacuum chambers that are really cold
in her lab and compares the two.
And so, there's a lot of effort there in, you know,
first of all instrumentation as far as, you know,
being able to measure these things.
And also in data analysis because you have to do a lot
of pattern recognition and sift through a lot of spectra
and compare whether they are the same or not.
So, this is what this stuff is actually used for in real life.
And here's the instrument that you need to do it.
So it's kind of exotic.
It doesn't come up much.
Its neat but it's not used all over the place.
Next time, we're going to talk about vibrational spectroscopy
which is used all the time in research labs, you know,
and you've probably all used it yourself in the context of IR
and maybe raman as well.
OK. Happy Martin Luther King Day on Monday.
And I'll see everyday on Wednesday.
[ Inaudible Discussions ] ------------------------------05c536bc3a2a--