Lecture 06. Rotational Spectroscopy Pt. II.
Martin, Rachel
 Principles of quantum mechanics with application to the elements of atomic structure and energy levels, diatomic molecular spectroscopy and structure determination, and chemical bonding in simple molecules. 
 Good morning. Let's get started. It's time. It is. It's time for PCAM. 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 them 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 if there's any sort of consensus. My schedule isn't completely flexible, but I will take people's preferences into account. So go ahead and if you have an opinion, please put it on the Facebook page. 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. Last time we looked at this picture, that's the big picture of spectroscopy. Let me see if I can fix the screen because that's going to be annoying. Much better. Right now, we are away in the bottom of the ground electronic state and we're all the way in the bottom of that well 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 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 because 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're 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 Hamiltonians for these things because I'm not going to go through and solve the Schrodinger 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 used to making that transition between these things. Okay, so here's the results. We can write our Hamiltonian in terms of the angular momentum operator L squared and here's what we get and so from that we can pull out our energy eigenvalues. So things that should look familiar. This Hamiltonian, the solution to the Schrodinger equation in this kind of a system and these energy eigenvalues. I also want to briefly talk about commutators because the commutation relationships of the angular momentum 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. An example that I know you've 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. Well, we're actually talking about a molecule rotating around and we're looking at its rotational states. But it's worth spending a little 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, C13 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. Okay, so if we look at our angular momentum operators in the x, y, and z directions, here's how they're defined. And again this should be familiar from last quarter but maybe we're looking at it in a little bit different context. The main thing that I want to point out here is well first of all just 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 these angular momentum operators and work them out, here's what you get for their commutators. The commutator of Lx and Ly is I Lz. The one for Lz and Lx is I Ly. And we call that a cyclic commutation relationship. So we have this set of three operators and their commutators are related to each other in this cyclic way. Question in the back. What's the capital D again? The capital D is the partial derivative with respect to whatever the subscript is. So it's defined down here on the right. Just it's a useful shorthand for later on. We're not going to have so much space. Okay 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 harmonic. 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 okay they don't commute with each other. Let's look at the eigenfunctions of L squared. So L squared is the total angular momentum operator. 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 online. 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 look like mathematically. Because one of the things that you're going to have to do in the practice problems is we're going to look at selection rules and we're going to say, okay 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 integrals with respect to these functions and say, alright 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're really great at drawing stuff really fast that's not going to work in the context of an 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 look like mathematically. So again we're going to represent our states, you know, you've seen them as y sub l m sub l. That just means they're described by these two quantum numbers. We're going to look at that in Dirac notation by just sticking those two quantum numbers that we have to keep track of in the, the ket. So the state zero zero looks like this. And then as we go through we can put in the rest of the Legendre polynomials. Again if you're frantically writing this down don't. You can, you can look it up. I just want to remind you that these, 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 integrals 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. Okay 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, LI 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? Okay. So you've seen it but haven't necessarily done it. Is that? Yeah, it's good practice. Should go ahead and do it. And there are also a bunch of extra practice problems. Many of them are from the book. Some of them are not. Some of them I made up. They're not posted yet but I'll do it as soon as I get back from lunch. So the practice problems that 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? I'm not going to post the answers. However, 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 up with anything like that. But I'm not going to just post the key. Okay, so we talked about where rotational spectroscopy fits in in the grand scheme of spectroscopy. It's a very humble modest little spectroscopy. It doesn't take pretty much energy to do it. We talked about some properties of angular momentum which are going to be important for a lot of different things. Let's get into the details of rotational spectroscopy. Okay, 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 spectra 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, alright, 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, 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 the type that we're talking about is something like a tenth of a wave number to ten 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 Planck's constant. So there is a really straightforward relationship between this and an energy. And you'll, you'll see people use that as a shorthand. Okay, somebody had a question over here. Yeah. What is the C in that equation? Speed of light. How did you get from the middle portion there to that portion? Because the Hr is going away, it's, you know, kind of like the H, I stir or whatever. That shouldn't be there. Okay. Thanks for pointing that out. So, yeah, that's relevant when we start 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 things, now we need to consider different moments of inertia. So if we look at our classical rotational kinetic energy, we've got these three moments of inertia. They're labeled A, B, and C just to emphasize that it's out in free space. You know, we could call them XYZ, 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 it's 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. Okay. 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 inter nuclear distance stays the same. And we also have to worry about the selection rules. So selection rules are just telling us by symmetry, which transitions can we observe in the spectrum. And the gross selection rule, sort of the, you can think of it as 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're only going to see anything if it has a dipole. So the molecule's 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 NB? 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. Okay. So we're only able to observe these transitions if the molecule actually has a permanent dipole moment. And of course, it's 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 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 one. And I'm not going to prove that to 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. Okay, 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'll see J plus one 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. Alright, 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 one. Again, just from the eigenvalues of the LZ operator. Which, you know, again, we have the result here of what it is in polar coordinates, which you're going to show in your homework. Okay, 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. Okay, 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. Okay, so as spectroscopics method go, this one's a little lame. It doesn't actually contain that much information. I mean, so a lot of times you're going to know the moment of inertia of 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 it is useful, which we're going to talk about a little later, but 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. Okay, 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. Okay, so E sub j is the energy of some rotational level j. And that's in normal good old energy units. F of j with a tilde over it 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 nu of j, of the transition j to j plus 1, so again, nu with a tilde over it is, 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, to know how to look at that and then back out all of this other stuff that tells us about the, 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 that's 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 that's been around for a long time. It's something that we just have to learn to read. Okay, so let's talk about our rotational energy levels in a little bit more detail. So we're back to talking about a diatomic molecule. And these are things that we've already seen. Yeah, we've already talked about that, so we don't need to go into more detail about it. Okay, 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. Okay, 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. Okay, so for 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, 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 the radius in nanometers. So if we take the one from, going from 3 to 4, that's the frequency it has. And here's the bond length that we get if we calculate it. And now if we take the higher energy states, the bond length starts to increase a little bit that we measure. And again, if we keep going, 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 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 is a very simple process. So it's important to be aware that a lot of times we're using approximations because that makes things easy to treat and 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 it's a very stiff kind of bond, then this isn't going to happen until much higher energy than it will for a floppier bond that can move around more. Okay, so let's talk about the types of rigid rotors. So again, there's a nice table in your book of what all the moments of inertia are. But I just want to talk conceptually about what they look like. Okay, 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 linear molecule, you know, we're looking at, 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 degenerate. There are more ways to generate that state than if you're in 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. Okay, 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 rotate it around the z axis or if I rotate it 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're not talking about, you know, we've spent all this time talking about rotations as symmetry operations. Here we're not talking about it in that sense. We're just talking about, 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 the 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. Okay, 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 it around x or we rotate it 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 constants. And they're called A and B. And they're defined as parallel and perpendicular. And 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. Okay, so for the spherical rotor, all the moments of inertia are the same. And that's all that is meant by spherical. So something like methane, an icosahedral 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. Okay, so let's look at this 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're calling them B and C. And 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. 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 eigenvalues 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 sets of rotational motion. And we have quantum numbers for all of those things. And so what that means is that if k equals 0, that means there's no rotation about the principal axis. So the molecule is in space and it's rotating 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. We're just talking about what direction is it quantized. And again, those are always quantized in increments of h bar. So 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 0. And we can have delta j being plus or minus 1. And then k also has to take these values up to an including plus and minus j question. It is h bar. Alright, 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 large really quickly. So if we have j equals 10, so we're only in the tenth state, there are 441 ways to get that. And this has some important consequences for what the spectra look like. So this is a simulated spectrum for FClO3. And it's at one kelvin. So it's really, really cold. This molecule is not rotating very much. So when it's really cold, you know, we're used to thinking 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 that is that the ground state is the lowest energy, sure, but there's only one way to get it. That state is non-degenerate. There are entropy considerations, if you will, to getting that. There's only one way to do it, so it's rare. Whereas the little bit higher energy states just have more ways to get 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 stat mech, 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 the most likely state for the first midterm is that everyone gets 100 because everybody's 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 at interstellar molecules. So here's a picture of this cloud of gas that has a bunch of molecules out in it that's out in space. And many of the molecules that are known to be out there were discovered near this feature. And 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 measuring these spectra using a radio telescope. And then they make mixtures of molecules in their lab. So you get these spectra that are a big mass. 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 data is from the lab of Professor Lucy Zerise who is at University of Arizona. I visited her lab a couple 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 she can log in 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're the same or not. So this is what the 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. It's 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. Okay, happy Martin Luther King Day on Monday, and I'll see you everybody on Wednesday.
 Good morning. Let's get started. It's time. It is. It's time for PCAM. 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 them 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 if there's any sort of consensus.  My schedule isn't completely flexible, but I will take people's preferences into account.  So go ahead and if you have an opinion, please put it on the Facebook page. 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. Last time we looked at this picture, that's the big picture of spectroscopy. Let  me see if I can fix the screen because that's going to be annoying. Much better. Right now,  we are away in the bottom of the ground electronic state and we're all the way in the bottom  of that well 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 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 because 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're 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 Hamiltonians for these things because I'm not going to  go through and solve the Schrodinger 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 used to making that transition between these things. Okay, so here's the  results. We can write our Hamiltonian in terms of the angular momentum operator L squared  and here's what we get and so from that we can pull out our energy eigenvalues. So things  that should look familiar. This Hamiltonian, the solution to the Schrodinger equation in  this kind of a system and these energy eigenvalues. I also want to briefly talk about commutators  because the commutation relationships of the angular momentum 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. An example that I know you've 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. Well, we're actually talking  about a molecule rotating around and we're looking at its rotational states. But it's  worth spending a little 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,  C13 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. Okay, so if we look at our angular momentum operators  in the x, y, and z directions, here's how they're defined. And again this should be  familiar from last quarter but maybe we're looking at it in a little bit different context.  The main thing that I want to point out here is well first of all just 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 these angular momentum operators  and work them out, here's what you get for their commutators. The commutator of Lx and  Ly is I Lz. The one for Lz and Lx is I Ly. And we call that a cyclic commutation relationship.  So we have this set of three operators and their commutators are related to each other  in this cyclic way. Question in the back.  What's the capital D again?  The capital D is the partial derivative with respect to whatever the subscript is. So it's  defined down here on the right. Just it's a useful shorthand for later on. We're not  going to have so much space. Okay 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  harmonic. 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 okay they don't commute with  each other. Let's look at the eigenfunctions of L squared. So L squared is the total angular  momentum operator. 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 online. 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 look like mathematically. Because one of the things that you're going  to have to do in the practice problems is we're going to look at selection rules and  we're going to say, okay 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 integrals with respect to these functions and say, alright  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're really great at drawing stuff really fast  that's not going to work in the context of an 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 look like mathematically. So again we're going to represent  our states, you know, you've seen them as y sub l m sub l. That just means they're described  by these two quantum numbers. We're going to look at that in Dirac notation by just sticking  those two quantum numbers that we have to keep track of in the, the ket. So the state  zero zero looks like this. And then as we go through we can put in the rest of the Legendre  polynomials. Again if you're frantically writing this down don't. You can, you can look it  up. I just want to remind you that these, 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 integrals 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. Okay 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, LI 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? Okay. So you've seen it but haven't necessarily done it. Is  that? Yeah, it's good practice. Should go ahead and do it. And there are also a bunch  of extra practice problems. Many of them are from the book. Some of them are not. Some  of them I made up. They're not posted yet but I'll do it as soon as I get back from  lunch. So the practice problems that 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? I'm not going  to post the answers. However, 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  up with anything like that. But I'm not going to just post the key. Okay, so we talked about  where rotational spectroscopy fits in in the grand scheme of spectroscopy. It's a very  humble modest little spectroscopy. It doesn't take pretty much energy to do it. We talked  about some properties of angular momentum which are going to be important for a lot of different  things. Let's get into the details of rotational spectroscopy. Okay, 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 spectra 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, alright, 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, 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 the type that we're talking about is something like a tenth of a wave number to ten 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 Planck's constant. So there  is a really straightforward relationship between this and an energy. And you'll, you'll see  people use that as a shorthand. Okay, somebody had a question over here. Yeah.  What is the C in that equation? Speed of light.  How did you get from the middle portion there to that portion? Because the Hr is going away,  it's, you know, kind of like the H, I stir or whatever.  That shouldn't be there. Okay. Thanks for pointing that out. So, yeah, that's relevant  when we start 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  things, now we need to consider different moments of inertia. So if we look at our classical  rotational kinetic energy, we've got these three moments of inertia. They're labeled  A, B, and C just to emphasize that it's out in free space. You know, we could call them  XYZ, 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 it's 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. Okay. 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 inter nuclear distance stays the same.  And we also have to worry about the selection rules. So selection rules are just telling  us by symmetry, which transitions can we observe in the spectrum. And the gross selection rule,  sort of the, you can think of it as 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're only going to see anything if  it has a dipole. So the molecule's 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 NB? 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. Okay. So we're only able to observe these transitions if the molecule  actually has a permanent dipole moment. And of course, it's 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 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 one. And I'm not going to prove that to 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.  Okay, 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'll see J plus one 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.  Alright, 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 one. Again, just from the eigenvalues  of the LZ operator. Which, you know, again, we have the result here of what it is in polar  coordinates, which you're going to show in your homework. Okay, 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. Okay, 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. Okay, so as spectroscopics method go, this  one's a little lame. It doesn't actually contain that much information. I mean, so a lot of  times you're going to know the moment of inertia of 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 it is useful, which we're going to talk about a little later,  but 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. Okay, 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. Okay, so E sub j is the energy of some rotational level j.  And that's in normal good old energy units. F of j with a tilde over it 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 nu of j, of the transition j to j plus 1, so again, nu with a tilde over it is,  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, to know how to look at that and then back out all of this  other stuff that tells us about the, 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 that's  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 that's been around for a long time.  It's something that we just have to learn to read. Okay, so let's talk about our rotational  energy levels in a little bit more detail. So we're back to talking about a diatomic  molecule. And these are things that we've already seen. Yeah, we've already talked about  that, so we don't need to go into more detail about it. Okay, 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. Okay, 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. Okay, so for 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,  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 the radius in nanometers. So if we take the one from, going from 3 to  4, that's the frequency it has. And here's the bond length that we get if we calculate  it. And now if we take the higher energy states, the bond length starts to increase a little  bit that we measure. And again, if we keep going, 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 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  is a very simple process. So it's important to be aware that a lot of times we're using  approximations because that makes things easy to treat and 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 it's a very stiff kind of bond, then this isn't going to happen until  much higher energy than it will for a floppier bond that can move around more.  Okay, so let's talk about the types of rigid rotors. So again, there's a nice table in  your book of what all the moments of inertia are. But I just want to talk conceptually about  what they look like. Okay, 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  linear molecule, you know, we're looking at, 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 degenerate. There are more ways to  generate that state than if you're in 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. Okay, 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 rotate it around the z axis or if I rotate it 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're not talking about, you know, we've spent all this time talking about  rotations as symmetry operations. Here we're not talking about it in that sense. We're  just talking about, 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 the  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.  Okay, 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 it around x or we rotate it 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 constants. And they're called A and B. And they're defined as parallel and  perpendicular. And 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. Okay, so for the  spherical rotor, all the moments of inertia are the same. And that's all that is meant  by spherical. So something like methane, an icosahedral 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. Okay, so let's look at this 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're calling them B and C. And 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.  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 eigenvalues 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 sets of rotational motion. And we have quantum numbers for all of those things. And  so what that means is that if k equals 0, that means there's no rotation about the principal  axis. So the molecule is in space and it's rotating 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. We're just talking about what direction is  it quantized. And again, those are always quantized in increments of h bar. So 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 0. And we  can have delta j being plus or minus 1. And then k also has to take these values up to  an including plus and minus j question. It is h bar. Alright, 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 large  really quickly. So if we have j equals 10, so we're only in the tenth state, there are  441 ways to get that. And this has some important consequences for what the spectra look like.  So this is a simulated spectrum for FClO3. And it's at one kelvin. So it's really, really  cold. This molecule is not rotating very much. So when it's really cold, you know, we're  used to thinking 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 that is  that the ground state is the lowest energy, sure, but there's only one way to get it.  That state is non-degenerate. There are entropy considerations, if you will, to getting that.  There's only one way to do it, so it's rare. Whereas the little bit higher energy states  just have more ways to get 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 stat mech, 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 the most likely state for the first midterm is that everyone gets 100  because everybody's 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 at interstellar  molecules. So here's a picture of this cloud of gas that has a bunch of molecules out in  it that's out in space. And many of the molecules that are known to be out there were discovered  near this feature. And 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 measuring these spectra using a radio telescope. And then they make mixtures  of molecules in their lab. So you get these spectra that are a big mass. 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 data is from the lab of Professor Lucy  Zerise who is at University of Arizona. I visited her lab a couple 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 she can log in 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're the same or not. So this is what the 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. It's 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. Okay, happy Martin Luther King Day on Monday, and I'll  see you everybody on Wednesday.
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Good morning. Let's get started. It's time. It is. It's time for PCAM. 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 them 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 if there's any sort of consensus. My schedule isn't completely flexible, but I will take people's preferences into account. So go ahead and if you have an opinion, please put it on the Facebook page. 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. Last time we looked at this picture, that's the big picture of spectroscopy. Let me see if I can fix the screen because that's going to be annoying. Much better. Right now, we are away in the bottom of the ground electronic state and we're all the way in the bottom of that well 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 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 because 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're 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 Hamiltonians for these things because I'm not going to go through and solve the Schrodinger 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 used to making that transition between these things. Okay, so here's the results. We can write our Hamiltonian in terms of the angular momentum operator L squared and here's what we get and so from that we can pull out our energy eigenvalues. So things that should look familiar. This Hamiltonian, the solution to the Schrodinger equation in this kind of a system and these energy eigenvalues. I also want to briefly talk about commutators because the commutation relationships of the angular momentum 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. An example that I know you've 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. Well, we're actually talking about a molecule rotating around and we're looking at its rotational states. But it's worth spending a little 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, C13 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. Okay, so if we look at our angular momentum operators in the x, y, and z directions, here's how they're defined. And again this should be familiar from last quarter but maybe we're looking at it in a little bit different context. The main thing that I want to point out here is well first of all just 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 these angular momentum operators and work them out, here's what you get for their commutators. The commutator of Lx and Ly is I Lz. The one for Lz and Lx is I Ly. And we call that a cyclic commutation relationship. So we have this set of three operators and their commutators are related to each other in this cyclic way. Question in the back. What's the capital D again? The capital D is the partial derivative with respect to whatever the subscript is. So it's defined down here on the right. Just it's a useful shorthand for later on. We're not going to have so much space. Okay 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 harmonic. 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 okay they don't commute with each other. Let's look at the eigenfunctions of L squared. So L squared is the total angular momentum operator. 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 online. 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 look like mathematically. Because one of the things that you're going to have to do in the practice problems is we're going to look at selection rules and we're going to say, okay 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 integrals with respect to these functions and say, alright 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're really great at drawing stuff really fast that's not going to work in the context of an 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 look like mathematically. So again we're going to represent our states, you know, you've seen them as y sub l m sub l. That just means they're described by these two quantum numbers. We're going to look at that in Dirac notation by just sticking those two quantum numbers that we have to keep track of in the, the ket. So the state zero zero looks like this. And then as we go through we can put in the rest of the Legendre polynomials. Again if you're frantically writing this down don't. You can, you can look it up. I just want to remind you that these, 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 integrals 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. Okay 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, LI 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? Okay. So you've seen it but haven't necessarily done it. Is that? Yeah, it's good practice. Should go ahead and do it. And there are also a bunch of extra practice problems. Many of them are from the book. Some of them are not. Some of them I made up. They're not posted yet but I'll do it as soon as I get back from lunch. So the practice problems that 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? I'm not going to post the answers. However, 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 up with anything like that. But I'm not going to just post the key. Okay, so we talked about where rotational spectroscopy fits in in the grand scheme of spectroscopy. It's a very humble modest little spectroscopy. It doesn't take pretty much energy to do it. We talked about some properties of angular momentum which are going to be important for a lot of different things. Let's get into the details of rotational spectroscopy. Okay, 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 spectra 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, alright, 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, 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 the type that we're talking about is something like a tenth of a wave number to ten 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 Planck's constant. So there is a really straightforward relationship between this and an energy. And you'll, you'll see people use that as a shorthand. Okay, somebody had a question over here. Yeah. What is the C in that equation? Speed of light. How did you get from the middle portion there to that portion? Because the Hr is going away, it's, you know, kind of like the H, I stir or whatever. That shouldn't be there. Okay. Thanks for pointing that out. So, yeah, that's relevant when we start 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 things, now we need to consider different moments of inertia. So if we look at our classical rotational kinetic energy, we've got these three moments of inertia. They're labeled A, B, and C just to emphasize that it's out in free space. You know, we could call them XYZ, 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 it's 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. Okay. 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 inter nuclear distance stays the same. And we also have to worry about the selection rules. So selection rules are just telling us by symmetry, which transitions can we observe in the spectrum. And the gross selection rule, sort of the, you can think of it as 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're only going to see anything if it has a dipole. So the molecule's 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 NB? 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. Okay. So we're only able to observe these transitions if the molecule actually has a permanent dipole moment. And of course, it's 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 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 one. And I'm not going to prove that to 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. Okay, 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'll see J plus one 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. Alright, 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 one. Again, just from the eigenvalues of the LZ operator. Which, you know, again, we have the result here of what it is in polar coordinates, which you're going to show in your homework. Okay, 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. Okay, 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. Okay, so as spectroscopics method go, this one's a little lame. It doesn't actually contain that much information. I mean, so a lot of times you're going to know the moment of inertia of 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 it is useful, which we're going to talk about a little later, but 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. Okay, 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. Okay, so E sub j is the energy of some rotational level j. And that's in normal good old energy units. F of j with a tilde over it 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 nu of j, of the transition j to j plus 1, so again, nu with a tilde over it is, 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, to know how to look at that and then back out all of this other stuff that tells us about the, 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 that's 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 that's been around for a long time. It's something that we just have to learn to read. Okay, so let's talk about our rotational energy levels in a little bit more detail. So we're back to talking about a diatomic molecule. And these are things that we've already seen. Yeah, we've already talked about that, so we don't need to go into more detail about it. Okay, 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. Okay, 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. Okay, so for 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, 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 the radius in nanometers. So if we take the one from, going from 3 to 4, that's the frequency it has. And here's the bond length that we get if we calculate it. And now if we take the higher energy states, the bond length starts to increase a little bit that we measure. And again, if we keep going, 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 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 is a very simple process. So it's important to be aware that a lot of times we're using approximations because that makes things easy to treat and 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 it's a very stiff kind of bond, then this isn't going to happen until much higher energy than it will for a floppier bond that can move around more. Okay, so let's talk about the types of rigid rotors. So again, there's a nice table in your book of what all the moments of inertia are. But I just want to talk conceptually about what they look like. Okay, 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 linear molecule, you know, we're looking at, 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 degenerate. There are more ways to generate that state than if you're in 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. Okay, 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 rotate it around the z axis or if I rotate it 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're not talking about, you know, we've spent all this time talking about rotations as symmetry operations. Here we're not talking about it in that sense. We're just talking about, 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 the 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. Okay, 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 it around x or we rotate it 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 constants. And they're called A and B. And they're defined as parallel and perpendicular. And 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. Okay, so for the spherical rotor, all the moments of inertia are the same. And that's all that is meant by spherical. So something like methane, an icosahedral 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. Okay, so let's look at this 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're calling them B and C. And 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. 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 eigenvalues 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 sets of rotational motion. And we have quantum numbers for all of those things. And so what that means is that if k equals 0, that means there's no rotation about the principal axis. So the molecule is in space and it's rotating 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. We're just talking about what direction is it quantized. And again, those are always quantized in increments of h bar. So 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 0. And we can have delta j being plus or minus 1. And then k also has to take these values up to an including plus and minus j question. It is h bar. Alright, 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 large really quickly. So if we have j equals 10, so we're only in the tenth state, there are 441 ways to get that. And this has some important consequences for what the spectra look like. So this is a simulated spectrum for FClO3. And it's at one kelvin. So it's really, really cold. This molecule is not rotating very much. So when it's really cold, you know, we're used to thinking 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 that is that the ground state is the lowest energy, sure, but there's only one way to get it. That state is non-degenerate. There are entropy considerations, if you will, to getting that. There's only one way to do it, so it's rare. Whereas the little bit higher energy states just have more ways to get 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 stat mech, 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 the most likely state for the first midterm is that everyone gets 100 because everybody's 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 at interstellar molecules. So here's a picture of this cloud of gas that has a bunch of molecules out in it that's out in space. And many of the molecules that are known to be out there were discovered near this feature. And 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 measuring these spectra using a radio telescope. And then they make mixtures of molecules in their lab. So you get these spectra that are a big mass. 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 data is from the lab of Professor Lucy Zerise who is at University of Arizona. I visited her lab a couple 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 she can log in 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're the same or not. So this is what the 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. It's 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. Okay, happy Martin Luther King Day on Monday, and I'll see you everybody on Wednesday.
fundamental property in quantum mechanics. And practice problems for rotational spectroscopy and let's look at what the energy levels look like. And in the case of diatomic molecules, we really lines show up in the spectrum with respect to the