 So, at this point we have a fairly complete understanding of the harmonic oscillator, the rigid rotor, the particle on a box, all of which can be used to describe different motions for a diatomic molecule. So, let's see what we can learn by putting those pieces together and describing the diatomic molecule as a whole. So, we know that the energies of a harmonic oscillator have this form, 1 half h nu plus one or two or three or four more copies of h nu for every excited energy level. The rigid rotor has different form for the energies, a rotational constant multiplied by the angular momentum quantum number L times L plus one. So, those are completely different states for the rotational energies and the vibrational energies. And remember that the sizes of these rotational constant and the vibrational constant, the vibrational, fundamental vibrational frequency are quite different from one another. If I convert both of those to the same units, if I convert them to units of wave numbers, remember that the fundamental vibrational frequency, that's typically on the order of several thousand inverse centimeters, whereas the rotational constant, that's somewhere in the vicinity of maybe a few tenths or maybe a few tens of inverse centimeters. So, the rotational constant is smaller by at least a couple of orders of magnitude compared to the vibrational energy. Vibrational photons that excite vibrational excitations were in the infrared photons that excite rotational excitations are in the microwave. So, real molecules of course both vibrate as well as rotate at the same time. So, rather than just treating them one at a time, we can consider the possibility that a molecule is vibrating and rotating simultaneously, in which case it'll have a vibrational quantum number, a rotational quantum number, angular momentum quantum number, and we can say that the total energy composed of the vibrational contribution to the energy and the rotational contribution to the energy, if we just sum those two together, we find that the energy will look like just the sum of these two terms. But where it will get interesting is when we see what those energy states look like. So, let's draw an energy ladder. So, here will be the energies of my molecules. So, these quantum numbers mean exactly the same thing they did for the harmonic oscillator and for the rigid rotor. The ground state, the ground vibrational state will be when n equals zero. So, let's think about what this would mean when we have n equals zero and l equals zero. n equals zero, the vibrational contribution is still going to be one-half h nu, l equals zero, with a zero here, the rotational contribution is just going to be zero. So, if I add these two pieces together, we predict that there's going to be an energy level for the molecule with an energy of one-half h nu. That's just the same zero point energy we had for the harmonic oscillator. There's no additional rotational energy, but it's certainly possible to also have a little bit of rotational energy in addition to that zero point vibrational energy. So, we'll put the molecule in the first excited rotational state. Give it l equals one, the first quantum's worth of rotational energy, and then the energy is going to be one-half h nu from the vibrational energy. And then the rotational constant times l times l plus one, that's a one and a two. So, we'll add twice the rotational constant as the energy of this state. Remember now that the rotational energy, when they're in the same units, either energy units or wave number units, the rotational energy is smaller by several orders of magnitude compared to the vibrational energy. So, I'm going to have a hard time drawing this to scale. I'll draw this level just barely above the ground state, the ground vibrational and rotational state. In fact, this extra amount of rotational energy should be several hundred or maybe even several thousand times smaller than the amount of vibrational energy. So, if I do it to scale, we wouldn't see this line. It'd be buried very close to the top of this line. So, I'll just draw a very small excitation due to the rotation. But also remember that these states are degenerate. The rigid rotor states have a degeneracy of two l plus one. So, there are, in fact, three states with l equals one slightly above the ground vibrational state. And if I continue up to the state beyond that, there's going to be one, two, three, four, five states with l equals two. And then I can continue to draw more states, seven and then nine stacked up here. The energy, let's see, the one half plus two b, that's the energy of this l equals one state. This state would have an energy of one half h nu plus six b. We've considered rotational excitations. What if I put the molecule in an n equals one state? So, somewhere up higher, we're going to have the n equals one, l equals something, l equals zero, l equals one. So, if this state is one half h nu, somewhere up here is three halves h nu. And then stacked on top of that vibrational excited state, we're going to have some rotationally excited states. A triply degenerate state with one quantum of rotational excitation, a five-fold degenerate state with l equals two, with more rotational energy. So, this would be the n equals one, l equals two state, and so on. So, we've got a whole family of rotational states stacked on top of the ground vibrational state, stacked on top of the first excited vibrational state, and then further on as we go up. That's what our energy ladder looks like. That's beginning to look more interesting and more complicated than either the harmonic oscillator or the rigid rotor on its own. Things get even more interesting when we remember that there are selection rules. The harmonic oscillator had a selection rule that required that we can change the vibrational quantum number by one in the upward direction or one in the downward direction, but no more than that. There was also the selection rule for the rotational quantum numbers. l had to change by plus or minus one, and the magnetic quantum number, the quantum number that gives rise to this degeneracy, that one was not allowed to change. So, let's say, for example, we have a molecule in this ground vibrational and ground rotational states. Which states is it allowed to make a transition to? Only the ones that obey these selection rules. So, we could, for example, go from n, l, m. If we start in the 0, 0, 0 state, ground vibrational state, ground rotational state, we could make a transition. We can increase n. We can increase l. m must stay the same. So, we're allowed to make a transition to the 1, 1, 0 state. So, that would be this transition starting here and going to this state. No, sorry, that's not correct. Going to this state, the one, the, the rotationally excited state. Notice that the energy of that transition, I'm going from energy of one-half h nu up to an energy of three-halves h nu plus a little bit of rotational energy, plus two rotational constants worth of energy. So, this, the delta e for this would be h nu for the vibrational excitation plus two times b for the rotational excitation. Some other examples, if I were starting, let's say, in the somewhat rotationally excited state. If I'm starting in, let's do m equals one for variety. Let's say I'm starting in the m equals one state. So, that would be starting here, let's say, ground vibration, one unit of excitation in the rotational energy, and then I'll call this one the m equals one state. I'm allowed to make a transition up one in n, up one in l, and I must stay the same in m. Or, I could make a transition up one in n, down one in l, and keeping m the same. So, I could make a transition either to there or to here, but this one doesn't exist. That one's not possible. There is no m equals one when l equals zero. That one violates the rules for how these quantum numbers need to be. The only state I could make a transition to from the zero one one state would be to the one two one state. If we think about what the energy of this transition would be, I'm gaining h nu for vibrational energy. I've gone from two b worth of rotational energy up to six b worth of rotational energy. So, I've gained a total of four b in rotational energy. As one final example, let's say I'm starting in the zero one zero state. I could make a transition increasing vibrationally, increasing rotationally, and keeping the same m. Or increasing vibrationally, decreasing rotationally, and keeping the same m. This time, both of these are in fact acceptable. This one doesn't break the rules for the values of m. This state can only go upwards. This state could go up to this state and have the same energy as the excitation we considered here. Or it could drop in rotational energy and then land in the vibrational excited but rotationally completely unexcited state. That would have an excitation energy. I've gained h nu worth of vibrational energy, but now I've lost some rotational energy. I started out a little bit excited. One quantum of excitation in the rotational energy. When I ended up, I had no rotational energy, so I've lost both of my b e units worth of rotational energy. So, we've begun to see here that the complicated structure of the vibrational and rotational energies combine together. So, we call those row vibrational states. Rotational energy stacked on top of vibrational energy. Those row vibrational states combined with the selection rules tell us specifically what energies the molecules allowed to absorb. And they've begun to look a little bit complicated. This is going to be the key to understanding specifically what energies or frequencies or wavelengths of light diatomic molecules are going to absorb in the infrared portion of the spectrum. So, we'll consider that next.