 All right, so what we know about the rigid rotor so far is once we've written down the rigid rotor Hamiltonian, we can write down solutions, wave functions that have the form, some normalization constant, a Legendre polynomial that describes the theta dependence, and then this exponential, imaginary exponential that describes the phi dependence. And we've seen a few cases of this wave function so far. So for example, the 1, 0 wave function, the one with quantum numbers, l equals 1 and m equals 0, was n cosine theta. And for every one of these wave functions that we write down, if it solves Schrodinger's equation, meaning the Hamiltonian acting on the wave function gives us back a constant multiplied by the wave function, we can write down the energy that's associated with that wave function. So in the two cases we've seen so far, when the wave function is n times cosine theta, the energy of that particular wave function is some constants that are just the constants in front of the Hamiltonian here multiplied by a 2. We can also write down the psi 1, 1 wave function, which looked like sine theta rather than cosine theta and an e to the i phi. And then that wave function has the energy also same collection of constants, h squared over 8 pi squared mu r squared multiplied by the same number 2. So those two wave functions have exactly the same energy as each other. But of course, not every wave function has exactly the same energy. We'll see that when l has a different value, the energy turns out to have a different value. So the easiest case that we can demonstrate that for, we've seen this wave function before, but we haven't talked about its energy. The simplest of the wave functions is the 0, 0 wave function. And that wave function is just a constant. If I insert the constant and normalization constant for the wave function, if I take these derivatives of that wave function, then I'll get back a constant times that wave function. And in fact, those derivatives are not hard to do. If this is just a constant n, when I take the phi derivative of the constant, I get 0. When I take the theta derivative of the constant, I get 0. So everything inside parentheses here is just 0. So the energy of the 0, 0 wave function is just 0. So that is a constant times the wave function. The constant is 0 times the wave function. There's other more complicated wave functions. When we write down the 2, 0 wave function, that one's a little more complicated. It involves a quadratic dependence on cosine theta, as we know because it's an L equals 2 function. And we'll skip the math. I think you've seen me evaluate Schrodinger's equation enough times on wave functions. If we take these various derivatives of this particular function and evaluate what we get, we're going to get back the same function we started with multiplied by a constant. And the constant in this case will be not surprisingly. Same collection of constants from here. And then another number multiplying that collection of constants. So things to stop and point out here are the energies that we get are always, so we can do this as many times as we have the patience for, in general, for the L, M wave function. The energy of that wave function is always going to be this particular collection of constants, 8 pi squared mu R squared multiplied by some number. That number is always going to be an integer. The integer might be 0. The integer might be 2. And in fact, that's what we're going to get every time L is equal to 1. Interger 0 is what we get when the O equals 0. For any one of the wave functions that has L equals 2, remember there were five of those, we're going to get the integers are 6. And the general pattern is the integer that we multiply these constants by is L times L plus 1. So we have not proven that statement. I'm just telling you that statement is true. As you remember, we have a recipe for if we know what the wave function looks like for a particular L, let's say L equals 2, we have a recipe in the recursion formula for the Legendre polynomials that tells us how to get the next one. So you can use that recursion formula essentially to figure out if I know the energy for the L equals 2 wave function, what would it be for the O equals 3? What would it be for the O equals 4? I won't show you those steps explicitly, but it turns out that any one of these wave functions, the energy of that wave function turns out to be this collection of constants and integers. So that's an equation we'll use several times, so I'll put that in a box. A couple of things to point out about that equation. Number one, we can just call that E sub L, not E sub L and M. If I want to know the energy of wave function two zero or zero zero, those have an index, a quantum number L and a quantum number M, but M doesn't show up anywhere in this energy. So it turns out the value of M is irrelevant. That's why the one zero and the one one wave functions both turned out to have the same energy as each other. Their M values were different, but the values of L were the same, so they end up having the same energy. So we normally just write that E sub L not E sub L M, because it ends up only depending on the value of L. The other important consequence is because multiple wave functions, the one zero, the one one, even the one negative one, wave function all have the same energy as each other, have all the same energy as each other. So multiple wave functions psi L M have the same energy E sub L. So that idea begins to sound like the concept of degeneracy that we've talked about previously, and the degeneracy of these wave functions turns out to have some pretty important consequences for the properties of the molecules that we treat with the rigid rotor approximation. So our next step will be to explore that degeneracy a little further.