 OK, so we have seen that the angular portion of the Schrodinger equation for the hydrogen atom, once we break it down into a radial piece and an angular piece, the angular piece of the Schrodinger equation looks like this. So we have this differential equation. Notice that we're not solving for a wave function anymore, we're solving for y, which is only the angular component of the hydrogen atom Schrodinger equation. But still we've got a differential equation to solve if we can figure out the function y that obeys this differential equation. If we take these derivatives of it and add it to these derivatives, we get back a particular constant. So that's what we're looking to do, is to find the solutions to this differential equation for the function y of theta and phi. So as a start, we can do the following. If we take this equation, and if we multiply by y to get rid of this term, and if we divide by r squared, for reasons I'll mention in just a second, then what this equation looks like is multiplying by y gets rid of this, dividing by r squared does that. And on this side, if I multiply by y and if I divide by r squared, then I've done the following to the equation. So that's what the differential equation looks like now. And the reason I've done those two particular things is because this differential equation might look very familiar. This, in particular, the theta and phi portions of the kinetic energy term and the lack of a potential energy term looks exactly like the differential equation that we solved when we were studying the rigid rotor. So in particular, if I notice that this alpha over r squared, if I identify that with the thing we called energy for the rigid rotor, and if I replace the mass of an electron with the reduced mass for a diatomic molecule, if I replace m with mu and if I replace alpha over r squared with e, that's exactly the same thing as the differential equation we solved when we were solving for the rigid rotor, rigid rotor wave function. So that means two things which are fairly convenient and somewhat remarkable. The first one is that for some reason the angular behavior of an electron around a hydrogen atom, the problem we're currently trying to solve. So we have an electron at some distance r away from a hydrogen nucleus, which I suppose is just a proton. That behaves exactly the same way quantum mechanically as if we have a diatomic molecule where we have one atom with mass one at some distance away from an atom with mass two. So drawing the diagrams makes it clear exactly why this is true. We just have two point masses interacting with no potential energy because we're only considering the right angular term with a fixed r. It doesn't matter whether we call them a proton and electron or whether we call them an atom and another atom. If we keep their distance fixed and leave the radial portion of the behavior for a different day, then at fixed r, a rigid rotor molecule behaves exactly the same as the angular behavior of an electron and a hydrogen atom. So we've already solved this problem as it turns out. And the solutions that we saw for the rigid rotor are going to be exactly the same as the solutions for this angular term in the hydrogen atom wave function. And what those looked like for the rigid rotor was some normalization constant, some polynomial that looks like a spherical harmonic. And then I guess I don't need the plus or minus. e to the i m phi m could be a positive number or a negative number. And that reminds us that there's a whole sequence of these solutions. We can have the l equals 0, l equals 1, l equals 2 solutions. Each l can have its own family of m's ranging from negative l to positive l. So these are exactly what we wrote down for the rigid rotor wave functions. They're also exactly the same as the hydrogen atom angular terms. And just as a reminder, the 0, 0 wave function was just some normalization constant with no angular dependence. The 1, 0 wave function was different. If we want to label these, we can label them the 0, 0 normalization constant or the 1, 0 normalization constant. The 1, 0 wave function had some angular dependence in terms of the theta variable. And then the 1, 1 and the 1, negative 1 wave functions for the rigid rotor looked like a normalization constant multiplied by sine theta. And then e to the i phi or e to the minus i phi, depending on whether I have a plus 1 or a minus 1 here. So there's an infinite number of those wave functions, any l and any m that we inserted in this equation. We've got a particular polynomial in cosine theta or sine theta and e to the sum number of i phi's and a normalization constant out front. And those are also the solutions for the angular behavior of the hydrogen atom. The only last thing we have to consider is that energy means something slightly different now than it did before. Previously, our energy now corresponds to this alpha over r squared. So this constant alpha that we wanted to know if I take these derivatives of the angular function, what I get back is an alpha. We're in this new equation. What I get is alpha over r squared times the original function. So the value of this alpha is r squared times what we used to think of as the energy of the rigid rotor. Those energies for the rigid rotor, the energy of the elf wave function was this collection of constants, h squared over 8 pi squared. Used to be a mu for reduced mass. Now it's me for the mass of an electron. r squared is in the denominator. And that all multiplies the quantum numbers l times l plus 1. Those were the energies of the rigid rotor. What that means now for our value of alpha, for the hydrogen atom, if I just take that energy and multiply by r squared, there's an r squared in the denominator. So the value of alpha just works out to be h squared over 8 pi squared mass of the electron. The r squared has gone away, and we still have an l times an l plus 1. So that's the significant conclusions we've got so far for the angular behavior of an electron in the hydrogen atom. It obeys any one of these functions for the angular component of the wave function. And the value of alpha that will need to solve the radial equation is this collection of constants times some quantum numbers.