 So we've gotten as far as developing a model for the geometry of a diatomic molecule and we've determined that rather than thinking about a molecule with two different masses and two different atoms at two different positions, we can get away with thinking about a molecule that's in the same orientation but only a single point with mass, a mu, the reduced mass, a distance r from the origin with again the same angles theta and phi. So the other advantage of using this coordinate system in addition to reducing the number of variables that we have to think about is it also helps us distinguish pretty clearly between the two different types of motions this molecule can undergo, vibrational motions and rotational motions. When the molecule vibrates, if it keeps the center of mass where it is and changes its bond length by vibrating, in this model that just corresponds to changing the length of this coordinate r, the distance away from the origin. On the other hand, if the molecule rotates, if it changes its orientational angle, then that's a change in theta or phi or both of those at the same time. So rotations are described by theta and phi, vibrations are described by changes in r. And we're actually going to treat those two problems separately. We'll postpone vibrations for a little while and let's think first only about the rotations. So the way we'll do that is we'll require that r be constant, r is not going to change. So when this molecule rotates, it does so with a fixed bond length. So in other words, it's a rigidly rotating molecule, it doesn't change it's not flexible but it's rigid, it doesn't change its bond length. So r is constant, we are going to allow theta and phi to change. So those two variables can change and therefore the way that we'll describe the state of this system, the two variables that can change are theta and phi and the state of this system is going to be described at least quantum mechanically by a wave function that depends on theta and on phi. So in order to describe this system quantum mechanically with a wave function, we're going to need to start talking about Schrodinger's equation. So let's remind ourselves of what Schrodinger's equation is. In general, Schrodinger's equation is a kinetic energy term which looks like some constants minus h squared over 8 pi squared mass, but remember now our mass, the mass of this particle that we're studying is this reduced mass mu times the second derivative in the form of this Laplacian acting on the wave function. Our wave function only depends on two variables theta and phi and then we've got a potential energy term multiplying the wave function. Kinetic energy plus potential energy must equal e total energy. So h psi, Hamiltonian acting on psi gives us energy multiplying psi. So that's Schrodinger's equation in its general form. In this particular case, if we think about the potential energy term, for this diatomic molecule this potential energy term is asking how different is the potential energy for this diatomic molecule if it has this orientation or this orientation or this orientation and if you think about a gas molecule changing its orientation, we don't expect its potential energy to be any different than any one of these orientations. So we will assume that the potential energy is equal to zero for this model that we're calling the rigid rotor model. So if we make that assumption, Schrodinger's equation simplifies quite a bit. Leaving out the variable, the coordinate dependence, kinetic energy term minus h squared over 8 pi squared mu del squared acting on psi is equal to e times psi. In other words, kinetic energy term is equal to e times psi. That looks an awful lot like what we had when we solved the 3D particle in a box problem. Really the only difference in the equation is I've got a mu here for a mass instead of an m, but there's one important difference that we need to solve this problem in spherical polar coordinates, theta and phi, whereas we solved the 3D particle in a box problem in Cartesian coordinates, x, y and z. So let me summarize so we don't get confused the properties of this rigid rotor model. So we've assumed that for our diatomic molecule the bond length is constant, it stays rigid as it rotates, that's why we call it a rigid rotor. We've assumed that the potential energy is zero, that's the same assumption we made for the particle in a box, potential energy is zero. That means we've only got two variables to worry about, theta and phi, so our wave function is going to depend on theta and phi. This is a reminder that we're doing this problem in spherical polar coordinates, theta and phi, rather than x, y and z. The other important difference with the 3D particle in a box model is the boundary conditions. Remember for the particle in a box we can find a particle in a box and the wave function had to reach zero as we got to the edges of the box. For this case there is no edge, the coordinates we're talking about are theta and phi. If I take the variable theta and I keep increasing it, the molecule bends down away from the z-axis. If I keep increasing it there's no edge of the box, the molecule can just spin, can I have any values of theta and phi, well I want, if I increase theta from zero to pi and keep going it just does circles. So rather than an edge of a box that we have to worry about the real periodic boundary conditions we have to worry about are the fact that in particular if I take phi, if I take this molecule, if I start increasing the value of phi, the molecule will spin around the z-axis and when I spin by a full two pi radians around the z-axis I get back to where I started. So what that means is if I evaluate the function either at an angle theta and phi like I've drawn it here or at an angle theta and phi where I've spun it by a full revolution around the z-axis and get back to where I started the wave function before and after I spin it around has to be exactly the same. So we're going to require that our wave function obey this boundary condition. The wave function at theta and phi has to have the same value if I rotate phi by an angle of two pi. So that's the requirement we're going to place on our wave function. So with those things in mind we need to rewrite this Schrodinger equation now not with del squared equal to what we used in Cartesian coordinates so just as a reminder we won't be using this right now but in Cartesian coordinates the Laplacian del squared is the second derivative operator d squared dx squared plus d squared dy squared plus d squared dz squared so that's what we did for the 3D particle in a box. In spherical polar coordinates del squared is still a second derivative operator second derivative with respect to these r and theta and phi coordinates but it looks a little bit different. It looks like let's see the second derivative with respect to r looks a little bit convoluted has the d dr separated by a little bit likewise there's a second derivative with respect to theta but the two d d theta derivatives are separated they have this sine theta sandwiched in between them and they have some terms out front and there's a third term one over r squared sine squared theta the second derivative with respect to phi does look like an ordinary second derivative. So those three terms all together you have probably seen in a calculus three class at some point that's what the Laplacian looks like that's the second derivative Laplacian in spherical polar coordinates it gets a little bit simpler for us because once we've made this assumption that r is constant the derivative with respect to r those terms go to zero so we don't have to worry about the d dr terms specifically because we're not allowing the value of the bond length to change so it can't have a derivative so what we're left with is just the second two terms if I then take this expression for the Laplacian and plug it into Schrodinger's equation and that'll be the last thing to write on the board for this lecture so the rigid rotor Schrodinger equation the one we'll want to solve is the Schrodinger equation with r constant so this term goes away with v equals zero so this term goes away and with the Laplacian in spherical polar coordinates so I use these two terms for the Laplacian so I've got minus h squared over eight pi squared mu let me notice also that both of these terms that I'm about to write down this term and this term they both have a one over r squared in them so let me go ahead and put the r squared here from both of those terms and then I've got one over sine theta d d theta sine theta d d theta that's the remaining terms from here and then I've also got one over sine squared theta d squared d phi squared those things acting on the wave function which remember as a function of theta and phi must equal energy times the wave function which is still a function of theta and phi all right so this expression that I've just written down that is the Schrodinger equation in the form we need to worry about it to describe a diatomic molecule so remember what that means is we're looking for some function some wave function psi of theta and phi that obeys this particular equation that when I do this stuff on the left to it I get back a constant times the same wave function that I started with so the next thing we'll tackle is trying to find wave functions that obey this Schrodinger's equation.