 So, here is the Schrodinger equation for the rigid rotor model, a model for diatomic molecules that allows the molecule to rotate with no potential energy. It's rigid so it does not allow the molecule to vibrate and its center of mass is fixed at a point in space. So, with that model for the rotating diatomic molecule, we've been able to write down Schrodinger's equation and our next task is to figure out what functions, what wave functions psi solve this differential equation, solve Schrodinger's equation. So we're looking for functions psi of theta and phi, functions that depend on the two angular coordinates that allow the orientation of a molecule to change that will solve this function. So, we can explore this Schrodinger equation a little bit more by using some trial and error to find the first few solutions of this wave function. As a start, we can say, let's start by saying, again just by trial and error, let's say, function needs to depend on some angles, let's try cosine of theta as a possible candidate solution for this function and see how it works. So we need to figure out if I take the theta derivative of that function, multiply by sine theta, take the theta derivative again, divide by sine theta, add that to the second phi derivative divided by sine squared theta, multiplied by some constants. Do I or do I not get back the same function I started with? So that's our task to figure out right now. So let's do that step by step. The function is simple enough. It only involves theta as it doesn't involve any phi. So this first derivative, taking the derivative of that function with respect to phi once or twice, gives us zero. So that derivative goes away completely. This entire term disappears. This term d psi d theta doesn't go away. So I need to take the derivative of cosine theta with respect to theta. That gives me a minus sine theta. What the equation tells me to do next is multiply that by a sine theta. So when I multiply negative sine theta by sine theta, I get minus sine squared theta. If I then take the derivative, the theta derivative of that minus sine squared theta, I get minus derivative of sine squared is 2 sine theta. And then I need to use the chain rule derivative of sine theta itself is cosine theta. So the derivative of minus sine squared is minus 2 sine times cosine. So now I'm up to including everything back through this theta derivative. What's left is to divide it by sine theta. So minus 2 sine theta, cosine theta divided by sine theta cancels the sine theta, and I've got minus 2 cosine theta. Now I'm done with taking the derivatives. And shortening of the equation, we want to know is this or is this not true for each candidate function that we stick into it? So is it or is it not true that minus these constants, h squared over 8 pi squared mu r squared times negative 2 cosine theta. This theta-related term in parentheses, the phi-related term disappears completely. So this whole term in parentheses is just minus 2 cosine theta. Is that equal to, on the right side of the equation, a constant that we call the energy. Also multiplying by our candidate wave function, cosine theta. Is it true that constants times cosine theta equals a different constant times cosine theta? Sure, as long as that constant e is equal to, after canceling these negative signs, 2h squared over 8 pi squared mu r squared. And what we've just determined is two things. Number one, this candidate wave function does in fact solve the rigid rotor shortening equation. When I plug this function into the shorting equation, it obeys the shorting equation. And we've also found that the energy of this wave function is this collection of constants, 2h squared over 8 pi squared mu r squared. So verifying a solution once we have a good guess at a solution that does in fact work is not that difficult. Not every guess we make will be a valid solution to Schrodinger's equation, of course. So I can, for example, try a different function. If cosine theta works, we can ask ourselves, does sine theta work? Will that also solve Schrodinger's equation? And instead of writing everything down the same way I did over here, we can just take the terms one by one. If this is psi, then the derivative with respect to, let's do the phi derivatives first because they're once again easier. Derivative of that function with respect to phi, again, there's no phi's in there, so that disappears. If I take the derivative with respect to theta, derivative of sine theta with respect to theta is cosine theta. So now I'm working on this term here. If I multiply that by sine theta, I've got cosine theta times sine theta. If I now take the theta derivative of that, so theta derivative of sine theta theta derivative, so I need the theta derivative of this piece. Now I need to use the product rule. Derivative of cosine is negative sine times sine makes sine squared. And using the other piece, if I leave cosine alone, take the derivative of sine, derivative of sine is cosine, which when multiplied by this cosine gives me a cosine squared. So I've got negative sine squared plus cosine squared. What I have next to do after that theta derivative is divide that whole thing by sine theta. So 1 over sine theta, theta derivative, if I divide all that by sine theta, it's not going to divide as cleanly as it did the first time around. So sine squared theta divides by theta just fine. And then the second term is going to look like cosine squared theta over sine theta. That's not canceling terribly cleanly. There's some trig identities we could use. We could rewrite cosine squared as 1 minus sine squared. But still, this is not going to simplify down to just constants multiplying my original wave function sine theta. So what that means is, even without proceeding all the way to asking ourselves, does the left side equal the right side? We can see that the derivatives we've taken on the left side leave me with something that's not just the original function. It's the original function plus some extra stuff that isn't going to end up canceling. So what that means is this candidate wave function does not solve the Schrodinger equation. So psi equals cosine theta solved the Schrodinger equation. Psi equals sine theta did not solve the Schrodinger equation. So like I said, we can use trial and error. We've got two trial cases. One of them worked. One of them didn't. Luckily, we don't have to proceed purely by trial and error. This particular differential equation, long before the Schrodinger equation was developed, mathematicians had been studying this equation. So the solutions to this equation are well understood, even before anybody knew that it was useful in quantum mechanical problems for the rigid rotor. So the collection of solutions, there's a whole bunch of solutions to this Schrodinger equation. One of them we've seen is cosine theta. And in fact, I can stick any constant I want in front of that cosine theta. If psi equals cosine theta solves that Schrodinger equation, n cosine theta will solve it as well, because each of these terms would just have an n in front of it. So if I have an n on the left side, I also will have an n on the right side, and n doesn't affect the solutions at all. And you won't be surprised to hear or to remember that that n is a normalization constant that we can solve for the value of. And we'll talk about that on a different day. But in addition to this solution, cosine theta, which we've validated, solves Schrodinger's equation. There's some other solutions as well. I've left room above that, because there's an even simpler expression that also solves Schrodinger's equation for the rigid rotor, which is just the constant n, a normalization constant all by itself. I'll skip the math of showing you that that's true. But these derivatives are not terribly difficult when I'm taking the derivative of a constant, so you can verify that for yourself. Just the wave function equal to a constant will solve Schrodinger's equation. Constant times cosine theta will solve Schrodinger's equation. We've seen that n times sine theta does not solve Schrodinger's equation. But if I add in some phi dependence, n times sine theta times e to the i phi, that will solve Schrodinger's equation as will n times sine theta times e to the negative i phi. So those functions, as well as many more. So we'll have occasion to talk about the rest of these wave functions coming up soon. But there's a long list of functions that solve this Schrodinger's equation, or this differential equation. And we'll explore those solutions in a little bit of detail soon. The one last thing I'll point out about these solutions for right now is in order to get sine theta to solve the rigid rotor Schrodinger equation, I had to stick this e to the i phi on the end of it. And the i in this expression is the imaginary constant, the square root of negative 1. So these last two solutions I've written down are actually complex solutions. They're not purely real valued. They involve the square root of negative 1 in them. So that seems a little bit confusing. And that has some important consequences that we'll talk about in the next video lecture.