 So, in thinking about the rigid rotor as a model for the diatomic molecules, we've got a Schrodinger equation that describes the rigid rotor, and we've seen a few solutions to the rigid rotor Schrodinger equation. There's a large number of these, but the first few, the simplest ones, are shown here. And we have the somewhat mysterious fact for right now that some of those solutions appear to have i, the square root of negative 1, in them. And that raises some questions. First of all, what is it doing there? Why would those solutions need to have a complex number in them? And does that present any problems for understanding what these wave functions mean? We're used to thinking about real numbers, but we have a little more difficulty thinking about what complex numbers mean in the real world. So, as a first step, let's first just confirm that these solutions with the i in them do in fact solve the rigid rotor Schrodinger equation. So that's the first thing we can do is take, we'll start with this one. Take this function and insert it into the Schrodinger equation for the rigid rotor and confirm that it does, in fact, present a solution to Schrodinger's equation. So we'll start by taking, so I don't need psi in there twice. So we'll start by taking this function and inserting it in here for psi and taking the derivatives that are indicated. So if I take the phi derivative, so this is, take the second derivative with respect to phi of my function. So if I take the derivative once, so I'll take the d d phi derivative one time. Only place phi shows up is in the exponential. So the derivative of e to the i phi just pulls down a factor of i and that first derivative is going to look like n times i times sine theta e to the i phi. If I do it again, if I take a second phi derivative, another derivative, the first derivative of this function will just pull down another factor of i and so I'll have n times i squared times sine theta e to the i phi. But i squared, i is the square root of negative one, so i squared is just negative one. So I've got minus one times n times sine theta e to the i phi. That's what I get for this term, actually for just the second derivative. If I then do what's indicated here, if I divide that result by sine squared, then I'll get minus n, sine divided by sine squared leaves me with one over sine times e to the i phi. So that's the result from the second term, the term involving phi derivatives. Next step is to take that same function and plug it in here and take the indicated derivatives with respect to theta. So we'll do that next. So let's start by first just taking the theta derivative, the first theta derivative of this candidate wave function. So derivative of sine theta is cosine. So I've got cosine theta e to the i phi with an n out front. That's the theta derivative. If I then multiply it by sine theta, what I have looks like this. And cosine sine times an exponential e to the i phi. Now I've got all of the terms including multiplying by sine theta. The next step is to take the theta derivative. So I need to take the theta derivative of cosine times sine. So product rule will help me here. Derivative of cosine is negative sine, multiplied by the other sine makes sine squared. And then if I leave the cosine part undifferentiated, take the derivative of sine. The second term, derivative of sine is cosine. So that cosine times the first cosine leaves me cosine squared. And I still have e to the i phi at the end of it all. So now I've gone as far as taking the theta derivative of everything that comes after it and I've got this result. The next step, last step inside the parentheses here is to divide by sine theta. So when I do all those things together and divide by sine theta, what I'll end up with is negative n sine squared divided by sine is sine theta. Cosine squared, I can write this as n cosine squared divided by sine theta. Or I can recognize that cosine squared is 1 minus sine squared. That looks like it'll make a little bit of progress. So cosine squared became 1 minus sine squared. I'm dividing all that by sine theta. I'm going to have some cancellation but the one is not going to be canceled. Still multiplying e to the i phi and to that whole result. So everything here is the result from doing these theta derivatives. The result here is what I got by doing the phi derivatives. If I combine those together, what I have here is the entire term within parentheses. So that looks pretty ugly but there's a whole lot of signs in here and if we look closely we can see there's some cancellation that's going to happen. I have, let's take a look at this last term, minus n 1 over sine times e to the i phi. That looks a lot like this n 1 over sine e to the i phi which has a positive sign out in front of it. So this term that wasn't going to completely cancel this 1 over sine theta does end up canceling the term that came from the phi derivative. So that's actually the reason we wanted to have this e to the i phi in here in the first place is the derivatives of e to the i phi twice brought in a negative sign which is what we needed to have here to cancel the inconvenient term in the theta derivative. So after canceling those pieces what we have left is a minus n sine theta e to the i phi and then a minus n sine squared over sine gives me another sine theta e to the i phi. So altogether that looks like n sine theta e to the i phi that shows up twice, once from this term and once from what's left of the second term. So the entire term in parentheses simplified down to minus 2n sine theta e to the i phi. If I now multiply by these constants minus h squared over 8 pi squared mu r squared, the question is is that or is that not equal to a constant times my original function? My original function is n sine theta e to the i phi and the cancellation that we can do now. The ends cancel on both sides as does the sine theta as does the e to the i phi. The two negative signs cancel each other and we found that on the right I now just have h squared over 8 pi squared mu r squared and the only thing left in the parentheses that hasn't canceled is this factor 2. So constants on the left, constants on the right, everything is okay. This function does in fact solve Schrodinger's equation. The energy of that wave function is this collection of constants 2h squared over 8 pi squared mu r squared. And we've seen this does solve Schrodinger's equation. The i's were in there out of necessity because we needed the sign here to be a negative. We needed that negative sign to come from an i that got squared. So the purpose of the i's in this wave function are in order to guarantee that this function solves Schrodinger's equation. If I had left the exponential term out, as we saw in the previous video lecture, a plane sine theta does not solve Schrodinger's equation. So the i's are necessary. The remaining question is does that cause problems for us? What does it mean to have a complex wave function? Not complex as in complicated, but complex as in not real. It's not a real function. It's got some imaginary terms in it, so it's a complex function. So what does it mean if our wave function is complex? Physically, remember what the wave function means is that the probability of finding a quantum mechanical system in some state is the square of the wave function, but squared in this way that's particular to these quantum mechanical systems. Squaring inside these vertical lines here means I take the complex conjugate of the wave function and I multiply by the wave function. So squaring in this particular sense, the only time we have to worry about the complex conjugate is in the cases where we do have i's showing up in the wave function. So remember what a complex conjugate means. That means a complex conjugate means take all the i's in a function and turn them into negative i's. That's the recipe given by taking the complex conjugate. So this is the wave function, n sine theta e to the i phi. The complex conjugate of that wave function would be n sine theta e to the minus i phi. So that's actually the function we have right here. These two functions are complex conjugates of each other. The good news is if I take a complex number and I multiply by its complex conjugate, I'm always going to end up with a real number. If it's been a little while since you saw the brief introduction that you might have had to complex numbers, let's just say I have some complex number that involves a real part and an imaginary part. Let's just say that's some complex number z. Then the complex conjugate of that number, I leave the real part alone. I turn the i into a minus i. So i has become minus i. So this is the complex conjugate of the number z. Notice now that if I take the complex conjugate times the original number, in other words, a minus bi, the complex conjugate, times a plus bi. This is now just algebra. If I take first, outer, inner, last, the first two terms give me a squared. The outer terms give me a plus bi or a times b times i. The inner terms give me a times b times i with a negative sign. And the last terms give me a negative b squared times i squared, but i squared is negative 1. So we have a cancellation between the two terms involving the square root of negative 1, the imaginary terms cancel. And all I'm left with is an a squared and a b squared. So what that means is any time I have a complex number times its complex conjugate, the result is going to be a real number. I've made the imaginary parts go away. So what that means is if I take a wave function that happens to be complex, multiply by its complex conjugate, we know that what we get when we do that is a probability. If we had found that the probability was a complex number, that would be a little worrisome. If I said the probability of finding a particle at this point in space or a molecule with this orientation was 3i, that doesn't really make sense. I need the probability to be a real number because it's a measurement about the system in the real world. But when I take the wave function times its complex conjugate, what I get is a real number. So it's OK, the probability, the thing we use wave functions for in part, measuring probability, that still comes out to be a real number. So it's actually not a problem that the wave functions are complex. The wave functions aren't a real world property, the molecules. They are used to generate measurements or predictions or probabilities that do end up having real values. So the complex, the imaginary numbers in these wave functions might be a little worrying to you at first, but they're there for a good reason, and they don't end up causing any problems with any real measurements that we make in the real world. So with that, we can say that we've almost completely understood these wave functions for the rigid rotor. But the way I've written them down here, we don't yet know what these normalization constants are. So that's the next thing we'll tackle is solving for the values of those normalization constants.