 Okay, so our wave functions for the rigid rotor that take this form a normalization constant polynomial in cosine of theta that is related to the Legendre or associated Legendre polynomials and then an exponential term involving phi that is complex. Those wave functions we have seen are normalized or at least we can choose the value of n to guarantee that they're normalized. They have one other property that's very handy which is that they're orthogonal and what I mean by that is every pair of wave functions in this full set of wave functions is orthogonal to each other. So we say that the set of wave functions are mutually orthogonal so any pair that I choose as long as they're not the same as each other they end up orthogonal. So just as a reminder of what these words mean the function is normalized if the integral of a wave function squared is equal to one and orthogonal means the integral of a wave function times a different wave function. So if I take this wave function size of lm multiplied by a different one l prime lm prime then that's not going to come out equal to one but it's going to come out equal to zero if either l prime is not m prime or if l is not equal to l prime or if m is not equal to m prime. So that's what it means for these wave functions to be orthogonal. We can do a quick example and make sure that's clear. Let's say take two of the simpler wave functions the psi one one wave function the psi zero zero wave function l equals one m equals one l equals zero m equals zero in this case they both the l's and the m's are different from each other but we can ask either are they orthogonal to each other or not or since I've already told you that they're going to be we can verify that they're going to be orthogonal. So we can say is it true that the integral of psi one one star times psi zero zero is going to come out to be zero if so then the two functions are orthogonal to one another. So to do this we need to be able to write down what the wave functions are and we would use this expression to write them down. We don't particularly care about the values of normalization constants when we're dealing with orthogonality either this is going to come out to be equal to zero or it's going to come out to be something that's not zero and the value of n won't change whether it's equal to zero or not so I'll leave the n's alone I won't bother to normalize these functions and find the values of n for either the one one or the zero zero wave function. The Legendre polynomial when l equals one and m equals one that's one we've seen before that's sin theta and then I've got a e to the m or e to the one times i times phi if I then take the complex conjugate it becomes negative. So that's the complex conjugate of the psi one one wave function psi zero zero that's the particularly simple one that's just a normalization constant the zero zero Legendre polynomial is one and e to the zero phi gives me one. So I want to do that integral over d theta and d phi with the sin theta integration factor. So I've got a theta integral and a phi integral that's again integral we want to know whether that does or doesn't come out equal to zero I'll go ahead and pull the constants out front those aren't going to make the function equal to zero I'll break the integrals up into first the theta part of the integral and then separately the phi part of the integral. The theta integral I've got a sin theta and a sin theta sin squared theta integrated from zero to pi that's already looking like not great news that's an integral we know how to do integral of sin squared theta from zero to pi that's going to be it's going to have a particular value the values pi over two but in particular it's not equal to one but it's also not equal to zero which is the more important fact that integral is not equal to zero if this integral is going to come out equal to zero it's not going to be because of the normalization constants it's not going to be because of this theta term if it's equal to zero it's only going to be because of this term so let's see if that works out the only phi dependence I have is this e to the minus i phi term. So I need to know what is the integral of e to the minus i phi when phi goes from zero to two pi so that integral is the one we should focus on because that's the only one that has any hope of making this thing equal to zero integral of an exponential is just an exponential where I need some constants out front just to double check if I take the derivative of this function derivative of this function pulls down a minus i the minus cancels the minus the one over i cancels the i and I get back this function so this is the correct integral of e to the minus i phi evaluated from zero to two pi that's looking a little weird one over i with the negative sign out front of it that's certainly not going to help me get to zero but if I plug in the upper and lower limits of this definite integral so this is going to become minus one over i e to the minus i times two pi minus using the lower limit e to the zero so the e to the zero we know what that is that's equal to one but e to the two pi i is also equal to one e to the minus two pi i also equal to one so the few small facts you need to understand about complex numbers in order to do quantum mechanics at this level are number one how to do how to take complex conjugates number one not to be too frightened when you see an eye showing up in equations and number three a fact that we frequently need to use is the fact that e to the two pi i equals one so that fact I've used in this expression e to the minus two pi i is just one over e to the two pi which is one over one which is still one so using that fact e to the minus two pi i is equal to one and now we see that one minus one is going to be zero so remember what we're calculating is the pi portion of this integral that's the work we've just done here the pi integral gives us zero so after all of this work we find that the full integral this overlap integral between the one one wave function and the zero zero wave function does turn out to be zero and we see that those two wave functions are indeed orthogonal to each other and in fact any time we have m not equal to m prime you can see exactly why that's going to work if I have any amount of e to the i pi that doesn't end up canceling in this five portion of the integral then the integral the five portion of the integral is going to come out equal to zero just like it did here so anytime m is not equal to m prime then the phi integral will work out to be zero if m does happen to equal to m prime then this integral won't be zero but the L part of the integral will be so anytime either one of these quantum numbers is different for the two wave functions then the overlap integral between them will work out to be zero and every pair of different wave functions for the rigid rotor it comes out orthogonal to each other so every one of these pairs of integrals is mutually orthogonal as a result if all of our wave functions are both normalized and the full set is more mutually orthogonal then the other way to say those two facts is to say that those wave functions form an orthonormal set so they're orthonormal probably better to say they are an orthonormal set of wave functions so that fact will be useful when dealing with rigid rotor wave functions occasionally and then now the next thing we have to do now that we have a more complete understanding of these wave functions is to use them again in the Schrodinger equation and figure out what the energies of these wave