 So we can continue our investigation of what it takes to make this transition dipole moment either zero or non-zero, making the transition forbidden or allowed. So far, we've determined that the phi portion of this integral, when we insert the rigid rotor wave functions into this integral, we get terms that involve Legendre polynomials and e to the i m phi's. The phi portion of the integral tells us that the magnetic quantum number is not allowed to change if the molecule is to absorb electromagnetic radiation to change its rotational state. The next question is for this more complicated looking integral, by how much is the angular momentum quantum number l allowed to change? So we'll focus on this theta integral. So one step we can immediately make to make this integral a little simpler. We see the Legendre polynomials in this cosine theta term, those all depend on cosine theta. The integration variable is sine theta d theta. So if we simplify this a little bit by using u substitution, or I'll actually use x substitution, if I just let x be shorthand for cosine theta, then dx is going to be minus sine theta d theta. So the sine theta d theta already appears in this integral, so I can rewrite the whole integral in the simpler form. Integral of p l m as a function of my shorthand x times the p l prime m prime. Also as a function of x, this cosine theta I can now just write as x. And then sine theta d theta I can write as dx. And this integral is going to go instead of 0 to pi. Cosine theta makes this go from 1 to negative 1. So that's at least a simpler way of writing down the integral. In order to make progress doing anything with that integral, we could certainly plug in specific values like the 0, 0, 1, 0 example we've considered previously. But to do anything with that in the general case, what we need to remember is that there's a recurrence relation between these Legendre polynomials. I've talked previously about a recursion relation for the plain Legendre polynomials, the ones with m equals 0 up as a superscript. I'll write down for you now what that recurrence relation looks like for the associated Legendre polynomials, the ones where m is allowed to be non-zero, as these very well could be. So what that tells us is the l plus first Legendre polynomial can be written in terms of x times the lm polynomial minus l plus m times the l minus first Legendre polynomial. So that's similar to but not exactly the same as the recurrence relation we've seen previously for the Legendre polynomials with the m equals 0. More importantly for our purposes, remembering that we don't really care about the value of constants, about the numerical value of what this integral comes out to be equal to. We just want to know if there's anything that's going to cause it to completely become 0 or not. So actually these constants are not terribly important. I'll just rewrite this equation as the l plus first associated Legendre polynomial is equal to x times the l minus the l minus 1th. And then recognizing that there's some constants out front that I'm not going to bother with. They might be, depending on the values of l and m, they might be 2s, or they might be 3s or 1s, but they're not going to be 0. And the specific values aren't going to affect whether the integral comes out equal to 0 or equal to some other value. So the next step is to recognize that since I have an x in this integral, an x times one of these Legendre polynomials looks like this x times the Legendre polynomial. So if I rearrange this equation a little bit, I can see that x times the lth Legendre polynomial is equal to the l plus first. And if I bring this l minus 1 over to the other side, I'm adding the l minus 1 Legendre polynomial. So x times the Legendre polynomial is some combination of the one above it with l plus 1 and the one below it with an l minus 1. Again, remembering that there's some constants out front of these numbers that I'm not bothering to write down. So that's enough for me to use in this expression. If I have x times a Legendre polynomial, either x times this one or x times this one, I can rewrite it as the one above it and the one below it. So let's do that. Let's take this x times the lm Legendre polynomial so that it's in exactly the form I've gotten written here. And those terms I've underlined in pink, I can rewrite as the one above it, pl plus 1, and the one below it, pl minus 1. That's rewriting x times the plm term. Those are still multiplying this pl prime m prime term. So the good news is I've made the x go away. If I rewrite, simplify this integral a little bit, distribute this product, this integral of a sum, I can write as the sum of two integrals. I've got the integral of pl plus 1 times l prime from this term multiplied by this term. And I've also got the one within l minus 1 from this term multiplying this term. So pl minus 1 multiplying pl prime. And now the last thing we have to remember is that these associated Legendre polynomials are mutually orthogonal to one another. If I have the integral of two different Legendre polynomials over the appropriate coordinates and the appropriate bounds, then that integral will come out to zero. If I integrate one with itself, then it won't come out to be equal to zero. But if I integrate two different ones, they're mutually orthogonal, so that integral will come out to be equal to zero. So this integral looks like I've got two different Legendre polynomials, an l plus 1 and an l prime. First of all, let me point out that from the magnetic quantum number, we know that these two numbers are in fact the same. The magnetic quantum number, this contribution to the integral will go to zero unless there's no change in the magnetic quantum number. So for any transition that's allowed, the m's are already going to be equal to each other. Same thing here. Delta m has to be zero. So the superscripts are going to be the same for any integral that matters. The question is what about the subscripts, l plus 1 and l prime? If those are different numbers from one another, this whole integral goes away. So this integral will be zero unless that number l prime is equal to l plus 1. Likewise, this integral will be zero unless this number l prime is equal to that number l minus 1. So the orthogonality of these Legendre polynomials guarantees that that will be zero unless these are both the same Legendre polynomials. So if l prime is equal to, in this case, l minus 1, then that will save this integral. So what that means is we've now determined the selection rule. Let's say we have l changing by, so this l changing l prime, let's say it changes by 5. In that case, since l prime is 5 higher than the starting value, it's not true that l prime is l plus 1. It's not true that l prime is l minus 1. So this integral is zero. This integral is also zero. And the entire theta integral will be zero. So if l changes by 5, both these integrals are zero and the transition's forbidden. The only way to make at least one of these integrals not equal to zero is if I'm changing l by plus 1. In this case, l prime can be 1 higher than l, or if I'm changing by negative 1. If the value of l is decreasing by 1, then this integral will be non-zero. So what that means is our selection rule for the angular momentum quantum number, l is allowed to change by plus 1 or by minus 1. If it changes by any other value, then this theta contribution to the integral will be zero. We had seen previously the selection rule for the magnetic quantum number is that the change in magnetic quantum number has to be zero. It's not allowed to change. So those two selection rules put together tell us which transitions are allowed or forbidden for a rigid rotor. These are the conditions that a transition has to make in order for the transition to be allowed. Anything else in the transition will be forbidden.