 So we've seen that the rigid rotor wave functions, which can be identified by a pair of quantum numbers, the angular momentum quantum number and the magnetic quantum number, can be written as a normalization constant, some polynomial dependence on trigonometric functions of theta multiplied by this complex term e to the i m phi. So this polynomial dependence on a trigonometric function of theta deserves a little more attention and there is a recipe for how to calculate exactly what that polynomial function looks like and that's what we'll talk about now and it's a, this polynomial is called a Legendre polynomial. So let's start with, in general, if I want to know the l m wave function, I need the polynomial with an l and an m as a simpler problem. We can talk first about what if m happens to be equal to zero. When l is some integer and m happens to be zero, then there's a recipe for this polynomial. This polynomial is called, I'll go ahead and write it down first. So if I take 1 over 2 to the l, divide that by l factorial and then take the l-th derivative of x squared minus 1 raised to the l-th power. So that's a pretty complicated recipe, but if I know the value of l, that's just a recipe for how to obtain some polynomial in x with those particular values of l. So that deserves an example to make sure we understand how that works. So let's say we want to calculate a particular polynomial and now I'll give you the name of this polynomial, that's called a Legendre polynomial. Let's say we want to know the l equals 2 version of the Legendre polynomial. So when l equals 2, we just follow this recipe. This recipe says take 1 over 2 to the l, 1 over 2 to the 2, l factorial is 2 factorial. This l says take the second derivative with respect to x of this quantity x squared minus 1 raised to the l-th power. So l's show up lots of places, one here, one here, one here, one here, and one here. That's now just a recipe for doing a little bit of calculus and a little bit of algebra. So 1 over 2 squared, 2 squared is 4, 2 factorial is 2. So before I take the derivative, let's figure out what x squared minus 1 quantity squared is, the thing we're taking the derivative of. The square of x squared minus 1 is x squared squared minus twice the inner terms minus 2x squared and then the last term squared plus 1. So that's the thing we want to take the x derivative of. So if I simplify this, 1 over 4 and 2 becomes a 1 over 8. Second derivative of this, we'll do them one at a time. So is the first derivative of the first derivative. First derivative of this function is 4x cubed minus 4x. Now the first derivative of that is, so I've still got a 1 over 8 out front, derivative of this term in parentheses is 12x squared minus 4. If I divide by 4 on the inside and use that to cancel the factor of 4 in the denominator, I'm left with 1 half of 3x squared minus 1. So that's about as simple as I can get it. That is the Legendre polynomial when L is equal to 2 as specified by this recipe. What does that have to do with our rigid rotor wave functions? Notice that the rigid rotor wave functions involve not the Legendre polynomial of x, but the Legendre polynomial of the argument cosine theta. So when I want to know the psi 2, 0 wave function, it's got a normalization constant and it's got some e to the 0 i phi from this value of m. But the theta dependence, I don't write this polynomial in x. Everywhere I have an x in the Legendre polynomial, I replace it with a cosine theta for the wave function. So this looks like 1 half instead of 3x squared minus 1. And I've got 3 cosine squared theta minus 1 and e to the i times 0 times phi goes away. And all I'm left with is 3 cosine squared theta minus 1 times some constants out front. And that's, we don't have it written out on the board, but that's exactly the wave function that we've seen previously for the L equals 2 and equals 0 wave function. So we can see that this is a recipe for getting the theta dependence of a wave function using these Legendre polynomials as long as m is equal to 0. I'll mention also that there's a somewhat easier way, this way works fine. There's an easier way to get these Legendre polynomials as well, using something called a recursion relation. And what that does is it writes one of these Legendre polynomials in terms of the prior 2. So if I want to know the L plus 1 Legendre polynomial, I can write it as some things multiplying the Lth Legendre polynomial and the L minus 1th Legendre polynomial. So again, the L plus 1 polynomial with some coefficients and x's out front is related to the Lth and the L minus 1th polynomial. So we can also see how that works. Let's say we want, again, to get the P2 0 function. If I want to know that, I have to know the two that come before it, the P1 and the P0. And so anytime you use a recursion relation like this, we need to start out with the first few values. So if we were to use this expression or if I were to just tell you that the 00 polynomial is just 1, the 10 polynomial is just x, those are enough now to bootstrap our way up to higher ones. 0 and 1 combined with this formula to give us the second one, the first and second combined to give us the third one, and we can just climb that ladder getting higher and higher values if we want. So to make sure it's clear how that works, let's do this example again using the recursion relation. So P2 0, if I just use this expression. So this, I want to do this with L plus 1 is equal to 2. So notice I'm solving for the L plus 1 polynomial. L plus 1 has to be equal to 2, so L has to be equal to 1. Out in front of this L plus 1 term, I have a factor of L plus 1. So that factor of L plus 1 looks like 2. On the right hand side, if I just insert an L equals 1, everyone in this expression, I've got twice L plus 1, that would be equal to 3 times x, times the Lth Legendre polynomial, L is 1, and I know that P1 polynomial is x. So then I subtract L, which is 1, times the L minus 1 polynomial, L minus 1 is 0, the 0th Legendre polynomial is 1. So the recursion relation tells me this formula. Twice P2 is equal to 3x squared minus 2. If I divide by 2 on both sides, I find that P2, after dividing by 2, is equal to 1 half of 3x squared minus 2. Which means I've got a typo. No, I don't have a typo. Yeah, the error is right here. 1 times 1 is 1. I think I must have been adding instead of multiplying. So what I'm taking 1 half of is 3x squared minus 1. And notice now I get the same answer here as I got here. And if you can do arithmetic correctly, using the recursion relation gives you the same result as using this definition of the Legendre polynomial. So either approach is fine. As you can see here, sometimes the recursion relation involves a little bit less arithmetic or calculus or algebra than using the definition. So far so good, we have two different ways of calculating Legendre polynomials as long as m is equal to 0. We often have to calculate these polynomials when m is not equal to 0, however. So to do that, I need to tell you how to calculate the polynomials when m is not equal to 0. And the recipe for doing that is the following. I take 1 minus x squared, and I raise it to a power, not m, but half of m. I multiply that by another derivative, the m-th derivative of the one we already know how to calculate the Legendre polynomial with l. So this recipe is for how to turn the l-th Legendre polynomial into one with an m up top. These are called the associated Legendre polynomials. And again, we have a recipe for calculating them. And so as our last example, let's say we want to calculate p22 just to make sure we understand how this expression works. Let's say I want to calculate, if I already know what p20 is, it's either this or this, depending on how I obtained it. If I want to calculate p22, I just follow this recipe. So I want to take 1 minus x squared raised to a power. What value of m am I using? I've set m equal to 2. So I want to follow this recipe with an m equals 2. So m over 2 is just 1. The derivative I'm taking, this m-th derivative, is the second derivative of the polynomial when m is equal to 0. So that's this polynomial, 1 half of a 3x squared minus 1. So to compute that, I'll leave the 1 minus x squared out front. Second derivative of 3x squared minus 1, you might be able to look at that and tell me what it is. But if I do that one derivative at a time, the first derivative of 1 half 3x squared minus 1 is 1 half times 6x. The 1 goes away. And then if I take another derivative, I'm going to get 1 minus x squared. Derivative of 1 half times 6x is just going to be 1 half times 6 or 3. So the l equals 2, m equals 2 associated Legendre polynomial is this function 3 times 1 minus x squared. If I ask myself, what does that have to do with a wave function like the psi 22 wave function, that's going to be a normalization constant. The 22 Legendre polynomial associated Legendre polynomial that we've just calculated, but using cosine theta as the argument. So not 1 minus x squared, but 1 minus cosine squared. But we know 1 minus cosine is sine squared. And then the last term, e to the i, m phi, when m is equal to 2, then I have e to the 2 i phi. So again, this expression, the 22 l equals 2, m equals 2 wave function, isn't on the board right now. But if you look back in your notes or watch the previous video where we listed many of these rigid rotor wave functions, you'll see that the 22 wave function that we listed was in fact constants times sine squared e to the plus 2 i phi. So now between Legendre polynomials and associated Legendre polynomials, we have a recipe for how specifically to calculate the theta dependence of these rigid rotor wave functions. And we'll use those to understand more about the rigid rotor.