 So, we'd like to know which transitions between states of the harmonic oscillator are allowed and which ones are forbidden. Can we make transitions from 0 to 1 or to 2 or to 3 and so on. And what we've determined so far is that this transition dipole moment that tells us whether those transitions are allowed if it's non-zero or forbidden if it's zero has broken down to look like this. There's a rate of change of the dipole moment of the harmonic oscillator with respect to the bond displacement. If that's zero then the molecule can't absorb any light that gives us the gross selection rule. What we're interested now is this integral and to do that we're going to have to recall that these wave functions for the harmonic oscillator take the form the nth wave function n or n prime whatever this harmonic oscillator quantum number is. Those look like some normalization constant which is different for every n. A Hermite polynomial which is a function of x, it's actually a function of the square root of some constants we call alpha multiplying x and then e to the minus alpha x squared. Those constants appear up in the exponent of the Gaussian as well. So that's the form of these wave functions. We won't have to get any deeper, we won't have to write down what any of these Hermite polynomials are in order to do these integrals today. But if we remember that then this transition dipole moment, at least the integral portion of the transition dipole moment ends up looking like integral of writing the complex conjugate of the nth wave function first. I've got n sub n complex conjugated Hermite polynomial. I don't have to worry about the complex conjugate because these are not imaginary. Those are real. So there's an h n term and there's an e to the minus alpha x squared, I'm sorry that's e to the minus alpha x squared over 2. When I combine the e to the minus alpha x squared over 2s for both of these pieces I'll go ahead and do that ahead of time. That's going to look like e to the minus alpha x squared over 2. And another one, when I combine those together I just get e to the minus alpha x squared not divided by 2. So these pieces are from the first wave function. I've also got an x and then I've got constants with n primes, Hermite polynomial with n prime, and the rest of the Gaussian, the other factor of e to the minus alpha x squared over 2 that combined to give us this term and that's all integrated over dx. So some of those are constants, we can pull out of the integral, the constants n, I'm not particularly interested in the values of those constants. All I want to know is whether this integral is going to work out to be zero. The constants themselves are not going to be zero, so they're not terribly interesting right now. The rest of the integral looks like h n times x times h n prime times a Gaussian. Now, the trick is to remember that we know something about these Hermite polynomials. We know there's a recursion relation that relates the Hermite polynomials to the ones just above or just below them. That recursion relation tells us that the n plus first polynomial is equal to, let me look this up and make sure I get the coefficients right. Two times x times the nth polynomial, and minus two times the quantum number. Times the one below that, h n minus one. So what I've got here is an x times a Hermite polynomial. Let me rearrange this equation to solve for x times a Hermite polynomial on one side of the equation, so x times the nth Hermite polynomial. If I bring this over to the left, everything over to the right and divide through by two to get rid of that factor of two. I'll have what will I have? One half h n plus one minus after dividing by two, that two goes away. And I've just got n times h n minus one. So that x times h n, I'll use that to replace this x times h n prime. So my integral has now become, I've got constants out front. Integral now looks like the nth Hermite polynomial multiplied by what I have written here, one half of the n plus first polynomial minus n prime times the n minus, n prime minus first polynomial with a Gaussian integrated over x. That's now turned into something we can use because if I, I'll go ahead and rewrite this one more time. So moving up to the top over here, that's gonna look like constants that I'm not particularly interested in, some one halves and ns and other numbers that I know are not zero and I'm not particularly interested in. And then I've got integral, so I've got an integral that looks like h n times h n prime plus one and then some halves and some, that's what I get by combining this term. And I'll go ahead and throw in the one half. I've also got the other term. So now I've got, so that minus sign and n prime and then the inside of the integral that I am interested in looks like an h n times h n prime minus one. So the next step is to say, this looks an awful lot like a normalization, not a normalization, an overlap integral between the nth and the n prime plus one wave function. And if, but I know that wave functions, different harmonic oscillator wave functions are orthogonal. The zeroth wave function is orthogonal to the first and to the second and to the third. If these are different wave functions, this integral is gonna disappear. If these are different wave functions, this integral is gonna disappear. So what that means is this whole integral is zero unless these are actually the same wave function. If the nth wave function is the same as the n prime plus one wave function. For example, if n is five and n prime is four, so that four plus one is the same thing as five. If I'm making a transition from the fifth to the fourth, that's okay. But any other transition is gonna cause this integral to be zero. Likewise, this integral will be zero unless these two permate polynomials have the same index. These two wave functions are the same as each other. So that's zero unless n is equal to n prime minus one. Again, maybe if n is equal to five, if n prime is equal to six, then these two sides of the equation are equal to each other. So I can make a transition between the fifth and the sixth energy level because this integral will be non-zero. I can make a transition between the fifth and the fourth energy level because this integral will be zero. But any other transition I make will be zero. My selection rule ends up being n must be equal to n prime plus or minus one, either plus one or minus one. Or if I want to write the change in energy level, n prime minus n. That has to be plus or minus one. My energy level has to go up by one, has to go down by one. And that's our selection rule for the harmonic oscillator. I am allowed to make, let's say this transition to up to three. I'm allowed to, if I have a harmonic oscillator that's in the second, the n equals two energy level. If I provide this much energy to it, if I provide a photon with exactly that much energy, I can induce it to absorb the light and make that transition. On the other hand, well, on the same hand, if I have it make a transition that drops by one, if the change in the energy level is negative one, it can make this transition. In that case, it will emit a photon. So instead of absorbing a photon to gain energy, it drops by one level and emits a photon, so it can emit light with that same frequency. Those are both allowed transitions. This transition, however, e2 down to e0, that will not happen. Likewise, e2 up to e4, that will not happen. The change in the quantum number must be plus one or minus one, plus one if we're absorbing light, minus one if we're emitting light. So that's the selection rule that tells us what transitions are allowed for the harmonic oscillator vibrational transitions.