 So the general form of the harmonic oscillator wave functions is this polynomial in x, the Hermite polynomial, multiplying a Gaussian function with some carefully chosen constants to make sure it's an actual solution to the Schrodinger equation. We haven't yet talked about the normalization constant out front of these wave functions. So as usual, once we obtain a solution to the Schrodinger equation, we can guarantee that that function is normalized, and we can also investigate whether the family of solutions is mutually orthogonal or not. So that's what we'll do next. So as always, normalization involves guaranteeing that if I integrate the probability of finding the particle over all positions, then that has to come out to 1, and probability is wave function squared. The complex conjugate, we don't have to worry about too much in this case because neither the Hermite polynomial nor this Gaussian involves any i's, any square roots of negative 1. So the complex conjugation is not needed for the harmonic oscillator wave functions because they're not complex. We'll do one example of this, the simplest one we can think of because normalization is not particularly fun, and watching someone normalize a wave function is certainly not fun. But it's worth doing one example just to make sure we know what variable we're integrating over, and because this wave function is a one-dimensional function that depends on x, the variable we integrate over is x, and remembering what that variable means, it's a bond displacement. It's how much I've stretched the bond, or how much I've compressed the bond. That displacement can go anywhere between negative infinity and infinity. Negative infinity, the infinite amount of compression might seem like it doesn't make sense because if you're thinking of the bond as, so here's potential energy as a function of bond length. If you're thinking of the bond as having this sort of potential energy, I can only compress the bond from its equilibrium position, where x equals 0, down to fully compressed until the two atoms are sitting on top of each other. But remember, we're not solving Schrodinger's equation for the correct potential energy function, we're solving it for an approximation where we've treated the bottom of this well as a parabola, and we acknowledge that we're making some mistake over in this region. We're also making a mistake over in this region, and this function can continue to negative values of the bond displacement. If we're treating the potential energy as a parabola, x can be all the way from negative infinity all the way up to positive infinity. Normalize the zeroth wave function, which remember is equal to some normalization constant that we have not determined yet. The Hermite polynomial is 1, and the Gaussian looks like e to the minus alpha x squared over 2 for this particular value of alpha. Normalization means I need for 1 to be equal to this integral, negative infinity to infinity of the wave function, and not e to the minus alpha x squared over 2, times the wave function again, so that's another factor of n naught, and another factor of e to the minus alpha x squared over 2, integrated over x. So the n's I can pull out of the function, the integral, the two Gaussians combine e to this exponent, and e to this exponent multiplied together is e to the sum of the two exponents, so one half of this quantity and one half of this quantity add up to just give one of that quantity, and luckily that's an integral that we know how to do. We've seen Gaussian integrals before. Integral of a Gaussian function from negative infinity to infinity is the square root of pi over the exponent of the Gaussian, the coefficient alpha. So we need for 1 to be equal to n naught squared times the square root of pi over alpha. Since we're solving for n naught, I'll rearrange this to say n naught squared must be square root of alpha over pi, or when I take the square root, n naught must be alpha over pi to the 1 fourth power. So that's the normalization constant. If we really care about the full details of the n equals 0 wave function, then the value of this n naught normalization constant is the fourth root of alpha over pi, where alpha is this collection of constants. Often we don't particularly care about those values, but when you do, here's how to do the normalization, the variable we integrate over and the ranges of that integral. We can also talk about the orthogonality of those wave functions. Are they orthogonal to one another? So here I can't ask, is psi naught orthogonal? Remember, I have to ask, is it orthogonal? Is one wave function orthogonal to another wave function? So I can ask, is, for example, psi naught orthogonal to psi 1, or to psi 2, or to psi 3, and so on, any pair of wave functions I want? In general, to answer that question, I would need, how about, let's say, the nth wave function orthogonal to some n prime, some different wave function? To answer that question, I'm asking, is it true that the overlap integral between n and n prime, this integral of psi n complex conjugated, if we need to, which we don't for harmonic oscillator, multiplied by n prime wave function, is that integral equal to 0? If it is for n's that are different from one another, then those two wave functions are orthogonal to one another. We can certainly solve one such problem mathematically. We can take the psi 0 wave function, the psi 1 wave function, insert them in an integral, do the calculation. It's a little bit easier to see how that works out if we do it visually. If I draw a graph of what the psi 0 wave function looks like, psi 0 looks like a Gaussian, constant times e to the minus alpha x squared, so that, ignoring the normalization constant, looks like, in fact, let's ignore all the constants and say that just looks like e to the minus x squared with some constants thrown in. The shape of the function is the shape of a Gaussian. If I am asking whether that's orthogonal to psi 1, or psi n, or some other wave function, I need to know what those look like also. So if I draw psi 1, that has the shape first order function, x, times the Gaussian, again with normalization constants up front, alpha over 2 up top. But the shape of that function generally looks like this. We know it looks like that because it's the same Gaussian we started with before, and we've identified when x, this axis is x, when we multiply by x equals 0, the function has the value of 0 because x is equal to 0. For small values of x, I'm multiplying my Gaussian by a small number that's either positive or negative if x is positive or x is negative. And then as x gets larger, I'm multiplying by a large number, but it doesn't blow up because the Gaussian is dying fairly quickly, faster than the value of x is growing. So it's negative when x is negative, positive when x is positive. Let me go ahead and shade the area under this integral to illustrate the fact that if I were to compute the overlap integral between the 0th and the first wave function, so that's my overlap integral between psi 0 and psi 1, multiply this wave function by this wave function. Both of these wave functions are symmetric. The first wave function is, in fact, symmetric about x equals 0, if I had drawn it as carefully as I should, what it looks like on the left or on the right matches what it looks like on the left. The psi 1 wave function is odd, remember the Hermite polynomial is an odd function, so the value on the left is negative of the value on the right. So when I multiply the two functions together, the areas where it's positive times positive exactly cancel the areas where it's positive times negative, and by symmetry we can see that that overlap integral will come out to be equal to 0. We can continue and do a very similar thing for one more case, the psi 2 function. Remember what psi 2 looks like. That's again ignoring the constants x squared minus a constant times the Gaussian. Second order Hermite polynomial, multiplying my Gaussian function, so it looks like a Gaussian not multiplied by a linear function that has 1, 0, that gives the wave function a single node at x equals 0, but now I've got a quadratic function which has 2, 0s, which is going to cause there to be two different nodes at x equal to positive and negative some collection of constants, so the wave function is going to go through 0 at two different locations. It's going to start out positive, dip negative, cross back positive again, and then die at either extreme because of the Gaussian. So again shading the area under the integral to illustrate where it's positive and where it's negative. If we compute the overlap integral between wave function 1 and wave function 2, again we can see because of the symmetry this is an even function. If I multiply it by an odd function, the places where it's positive and negative will cancel out the places where it's negative and negative, where negative positive will cancel out positive, positive. So by symmetry these two are orthogonal to one another. It gets a little more complicated when I talk about the overlap between function 0 and function 2. So here I can't use symmetry in quite the same way. This is an even function, this is also an even function, there's no guarantee that an even function times an even function will integrate to 0, but I can see in this case there are some regions where positive multiplies positive. Over here more regions where positive multiplies positive, and in the middle there are some regions where positive multiplies negative. And it just happens to be the case because of the features of this Hermite polynomial, essentially because of the recursion relation between different Hermite polynomials, that the negative areas of this product will exactly cancel the positive areas of this product, and this overlap integral will in fact be 0, although it's not obvious that that has to be the case from the pictures we've drawn here. So in fact it is true that every pair of different harmonic oscillator wave functions is 0. The overlap integral between them is 0, so those two wave functions are orthogonal. The full set of harmonic oscillator wave functions is indeed mutually orthogonal. So that has summarized both the normalization and the orthogonality of these integrals since the wave functions with these choices of normalization constant are normalized and they're mutually orthogonal to one another, we say that the harmonic oscillator wave functions form an orthonormal set of functions. So now that we know that, the next step is to go back to the wave function and see what we can learn about the energies of the wave functions.