 So, we know one function, this psi of x e to the minus x squared with some constants, that solves the harmonic oscillator Schrodinger equation. When we insert this equation for psi into the Schrodinger equation, the left side is equal to the right side, and the primary reason that happened is because when we take the second derivative, it pulls down two extra powers of x, which end up canceling the x squared times the wave function in the potential energy term. So, that feature, that pulling down two extra powers of x, is just a feature of this e to the x squared term, and that would happen with a bunch of other mathematical functions also. So, for example, if I write down a function that looks like n times x times e to the minus alpha x squared over 2, it's also true. We won't do the derivatives, but perhaps you can look at that and convince yourself it's true, or perhaps you can pause the video and take the derivatives yourself and convince yourself that if I take derivatives of this function, despite the fact that there's some x's out front and some x's in the exponent, when I take the second derivative, I will end up with terms that have pulled down two extra powers of x into that polynomial pre-factor term. So, those we can make cancel with the two powers of x multiplying the wave function in the potential energy term, and this can also be a valid solution to Schrodinger's equation. Likewise, there are functions that look like x squared times e to the minus alpha x squared over 2 with normalization constants out front. This one gets a little more complicated because there's now some cancellation that needs to happen between the various different product rule terms involved in taking this derivative. So, because of that extra cancellation, if I use the 4 alpha and the x squared and also a constant term, that's a function that, again, we won't do ourselves, but we can convince ourselves if we insert it into Schrodinger's equation, this also solves the harmonic oscillator Schrodinger equation. So, as you can see now, there's a whole series of solutions, one with just a constant multiplying this Gaussian, one with a linear term in x multiplying the Gaussian, one with a quadratic function of x multiplying the Gaussian, there's also a cubic and a quartic and so on. So, in general, we can give these solutions numbers. This is, we call this the zeroth solution because the polynomial in front of the Gaussian is a zeroth order polynomial. This linear function or first order polynomial is the n equals one solution. There's an n equals two solution. In general, for any n, I can write a solution that looks like a normalization constant times a polynomial, and we'll talk more about the features of that polynomial in just a second. In fact, I'll write that polynomial as not just a polynomial of x, but you'll notice every time we see an x, it's paired with some alphas. So this x squared is paired with an alpha, this x squared is paired with an alpha, these x squared are paired with an alpha. Every time we see an x, it's paired with an alpha to the one half. So, this polynomial is a function of alpha to the one half times x, and all multiplying this Gaussian term, e to the minus alpha x squared over two, that's the same for every one of these solutions. So the polynomial term is different for every case. The normalization constant will end up different for every case as well. And these are the general form of these solutions to the harmonic oscillator Schrodinger equation. In order to be able to write down any particular solution, there's a couple things you could do. You could write out a polynomial if I want to know the n equals 5 solution. What fifth order polynomial should I stick in front of this Gaussian to solve Schrodinger's equation? We could do it the hard way. We could write down a fifth order polynomial, plug it into Schrodinger's equation, figure out what constants need to cancel other constants, or now that I've written the general form of this polynomial, or at least the name of this polynomial, that polynomial is called a Hermite polynomial. As usual, that's a function that was well-studied and understood by mathematicians before anybody knew it was useful in quantum mechanical problems. That function, so notice that here I've written it a Hermite polynomial as a function of some arbitrary argument. Normally, I would say h of x, but for the harmonic oscillator problem, x has another very specific meaning that's the bond displacement. So here I'm just writing h as a function of some argument. We can write down the first few of these Hermite polynomials and get a sense for what they look like. So the zeroth polynomial, as we've already seen, is just a constant, and the constant value is 1. h1, I'll have to double check the constants. That is twice alpha to the 1 half times, oh no, that's not true. Twice the argument of the function. So what that means is if I insert alpha to the 1 half x, as we've said here, the n equals 1 polynomial whose argument is alpha to the 1 half times x, that will be 2 alpha to the 1 half x, because the argument of the function is alpha to the 1 half times x. The n equals 2 polynomial is, as we see here, let's see, that's going to be 4 times c squared minus 2. Or in the case where the argument is alpha to the 1 half times x, the squiggle squared is c squared equals alpha x squared. We can write one more, we can write a whole lot more, but I'll write one more that is not already on the board, which is the n equals 3 version of this polynomial, which is 8 times the argument cube minus 12 times the argument. And again, this series goes on. That's only useful. These examples are useful if we never want to go beyond n equals 3. What's useful is to be able to predict what the nth polynomial is, and we can do that with a recursion formula. If we want to know the n plus 1th polynomial, that's going to be twice the variable that we're talking about, multiplying by the previous polynomial, plus or minus. It's minus twice this number n times the polynomial prior to that. So this recursion formula tells me the n plus first in terms of the nth and the n minus first. So if I want to know the fourth one, I can plug in the second one and the third one here and here and use that to solve for the n equals 4 polynomial. The other thing that's worth noting about these polynomials is their parity. So if we look carefully, we can see that let's say that n equals 1 and that n equals 3 polynomial. This is n equals 1 polynomial has its argument raised to the first power, n equals 3 polynomial has the argument raised to the third power, but also the argument raised to the first power. So it's only the odd powers of the argument that show up in n equals 1 and n equals 3. So those functions have odd parity. They only have odd powers of their argument, whereas the even values of n, 0 and 2 and continuing with 4 and 6 and so on, those have only even powers of their argument. So h2 has not only the argument squared term, but also an argument to the 0 term, but it doesn't include an argument to the first term. So when n is an odd number, the polynomial includes only odd, when n is an even number, it includes only even powers of the argument. When n is an odd number, it includes only odd powers of the argument. And that's going to be important when we get a little further into talking about the parity of the wave functions because the Hermite polynomial, knowing what the Hermite polynomial is, knowing how to predict it for any arbitrary value of n, is now going to allow us to write down the wave function for any arbitrary value of n as well.