 So now it's time to talk about which wave functions actually solve the harmonic oscillator Schrodinger equation. What functions are there that we can take the second derivative of, add them to x squared times the same function with various constants thrown in, and get back the function that we started with. So we can consider at least some trial and error solutions to the Schrodinger equation and learn something about what type of functions will solve the Schrodinger equation. So we can start with a relatively simple equation. Let's just start with a polynomial. Let's just take a function like x cubed or x to the fifth or x to some power. We could also consider polynomial with multiple terms, but if we start simple, let's just take one term. We can throw a constant in front of that. And then the first derivative of that function, we're going to need the second derivative of that function in order to plug it into Schrodinger's equation. So taking them one at a time, the first derivative of that function is going to look like 3ax squared. And then the second derivative will pull down the x squared down to just an x and pull down the 2 into the pre-factors. Now I've got 6a times x is the second derivative. If we go to plug those into Schrodinger's equation, so I've got some constants and throughout this whole derivation, I'll not worry about the constants. We can worry later about whether these constants happen to cancel some other constants. But just to see if the function is going to work, I've got some constants multiplying the second derivative of the wave function. So that looks like more constants times x. And to that, I want to add some constants, 1 half k, so constants, multiplying x squared times the wave function. So x squared times the original x cubed gives me a total of x to the fifth. So constants times x, constants times x to the fifth. Does that equal constants times the original function, which is x cubed? So is it true that a linear function plus a fifth order term equals a cubic term? Definitely not. The x terms don't match in this expression. So that means that this guess at a solution failed. If we start with a function like x cubed, the reason it didn't work is because the second derivative knocked the power of x down by 2 to just x. This x squared raises it by 2 to x to the fifth. Neither of those matches x cubed that we had in our original function. So we're going to have to try something different. Just an x to some power is not going to work. So we can try something else. Part of this problem was that the second derivative kept knocking down the power of x, and we wanted eventually to be equal to the original function. We certainly know some functions that don't get knocked down in powers as we take their derivatives. Those are functions like e to the, let's say, a times e to the bx, exponentials. If I take derivatives of an exponential, the first derivative just pulls down the derivative of the exponent, pulls down to b. The second derivative is going to pull down another factor of b, so why don't I just say psi double prime, the second derivative is equal to ab squared, pulls down another factor of b. And if I plug those into Schrodinger's equation to see if they work. Second derivative, so again, constants, I won't worry about the specific values of. The second derivative looks like e to the bx. On the right side, I've got constants times the original function e to the bx, so that's good news. This term is matching this term, that looks good so far. But when I include this potential energy term, one-half kx squared times the wave function, I've got constants, the x squared, I can't lump it in the constants, that's part of the x dependence. So I've got x squared times e to the bx. So that's better than we had in our first try. This e to the bx matches this e to the bx, but the x squared, the potential energy term, is causing problems. So we're going to have to work a little bit harder to figure out a function that has a chance of working. If we think about what went right and what went wrong in this solution, it was good that our function didn't get knocked down any powers of x by taking the derivatives. But what we really need is something whose second derivative gets knocked up by a couple of powers of x, so that these two terms have some terms that match each other or perhaps cancel each other. And if we think about it a little bit, we can find a function that does that. So what I'm looking for is a function that if I take derivatives, pulls down not just extra constants in front of the function, but extra factors of x in front of the original function. So I can do that with a function like an exponential again, but now instead of e to the bx, I'm going to have e to the b times x squared. I'm going to have more than one power of x up in the exponent so that when I take the derivative of the exponent, it's going to pull down some factors of x out front. In particular, the first derivative is going to be a times the exponential e to the bx squared times the derivative of the exponent. The derivative of bx squared is 2 times b times x. So that's doing what I wanted it to. The first derivative has pulled down a factor of x. The second derivative gets a little more complicated because now I've got two factors of x to worry about the x in front of the exponential and the x squared in the exponential. The product rule tells me the derivative of this function is going to look like 2ab e to the bx squared. That's taking the derivative with respect to x of this polynomial term in front of the exponential and 2ab times x times e to the bx squared times the derivative of the exponent. So I'm going to pull down another factor of 2bx. So the 2 becomes a 4 and the b becomes a b squared. So that's what I get for my second derivative of this function. I'm sorry, pulls down a b and pulls down another factor of x. Yes, so 2 times b times x in addition to the original 2 and b at x. So this is the second derivative of this function and it's done what I claimed we wanted to happen when I take the second derivative. I get some x squared factors pulled down that are going to match this term and I also have something that looks like the original function that's going to match the term on the right side. So plugging those all into the Schrodinger equation, again we've got constants times second derivative of wave function gives me e to the bx squared and also some constants times x squared e to the bx squared. So that's the first term, the kinetic energy term. The potential energy term just looks like constants times the original wave function. So I'm adding in constants times x squared from here times e to the bx squared from the original wave function. And that is supposed to be equal to energy or some constants times the original wave function. So energy times e to the bx squared. Is it true that e to the bx squared with some constants in front plus an x squared e to the bx squared and an x squared e to the bx squared is equal to e to the bx squared? At first glance, that might look like it's not true. I have some terms on the left, this x squared e to the bx squared, that don't match any terms on the right. But depending on what these constants are, if the value of this constant, that's not a 0, that's just some particular value. If the value of this constant and the value of this constant when I add them together, when they cancel each other, if there's some cancellation that occurs and causes these two terms to cancel each other, then I'm left with something that can work. So at least if we choose our constants right, if we choose the value of this b constant well, this function can solve Schrodinger's equation, whereas these first two cases that we talked about cannot solve Schrodinger's equation for the harmonic oscillator. So this is our candidate solution. Possibly functions that look like e to the bx squared are going to solve Schrodinger's equation. What does that mean physically? If I draw a graph of what that wave function looks like, what would the wave function look like? So e to the, some constant times e to the bx squared. We don't know what the constant is yet, but certainly we know what the function e to the x looks like or e to the x squared looks like. If it's e to the x squared, then the function is going to look like this. The larger x gets, the larger e to that x squared gets, and the function is going to rise pretty quickly as I move away from x equals 0. And that begins to look a little strange. Remember what the wave function means. In order for us to interpret what the wave function means, remember that the probability of finding the particle at some point on the x-axis is proportional to the wave function squared. So that means I get more and more likely to find the wave function, find the particle at x values as x gets either very large or very small. And that probability reaches infinity as the value of x gets approaches infinity. So that sounds like it doesn't make any physical sense. I can't have an infinite probability of finding the particle at any location. But the good news is I haven't decided what this constant b is yet. It doesn't have to be a positive constant. So this graph is what the function would look like if b were a positive number. If b is a negative number, then this is going to look like e to the negative some constant times x squared. That sounds like a Gaussian function or a bell curve. So we know what Gaussian functions look like. So if b is less than 0, b is a negative constant, then the wave function is going to look like this. And now the good news is the further x gets away from the origin, the smaller the wave function gets. So the smaller the probability of finding the particle at that distance. And likewise the area under this curve or the area under this psi squared curve is bounded and finite. So that we have a chance of normalizing this function and making the probability equal to 1 if I ask what's the probability of finding it anywhere under this curve. So what that leads us to conclude is that a possibly valid version of the wave function would be some constant times e to the minus some other constant squared. So if I rename my constants instead of calling them a and b, I'll go ahead and start using n for the normalization constant out front. e to the negative some constant times x squared. And for my constant, I've made it negative, so I've changed its meaning. So b is equal to negative some other constant. I'll change its name to alpha. And in particular, so that we don't end up with this problem of these twos proliferating every time I take a derivative x squared causes some twos to fall down out of the exponent. Either one or more of those factors of two as I take derivatives. So to avoid the arithmetic complications when those factors of two show up, I'm going to make the exponent half as big as I had in the original definition. So still constant times e to the constant times x squared. My constant is just now negative this thing I'm calling alpha divided by two. So this is at least a reasonable guess for a function that will solve the Schrodinger equation for the harmonic oscillator. To find out what there does, in fact solve Schrodinger's equation and what this value of alpha needs to be, what the value of n needs to be. We need to do a little more work and actually plug this into Schrodinger's equation and not be so cavalier with the constants that we cancel. And so that's what we'll do next.