 So we've seen Schrodinger's equation in one dimension, or as it's written in here in three dimensions, which is an equation that the wave function needs to satisfy. What we'll point out now is that the Schrodinger equation is full of operators, something called operators. An operator is something you may not sound familiar to you, but you're familiar with operators and how operators work, so I'll convince you that they're nothing terribly new. An operator is just a mathematical action that acts upon something else, operates upon something else. So for example, I can define an operator. I'll say let's let the D operator, and we'll often use a hat to describe an operator. So if I put the D with a hat on it, it means I'm talking about an operator called D. This operator D that I'm going to define is the take the first derivative of operator. So derivative is an action that doesn't mean anything by itself. I can't tell you what the derivative is until you tell me what you want to take the derivative of. So it acts upon some function. So another way to define this function would be to say the derivative operator when it's acting on some function gives me the derivative of that function. So we can do a few examples. Derivative operator acting on the function sine x will give me cosine x. On the other hand, the derivative operator acting on a different function log of x gives me derivative of log x is 1 over x. So the operator is the same in both cases, but the result I get depends on what it is that I'm acting on. So that's a moderately interesting operator. Operators don't have to be complicated or interesting. They can be very simple. In fact, let's define an operator, the multiply by five operator. So notice I've put a hat on the five here, meaning the five operator. I'm defining to be tell me what you get when you multiply the thing by five. And again, you can't tell me what this operator does until I tell you what it's acting on. But if I act with the multiply by five operator on some function like 3x, then what I'll get is five times that function or 15x. So multiplying by five is not a terribly complicated operation. You've probably never thought of that as being an operator before, but we can think of multiplication, differentiation. We can think of most mathematical operations as operators. One more important thing to understand about operators is they connect sequentially. So let's see, we've got this differentiation operator. I'll define another operator, the squaring operator. So if I take the squaring operator s, s hat, act on a function, what I get back is the function squared. So, so far so good. If I act with the squaring operator on some function, let's say 3x squared, then what I'll get is the square of 3x squared. I'll get nine times x to the fourth. But notice that I can do more than one operator back to back. So if I say take the squaring operator and the differentiation operator acting on some f, in fact, let's not act on f, let's act on this 3x squared. So the way I've written this down, what this means is first I need to take the d operator acting on the function, and then when I'm done with that, I take the squaring operator acting on what's left. So notice that the operators proceed right to left. First act with the one closest to the function, and then act on the result with the one further away. So in order to understand what this means, d derivative of 3x squared is 6x. So what I'm asking here is what is the squaring operator acting on 6x. So that's going to give me 36x squared. So far so good. That tells us what happens when I sequentially first take the derivative operator and then take the squaring operator acting on the result. I could also have asked what's the result when I do those in the opposite order on the same substrate, on the same function, first square it and then take the derivative. So if I square the function 3x squared like we saw just above, squaring 3x squared gives me 9x to the fourth. If I take the derivative of 9x to the fourth, 9 times 4 is 36. So that part looks the same, but x fourth gets knocked down to x cubed. And I see that what I got by doing the square and then the derivative is not the same as I get by taking the derivative and then squaring it. So those two results are not the same as each other. So in general, it will not be true that one operator followed by another is the same as the two operators in the opposite order. In other words, what we've discovered is that these two operators, s and d, don't commute. So commutativity, the commutative property is something you learned about in elementary school when we taught you that addition is a commutative property, multiplication is a commutative property, a times b is the same as b times a. In third grade it probably didn't make much sense why we would bother teaching you this terminology because of course multiplication is commutative. But we taught it to you in third grade so that you could one day do quantum mechanics because sometimes we run across properties like these operators that don't commute with each other. Operations are not necessarily commutative. Sometimes they will be, but they're not guaranteed to be like these are not. So what does any of this have to do with Schrodinger's equation? I told you Schrodinger's equation is full of operators. Fairly, clearly, just like the differentiation operator is an operator, the Laplacian is also an operator. Remember what's hiding inside this Laplacian is d squared dx squared plus d squared dy squared plus d squared dz squared. So this is an operator. I could easily stick a hat on top of del squared and that is an operator that's acting on psi. I don't know what I get until I tell you what this function is psi that I'm taking derivatives of. And also just like five, the five hat operator is five times something. We can think of either e times or potential energy times as being operators. They're fairly simple operators. They're just multiply by operators. But what that tells us is the e times is the energy operator. We know potential energy is potential energy. We talked briefly before about how this whole term, this collection of constants, times the Laplacian all acting on a wave function that gives us the kinetic energy. So that's the kinetic energy operator. So we can just like we name the differentiation operator, call it d with a hat on it. We can give a name to the kinetic energy operator. We could call that, for example, t with a hat on it is a common name. You might also see some people call it e sub k for kinetic energy with a hat on it. The potential energy operator is often called v, capital V with a hat on it. And then I won't actually write this one as an operator. But what that means is I could rewrite the Schrodinger equation if I want to say this whole kinetic energy operator acting on the wave function if I add that to the potential energy operator acting on the wave function gives me back the total energy or if I wanted the total energy operator acting on the wave function or multiplying the wave function. And in fact, this whole process, take these constants times the Laplacian acting on a wave function, add that to the potential energy multiplying the wave function. I can combine all those things into one big operator. So if I rewrite potential energy acting on psi and kinetic energy acting on psi and I combine those both into a single operator, that operator we call the Hamiltonian. So that Hamiltonian operator t plus v acting on psi tells us to do exactly this. The definition of the Hamiltonian operator is minus h squared over 8 pi squared m times the Laplacian of the wave function added to potential energy times the wave function. That's the definition of the Hamiltonian. But yet another way of writing down Schrodinger's equation is the Hamiltonian operator acting on a wave function gives us back the energy times the wave function. So what we've seen is we've got operators hiding inside Schrodinger's equation and those give us a couple of new ways, simpler ways to write down Schrodinger's equation. The next thing we have to do is to understand how it is we're going to solve equations like this one, equations that involve operators, how to solve operator equations for the functions that they act on. And so that's coming up next.