 Okay, so we've talked about the solutions to the Schrodinger equation, the one-dimensional version of the Schrodinger equation with no potential energy, the problem we called the free particle problem. So for the free particle Hamiltonian, some of the solutions to Schrodinger's equation are these sine functions, sine of some constant times x multiplied by some other constant. That function, that wave function will solve Schrodinger's equation and those wave functions have energies of this particular equation of constants involving the k squared where k is the constant inside the sine wave. Turns out there's a problem with this solution to Schrodinger's equation. To see what the problem is, let's remind ourselves of what the wave function means. This wave function, if I were to graph the wave function constant times sine of kx as a function of x, sine wave just oscillates up and down, goes on in the positive x direction as well as the negative x direction that continues on in both directions, right? What does that mean that the wave function is a sine wave? Remember what the physical meaning of the wave function is when you want to ask yourself what the wave function means. The thing to remember is that the probability of the wave function, the probability, sorry, of the particle being somewhere is equal to the wave function squared. Maybe with a complex conjugate, but there's no complex numbers in this version of the wave function, so I'll just square the wave function. When I square a sine wave, so a squared sine squared of kx, that just makes the positive numbers bigger. It makes the negative numbers positive. This is poorly drawn, but the green curve here is what sine squared looks like, and again it goes on and on in both the positive and negative directions. In particular, what I want to be true is that if I integrate, if p is the probability of finding the particle at some position x, so what this green function tells us is I'm likely, somewhat likely to find the particle here, I'm much less likely to find it where the green curve is small, and there are some places where the curve hits zero where I'm not likely to find it at all. Sine squared is telling us how likely I am to find the particle at various places, so this particular wave function generates this sort of washboard probability function for finding the particle at different places. The problem with this function is if I integrate the probability of finding the particle at a position x from x equals negative infinity all the way up to x equals positive infinity, positive infinity, what I should get is one. There's a hundred percent of probability of finding the particle somewhere between negative infinity and positive infinity. In fact, when I calculate the area under this green curve, calculate the area under this lump, add it to the area under this lump, and the next one and the next one all the way out to positive infinity, that area without working out the integral in detail, that integral is clearly going to be infinite, so we need the probability to integrate to one. In fact, for this wave function, it doesn't integrate to one, at least not if a is a number greater than zero. That's the problem we have with this wave function. It turns out one of the requirements we're going to impose on wave functions is that when we square them, their integral has to be finite. It cannot be infinite. Right now, we'll just say something went wrong, that's not a good wave function, we'll have to figure out how to deal with that. One thing we can do to deal with that is to recognize that in the real world particles can't extend everywhere. This free particle problem that we've solved Schrodinger's equation for with the potential energy equal to zero, the potential energy term is missing here. The only way for that to be true is for there to be no particles anywhere, not all the way out to positive infinity or negative infinity for the original particle to interact with. If in fact this was the only particle in the entire universe and there was nothing else around it, then the free particle would be an appropriate equation to solve. But in the real world, we can't have an infinite volume that our particle occupies. We can put a particle in an empty box, but there's always going to be some walls on the box for the particle to remain within. So we're never going to have the probability of finding the particle all the way out to positive infinity and negative infinity. In fact, what we more typically have, as you might have guessed from the title of this lecture, is a particle confined to a box. So my wave function sine wave might be fine as a wave function for this particle, but let's say I confine the particle to somewhere between zero and A. So I don't allow the particle to exist anywhere outside of a box. The box ranges from x equals zero up to x equals A. If x equals some negative number, then I'm not going to allow the particle to be there. If x is a positive number bigger than A, I'm not going to allow the particle to be there. So the wave function is zero, the probability is zero outside of the box. So I've confined my particle to a box. The problem with this, however, is if I draw my sine wave wave function, I've got a sine wave that must be non-zero inside the box. It's this oscillating sine wave inside the box. I'm requiring it to be zero outside the box. In other words, I'm requiring that the probability of finding the particle was zero outside the box. So my wave function might look like a sine of chi x k x when I'm inside the box, when x is between zero and A, but it's going to be zero when either x is negative or when x is too big, when x is bigger than A. The problem here, so I've drawn that function here, it's zero for negative values, oscillating inside the box, zero for values greater than A, but notice here the function is discontinuous. My wave function has changed from some finite value to zero at the edge of the box. So that's not good. The particle can't both have a high probability of being at the edge of the box and a zero probability of being at the edge of the box when I approach that limit from different sides. So we're going to require that our wave functions be continuous. So the function I've drawn here is not a valid wave function because it's not continuous. A better version of that wave function that is allowed would be to have it be zero and have it oscillate. Let me keep the box size the same. So if the edge of the box is here, if my wave function hits the edge of the box exactly, then everything's okay. That wave function that I've just drawn is continuous and that's good. So I need the wave function to be zero at the left edge of the box and I need the wave function to be zero at the right edge of the box no matter which side I approach it from. So what that means is I'm only certain wavelengths of the wave inside this box, this sine wave, only certain constants k for that sine wave are going to be allowed. If the sine wave is also letting it adjust the right speed to hit the edge of the box and reach zero, then everything's okay. If it's oscillating at a different speed where it hits the edge of the box and it's non-zero, then the wave function will be not continuous and I don't want that solution. So what I actually require is that the wave function at the left edge of the box when x equals zero has to be zero and the wave function at the right edge of the box also has to be zero. So those are the conditions I'm imposing on the wave function in order for it to be continuous. This first condition is no problem. When I plug x equals zero, so all I've done here is say evaluate the wave function when x equals zero. When x equals zero, this wave function, if I put a zero into the wave function, sine of zero is zero. So that's fine. Sine waves always start from zero and head upwards before they start oscillating back downwards. So at the left edge of the box, everything is fine. Here at the right edge of the box where x is equal to a is where I have a problem or a potential problem. My wave function sine of kx, when I insert an a, sine of kx becomes sine of ka and I need that to be zero. A sine of some argument is going to be zero only when the argument is zero or maybe the argument is pi or 2pi. If I take the sine of zero, sine of pi, sine of 2pi, sine of 3pi, sine of 4pi, sine of all those integer multiples of pi do come out to zero but those are the only values I can take sine of and get a zero. So what I want when I say sine of something must be zero is that something must be some integer multiple of pi. So I'm requiring that ka, the inside of that sine function, be equal to an integer times pi. Now let's step back and ask ourselves what we're solving for. I want to know what sine waves are the special sine waves that will oscillate and hit the edge of the box reaching zero right at the edge of the box. In other words I want to know which values of k make this sine wave oscillate just fast enough to hit the edge of the box. So if I solve for k then that's what I've learned. Those values of k are the ones that provide good continuous wave functions that can be confined within a box and remain continuous at the edge of the box. So instead of saying my wave function is any old sine wave inside the box in zero outside, a better description of the wave function that generates only valid continuous wave functions would be to say it's a times, capital A times the sine of kx, but k has to be n pi over a. So sine of n pi over a, all multiplying an x when x is inside the box and it's zero when x is outside the box, either when x is negative or when x is bigger than a. And that's describing wave functions that are confined within the box and continuous. So let me point out that just like our original function this wasn't just one solution to Schrodinger's equation this was a whole family of solutions. Any value of k solves Schrodinger's equation although it didn't generate a wave function with one hundred percent probability of being found somewhere. When we can find the particle to a box, we can't just use any value of k. We have to use some integer multiple of pi over a as our value of k. We still have an infinite number of solutions. The n values I can use here are some integer one or two or three. You might ask yourself why don't I also use n equals zero? Sine of zero times pi is zero. So that obeys the boundary conditions. But if we think about what the wave function looks like when I have sine of zero times pi times x, that always equals zero. So the wave function itself is zero. When n equals zero the wave function is just a flat line that doesn't exist anywhere. So again that wave function doesn't have a hundred percent probability of existing at some point in space. That is a wave function of a particle that doesn't exist anywhere. So only these values of one, two, three and so on generate useful values of the wave function. But still I have an infinite number of n's that I can insert into this wave function. A sine of one pi x over a. A sine of two pi x over a. A sine of three pi x over a. That infinite family of solutions can each be labeled with this letter n. So I've got a solution number one, solution number two, solution number three, each corresponding to one of these values of n. So I have an infinite number of solutions. But notice something is important, something important has happened in changing from this expression to this expression. Here k could be any value. At this point n can only be certain discrete specific values, certain quantized values. So the wave function is only allowed to be certain things, it's not allowed to be other things. So the act of confining the particle inside this box has quantized the wave function. And that's going to be a key feature that we run across frequently in quantum mechanical problems is once you confine a particle to a finite volume, you quantize the wave function. The energies, we haven't talked about energies yet, for the free particle the energies were collection of constants times k squared. Now we know that k has to be these particular values, n pi over a. The energies are still h squared over 8 pi squared mass times k squared. But with this value of k, if I insert n pi over a for k and if I square it, then there's some simplification that happens. I have pies that cancel. And so the energy ends up looking like h squared over 8 mass box length squared, a squared, and I still got an n squared. So the energy of this nth wave function, if I stick n equals one into the wave function that describes one of my wave functions. That wave function has an energy e sub one where I take these constants multiplied by one squared. I could take the n equals two or the n equals three solution. The energies of those are this constant times two squared or times three squared. So if I draw that graphically on an energy ladder, so here's an energy of zero. The lowest energy I can give my particle that's confined to a box is this n equals one solution. When I multiply by one squared, the particle has an energy e one, which is h squared over 8 m a squared times one. If I insert n equals two, the energy is four times bigger. Two squared is four, so this energy is e two, which is the same collection of constants times two squared or four. The next one, the n equals three solution is going to be even higher. So notice number one, the energies are discrete. They're quantized. Some energies are allowed. Some energies are not allowed. I can have an energy of one times h squared over 8 m a squared or four times those constants or nine times those constants. I can't have energies in between. So those quantized energy levels come from the discrete individual solutions to Schrodinger's equation, which happened when we confined the particle inside a box. Notice also this will be relevant later on when we talk about real world systems that we can solve with this particle in a box example. These energies are increasing proportionally to n squared. So as n gets bigger, the energies are getting bigger even faster as n squared. We could also point out that the energies, those energies are proportional to one over the box length squared. So h and m, we don't usually have much control over. A particle has the mass that it has and Planck's constant has the value it has. But if I've confined my particle to a box, I can choose, do I confine it to a small box or do I confine it to a large box? And that has a pretty big effect on the energies. When I confine a particle to a very small box, then that small number in the denominator makes the energies pretty large. On the other hand, if I confine the particle to a large box, the opposite is true, dividing by a large number squared in the denominator keeps the energies low. So that's again going to be an important feature that recurs over and over as we treat quantum mechanical problems. When we confine quantum mechanical systems to boxes, it quantizes their energy levels. And the smaller the box we put them in, usually the higher that makes their energies. All right. So that's all we have to say about the particle in a box with one exception. We've described these wave functions that solve the particle in a box problem. We've figured out what the value of k has to be in order for this particle to solve the particle in a box problem. But we haven't said anything at all yet about this number a, the constant a that multiplies the wave function out front. So that's the next thing we'll talk about is the meaning of this constant a and how to determine the value of it.