 We've seen that the one-dimensional particle in a box is a problem that we can solve Schrodinger's equation for. So here's a reminder of what Schrodinger's equation looks like for the one-dimensional particle in a box. And we got a little bit of mileage out of solving that problem. We discovered after solving it that it's a useful model to be able to describe things like an electron in a linear conjugated hydrocarbon molecule. And it may be useful for some small number of other things as well. But in general, problems in the three-dimensional world are not well described by a one-dimensional particle in a box. So we'll get a lot more mileage out of solving the same particle in a box problem in three dimensions rather than in just one dimension. So that's our next task. So as a reminder of how we proceeded for the one-dimensional particle in a box, because the 3D particle in a box is going to be very similar, we wrote down Schrodinger's equation. The potential energy term was zero, because we assume the particle has no potential energy. So that's our 1D particle in a box Schrodinger's equation. We could then write down or solve Schrodinger's equation to write down the wave functions. Anything that looks like a sine wave is going to have a second derivative that's the negative of the original function. So these are all going to be solutions to this Schrodinger's equation. With the extra caveat that because we require that the solutions be zero at the edges of a box, I can only have that sine wave or that sine wave or that sine wave, only sine waves that end at the edges of the box. So solving the Schrodinger's equation gives us one set of solutions obeying the boundary conditions, making those sine waves end at particular places defined by the edges of our box, places some additional restrictions, and we found that the one-dimensional particle in a box wave function has to look like not just sine of any constant times x, but sine of an integer times pi x divided by the box length. So that's a real quick recap of what we did for our one-dimensional particle in a box. And remember this step, we required that the, at the two edges of the box, when I plug in x equals zero or when I plug in x equals a, I have to get zero for the value of the wave function. That's how this k became an n pi over a. So we're going to now do the exact same thing for the three-dimensional particle in a box. We can start by writing down Schrodinger's equation. So it's still a particle in a box. We're still going to assume that when the molecule is inside the box, there's no potential energy and it's not allowed to be outside the box. So we still have just a kinetic energy operator. The only difference is that instead of having d squared dx squared, we need the full Laplacian, which is just a fancy way of saying we need the derivatives not just with respect to x, but also with respect to y and with respect to z. So those are the derivatives that we take of the wave function and sum them together multiplying by the constants out front to obtain a constant times the wave function. So now we've written the three-dimensional version of the Schrodinger equation, just like the one-dimensional version, but we have a y and a z. Also I suppose I should point out this wave function itself is now a function of x and y and z instead of just a function of x as it was for the one-dimensional problem. So if we were to solve this equation formally, for example, as you would have done in the differential equations class, we would invoke separation of variables. I've got some x derivatives that are separate from the y derivative separate from the z derivatives. So there's some steps we could take that will assure us that this wave function is going to be a product of some parts of x, some parts with y, and some parts of z. Being quite so formal about it, what we can do is I'll write down what the solution should look like and then we'll be able to confirm that those solutions are, in fact, solutions to the Schrodinger equation. So again, because we know how to solve the problem, if we only had the x derivative, if all we had was this piece, the solutions would look like this. So that's a good starting place. But since we also have a y portion and a z portion, because these derivatives sum together, the wave function, it turns out, multiplies x, y, and z portions together. And there's no difference between d squared, dx squared, d squared, dy squared, d squared, dz squared, other than the variable names. So the things we multiply together are sine of k times x, sine of k times y, and sine of k times z. And in fact, there's no necessity that these k's all be the same value as each other. I could have a different constant case of x multiplying x and a different constant case of y that multiplies y and a different constant case of z that multiplies z. There's possibly three different constants in this expression. So that's the equivalent of the equation I have here, where we've solved Schrodinger's equation. If we took this expression, plugged it back into Schrodinger's equation, we could confirm that taking the second derivatives of this function do, in fact, give us back the same function we started with, multiplied by some constants. So that's step number one. Step number two proceeds very much like step number two in one dimension, again, keeping in mind that we're in 3D now instead of 1D. So instead of having a one-dimensional region that we've confined particle two, to keep it in the box, we have a three-dimensional, so I've got a three-dimensional box. That the particle is confined to, and the potential energy is zero inside this box, and the particle is not allowed to be outside the box. So what the equivalent of these statements, the wave function, has to be zero at the edges of the one-dimensional box. The same thing has to be true. The edges of the box in x, the box may range from zero to a. In y, we can choose a different box length. Let's let b be the length of the box in the y direction. And then let's let c be the height of the box in the z direction. So what that means is my new boundary conditions. I want whenever the particle is at the left edge of the box, whenever x is equal to zero, I want the wave function to be zero. Whenever x is at the right edge of the box, I want the wave function to be zero. That's like saying, the second condition in particular is like saying anytime the particle is at the right face of the box, the whole right side of the box, I need the wave function to be zero there. The first condition is like saying I need it to be zero on this left face of the box. There's an equivalent boundary condition for the top of the box and the bottom of the box, and the front of the box, and the back of the box. So I also need the wave function to be zero when y is equal to zero. Or when y is equal to b, or when z is equal to zero. Or when z is equal to c. So those are the six conditions for the six different sides of the box, not just left and right, because we're in three dimensions. Also front and back and top and bottom as well. So after requiring, so these two conditions place a restriction on x and that works exactly the same way it did for the one dimensional particle in a box that's going to turn this k into some integer number of pi over a's. The wave function has to oscillate an integer number of times when it gets from the left edge of the box to the right edge of the box. But the conditions on y are going to place some restrictions on k sub y. The restrictions on z are going to place some restrictions on k sub z. So by the time we apply all those, the result is going to look very analogous to the solution in 1D. And we're going to find that the wave function looks like sine of n pi x over box length nx multiplied by sine of n pi y over the box length in y. Multiplied by sine of n pi z over the box length in z. But again, these constants don't need to be equal to each other. The number of oscillations in the x direction doesn't have to match the number of oscillations in the y direction. It doesn't have to match the number of oscillations in the z direction. So this n sub x, n sub y, and sub z may be or allowed to be different values. And just like in the one dimensional problem when we applied the boundary conditions, we went from an infinite number of solutions with any possible value of k to a discretized set of solutions where these ends had to be integers. And we started labeling the wave function size of n. Same thing here. We're going to label these wave functions not just by a single n, but by the three n's, nx, ny, and z that we need to be able to evaluate it. So these equations that I've written down here for any choice of nx, ny, and z, nx is any positive integer as well as ny, nnz. Each one must be a positive integer. For any choice of positive integers, I can write down a wave function that will obey Schrodinger's equation. And just to, in fact, confirm that these wave functions do solve the three dimensional particle in a box wave Schrodinger equation. Let's go ahead and convince ourselves that that's true. If this wave function solves the Schrodinger equation, if I take the second x derivative, add it to the second y derivative, add it to the second z derivative, multiplied by these constants, I'll get back to the original wave function. So I need to know what is the second x derivative of my wave function. The x's only show up in this sine term, right? y term and z term don't have any x's in them. So if I take the x derivative of sine, I get cosine. If I take another derivative, I'll get back to negative sine. Every time I take a derivative, it's going to pull out some constants, the n pi over a. So I'm going to have negative n pi over a. I've done that twice. I've still got an a, I've still got the sine that's turned into the negative sine. I've still got the sine y and sine z. So that's all multiplying the original function. So taking the original, taking the second derivative of the function, gives me back the same function multiplied by these negative constants. Likewise, for y and z, the second y derivative is going to pluck out some negative constants and y squared, pi squared over the box length in y squared. And likewise for z, it's minus n sub z squared pi squared over c squared. So those are all the building blocks I need to be able to evaluate the Schrodinger equation. So if I take that and let's go ahead and plug our wave function now into Schrodinger's equation. So Schrodinger's equation tells us minus h squared over 8 pi squared m, second derivative with respect to x gives me minus nx squared pi squared over a squared times psi. Second y derivative gives me minus ny squared pi squared over b squared times psi. The z derivative gives me minus nz squared pi squared over c squared all times psi. The claim in Schrodinger's equation is that that collection of things is equal to some constants times the wave function and it's true. Everything I have here, everything in parentheses, everything in front of the parentheses, those are all constants. So that's all equal to constants times the wave function and that's true. That also lets us evaluate what is the energy of a three dimensional particle in a box. This particular particle in a box with choice of nx, ny, and z as the integers gives me this result for the energy. There's some cancellation that happens. These pi squareds all cancel. The negative signs all cancel each other. So after that simplification I can say that the energy, everything that precedes the psi, the energy for the three dimensional particle in a box is h squared over 8 times mass. And then I've got some n's over some box lengths in each dimension. So it's nx squared over a squared, ny squared over b squared, and nz squared over c squared. That is the result for a three dimensional particle in a box. We now know what the wave function, sorry, this one, what the wave function is for a three dimensional particle in a box. Given an n for each of the x, y, and z directions, and we can also calculate what the energy of that three dimensional particle in a box is. The last thing we need to do is the very common case where we happen to have a cubic box. Notice I didn't assume, even though I drew it somewhat like a cube, I haven't assumed that the length of the box in A and B and C are equal to each other. But if I now do assume that A is equal to B is equal to C, this constant, this constant, this constant are all equal to each other. A squared and B squared and C squared are all the same number. In that special case, I can write the energy of the wave function A squared, B squared, C squared. I can write each one of them as A squared if I prefer, and I can pull that out of the parentheses. So I can just write A squared over 8ma squared times the sum of these integers squared, nx squared plus ny squared plus nz squared. So in the fairly common case where we have a cubic box, there's a somewhat simpler expression for calculating the entropy. Non-cubic box, I would use this expression. A cubic box, I would use this expression. So we're almost done with writing down the wave functions for the 3D particle in a box, but let me just point out that I haven't yet told you what this constant A is, or we haven't figured out what that constant A is. So we still have to normalize the wave function, and that's the next step that we'll do.