Hello, next, we're going to start solving the stationary wave equation. And we're going to start with the simplest problem in quantum mechanics, and that's the particle in a box. The particle in a box is the case where you have a single particle and it is just translating back and forth between the walls of the box. If one assumes that the motion in the three coordinate directions is independent of each other, which is the case if there is no external field acting on the particle, no force acting on the particle except at the walls, then you can decompose the problem into the three coordinate directions x, y, and z. So we'll solve the x direction, and then the solution will essentially be identical for the y and z motion, okay? So the rules are the motion is free in the box except at the walls where the potential energy goes to a large enough value that the particle cannot penetrate the wall. Think of two delta functions at the left side and the right side. As a result, the particle cannot exist beyond the wall, or at the wall, essentially. And so, the boundary conditions for this problem is that psi has to be 0 at both the left-hand side where x is equal 0, and the right-hand side where x is equal to L. And the one-dimensional version of the stationary wave equation for external motion is given there. It's h-bar squared over 2m, d-square psi dx-squared, is equal to minus Ex psi, where Ex is the x direction translational energy. Well, this is just a very simple second order differential equation with sines and cosines as the solution. And so, we can write the psi of x is equal to a coefficient A, yet to be determined, times the sine of the square root of 2m epsilon over h-bar squared times x, plus b cosine, the same square-root of 2mex divided by h-bar squared, all times x. So to get the coefficients A and B, we have to apply the boundary condition. Well, the first boundary condition that is at psi at 0 must be 0 immediately sets B equal to 0. Because the cosine of 0 is not 0. In order to force that boundary condition to hold, you have to set B equal to 0. The second boundary condition where psi of L is equal to 0 turns out to set the first coefficient, A, by this expression at 0 equals A sine square-root of 2me over h-bar squared, times L. And for that to be satisfied, it can only be satisfied if e is equal to h-bar squared, pi-squared over 2mL times and positive integer n sub x-squared, okay? So what this means is that not all energies are allowed as stationary solutions of the wave equation. And hence, that is why the term quantum is used, because only certain quanta of energy are allowed. If we apply postulate two, which has to do with the normalization, then we can get the absolute value of A. If we integrate from 0 to L psi-squared dx, that has to be equal to 1. That forces A to be equal to the square-root of 2 over L. So our final solution for the x direction wave function is square-root of 2 over L sine of Pi n sub x X over L where n sub x can take on the values 0, 1, 2 etc. So in three dimensions, putting the three solutions together, which are essentially identical except that the subscripts have to change with coordinate direction, we get the total energy. Which is equal to the sum of the energies in each of the three coordinate directions is equal to h-squared over 8mV to the two-thirds, times n sub x-squared plus n sub y-squared plus n sub z-squared. And the wave function becomes equal to the square root of 8 over L-cubed, square root of 8 over L-cubed, times the sine of pi nx X over L, etc., for the other two coordinate directions. So this solution leads to an interesting situation. And that is that for a given energy, there can be more than one solution to the wave equation. And we can see this by looking at this table where I've listed nx, ny, and nz. And then, I've listed the sum of the squares of those, which is by the previous slide proportional to the total energy, okay? So if nx, ny, and nz are each equal to 1, then the sum of the squares is 3. However, if I set nx to 2 and leave ny and nz 1, then the sum of the squares is 6. Suppose then, I set nx back to 1, set ny equal to 2, nz is 1, the sum is still 6. And again, if I set nx and ny to 1, and nz to 2, the sum is 6. So here, I have three states which are not exactly the same. They have the same total energy, but not the same component energies. And that is called degeneracy. In other words, when you can have more than one quantum state that contribute to the same energy, that is called degeneracy. And pretty universally in science, degeneracy is given the symbol g. So the first row, the degeneracy is 1. There's no other way to make that sum of squares 3. But there are three ways to make the sum of squares 6. And for large values of n, g could become very large. And we will need to explore that later on. At this point, it's worth thinking about some numbers. The mass of a proton is 1.67 times 10 to the minus 27 kilograms. h, Planck's constant, is 6.262 times 10 to the minus 34 joule-second. k, which is Boltzmann's constant, which is the gas constant per molecule, is 1.38 times 10 to the minus 23 joules for the Greek K. So if I have a 1 centimeter box, then h-squared over 8mMV to the two-thirds is equal to 3.28 times 10 to the minus 37 jewels. Now, later, when we talk about ideal gases, we'll learn that the average or expectation value of the translational energy for an ideal gas is three-halves kT. So if I put that all together and calculate a typical value of n as the square-root of the sum of the squares of the three components, then that turns out to be 1.38 times 10 to the 8th. So in other words, a typical value of n is a very, very large number, and that will turn out to result in a very, very large degeneracy. The other consequence of this is basically that translational energy levels are very closely spaced to each other. And as a result, in some instances dealing with translational energy, we'll be able to treat things in a continuous fashion rather than a quantized fashion. Finally, on this last slide is a plot of the x direction or one direction wave function for increasing values of n. The simplest curve, which just goes to 0 at the ends and peaks in the middle, is for n equals 1. And then, increasing ns. And it's easy to imagine that as the quantum number increases, that these solutions become more and more oscillatory to the point where you could hardly tell they were oscillating. So that's it for this video. And have a great day.
