 So, one important feature to be aware of in the energy levels for a particle in a box, a 3D particle in a box, is the degeneracy of those energy levels. So to understand that, let's pay a little bit closer attention to this expression we have for what the energy levels are for any individual energy level for a particle in a box. If I choose an n sub x and sub y n sub z, some integer values, I can calculate the energy of that particle in that box, in that three-dimensional box. So, let's draw an energy ladder. So here are the energy levels for the states of that particle in a box. And like I said, for any integers n, x, n, y, n, z, I can calculate the energy and place that energy on this ladder. So the lowest possible energy, the ground state for that particle in a box, smallest I can make n squared with a positive integer is by making n equal to 1. So if I choose nx equal to 1 and y equal to 1 and z equal to 1, that'll be the lowest energy level. So e sub 1, 1, 1, I've chosen n sub x equal to 1 and sub y equal to 1 and sub z equal to 1. That's going to be equal to h squared over 8m 1 squared over a squared, 1 squared over b squared, and 1 squared over c squared. If I ask myself what the next step up on this energy ladder is, I can take one of these ones and make it into a two. And which one I make into a two, I can make any one of them into a two. Which one is higher in energy than the others depends on what the box sizes are. So if one box size is relatively large, then it doesn't cost me much energy to make that n larger. But in general, without knowing anything about which box sizes, box lengths are bigger than the others, somewhere above the 111 energy level, there's going to be a 112, a 121, a 211 energy level. Unless the box sizes happen to be equal to each other, there's no reason that the energy for the 112, there's no reason this ratio should be the same as this ratio or this ratio because the box links in general won't be the same as each other. I can continue and add more levels on this tree. I can increase some of the twos to threes or I can have more than one of them be two at the same time. But before we do that, let's think about what this energy level diagram looks like when I do have a cubic box. When A, the box length in X and B, the box length in Y and C and the box length in Z are all equal to each other. So in that case, the energies simplify a little bit. Because A squared and B squared and C squared are all the same number. I can factor them out of this expression. And I just compute the sums of these n squareds, nx squared, ny squared, and z squared, sum those together, multiply it by the constants that just include this one box length. So if I draw what the energy ladder looks like for that case, again, for the ground state, the lowest this energy can be is if I use ones for each of those constants, and then one squared plus one squared plus one squared is just three. So there's the one, one, one energy level. The next step up on the tree, I can either make n sub z equal to two, I can make n sub y equal to two, I can make n sub x equal to two. Doesn't matter which one I make equal to two, I get the same answer in any case, two squared plus one squared plus one squared is the same as one squared plus two squared plus one squared or one squared plus one squared plus two squared. So what that means is somewhere higher up on this energy ladder, there's three degenerate energy levels. We can call those E 112, E 121, E 211. They all have the same energy as each other. And that energy is gonna be two squared is four, one squared is one, one squared is one, four plus one plus one is six. That energy is going to be six times those constant, h squared over eight ma squared. So twice as high as the ground state, but triply degenerate. There's a degeneracy of three. Let me go ahead and label these states as having a degeneracy of three, because there's three states, the x, y, and z excitations, that all have the same energy as each other. So that's something that didn't occur in the case where the box lengths were unequal does occur where the box lengths are equal to each other. So if I do continue climbing this energy ladder and I ask myself, what's the next highest energy level? There's a couple of things I could do. I could either make two of the, I can make, let's go ahead and write down states like E 221. So I've made not just the x, n sub x equal to two, but I've also made n sub y equal to two, but I've left n sub z equal to one. That's gonna give me an energy, in this case, two squared plus two squared plus one squared is going to be four plus four plus one is nine. So my energy is gonna be nine times some constants, nine times h squared over eight ma squared. So that's the math I did to get that result. And this state is also triply degenerate because the 212 and the 122 states all have the same energy as each other. It doesn't matter what order I sum those ends, they give me the same result. Let me do one more of these. Let's say we bring the third of them to zero. So our three ends are gonna be two squared and two squared and two squared. Four plus four plus four is 12. So that energy is going to be 12 h squared over eight ma squared. And that's what I get for the 222 energy level. So notice this one is not degenerate. nx equal to ny equal to nz equal to there's no different permutation of those numbers there's only the single 222 state. There's three different ways I can permute these numbers 212, 221, 122 to get different energy levels, but there's only one 222 level. So that state is only singly degenerate. Some of the states are degenerate and some are not. So that's one interesting feature of the particle in a box energy levels. Some of the energy levels are not degenerate. Some of the energy levels are degenerate. Another interesting feature comes if I ask myself, what if I hadn't just used twos and ones? What if I had used let's say threes? Let's say one of the dimensions I lift the excitation up to the n equals three state. And I'll do the other two in the n equals one state. So if I have, for example, nx equals three and y equals one and z equals one. Three squared plus one squared plus one squared equals only 11. Nine plus one plus one. So there's an energy level down here at, so let's see that one is 12 h squared over 8ma squared. This one that I've drawn here is only 11 h squared over 8ma squared. That's a triply degenerate energy level. That's going to include not only the 311 but also the 131 and the 113 energy level. That's a triply degenerate state. So that points out a somewhat annoying thing about the energy levels for their particle in a box. And that it's a little bit difficult to figure out without going to the effort of doing these calculations, which energy level is going to be next? I want to know, is my next state going to be the 321 state? Or one of the other excitations? I can't know for sure without pointing the numbers and calculating and then placing them on this tree in whatever order they come out in. So it's a little bit hard to predict without calculating all the energies, what the order of these states is going to be. One last feature I'll point out, we've seen so far that we have singly degenerate states, we have triply degenerate states. If I end up with a state like let's say nx equals 1 and y equals 2 and z equals 3, so the energy of that state is going to be square those numbers and add them together, 1 plus 4 plus 9 is 14. So there's going to be an energy level slightly higher than this 12 energy level. Has an energy h squared over a ma squared times 14. So that's this energy level right there. Ask yourself, what is the degeneracy of this level? The one that I've just told you I calculated is the energy of the 1, 2, 3 and x equals 1 and y equals 2 and z equals 3 energy level. How many degenerate states are there? What's the degeneracy of this state? That's just the number of ways I can shuffle the numbers 1, 2 and 3. The number of permutations of that number because no matter what order I write them in or how I permute them, their sum of their squares is still going to be 14. So it turns out there's three factorial. There's six different ways of writing 1, 2 and 3 in different order. I could write 1, 2, 3 or 1, 3, 2, 2, 1, 3 or 2, 3, 1, 3, 1, 2 or 3, 2, 1. So maybe I won't bother to label all those but I need to write down degenerate six different states that all have the same energy when I have three different indices. So I can have degeneracy of 1, degeneracy of 3, or sometimes degeneracy of 6 for different states. So it turns out there's multiple different degeneracies that these states can have. So each of these features of the 3D particle in a box are features that will show up in other types of problems that we solve quantum mechanically. But this is the first time we've run across cases where we have both degenerate and non-degenerate states in a mix of different degeneracies and some difficulty predicting what order those states will show up in.