 All right, so we've seen a couple of quantum mechanical problems so far, the free particle and the 1D particle in a box. Now let's consider a third problem, which turns out to be very similar to the particle in a box, but with some important differences. So that's going to be called particle on a ring for reasons that will become clear shortly. But mainly so that we can highlight the differences between the particle in a box and the particle in a ring. Let me just summarize a few things we know about the particle in a box, namely that if we take a particle and confine it to a one-dimensional box between zero and A on some axis, then the particle has some wave function. There's a lot of different wave functions I could have drawn. Each of those wave functions has to hit zero at the edges of the box. Those wave functions had a particular mathematical form and they have a particular energy and again each one of those different wave functions, n equals one or two or three, has a different wave function, has a different energy. So that's what we know about the particle in a box. Now let's consider a different problem. Let's consider still confining our particle to one-dimensional line, but now let's take that line and let's bend it. Bend that line around all the way until it forms a circle. So I could say still length A, but now that length is the circumference of the circle. So if I want to describe where the particle is, instead of describing where it is on the x-axis, I'll just describe at what angle I find it. So this circle has some radius r and I can use theta to describe the angle to describe where I am around that circle. I'm still going to have a wave function. I'm going to have a wave function that depends on the coordinates of the particle. Now because the particle can be anywhere in the plane of the board, it feels like a two-dimensional problem, but because if I'm confining the particle to live on the circumference of the circle, r has to be constant. So the only variable that's allowed to change is theta. So it's still a one-dimensional problem, but now I'm describing where the value of the wave function is a function of theta rather than as a function of x. So that's one of the first differences between this problem and the particle on a box problem. We can see now why it's called particle on a ring. Particle is confined to this ring. We could have called it particle on a circle, or if we've got this string that we've now wrapped around into a circle, we could call it particle on a rope. There's a lot of different names you could give to this particle. We'll call it particle on a ring. Some people call it particle in a ring. So we're going to have a wave function that depends on theta. The boundary conditions, the thing that for particle on a box required the wave function to hit zero at the edges of the box, that's no longer required. The wave function doesn't have to hit zero at the edge, because there is no edge of this box. If I have a wave function that starts somewhere and oscillates as it goes around the circle, all I need is that when I get back to where I started, notice I made these two curves join up. So really what I need to have my boundary condition is that when I get back after one full lap, back to where I started, plus one full revolution of the circle, that needs to have the same value as when I started. Otherwise, if I came back and missed, if I came back at a different value than when I started, I wouldn't have just a single function. I need to have the same value on every trip around this loop. So that's the boundary condition that's going to replace the zero at the edges boundary condition for particle on a box. So that's another difference between this problem and the particle on a box problem. And I'm not going to go through the solution in detail, I'll skim through the details of this particle in a ring problem, because we've already seen something very much like it in particle on a box. But if we solve this problem, we still need to write down the Schrodinger equation. I'll write that down for us, and then I'll explain the differences between it and the one for particle on a box. So this looks a lot like the particle on a box, 1D particle on a box Schrodinger equation. In that problem, our second derivative was d squared dx squared. Here the variable we're interested in is theta. I don't just take d squared d theta squared, I take 1 over r squared d squared d theta squared. You can take my word for it that that's what the Schrodinger equation, the little plus sign looks like for this problem, or if you want to double check me, what this is is if you take for this two dimensional problem, if I write down d squared dx squared plus d squared dy squared, because I have two different dimensions, if I convert those x's and y's to polar coordinates, and then I require r to be constant what survives is just 1 over r squared d squared d theta squared. If that sounds like fun, pause the video and do that little exercise right now. If that doesn't sound like fun, like I said, you can just take my word for it. That's the Schrodinger equation for the particle on a ring problem. So again, it basically boils down to the same problem. Just like particle on a box, we're going to be looking for some function that if I take the second derivative of it, I'm going to get back the same function, but with some constants and a negative sign. So, like before, function, a function whose second derivative is the negative of itself is trigonometric functions like sine. So sine of some constant time theta is still going to be a good solution. Unlike the previous case, cosine also works. Cosine didn't work before because we needed something that would start at zero, right, the wave function had to be zero at the edge of the box. Sine is zero at x equals zero, cosine is not. So, sine was a good solution for particle on a box. Cosine's also a good solution for the particle on a ring. If you're not too terrified by complex numbers, we can also use e to the i times a constant times theta. That'll also solve this equation perfectly well. If you don't like complex numbers, we can put off thinking about complex numbers until a little bit later in the course. And you can pretend I never wrote that down. For the moment, let's just stick with constant times cosine constant times theta. The next step would be figuring out what that value of k needs to be in order for this to be true. So, whether you're thinking about it graphically or visually, what does that k have to be so that after I complete a full loop, I get back to where I started? Or whether you think about it mathematically and think about how to obey this boundary condition, we can convince ourselves. Cosine of any integer times theta. So cosine of theta, cosine of 2 theta, cosine of 3 theta. Or in fact, cosine of 0 theta, or cosine of negative 1 theta, or cosine of negative 2 theta, those are all fine. So any integer, cosine of an integer times theta is a perfectly valid solution to this Schrodinger equation. And again, that just means the sine or the cosine needs to oscillate one full round trip as it goes around the circle. That would be cosine theta, or two full oscillations as it goes around the circle, cosine of 2 theta, or and so on. Or for 0, cosine of 0, then the wave function is just constant. It remains constant without oscillating as it goes around the circle. So any of those solutions are perfectly good. The next step, so each of these different solutions with a different value of n is a different wave function that we can call psi sub n, just like we did for particle on a box. If I want to know what the energy of those solutions are so that I can compare them to the particle on a box solution, we just need to plug these wave functions into the Schrodinger equation. So plugging into the Schrodinger equation, I can say minus h squared over 8 pi squared mass, 1 over r squared. Second derivative of the function, second derivative of cosine is negative cosine and I'll pull out an n twice. So I've got capital A and I've got an n squared that's been pulled out, cosine n theta. That's on the left, on the right I've got, that must be equal to e energy times the function, a times cosine n theta. Now this is just, I'm sorry, second derivative was negative a n squared cosine n theta so that pulled out an extra negative sign. I've got an A and a cosine n theta that cancels on both sides so what I'm left with is that the energy of this psi sub n wave function is equal to this collection of constants which looks like h squared over 8 pi squared mass r squared and an n squared. So again that's similar, this feels very similar to what we did for particle in a box. If I compare these directly they're not the same. I no longer have a box length to worry about. Instead I have an r squared in the denominator down here and a pi squared that survives that didn't previously. But aside from that the energies do depend on n squared just like they did before. All right so, and I guess the last step would be if I want to know what the value of this normalization constant is to normalize this function since we've not done that too many times. Normalizing the function requires that I take the wave function times itself or its complex conjugate and that has to equal one if I integrate everywhere. What that means for us is that if I integrate theta from zero to two pi, I'm sorry, if I integrate over theta from zero to two pi the wave function squared is going to look like a squared cosine squared of n theta. If I care about what the normalization constant is I would have to complete this integral solve for a and that tells me what the normalization constant is. It doesn't end up being squared of two over a as it did for particle in a box. What we get, skipping the details of doing that integral, this integral turns out to be a squared times pi so a results in one over squared of pi. So that's the normalization constant. The last thing worth mentioning about this particle on a ring model is why we should care. Why did I say take this straight line and bend it around into a circle? What in the world would we ever use this particle on a ring model for? The answer is if particle in a box was a useful model for simulating conjugated hydrocarbons that had the shape of a straight line if I take one of those conjugated hydrocarbons and bend it around into a circle then the particle on a ring model will be a good model for those molecules. For example, a conjugated hydrocarbon bent into not a circle but a hexagon would be benzene and there's a variety of other cyclical molecules that we could use as applications for this particular model. The reason it's worth talking about benzene as an example in particular, let's go back and look at what these energies are. If I draw a graph, a chart of the energies of these molecules, the n equals zero solution has an energy of zero times some constants. So the energy of the ground states down here, there's two states n equals one and n equals minus one that have the same energy as each other and then somewhere up higher there's two more states n equals two n equals negative two that have the same energy as each other so some of our states are degenerate, there's two states with the same energy at this level and at this level but the ground state is not degenerate, there's only one state. If I were to start filling these states with electrons notice that I fill up the first level after putting six electrons in, not coincidentally benzene has six pi electrons. So the six pi electrons in benzene occupy these lowest wave function energy levels that correspond at least approximately to these particle and a ring energy levels. So the reason benzene is a very stable aromatic compound is because it fills up the ground and first excited state. You may remember from an organic chemistry class, Huckel's rules about aromaticity specify that if you have four n plus two electrons a molecule will be aromatic and that is a consequence of this particle on a ring model. Two electrons per energy level, four n plus two electrons is like two n plus one energy levels so if I fill up to here six electrons gives me an aromatic benzene molecule, if I fill one level higher I would need 10 electrons to make the next aromatic ring size. So Huckel's rule about aromaticity is really an example of the particle on a ring quantum mechanical model.