Lecture 10. Particles on Rings and Spheres... a Prelude to Atoms
Shaka, Athan J.
 This course provides an introduction to quantum mechanics and principles of quantum chemistry with applications to nuclear motions and the electronic structure of the hydrogen atom. It also examines the Schrödinger equation and study how it describes the behavior of very light particles, the quantum description of rotating and vibrating molecules is compared to the classical description, and the quantum description of the electronic structure of atoms is studied. 
 Welcome back to chemistry 131A. Last time we talked about more realistic oscillators. We talked about the 612 potential. We talked about the Morse potential. And then we got away from one dimensional system, systems with just one variable, which we were calling either X or R, whatever. And we did a particle in a box in two dimensions. And we discovered that under certain conditions there was degeneracy if the lengths of the box were the same and so forth. And three dimensions is the same. And the main weapon we had was that we took a two-dimensional equation that had two variables in it. And we separated it into two one-dimensional equations, both of which we already knew the solution to. And then we just substituted in the solution. Today, rather than talking about a particle in a box, I want to talk about a particle confined to a ring, first of all, which will be an interesting thing to solve. And particle confined to the surface of a sphere. And both these things will be important as a prelude to atoms. But both of them are completely artificial because if you say, well, I've got a particle confined to a ring, you have to ask yourself what kind of potential does that have that the particle stays just on a circular ring? And that's not a very realistic looking potential that if you just get off, if you're on the ring, you're just there with kinetic energy. And then if you're off the ring by epsilon, it's infinite or something like that. And so we could get some kind of anomalous looking behavior. But really what it means is we're going to set up the equation in two dimensions and then we're going to freeze one dimension and solve the other, which will be an angular variable. And we'll leave the radius of the ring as something we fix at the outset and then we go from there. And we'll leave the radius of the sphere as the same thing. And luckily when we do real atoms, at least for simple ones, where we don't have too many electrons, just one being the right amount, we can actually factorize the thing. And so we can use these solutions for the ring and the sphere and then just paste them together exactly like an onion, like growing an onion in shells. We argued early on that a particle confined to a ring, which we could think of as sort of like an electron in a classical orbit, would have to have a wavelength that fit. And the idea is that every time it goes around, it has to match up perfectly because if it doesn't, it cancels out. And in fact, there's nothing left for the probability because it's basically minus itself half the time. And so that was an argument that the de Broglie wavelength had to match. And our qualitative condition then is that we have to have an integral number of wavelengths and lambda has to equal the circumference of the ring, which is 2 pi r. If that condition is met, then we can go around like that and we can come back and we'll be at the same place. And that means that we've got a stable standing wave, a probability pattern that's not changing in space or time. Now, the de Broglie wavelength is related to h divided by the momentum, therefore we can just substitute for that n times h upon p is equal to 2 pi r. And then we can write that in a very suggestive way. We can divide both sides by 2 pi, turn the h into an h bar, and we can multiply both sides by p. And then we have n h bar is equal to r times p. That's interesting because r times p is going to have something to do with angular momentum. And if we have a particle on a ring, we automatically think of angular momentum. Whenever we've got anything going in a circle or confined like that, we think of the angular momentum of such a particle. We learned about that in classical physics. Let's have the ring oriented in the x, y plane. Then that means that the particle has an angular momentum, vector j is equal to r times the vector cross product of p. In this case, r points to the ring and p is the way the particle is going. So r and p are always at right angles. And in fact, j is equal to r cross p, which points in the z direction if we've got the particle in the x, y plane. And so we can take the z component of the angular momentum and relate it to the particle. And that means that we've got jz, which turns out to be r times p times sine theta, sine theta sine 90 degrees, or sine pi over 2. That's 1. So jz is equal to rp and that's equal to n h bar by the condition that the integral number of wavelengths fit in. And now sort of by this backdoor root, we have that angular momentum seems to be quantized. Photons came in units h nu, angular momentum comes in units of h bar. Now our time dependent Schrodinger equation for the particle on a ring is the following. It's the kinetic energy with respect to x plus the kinetic energy with respect to y, second partial derivative. Same thing as the particle in a two dimensional box is equal to e times psi of x, y. And the problem is x and y in this problem are very, very, very awkward variables. And that's because r is the square root of x squared plus y squared and so on and so forth. And if you just try to just bully your way through this equation using Cartesian coordinates which are set up for square problems, you'll never get anywhere. What you first have to do is you first have to change your variables so that you're in the same kind of funhouse mirror as the problem is. And then the problem will seem very easy. And that's what we're going to do. We've got a circular problem. We're going to use polar coordinates. So we're going to set up two new variables rather than x and y. We're going to set up r. Y because r is constant. That's the perfect thing to have. Then I get a derivative with respect to r, zero. So r is constant. And then the other variable is just where I am on the ring. Let's call that phi. Then I have a relationship that x is equal to r times cosine phi. That's this leg of the triangle. And y is equal to r sine phi. That's that leg of the triangle. And of course, r squared is equal to x squared plus y squared which is always constant. Now we have to make unfortunately a transformation not only of the variables x and y but of the derivatives. And that's trickier. So we need some actual multivariable calculus to do that transformation. And I'm not going to go through the transformation over and over and over because it gets very tedious to go through it. But I want you to see one time exactly how you do it so that you'll understand where these terms come from. To do this, then here's what we have to do. We have to use the chain rule from calculus. And we have to understand that if we have a function of more than one variable and I change something, I want to figure out the change. I should figure out the slope that it does with respect to the x direction if I change x. And then the slope it does with respect to the y direction if I change y and that will give me the total change in the function. I need to take two things and add them up. And for each one of them, if I use a different variable, I can use the chain rule. So here we go. The derivative of psi, the wave function with respect to r has to be psi dx times dx dr. You can think of these things just like fractions where you just cancel out the dx. That's how you can remember the chain rule. Deep psi dr is deep psi dx dx dr. That's the one direction. But I have to, it's a function of two variables. So I have to add deep psi dy, dy dr. And now I have a formula for y with respect to r. And so what I get is I get deep psi dx cosine phi plus deep psi dy sine phi. The second derivative is the derivative of this thing. And it gets messy fast, but it's a really good exercise in thinking clearly. Because what you do is you just take each of these terms and call deep psi dx some other thing. And then put that into the formula and then expand it back out very carefully. And you'll see. So don't skip any steps. So let's take another derivative. The second derivative of psi with respect to r squared is just d by dr of what we got, d psi dx cosine phi d dy sine phi. And I can put that in by factoring out a cosine phi and get the second derivative of psi dx squared dx dr. Plus the second derivative of psi with respect to x and then with respect to y dy dr. And you might say, well, how, what about the order of the x and y? And the answer is for wave functions and nice things that we deal with, any kind of function we have, we don't really care about the order. If we take the derivative, partial derivative with respect to y first or the partial derivative with respect to x first, those are equal. So we don't worry too much about that. And then we have sine phi times, again, two terms, both of them with a second derivative, dx dr dy dr. And if we add all that up, we get cos squared phi times the second derivative of psi with respect to x squared plus sine squared phi times the second derivative of psi with respect to y squared plus two sine phi cos phi times the mixed partial derivative. The derivative with respect to phi, I'm going to leave as a problem. We do it the same way. We start with d psi d phi and we write it as the chain rule. And then whenever we have x over r or something like that, we express it in terms of cosine phi and sine phi. And what you'll find is that you get a little bit longer expression here. Actually about the same length. But you get minus r times d psi dr plus r squared times three terms, which look very similar to the other three except slightly different ordering. And now we can get the relationship we need between the Cartesian and the polar second derivatives because now we have these second derivatives with respect to r and with respect to phi. And we know what they are in terms of with respect to x and y. It's still quite a bit of algebra. And again, I'll leave that for you. I'll quote the result so you can see what it is. But there's still quite a bit of algebra to do. And what you find out then is that the second derivative with respect to psi with respect to x squared plus the second derivative of psi with respect to y squared is equal to the second derivative of psi with respect to r squared plus 1 over r times d psi dr plus 1 over r squared times the second derivative of psi with respect to phi with respect to the angle. And we can check that we haven't made any mistakes because we can put in units. These all have units of length squared on the bottom. Forget about the wave function units for the time being. We have second derivative dr squared on the bottom, 1 over r dr, 1 over r squared d phi, phi doesn't have any length. The angle is just a ratio of things because it's in radians. And now at this point, we make the totally artificial assumption. We just say, hey, particles on a ring, let's just freeze r. And so wherever we see a derivative with respect to r, let's just say it's 0 because r's not changing. And that's great because now we just have 1 over r squared times the second derivative with respect to phi and that's looking except for phi instead of x, that's looking awfully similar to things we've already done. Therefore, the Schrodinger equation simplifies to this. Instead of having the Cartesian coordinates, we have minus h bar squared over 2m times 1 over r squared times the second derivative with respect to phi of psi, which is now only a function of phi because r's fixed. That's a constant. And that's equal to e times psi of phi. This is beginning to look really good and here's why. We know that m times r squared is the moment of inertia of a classical particle orbiting around an orbit of radius r. And we're doing a rotational problem by keeping the particle on a ring and we found that we just got the rotational constant, the moment of inertia coming into the problem, just naturally, just by the way it worked out. The differential equation then, if we write it in terms of the moment of inertia times the energy, is just the second derivative with respect to phi is equal to minus i e upon h bar squared times psi. And that the solution of an equation like that, not surprisingly, is an exponential. So we can write down the solution and we can write down, look, it's a e to the i mu phi plus b e to the minus i mu phi, mu is the square root of 2i e upon h bar. We can verify if we shove that in that we solve the Schrodinger equation, we get the answer. We need to have i because we have a minus. After we take the derivative twice, we get minus. That could be i squared is minus 1 or minus i squared. You could go around the other side to minus 1. But we can't have any real exponentials. And so these functions are things that are corkscrewing around one way or the other. And when they corkscrew around, they come back and meet. And they could go the other way, but they meet. The quantization arises because the wave function has to meet itself on the way around. And that means that if we add 2pi to the wave function, it has to be the same thing. If that weren't true, the wave function wouldn't be single valued in space. We could, depending what variable we pick to call it, it would have a different value. But it's the same point on the ring. So it has to have the same value. So that means it has to exactly come around. And that means that mu times 2pi is equal to 2pi n. So that n is an integer. So if we add 2pi to the angle, it has to be 2pi times an integer. And conventionally, rather than using n because n gets used for other things to do with energy, we use m, where m is an integer. And m is called the magnetic quantum number. Why is it magnetic? Because if we've got a charge on a ring, and we imagine it's moving around, a charge on a ring then is a current loop. And a current loop makes a magnetic field. That's exactly how you make an electromagnet. You take a ton of wire and you wind it around a core. And then you put some current through it. And you can pick up all kinds of stuff. And that's quite a fun thing to do when you're in elementary school. And I remember spending considerable time doing exactly that and seeing what I could and couldn't pick up. In fact, we can make another connection with classical mechanics. So the angular momentum, Lz, is just r cross Pz. And we can put in our quantum operators. We can put in x hat, Py hat, minus y hat, Px hat. That's what the z component of r cross P is if we set it up. And that's equal to minus ih bar times x, the derivative with respect to y, minus y, the derivative with respect to x. So that's our operator. And we can take that and we can convert that to polar coordinates by exactly the same tricks as what we did before. And if you take that particular combination and you're very careful and you convert to polar coordinates, you find out it comes out to be this real simple thing. Minus ih bar d by d phi, just the first derivative with respect to the angle. That's what we get. Well, that's really interesting because what that means is that when we did the energy, we said, well, we could have e to the im phi or e to the minus im phi. And we could have a of that plus b of that. We could have any amount of each. But if we want the particle to be in an eigenstate not only of energy but of angular momentum, then that means it's one corkscrew. It's either e to the plus im phi going one way or e to the minus im phi going the other way. And so what we do when we set up these problems for neatness is we either set a equal to zero and say it's negative m or we set b equal to zero and we say it's positive m. And so m can vary from minus some value to plus some value anywhere in between, including apparently zero. And the interpretation then is that the energy is equal because a particle going this way at some rate and a particle going the other way at some rate while they have the same kinetic energy. This then is our final solution, e to the im phi, m is any positive or negative integer. And apparently zero could be zero. Why not? That solves it as well. And that certainly doesn't have a problem. If there's no twist at all, if it's just flat, of course it meets up. If we normalize the wave function over the ring, that means that the probability that the particle is somewhere on the ring is one. If we do the integral since e to the im phi times e to the minus im phi is one, the integral is 2 pi. We don't integrate over r because r is not in the problem any longer. It's fixed. Just integrate over phi. Then we can get our normalized wave function for the particle on the ring. One over the square root of 2 pi times e to the im phi. The lowest energy here is m is equal to zero, which is zero. And this seems to run counter to what I've been saying, which is that whenever you have a confined particle, you should have some zero point energy. Why? Because you want to satisfy the uncertainty principle. And the reason why this seems to violate the uncertainty principle is just kind of glossed over in the book. But the reason why it seems to violate the uncertainty principle is first of all, we just threw r out. We really had a two-dimensional problem, but we froze one of the variables. Who says you can freeze one of those variables exactly like that? That's point one. And point two is that phi seems to have an artificial range. We say phi goes from zero to 2 pi, but it would be the same if phi went from zero to infinity because it keeps wrapping around over and over and over. And so if you argue well, you don't know phi. It could be anywhere between minus infinity and infinity, and it would still be somewhere on the ring. Then you could have the momentum be zero. Of course, you have to have the momentum be zero if you have the energy zero. And you could still have the position in terms of the actual value of phi be indeterminate. And it's kind of a mathematical dodge. But you have to be careful if you set up a problem and then impose a constraint that might not be physical. Say it has to be exactly on the ring. And then get in a tizzy that something doesn't seem to be quite right because the arguments may be quite subtle at that point. The probability density for an energy and an angular momentum eigenstate is independent of the angular variable phi. The probability density is flat. And of course, it would have to be because there's no reason why I should expect to find the particle more on one side of the ring than the other side of the ring when there's no difference between the sides of the ring. So it's got to be flat, and that makes perfect sense. If there's nothing to distinguish the two sides, how are we going to tell? And quantization again arises from the fact that the wave function has to match up. Has to be single valued. Now, the next step is to expand to a sphere. And a sphere, you could argue a sphere is a bunch of rings. And that might be a good way to look at it, in fact, is to pile up rings. It's artificial again to assume that the particle can go anywhere on the sphere but can't move radially at all. But nevertheless, it's a really good stepping stone to getting to the point where we can write down atomic orbitals and figure out what's going on. We've got a sphere. We know if we write down the kinetic energy, and because we've got a sphere and we're keeping it on the sphere, we've got no potential energy except this completely arbitrary potential energy that's keeping the particle right on the sphere. We have to convert to spherical polar coordinates. And the coordinate system here is x is equal, we have two variables. We have phi which goes around the z-axis, changes x to y and so forth. And we have theta which starts here and goes from the north pole where theta is 0 to the south pole where theta is pi or 180 degrees. But we don't go around again because every time we go here, we make a ring like a tree ring. And then we go here, we make another ring like a tree ring, and we go here to the equator, we do that one, and we go down here and then we finish down here. And if we went around again, we'd be counting all the rings again twice. So, phi varies from 0 to 2 pi or 360 degrees around. And the other one, theta varies from just 0 to 180. We adopt a right-handed coordinate system and that means that x cross y, your thumb points in the plus z direction. And that's by far the easiest way to do it because if somebody draws a figure where x is shooting this way and y is that way and you try to reorient it in your mind so x is to the right and y is into the paper, it takes a long time mentally. It's quite a gymnastic. But you just grab your hand and do it, you can figure it out. Be careful though because what you tend to do when you're doing problems is you have the pen, if you're right-handed, you have the pen in your right hand. And you want to figure out which way something's going, you use your free hand. And that's the wrong hand. And if you do that on an exam in physics especially, you just get marked wrong because you didn't use a right-handed coordinate system. And I've watched that happen to people at various stages of my career. So here's a right-handed coordinate system, theta and phi are set out. And our position is given by three numbers, r, the distance out, theta, the distance down from the north pole, and phi, the distance around from the x-axis. What's very confusing is that if you take a course in math for some reason that I cannot fathom, the variables theta and phi are swapped. And so theta is the one this way and phi is the one this way. And I think that's because in math when you do two-dimensional problems, you always call it theta and they just like to keep theta for the same variable and use phi for the other one. But in physics, it's phi around the z-axis and theta away from the z-axis. And you have to keep them straight because if you open the wrong book, you'll get things backwards and you'll get a terrible mess then. Let's try a practice problem. Let's consider a volume element. dv is equal to dx dy dz in Cartesian coordinates. It's a little cube. What's the volume element for spherical polar coordinates? So let's take a radius r, angle theta from the z-axis. What's the formula for the volume? Well, we don't care what phi is because it's always the same here. But we do care what theta is and here's why. The size of the onion ring here at the top near the north pole is tiny. When theta is small, as theta gets smaller, the total size of the ring gets smaller and smaller. As we go toward the equator, the same ring around 360 degrees in phi is much bigger in terms of the volume that it'll hold if we take a little slice in r. And so we have to weight the volume elements by how close we are to the north pole. So like imagining you can walk around the north pole in a little circle and you've walked around all possible longitudes there. But if you try to do it at the equator, it's a very long walk. Same thing. So here's the beauty of calculus is that we actually draw this thing as a wedge because r is changing and so the inner part is smaller than the outer part and it's like a little cone. But when we have just very tiny differences, then it's like a cube. And all we need to do is take the size of the cube. And that's the beauty of taking very small things, dx's, that no matter how curvy something is, if you take it small enough, it's a straight line. That's why calculus is so great. And so we can figure out the distance here out is r sine theta. And so the distance along if I move by d phi is this distance here is r sine theta d phi. And the distance the other way is just r because that's a full distance times d theta. And the distance in the third direction is just dr. Neither theta nor phi changes. And so we can take those and multiply them together. And we get the volume is r squared sine theta dr d theta d phi. That's important to know because you're going to have integrals to do with psi and so forth in them. And they're going to have dv because you have to integrate over all space. And you need to know what dv is in terms of these variables and now you know what it is. And exactly the same way that we did before but I don't want to take an hour to go through it carefully. But I'll just leave it to you if you're interested to work it out once. We can take the Cartesian second derivatives, second derivative with respect to x, second derivative with respect to y plus the second derivative with respect to z. And we can cast it in the following form. Second derivative with respect to r plus 2 over r times the first derivative with respect to r plus 1 over r squared times this operator I've called lambda squared. And lambda squared has nothing in it except theta and phi. It has some sine squared theta, second derivative with respect to phi. And the second term which is written in a very funny way, 1 over sine theta d by d theta of sine theta d by d theta. Unless you're used to dealing with operators, this is written in a very compact, very nice way so you don't have a lot of terms. But you have to be quite careful when you actually put it on a wave function because you only put the wave function on the right of the operator. You don't start inserting it in between things. You just put it on the right and then you go sequentially. Right, take the derivative with respect to theta. Malt by by sine theta. Take the derivative again and so forth. But if you don't understand the operator notation, then you're very likely to get things wrong because you may stick a psi in wherever you think there's a blank. And that's not correct. This thing, lambda squared is called the Legendrian. It's very famous, the Legendre polynomials and so forth. And as I said, the operator takes a bit of getting used to. Only put the wave function on the right. Be careful about that. Don't put psi in front of both those derivatives. The operator lambda squared, the Legendrian, has all the angular energy in the Hamiltonian because the other operators had derivatives with respect to r. If we freeze r, we don't allow any change in r. There's no energy that way. That means that this thing, lambda squared, is what we want to focus on. And it's just the energy to do with all the possible angular motions of things on a sphere. So let's fix r. Throw that out again. Same way as we did with the particle on a ring. And now we've got this new equation to solve. Minus h bar squared over 2mr squared times lambda squared on the wave function, which is a function of theta and phi, is equal to some energy times a function of theta and phi. And if we can find the wave functions that solve that eigenvalue equation and the eigenvalues, then we know the energy and we know the possible wave functions on a sphere. And of course, we expect that it's going to be quantized and so on because it's a trapped thing again. And as we go around in theta and phi, it's much more complicated this time because there's two of them, so it's harder to see. But it's got to be similar as what we had before. There's no potential energy here. The only requirement on the quantization is just that the wave function fit into the space. Okay, let's do a practice problem here, then practice problem 14. Let's show that the angular Schrodinger equation is separable. And what does that mean? That means that whatever this wave function in theta and phi is, we can write it as a product of something that's only a function of theta and something else that's only a function of phi. And if we can do that, then we'd guess that the something else that's only a function of phi is what we had before because last time when we did two-dimensional particle on a box, then okay, it was a product and the X was the same one as what we had before. And so theta is different from phi because phi goes around in 2 pi and theta only stops here. So we wouldn't expect it's going to be so easy as that because theta could be different than phi. But still, phi should be the same as what it was before. And so that we got that, e to the i m phi stuff, that saves us a lot of work. Okay, here's what we got to show. It's separable if we can write the solution as a product and if we want to prove that, what we have to do is we have to rearrange the equation so that we have two terms, one of which only depends on theta plus or minus something. Another term that only depends on phi is equal to a constant. And then we make the same argument. If we fix theta and change phi, if it's a constant, that means that both of them are constant. Otherwise, that wouldn't work. And that means that we can do a one-dimensional equation for each one. So let's try the trial product solution. Again, we used capital X of X. Let's use capital theta of theta. It's just some function. We don't know what it is, but we don't care at this point. We use capital phi of phi and substitute it in. So we've got the Legendrean on this thing. And now we get a number and here epsilon, this number minus epsilon is just 2iE over h bar squared and i is the moment of inertia. So that just cleans it up so that we don't have to write a lot of extra terms. Now, if we substitute this in, here's what we find. We have 1 over sine squared, second derivative with respect to phi of this product. Plus 1 over sine theta, first derivative with respect to theta of sine theta, first derivative with respect to theta of the product is equal to minus epsilon times the product. I do the same thing. I divide both sides by the product. And first, I say, haha, I've got the product of two things. Second derivative of, with respect to phi of some function of theta and a function of phi. The function of theta is a constant so I can pull it out because it doesn't matter where it is. That's what the partial derivative means. In the second term where the derivative is with respect to theta, I can pull the function big phi out in front as a constant because it's just a bunch of derivatives with respect to theta and we treat phi as a constant. So let's pull it out just like the same way we pull out a constant in a derivative. And if we pull those out in front, that makes it much easier to see. Now we've got big theta out in front, sine theta squared, second derivative with respect to phi squared and then the rest that you can see is equal to minus epsilon times the product. And now if I divide by the product on both sides, the theta goes away and the term with phi and the phi goes away and the term with theta. But we still have this sine squared so we have to multiply through the whole equation by sine squared. And then if we do that, we finally find the following. We have 1 over phi times the second derivative with respect to phi. That's one term. Plus a bunch of gobbledygook but it doesn't matter what it is because it doesn't have phi. It's all theta and it has an epsilon sine squared theta. That's equal to zero. That's good enough because I have this thing over here which is just phi and this thing over here which is just theta and so I've got two equations. One of them is just phi, the other one is just theta and I'm in. And we can switch, once we've got it down to one variable, we can switch to the regular D instead of the funny D because it's the same thing and we can solve the differential equation. So the first term, as I said, is going to give us exactly what we had before in the particle on a ring because it's the same equation basically as the particle on a ring equation. It's what it's actually going to be will depend on what theta happens to be but that doesn't bother us. We have e to the i m phi and then whatever theta happens to be is going to be some other function that's going to be our business to solve in the second part of the equation which is a different equation to do. And so we can break this very complex problem up into this series of problems and solve them and it turns out for a real problem with an atom, what we're going to do is we're going to first break up the problem into the particle on a sphere and then we're just going to let R be the other variable and not surprisingly we're going to try a product, we're going to say G. I think that the wave function is a product of some function of R times some function of theta times some function of phi and see if that doesn't work and then that gives us the clue then of how to factorize the thing and see that it works. One will that fail? Well if it'll fail right away unfortunately if we have two electrons because if we have two electrons then where each of them happens to be and they're repelling each other and they're attracting to the nucleus, it gets too difficult. We can't separate the equation and so we run into problems. In that case what we do is we treat each electron separately, solve it and then take the electron, electron repulsion part that we couldn't handle and we couldn't separate and we treat that as a perturbation. How good that will be will depend a lot on how close the electrons are getting. You can imagine that if the electrons happen to get pretty close in space that the potential could get quite high and so the electrons will tend to avoid each other and that means that their motion is correlated. They're sort of like a cat chasing its tail. When one starts going this way the other may go that way and so forth and so the electrons may not be independently moving around and that kind of electron correlation is a very important aspect of multi-electron atoms and higher dimensional systems but we'll touch on that later on in the course and for now we'll leave it there. Please take time and look through these transformations and spend a little time in a quiet room with a pencil and a piece of paper and just methodically go through and take each step and at each step say what does it mean? What am I doing? Why can I do that and go through it and see if you can't figure out why these things have the structure they do. The d by d theta sine theta stuff with the one sine theta, one over sine theta out in front, that's going to be a little bit tricky to get but you can get that too if you work on it. And next time what we'll do is we'll pick up our solution. We can guess the solution in phi but we can't figure out the solution in theta yet because that's a different differential equation with sine theta in it and we haven't figured out anything to do with that equation yet. So that's a separate equation for us to solve and we'll have to figure out what kind of techniques we need to solve that equation and then figure out what these functions are and then hopefully we can get some idea of what these functions on a sphere actually look like. So we'll leave it there and then pick it up next time to figure out the actual wave functions for a particle on a sphere.
 Welcome back to chemistry 131A.  Last time we talked about more realistic oscillators.  We talked about the 612 potential.  We talked about the Morse potential.  And then we got away from one dimensional system,  systems with just one variable,  which we were calling either X or R, whatever.  And we did a particle in a box in two dimensions.  And we discovered that under certain conditions there was  degeneracy if the lengths of the box were the same  and so forth.  And three dimensions is the same.  And the main weapon we had was that we took a two-dimensional  equation that had two variables in it.  And we separated it into two one-dimensional equations,  both of which we already knew the solution to.  And then we just substituted in the solution.  Today, rather than talking about a particle in a box,  I want to talk about a particle confined to a ring, first of all,  which will be an interesting thing to solve.  And particle confined to the surface of a sphere.  And both these things will be important as a prelude to atoms.  But both of them are completely artificial because if you say,  well, I've got a particle confined to a ring,  you have to ask yourself what kind of potential does  that have that the particle stays just on a circular ring?  And that's not a very realistic looking potential  that if you just get off, if you're on the ring,  you're just there with kinetic energy.  And then if you're off the ring by epsilon,  it's infinite or something like that.  And so we could get some kind  of anomalous looking behavior.  But really what it means is we're going to set up the equation  in two dimensions and then we're going to freeze one dimension  and solve the other, which will be an angular variable.  And we'll leave the radius of the ring as something we fix  at the outset and then we go from there.  And we'll leave the radius of the sphere as the same thing.  And luckily when we do real atoms, at least for simple ones,  where we don't have too many electrons,  just one being the right amount, we can actually factorize the thing.  And so we can use these solutions for the ring and the sphere  and then just paste them together exactly like an onion,  like growing an onion in shells.  We argued early on that a particle confined to a ring,  which we could think of as sort of like an electron  in a classical orbit, would have to have a wavelength that fit.  And the idea is that every time it goes around,  it has to match up perfectly because if it doesn't,  it cancels out.  And in fact, there's nothing left for the probability  because it's basically minus itself half the time.  And so that was an argument that the de Broglie wavelength had  to match.  And our qualitative condition then is  that we have to have an integral number of wavelengths  and lambda has to equal the circumference of the ring,  which is 2 pi r.  If that condition is met, then we can go around like that  and we can come back and we'll be at the same place.  And that means that we've got a stable standing wave,  a probability pattern that's not changing in space or time.  Now, the de Broglie wavelength is related to h divided  by the momentum, therefore we can just substitute  for that n times h upon p is equal to 2 pi r.  And then we can write that in a very suggestive way.  We can divide both sides by 2 pi, turn the h into an h bar,  and we can multiply both sides by p. And then we have n h bar is  equal to r times p. That's interesting  because r times p is going to have something to do  with angular momentum.  And if we have a particle on a ring,  we automatically think of angular momentum.  Whenever we've got anything going in a circle or confined  like that, we think of the angular momentum  of such a particle.  We learned about that in classical physics.  Let's have the ring oriented in the x, y plane.  Then that means that the particle has an angular momentum,  vector j is equal to r times the vector cross product of p.  In this case, r points to the ring  and p is the way the particle is going.  So r and p are always at right angles.  And in fact, j is equal to r cross p,  which points in the z direction if we've got the particle  in the x, y plane.  And so we can take the z component of the angular momentum  and relate it to the particle.  And that means that we've got jz, which turns out to be r times p  times sine theta, sine theta sine 90 degrees,  or sine pi over 2.  That's 1. So jz is equal to rp and that's equal to n h bar  by the condition that the integral number  of wavelengths fit in.  And now sort of by this backdoor root,  we have that angular momentum seems to be quantized.  Photons came in units h nu, angular momentum comes  in units of h bar.  Now our time dependent Schrodinger equation  for the particle on a ring is the following.  It's the kinetic energy with respect to x plus the kinetic  energy with respect to y, second partial derivative.  Same thing as the particle  in a two dimensional box is equal to e times psi of x, y.  And the problem is x and y in this problem are very,  very, very awkward variables.  And that's because r is the square root of x squared plus y  squared and so on and so forth.  And if you just try to just bully your way  through this equation using Cartesian coordinates  which are set up for square problems,  you'll never get anywhere.  What you first have to do is you first have  to change your variables so that you're in the same kind  of funhouse mirror as the problem is.  And then the problem will seem very easy.  And that's what we're going to do.  We've got a circular problem.  We're going to use polar coordinates.  So we're going to set up two new variables rather  than x and y. We're going to set up r.  Y because r is constant.  That's the perfect thing to have.  Then I get a derivative with respect to r, zero.  So r is constant.  And then the other variable is just where I am on the ring.  Let's call that phi.  Then I have a relationship that x is equal  to r times cosine phi.  That's this leg of the triangle.  And y is equal to r sine phi.  That's that leg of the triangle.  And of course, r squared is equal to x squared plus y squared  which is always constant.  Now we have to make unfortunately a transformation not  only of the variables x and y but of the derivatives.  And that's trickier.  So we need some actual multivariable calculus  to do that transformation.  And I'm not going to go through the transformation over  and over and over because it gets very tedious to go  through it.  But I want you to see one time exactly how you do it  so that you'll understand where these terms come from.  To do this, then here's what we have to do.  We have to use the chain rule from calculus.  And we have to understand that if we have a function  of more than one variable and I change something,  I want to figure out the change.  I should figure out the slope that it does with respect  to the x direction if I change x.  And then the slope it does with respect to the y direction  if I change y and that will give me the total change  in the function.  I need to take two things and add them up.  And for each one of them, if I use a different variable,  I can use the chain rule.  So here we go.  The derivative of psi, the wave function with respect  to r has to be psi dx times dx dr.  You can think of these things just like fractions  where you just cancel out the dx.  That's how you can remember the chain rule.  Deep psi dr is deep psi dx dx dr.  That's the one direction.  But I have to, it's a function of two variables.  So I have to add deep psi dy, dy dr.  And now I have a formula for y with respect to r.  And so what I get is I get deep psi dx cosine phi plus deep psi dy sine phi.  The second derivative is the derivative of this thing.  And it gets messy fast, but it's a really good exercise  in thinking clearly.  Because what you do is you just take each of these terms  and call deep psi dx some other thing.  And then put that into the formula and then expand it back out very carefully.  And you'll see.  So don't skip any steps.  So let's take another derivative.  The second derivative of psi with respect to r squared is just d by dr  of what we got, d psi dx cosine phi d dy sine phi.  And I can put that in by factoring out a cosine phi  and get the second derivative of psi dx squared dx dr.  Plus the second derivative of psi with respect to x  and then with respect to y dy dr.  And you might say, well, how, what about the order of the x and y?  And the answer is for wave functions and nice things that we deal with,  any kind of function we have, we don't really care about the order.  If we take the derivative, partial derivative with respect to y first  or the partial derivative with respect to x first, those are equal.  So we don't worry too much about that.  And then we have sine phi times, again, two terms, both of them  with a second derivative, dx dr dy dr.  And if we add all that up, we get cos squared phi times the second derivative  of psi with respect to x squared plus sine squared phi times the second derivative  of psi with respect to y squared plus two sine phi cos phi times the mixed partial derivative.  The derivative with respect to phi, I'm going to leave as a problem.  We do it the same way.  We start with d psi d phi and we write it as the chain rule.  And then whenever we have x over r or something like that, we express it  in terms of cosine phi and sine phi.  And what you'll find is that you get a little bit longer expression here.  Actually about the same length.  But you get minus r times d psi dr plus r squared times three terms,  which look very similar to the other three except slightly different ordering.  And now we can get the relationship we need between the Cartesian  and the polar second derivatives because now we have these second derivatives with respect  to r and with respect to phi.  And we know what they are in terms of with respect to x and y.  It's still quite a bit of algebra.  And again, I'll leave that for you.  I'll quote the result so you can see what it is.  But there's still quite a bit of algebra to do.  And what you find out then is that the second derivative with respect to psi  with respect to x squared plus the second derivative of psi with respect to y squared is  equal to the second derivative of psi with respect to r squared plus 1  over r times d psi dr plus 1 over r squared times the second derivative  of psi with respect to phi with respect to the angle.  And we can check that we haven't made any mistakes  because we can put in units.  These all have units of length squared on the bottom.  Forget about the wave function units for the time being.  We have second derivative dr squared on the bottom, 1 over r dr,  1 over r squared d phi, phi doesn't have any length.  The angle is just a ratio of things because it's in radians.  And now at this point, we make the totally artificial assumption.  We just say, hey, particles on a ring, let's just freeze r.  And so wherever we see a derivative with respect to r, let's just say it's 0  because r's not changing.  And that's great because now we just have 1 over r squared times the second derivative  with respect to phi and that's looking except for phi instead of x,  that's looking awfully similar to things we've already done.  Therefore, the Schrodinger equation simplifies to this.  Instead of having the Cartesian coordinates, we have minus h bar squared over 2m times 1  over r squared times the second derivative with respect to phi of psi,  which is now only a function of phi because r's fixed.  That's a constant.  And that's equal to e times psi of phi.  This is beginning to look really good and here's why.  We know that m times r squared is the moment of inertia  of a classical particle orbiting around an orbit of radius r.  And we're doing a rotational problem by keeping the particle on a ring and we found  that we just got the rotational constant, the moment of inertia coming into the problem,  just naturally, just by the way it worked out.  The differential equation then, if we write it in terms of the moment  of inertia times the energy, is just the second derivative with respect to phi is equal  to minus i e upon h bar squared times psi.  And that the solution of an equation like that, not surprisingly, is an exponential.  So we can write down the solution and we can write down, look, it's a e to the i mu phi  plus b e to the minus i mu phi, mu is the square root of 2i e upon h bar.  We can verify if we shove that in that we solve the Schrodinger equation, we get the answer.  We need to have i because we have a minus.  After we take the derivative twice, we get minus.  That could be i squared is minus 1 or minus i squared.  You could go around the other side to minus 1.  But we can't have any real exponentials.  And so these functions are things that are corkscrewing around one way or the other.  And when they corkscrew around, they come back and meet.  And they could go the other way, but they meet.  The quantization arises because the wave function has to meet itself on the way around.  And that means that if we add 2pi to the wave function, it has to be the same thing.  If that weren't true, the wave function wouldn't be single valued in space.  We could, depending what variable we pick to call it, it would have a different value.  But it's the same point on the ring.  So it has to have the same value.  So that means it has to exactly come around.  And that means that mu times 2pi is equal to 2pi n.  So that n is an integer.  So if we add 2pi to the angle, it has to be 2pi times an integer.  And conventionally, rather than using n because n gets used for other things to do with energy,  we use m, where m is an integer.  And m is called the magnetic quantum number.  Why is it magnetic?  Because if we've got a charge on a ring, and we imagine it's moving around,  a charge on a ring then is a current loop.  And a current loop makes a magnetic field.  That's exactly how you make an electromagnet.  You take a ton of wire and you wind it around a core.  And then you put some current through it.  And you can pick up all kinds of stuff.  And that's quite a fun thing to do when you're in elementary school.  And I remember spending considerable time doing exactly that and seeing what I could  and couldn't pick up.  In fact, we can make another connection with classical mechanics.  So the angular momentum, Lz, is just r cross Pz.  And we can put in our quantum operators.  We can put in x hat, Py hat, minus y hat, Px hat.  That's what the z component of r cross P is if we set it up.  And that's equal to minus ih bar times x, the derivative with respect to y,  minus y, the derivative with respect to x.  So that's our operator.  And we can take that and we can convert that to polar coordinates  by exactly the same tricks as what we did before.  And if you take that particular combination and you're very careful  and you convert to polar coordinates, you find out it comes  out to be this real simple thing.  Minus ih bar d by d phi, just the first derivative with respect to the angle.  That's what we get.  Well, that's really interesting because what that means is that when we did the energy,  we said, well, we could have e to the im phi or e to the minus im phi.  And we could have a of that plus b of that.  We could have any amount of each.  But if we want the particle to be in an eigenstate not only of energy  but of angular momentum, then that means it's one corkscrew.  It's either e to the plus im phi going one way or e to the minus im phi going the other way.  And so what we do when we set up these problems for neatness is we either set a equal to zero  and say it's negative m or we set b equal to zero and we say it's positive m.  And so m can vary from minus some value to plus some value anywhere in between,  including apparently zero.  And the interpretation then is that the energy is equal because a particle going this way  at some rate and a particle going the other way at some rate  while they have the same kinetic energy.  This then is our final solution, e to the im phi, m is any positive or negative integer.  And apparently zero could be zero.  Why not?  That solves it as well.  And that certainly doesn't have a problem.  If there's no twist at all, if it's just flat, of course it meets up.  If we normalize the wave function over the ring, that means that the probability  that the particle is somewhere on the ring is one.  If we do the integral since e to the im phi times e to the minus im phi is one, the integral is 2 pi.  We don't integrate over r because r is not in the problem any longer.  It's fixed.  Just integrate over phi.  Then we can get our normalized wave function for the particle on the ring.  One over the square root of 2 pi times e to the im phi.  The lowest energy here is m is equal to zero, which is zero.  And this seems to run counter to what I've been saying,  which is that whenever you have a confined particle,  you should have some zero point energy.  Why? Because you want to satisfy the uncertainty principle.  And the reason why this seems to violate the uncertainty principle is just kind  of glossed over in the book.  But the reason why it seems to violate the uncertainty principle is first of all,  we just threw r out.  We really had a two-dimensional problem, but we froze one of the variables.  Who says you can freeze one of those variables exactly like that?  That's point one.  And point two is that phi seems to have an artificial range.  We say phi goes from zero to 2 pi, but it would be the same  if phi went from zero to infinity because it keeps wrapping around over and over and over.  And so if you argue well, you don't know phi.  It could be anywhere between minus infinity and infinity,  and it would still be somewhere on the ring.  Then you could have the momentum be zero.  Of course, you have to have the momentum be zero if you have the energy zero.  And you could still have the position in terms  of the actual value of phi be indeterminate.  And it's kind of a mathematical dodge.  But you have to be careful if you set up a problem and then impose a constraint  that might not be physical.  Say it has to be exactly on the ring.  And then get in a tizzy that something doesn't seem to be quite right  because the arguments may be quite subtle at that point.  The probability density for an energy and an angular momentum eigenstate is independent  of the angular variable phi.  The probability density is flat.  And of course, it would have to be because there's no reason why I should expect  to find the particle more on one side of the ring than the other side of the ring  when there's no difference between the sides of the ring.  So it's got to be flat, and that makes perfect sense.  If there's nothing to distinguish the two sides, how are we going to tell?  And quantization again arises from the fact that the wave function has to match up.  Has to be single valued.  Now, the next step is to expand to a sphere.  And a sphere, you could argue a sphere is a bunch of rings.  And that might be a good way to look at it, in fact, is to pile up rings.  It's artificial again to assume that the particle can go anywhere on the sphere  but can't move radially at all.  But nevertheless, it's a really good stepping stone to getting to the point  where we can write down atomic orbitals and figure out what's going on.  We've got a sphere.  We know if we write down the kinetic energy, and because we've got a sphere  and we're keeping it on the sphere, we've got no potential energy except this completely  arbitrary potential energy that's keeping the particle right on the sphere.  We have to convert to spherical polar coordinates.  And the coordinate system here is x is equal, we have two variables.  We have phi which goes around the z-axis, changes x to y and so forth.  And we have theta which starts here and goes from the north pole where theta is 0  to the south pole where theta is pi or 180 degrees.  But we don't go around again because every time we go here, we make a ring like a tree ring.  And then we go here, we make another ring like a tree ring, and we go here to the equator,  we do that one, and we go down here and then we finish down here.  And if we went around again, we'd be counting all the rings again twice.  So, phi varies from 0 to 2 pi or 360 degrees around.  And the other one, theta varies from just 0 to 180.  We adopt a right-handed coordinate system and that means that x cross y,  your thumb points in the plus z direction.  And that's by far the easiest way to do it because if somebody draws a figure  where x is shooting this way and y is that way and you try to reorient it in your mind  so x is to the right and y is into the paper, it takes a long time mentally.  It's quite a gymnastic.  But you just grab your hand and do it, you can figure it out.  Be careful though because what you tend to do when you're doing problems is you have the pen,  if you're right-handed, you have the pen in your right hand.  And you want to figure out which way something's going, you use your free hand.  And that's the wrong hand.  And if you do that on an exam in physics especially, you just get marked wrong  because you didn't use a right-handed coordinate system.  And I've watched that happen to people at various stages of my career.  So here's a right-handed coordinate system, theta and phi are set out.  And our position is given by three numbers, r, the distance out, theta,  the distance down from the north pole, and phi, the distance around from the x-axis.  What's very confusing is that if you take a course in math for some reason  that I cannot fathom, the variables theta and phi are swapped.  And so theta is the one this way and phi is the one this way.  And I think that's because in math when you do two-dimensional problems,  you always call it theta and they just like to keep theta for the same variable  and use phi for the other one.  But in physics, it's phi around the z-axis and theta away from the z-axis.  And you have to keep them straight because if you open the wrong book,  you'll get things backwards and you'll get a terrible mess then.  Let's try a practice problem.  Let's consider a volume element.  dv is equal to dx dy dz in Cartesian coordinates.  It's a little cube.  What's the volume element for spherical polar coordinates?  So let's take a radius r, angle theta from the z-axis.  What's the formula for the volume?  Well, we don't care what phi is because it's always the same here.  But we do care what theta is and here's why.  The size of the onion ring here at the top near the north pole is tiny.  When theta is small, as theta gets smaller,  the total size of the ring gets smaller and smaller.  As we go toward the equator, the same ring around 360 degrees in phi is much bigger  in terms of the volume that it'll hold if we take a little slice in r.  And so we have to weight the volume elements by how close we are to the north pole.  So like imagining you can walk around the north pole in a little circle  and you've walked around all possible longitudes there.  But if you try to do it at the equator, it's a very long walk.  Same thing.  So here's the beauty of calculus is that we actually draw this thing as a wedge  because r is changing and so the inner part is smaller than the outer part  and it's like a little cone.  But when we have just very tiny differences, then it's like a cube.  And all we need to do is take the size of the cube.  And that's the beauty of taking very small things, dx's,  that no matter how curvy something is, if you take it small enough, it's a straight line.  That's why calculus is so great.  And so we can figure out the distance here out is r sine theta.  And so the distance along if I move by d phi is this distance here is r sine theta d phi.  And the distance the other way is just r because that's a full distance times d theta.  And the distance in the third direction is just dr.  Neither theta nor phi changes.  And so we can take those and multiply them together.  And we get the volume is r squared sine theta dr d theta d phi.  That's important to know because you're going to have integrals to do with psi and so forth in them.  And they're going to have dv because you have to integrate over all space.  And you need to know what dv is in terms of these variables and now you know what it is.  And exactly the same way that we did before but I don't want to take an hour to go through it carefully.  But I'll just leave it to you if you're interested to work it out once.  We can take the Cartesian second derivatives, second derivative with respect to x,  second derivative with respect to y plus the second derivative with respect to z.  And we can cast it in the following form.  Second derivative with respect to r plus 2 over r times the first derivative  with respect to r plus 1 over r squared times this operator I've called lambda squared.  And lambda squared has nothing in it except theta and phi.  It has some sine squared theta, second derivative with respect to phi.  And the second term which is written in a very funny way,  1 over sine theta d by d theta of sine theta d by d theta.  Unless you're used to dealing with operators, this is written in a very compact, very nice way  so you don't have a lot of terms.  But you have to be quite careful when you actually put it on a wave function  because you only put the wave function on the right of the operator.  You don't start inserting it in between things.  You just put it on the right and then you go sequentially.  Right, take the derivative with respect to theta.  Malt by by sine theta.  Take the derivative again and so forth.  But if you don't understand the operator notation, then you're very likely  to get things wrong because you may stick a psi in wherever you think there's a blank.  And that's not correct.  This thing, lambda squared is called the Legendrian.  It's very famous, the Legendre polynomials and so forth.  And as I said, the operator takes a bit of getting used to.  Only put the wave function on the right.  Be careful about that.  Don't put psi in front of both those derivatives.  The operator lambda squared, the Legendrian, has all the angular energy in the Hamiltonian  because the other operators had derivatives with respect to r.  If we freeze r, we don't allow any change in r.  There's no energy that way.  That means that this thing, lambda squared, is what we want to focus on.  And it's just the energy to do with all the possible angular motions of things on a sphere.  So let's fix r. Throw that out again.  Same way as we did with the particle on a ring.  And now we've got this new equation to solve.  Minus h bar squared over 2mr squared times lambda squared on the wave function,  which is a function of theta and phi, is equal to some energy times a function of theta and phi.  And if we can find the wave functions that solve that eigenvalue equation  and the eigenvalues, then we know the energy and we know the possible wave functions on a sphere.  And of course, we expect that it's going to be quantized and so on  because it's a trapped thing again.  And as we go around in theta and phi, it's much more complicated this time  because there's two of them, so it's harder to see.  But it's got to be similar as what we had before.  There's no potential energy here.  The only requirement on the quantization is just that the wave function fit into the space.  Okay, let's do a practice problem here, then practice problem 14.  Let's show that the angular Schrodinger equation is separable.  And what does that mean?  That means that whatever this wave function in theta and phi is,  we can write it as a product of something that's only a function of theta  and something else that's only a function of phi.  And if we can do that, then we'd guess that the something else that's only a function  of phi is what we had before because last time when we did two-dimensional particle  on a box, then okay, it was a product and the X was the same one as what we had before.  And so theta is different from phi because phi goes around in 2 pi  and theta only stops here.  So we wouldn't expect it's going to be so easy as that  because theta could be different than phi.  But still, phi should be the same as what it was before.  And so that we got that, e to the i m phi stuff, that saves us a lot of work.  Okay, here's what we got to show.  It's separable if we can write the solution as a product and if we want to prove that,  what we have to do is we have to rearrange the equation so that we have two terms,  one of which only depends on theta plus or minus something.  Another term that only depends on phi is equal to a constant.  And then we make the same argument.  If we fix theta and change phi, if it's a constant,  that means that both of them are constant.  Otherwise, that wouldn't work.  And that means that we can do a one-dimensional equation for each one.  So let's try the trial product solution.  Again, we used capital X of X. Let's use capital theta of theta.  It's just some function.  We don't know what it is, but we don't care at this point.  We use capital phi of phi and substitute it in.  So we've got the Legendrean on this thing.  And now we get a number and here epsilon, this number minus epsilon is just 2iE  over h bar squared and i is the moment of inertia.  So that just cleans it up so that we don't have to write a lot of extra terms.  Now, if we substitute this in, here's what we find.  We have 1 over sine squared, second derivative with respect to phi of this product.  Plus 1 over sine theta, first derivative with respect to theta of sine theta,  first derivative with respect to theta of the product is equal  to minus epsilon times the product.  I do the same thing.  I divide both sides by the product.  And first, I say, haha, I've got the product of two things.  Second derivative of, with respect to phi of some function of theta and a function of phi.  The function of theta is a constant so I can pull it out  because it doesn't matter where it is.  That's what the partial derivative means.  In the second term where the derivative is with respect to theta,  I can pull the function big phi out in front as a constant  because it's just a bunch of derivatives with respect to theta  and we treat phi as a constant.  So let's pull it out just like the same way we pull out a constant in a derivative.  And if we pull those out in front, that makes it much easier to see.  Now we've got big theta out in front, sine theta squared,  second derivative with respect to phi squared and then the rest  that you can see is equal to minus epsilon times the product.  And now if I divide by the product on both sides, the theta goes away  and the term with phi and the phi goes away and the term with theta.  But we still have this sine squared so we have to multiply  through the whole equation by sine squared.  And then if we do that, we finally find the following.  We have 1 over phi times the second derivative with respect to phi.  That's one term.  Plus a bunch of gobbledygook but it doesn't matter what it is  because it doesn't have phi.  It's all theta and it has an epsilon sine squared theta.  That's equal to zero.  That's good enough because I have this thing over here which is just phi  and this thing over here which is just theta and so I've got two equations.  One of them is just phi, the other one is just theta and I'm in.  And we can switch, once we've got it down to one variable,  we can switch to the regular D instead of the funny D  because it's the same thing and we can solve the differential equation.  So the first term, as I said, is going to give us exactly what we had before in the particle  on a ring because it's the same equation basically  as the particle on a ring equation.  It's what it's actually going to be will depend on what theta happens to be  but that doesn't bother us.  We have e to the i m phi and then whatever theta happens to be is going  to be some other function that's going to be our business to solve  in the second part of the equation which is a different equation to do.  And so we can break this very complex problem up into this series  of problems and solve them and it turns out for a real problem with an atom,  what we're going to do is we're going to first break up the problem  into the particle on a sphere and then we're just going to let R be the other variable  and not surprisingly we're going to try a product, we're going to say G. I think  that the wave function is a product of some function of R times some function of theta times some  function of phi and see if that doesn't work and then that gives us the clue then  of how to factorize the thing and see that it works.  One will that fail?  Well if it'll fail right away unfortunately if we have two electrons  because if we have two electrons then where each of them happens to be  and they're repelling each other and they're attracting to the nucleus,  it gets too difficult.  We can't separate the equation and so we run into problems.  In that case what we do is we treat each electron separately, solve it and then take the electron,  electron repulsion part that we couldn't handle and we couldn't separate  and we treat that as a perturbation.  How good that will be will depend a lot on how close the electrons are getting.  You can imagine that if the electrons happen to get pretty close in space  that the potential could get quite high and so the electrons will tend to avoid each other  and that means that their motion is correlated.  They're sort of like a cat chasing its tail.  When one starts going this way the other may go that way and so forth  and so the electrons may not be independently moving around and that kind  of electron correlation is a very important aspect of multi-electron atoms  and higher dimensional systems but we'll touch on that later on in the course  and for now we'll leave it there.  Please take time and look through these transformations and spend a little time  in a quiet room with a pencil and a piece of paper and just methodically go through  and take each step and at each step say what does it mean?  What am I doing?  Why can I do that and go through it and see if you can't figure  out why these things have the structure they do.  The d by d theta sine theta stuff with the one sine theta, one over sine theta out in front,  that's going to be a little bit tricky to get but you can get that too  if you work on it.  And next time what we'll do is we'll pick up our solution.  We can guess the solution in phi but we can't figure out the solution in theta yet  because that's a different differential equation with sine theta in it  and we haven't figured out anything to do with that equation yet.  So that's a separate equation for us to solve and we'll have to figure  out what kind of techniques we need to solve that equation  and then figure out what these functions are  and then hopefully we can get some idea of what these functions  on a sphere actually look like.  So we'll leave it there and then pick it up next time to figure  out the actual wave functions for a particle on a sphere.
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Welcome back to chemistry 131A. Last time we talked about more realistic oscillators. We talked about the 612 potential. We talked about the Morse potential. And then we got away from one dimensional system, systems with just one variable, which we were calling either X or R, whatever. And we did a particle in a box in two dimensions. And we discovered that under certain conditions there was degeneracy if the lengths of the box were the same and so forth. And three dimensions is the same. And the main weapon we had was that we took a two-dimensional equation that had two variables in it. And we separated it into two one-dimensional equations, both of which we already knew the solution to. And then we just substituted in the solution. Today, rather than talking about a particle in a box, I want to talk about a particle confined to a ring, first of all, which will be an interesting thing to solve. And particle confined to the surface of a sphere. And both these things will be important as a prelude to atoms. But both of them are completely artificial because if you say, well, I've got a particle confined to a ring, you have to ask yourself what kind of potential does that have that the particle stays just on a circular ring? And that's not a very realistic looking potential that if you just get off, if you're on the ring, you're just there with kinetic energy. And then if you're off the ring by epsilon, it's infinite or something like that. And so we could get some kind of anomalous looking behavior. But really what it means is we're going to set up the equation in two dimensions and then we're going to freeze one dimension and solve the other, which will be an angular variable. And we'll leave the radius of the ring as something we fix at the outset and then we go from there. And we'll leave the radius of the sphere as the same thing. And luckily when we do real atoms, at least for simple ones, where we don't have too many electrons, just one being the right amount, we can actually factorize the thing. And so we can use these solutions for the ring and the sphere and then just paste them together exactly like an onion, like growing an onion in shells. We argued early on that a particle confined to a ring, which we could think of as sort of like an electron in a classical orbit, would have to have a wavelength that fit. And the idea is that every time it goes around, it has to match up perfectly because if it doesn't, it cancels out. And in fact, there's nothing left for the probability because it's basically minus itself half the time. And so that was an argument that the de Broglie wavelength had to match. And our qualitative condition then is that we have to have an integral number of wavelengths and lambda has to equal the circumference of the ring, which is 2 pi r. If that condition is met, then we can go around like that and we can come back and we'll be at the same place. And that means that we've got a stable standing wave, a probability pattern that's not changing in space or time. Now, the de Broglie wavelength is related to h divided by the momentum, therefore we can just substitute for that n times h upon p is equal to 2 pi r. And then we can write that in a very suggestive way. We can divide both sides by 2 pi, turn the h into an h bar, and we can multiply both sides by p. And then we have n h bar is equal to r times p. That's interesting because r times p is going to have something to do with angular momentum. And if we have a particle on a ring, we automatically think of angular momentum. Whenever we've got anything going in a circle or confined like that, we think of the angular momentum of such a particle. We learned about that in classical physics. Let's have the ring oriented in the x, y plane. Then that means that the particle has an angular momentum, vector j is equal to r times the vector cross product of p. In this case, r points to the ring and p is the way the particle is going. So r and p are always at right angles. And in fact, j is equal to r cross p, which points in the z direction if we've got the particle in the x, y plane. And so we can take the z component of the angular momentum and relate it to the particle. And that means that we've got jz, which turns out to be r times p times sine theta, sine theta sine 90 degrees, or sine pi over 2. That's 1. So jz is equal to rp and that's equal to n h bar by the condition that the integral number of wavelengths fit in. And now sort of by this backdoor root, we have that angular momentum seems to be quantized. Photons came in units h nu, angular momentum comes in units of h bar. Now our time dependent Schrodinger equation for the particle on a ring is the following. It's the kinetic energy with respect to x plus the kinetic energy with respect to y, second partial derivative. Same thing as the particle in a two dimensional box is equal to e times psi of x, y. And the problem is x and y in this problem are very, very, very awkward variables. And that's because r is the square root of x squared plus y squared and so on and so forth. And if you just try to just bully your way through this equation using Cartesian coordinates which are set up for square problems, you'll never get anywhere. What you first have to do is you first have to change your variables so that you're in the same kind of funhouse mirror as the problem is. And then the problem will seem very easy. And that's what we're going to do. We've got a circular problem. We're going to use polar coordinates. So we're going to set up two new variables rather than x and y. We're going to set up r. Y because r is constant. That's the perfect thing to have. Then I get a derivative with respect to r, zero. So r is constant. And then the other variable is just where I am on the ring. Let's call that phi. Then I have a relationship that x is equal to r times cosine phi. That's this leg of the triangle. And y is equal to r sine phi. That's that leg of the triangle. And of course, r squared is equal to x squared plus y squared which is always constant. Now we have to make unfortunately a transformation not only of the variables x and y but of the derivatives. And that's trickier. So we need some actual multivariable calculus to do that transformation. And I'm not going to go through the transformation over and over and over because it gets very tedious to go through it. But I want you to see one time exactly how you do it so that you'll understand where these terms come from. To do this, then here's what we have to do. We have to use the chain rule from calculus. And we have to understand that if we have a function of more than one variable and I change something, I want to figure out the change. I should figure out the slope that it does with respect to the x direction if I change x. And then the slope it does with respect to the y direction if I change y and that will give me the total change in the function. I need to take two things and add them up. And for each one of them, if I use a different variable, I can use the chain rule. So here we go. The derivative of psi, the wave function with respect to r has to be psi dx times dx dr. You can think of these things just like fractions where you just cancel out the dx. That's how you can remember the chain rule. Deep psi dr is deep psi dx dx dr. That's the one direction. But I have to, it's a function of two variables. So I have to add deep psi dy, dy dr. And now I have a formula for y with respect to r. And so what I get is I get deep psi dx cosine phi plus deep psi dy sine phi. The second derivative is the derivative of this thing. And it gets messy fast, but it's a really good exercise in thinking clearly. Because what you do is you just take each of these terms and call deep psi dx some other thing. And then put that into the formula and then expand it back out very carefully. And you'll see. So don't skip any steps. So let's take another derivative. The second derivative of psi with respect to r squared is just d by dr of what we got, d psi dx cosine phi d dy sine phi. And I can put that in by factoring out a cosine phi and get the second derivative of psi dx squared dx dr. Plus the second derivative of psi with respect to x and then with respect to y dy dr. And you might say, well, how, what about the order of the x and y? And the answer is for wave functions and nice things that we deal with, any kind of function we have, we don't really care about the order. If we take the derivative, partial derivative with respect to y first or the partial derivative with respect to x first, those are equal. So we don't worry too much about that. And then we have sine phi times, again, two terms, both of them with a second derivative, dx dr dy dr. And if we add all that up, we get cos squared phi times the second derivative of psi with respect to x squared plus sine squared phi times the second derivative of psi with respect to y squared plus two sine phi cos phi times the mixed partial derivative. The derivative with respect to phi, I'm going to leave as a problem. We do it the same way. We start with d psi d phi and we write it as the chain rule. And then whenever we have x over r or something like that, we express it in terms of cosine phi and sine phi. And what you'll find is that you get a little bit longer expression here. Actually about the same length. But you get minus r times d psi dr plus r squared times three terms, which look very similar to the other three except slightly different ordering. And now we can get the relationship we need between the Cartesian and the polar second derivatives because now we have these second derivatives with respect to r and with respect to phi. And we know what they are in terms of with respect to x and y. It's still quite a bit of algebra. And again, I'll leave that for you. I'll quote the result so you can see what it is. But there's still quite a bit of algebra to do. And what you find out then is that the second derivative with respect to psi with respect to x squared plus the second derivative of psi with respect to y squared is equal to the second derivative of psi with respect to r squared plus 1 over r times d psi dr plus 1 over r squared times the second derivative of psi with respect to phi with respect to the angle. And we can check that we haven't made any mistakes because we can put in units. These all have units of length squared on the bottom. Forget about the wave function units for the time being. We have second derivative dr squared on the bottom, 1 over r dr, 1 over r squared d phi, phi doesn't have any length. The angle is just a ratio of things because it's in radians. And now at this point, we make the totally artificial assumption. We just say, hey, particles on a ring, let's just freeze r. And so wherever we see a derivative with respect to r, let's just say it's 0 because r's not changing. And that's great because now we just have 1 over r squared times the second derivative with respect to phi and that's looking except for phi instead of x, that's looking awfully similar to things we've already done. Therefore, the Schrodinger equation simplifies to this. Instead of having the Cartesian coordinates, we have minus h bar squared over 2m times 1 over r squared times the second derivative with respect to phi of psi, which is now only a function of phi because r's fixed. That's a constant. And that's equal to e times psi of phi. This is beginning to look really good and here's why. We know that m times r squared is the moment of inertia of a classical particle orbiting around an orbit of radius r. And we're doing a rotational problem by keeping the particle on a ring and we found that we just got the rotational constant, the moment of inertia coming into the problem, just naturally, just by the way it worked out. The differential equation then, if we write it in terms of the moment of inertia times the energy, is just the second derivative with respect to phi is equal to minus i e upon h bar squared times psi. And that the solution of an equation like that, not surprisingly, is an exponential. So we can write down the solution and we can write down, look, it's a e to the i mu phi plus b e to the minus i mu phi, mu is the square root of 2i e upon h bar. We can verify if we shove that in that we solve the Schrodinger equation, we get the answer. We need to have i because we have a minus. After we take the derivative twice, we get minus. That could be i squared is minus 1 or minus i squared. You could go around the other side to minus 1. But we can't have any real exponentials. And so these functions are things that are corkscrewing around one way or the other. And when they corkscrew around, they come back and meet. And they could go the other way, but they meet. The quantization arises because the wave function has to meet itself on the way around. And that means that if we add 2pi to the wave function, it has to be the same thing. If that weren't true, the wave function wouldn't be single valued in space. We could, depending what variable we pick to call it, it would have a different value. But it's the same point on the ring. So it has to have the same value. So that means it has to exactly come around. And that means that mu times 2pi is equal to 2pi n. So that n is an integer. So if we add 2pi to the angle, it has to be 2pi times an integer. And conventionally, rather than using n because n gets used for other things to do with energy, we use m, where m is an integer. And m is called the magnetic quantum number. Why is it magnetic? Because if we've got a charge on a ring, and we imagine it's moving around, a charge on a ring then is a current loop. And a current loop makes a magnetic field. That's exactly how you make an electromagnet. You take a ton of wire and you wind it around a core. And then you put some current through it. And you can pick up all kinds of stuff. And that's quite a fun thing to do when you're in elementary school. And I remember spending considerable time doing exactly that and seeing what I could and couldn't pick up. In fact, we can make another connection with classical mechanics. So the angular momentum, Lz, is just r cross Pz. And we can put in our quantum operators. We can put in x hat, Py hat, minus y hat, Px hat. That's what the z component of r cross P is if we set it up. And that's equal to minus ih bar times x, the derivative with respect to y, minus y, the derivative with respect to x. So that's our operator. And we can take that and we can convert that to polar coordinates by exactly the same tricks as what we did before. And if you take that particular combination and you're very careful and you convert to polar coordinates, you find out it comes out to be this real simple thing. Minus ih bar d by d phi, just the first derivative with respect to the angle. That's what we get. Well, that's really interesting because what that means is that when we did the energy, we said, well, we could have e to the im phi or e to the minus im phi. And we could have a of that plus b of that. We could have any amount of each. But if we want the particle to be in an eigenstate not only of energy but of angular momentum, then that means it's one corkscrew. It's either e to the plus im phi going one way or e to the minus im phi going the other way. And so what we do when we set up these problems for neatness is we either set a equal to zero and say it's negative m or we set b equal to zero and we say it's positive m. And so m can vary from minus some value to plus some value anywhere in between, including apparently zero. And the interpretation then is that the energy is equal because a particle going this way at some rate and a particle going the other way at some rate while they have the same kinetic energy. This then is our final solution, e to the im phi, m is any positive or negative integer. And apparently zero could be zero. Why not? That solves it as well. And that certainly doesn't have a problem. If there's no twist at all, if it's just flat, of course it meets up. If we normalize the wave function over the ring, that means that the probability that the particle is somewhere on the ring is one. If we do the integral since e to the im phi times e to the minus im phi is one, the integral is 2 pi. We don't integrate over r because r is not in the problem any longer. It's fixed. Just integrate over phi. Then we can get our normalized wave function for the particle on the ring. One over the square root of 2 pi times e to the im phi. The lowest energy here is m is equal to zero, which is zero. And this seems to run counter to what I've been saying, which is that whenever you have a confined particle, you should have some zero point energy. Why? Because you want to satisfy the uncertainty principle. And the reason why this seems to violate the uncertainty principle is just kind of glossed over in the book. But the reason why it seems to violate the uncertainty principle is first of all, we just threw r out. We really had a two-dimensional problem, but we froze one of the variables. Who says you can freeze one of those variables exactly like that? That's point one. And point two is that phi seems to have an artificial range. We say phi goes from zero to 2 pi, but it would be the same if phi went from zero to infinity because it keeps wrapping around over and over and over. And so if you argue well, you don't know phi. It could be anywhere between minus infinity and infinity, and it would still be somewhere on the ring. Then you could have the momentum be zero. Of course, you have to have the momentum be zero if you have the energy zero. And you could still have the position in terms of the actual value of phi be indeterminate. And it's kind of a mathematical dodge. But you have to be careful if you set up a problem and then impose a constraint that might not be physical. Say it has to be exactly on the ring. And then get in a tizzy that something doesn't seem to be quite right because the arguments may be quite subtle at that point. The probability density for an energy and an angular momentum eigenstate is independent of the angular variable phi. The probability density is flat. And of course, it would have to be because there's no reason why I should expect to find the particle more on one side of the ring than the other side of the ring when there's no difference between the sides of the ring. So it's got to be flat, and that makes perfect sense. If there's nothing to distinguish the two sides, how are we going to tell? And quantization again arises from the fact that the wave function has to match up. Has to be single valued. Now, the next step is to expand to a sphere. And a sphere, you could argue a sphere is a bunch of rings. And that might be a good way to look at it, in fact, is to pile up rings. It's artificial again to assume that the particle can go anywhere on the sphere but can't move radially at all. But nevertheless, it's a really good stepping stone to getting to the point where we can write down atomic orbitals and figure out what's going on. We've got a sphere. We know if we write down the kinetic energy, and because we've got a sphere and we're keeping it on the sphere, we've got no potential energy except this completely arbitrary potential energy that's keeping the particle right on the sphere. We have to convert to spherical polar coordinates. And the coordinate system here is x is equal, we have two variables. We have phi which goes around the z-axis, changes x to y and so forth. And we have theta which starts here and goes from the north pole where theta is 0 to the south pole where theta is pi or 180 degrees. But we don't go around again because every time we go here, we make a ring like a tree ring. And then we go here, we make another ring like a tree ring, and we go here to the equator, we do that one, and we go down here and then we finish down here. And if we went around again, we'd be counting all the rings again twice. So, phi varies from 0 to 2 pi or 360 degrees around. And the other one, theta varies from just 0 to 180. We adopt a right-handed coordinate system and that means that x cross y, your thumb points in the plus z direction. And that's by far the easiest way to do it because if somebody draws a figure where x is shooting this way and y is that way and you try to reorient it in your mind so x is to the right and y is into the paper, it takes a long time mentally. It's quite a gymnastic. But you just grab your hand and do it, you can figure it out. Be careful though because what you tend to do when you're doing problems is you have the pen, if you're right-handed, you have the pen in your right hand. And you want to figure out which way something's going, you use your free hand. And that's the wrong hand. And if you do that on an exam in physics especially, you just get marked wrong because you didn't use a right-handed coordinate system. And I've watched that happen to people at various stages of my career. So here's a right-handed coordinate system, theta and phi are set out. And our position is given by three numbers, r, the distance out, theta, the distance down from the north pole, and phi, the distance around from the x-axis. What's very confusing is that if you take a course in math for some reason that I cannot fathom, the variables theta and phi are swapped. And so theta is the one this way and phi is the one this way. And I think that's because in math when you do two-dimensional problems, you always call it theta and they just like to keep theta for the same variable and use phi for the other one. But in physics, it's phi around the z-axis and theta away from the z-axis. And you have to keep them straight because if you open the wrong book, you'll get things backwards and you'll get a terrible mess then. Let's try a practice problem. Let's consider a volume element. dv is equal to dx dy dz in Cartesian coordinates. It's a little cube. What's the volume element for spherical polar coordinates? So let's take a radius r, angle theta from the z-axis. What's the formula for the volume? Well, we don't care what phi is because it's always the same here. But we do care what theta is and here's why. The size of the onion ring here at the top near the north pole is tiny. When theta is small, as theta gets smaller, the total size of the ring gets smaller and smaller. As we go toward the equator, the same ring around 360 degrees in phi is much bigger in terms of the volume that it'll hold if we take a little slice in r. And so we have to weight the volume elements by how close we are to the north pole. So like imagining you can walk around the north pole in a little circle and you've walked around all possible longitudes there. But if you try to do it at the equator, it's a very long walk. Same thing. So here's the beauty of calculus is that we actually draw this thing as a wedge because r is changing and so the inner part is smaller than the outer part and it's like a little cone. But when we have just very tiny differences, then it's like a cube. And all we need to do is take the size of the cube. And that's the beauty of taking very small things, dx's, that no matter how curvy something is, if you take it small enough, it's a straight line. That's why calculus is so great. And so we can figure out the distance here out is r sine theta. And so the distance along if I move by d phi is this distance here is r sine theta d phi. And the distance the other way is just r because that's a full distance times d theta. And the distance in the third direction is just dr. Neither theta nor phi changes. And so we can take those and multiply them together. And we get the volume is r squared sine theta dr d theta d phi. That's important to know because you're going to have integrals to do with psi and so forth in them. And they're going to have dv because you have to integrate over all space. And you need to know what dv is in terms of these variables and now you know what it is. And exactly the same way that we did before but I don't want to take an hour to go through it carefully. But I'll just leave it to you if you're interested to work it out once. We can take the Cartesian second derivatives, second derivative with respect to x, second derivative with respect to y plus the second derivative with respect to z. And we can cast it in the following form. Second derivative with respect to r plus 2 over r times the first derivative with respect to r plus 1 over r squared times this operator I've called lambda squared. And lambda squared has nothing in it except theta and phi. It has some sine squared theta, second derivative with respect to phi. And the second term which is written in a very funny way, 1 over sine theta d by d theta of sine theta d by d theta. Unless you're used to dealing with operators, this is written in a very compact, very nice way so you don't have a lot of terms. But you have to be quite careful when you actually put it on a wave function because you only put the wave function on the right of the operator. You don't start inserting it in between things. You just put it on the right and then you go sequentially. Right, take the derivative with respect to theta. Malt by by sine theta. Take the derivative again and so forth. But if you don't understand the operator notation, then you're very likely to get things wrong because you may stick a psi in wherever you think there's a blank. And that's not correct. This thing, lambda squared is called the Legendrian. It's very famous, the Legendre polynomials and so forth. And as I said, the operator takes a bit of getting used to. Only put the wave function on the right. Be careful about that. Don't put psi in front of both those derivatives. The operator lambda squared, the Legendrian, has all the angular energy in the Hamiltonian because the other operators had derivatives with respect to r. If we freeze r, we don't allow any change in r. There's no energy that way. That means that this thing, lambda squared, is what we want to focus on. And it's just the energy to do with all the possible angular motions of things on a sphere. So let's fix r. Throw that out again. Same way as we did with the particle on a ring. And now we've got this new equation to solve. Minus h bar squared over 2mr squared times lambda squared on the wave function, which is a function of theta and phi, is equal to some energy times a function of theta and phi. And if we can find the wave functions that solve that eigenvalue equation and the eigenvalues, then we know the energy and we know the possible wave functions on a sphere. And of course, we expect that it's going to be quantized and so on because it's a trapped thing again. And as we go around in theta and phi, it's much more complicated this time because there's two of them, so it's harder to see. But it's got to be similar as what we had before. There's no potential energy here. The only requirement on the quantization is just that the wave function fit into the space. Okay, let's do a practice problem here, then practice problem 14. Let's show that the angular Schrodinger equation is separable. And what does that mean? That means that whatever this wave function in theta and phi is, we can write it as a product of something that's only a function of theta and something else that's only a function of phi. And if we can do that, then we'd guess that the something else that's only a function of phi is what we had before because last time when we did two-dimensional particle on a box, then okay, it was a product and the X was the same one as what we had before. And so theta is different from phi because phi goes around in 2 pi and theta only stops here. So we wouldn't expect it's going to be so easy as that because theta could be different than phi. But still, phi should be the same as what it was before. And so that we got that, e to the i m phi stuff, that saves us a lot of work. Okay, here's what we got to show. It's separable if we can write the solution as a product and if we want to prove that, what we have to do is we have to rearrange the equation so that we have two terms, one of which only depends on theta plus or minus something. Another term that only depends on phi is equal to a constant. And then we make the same argument. If we fix theta and change phi, if it's a constant, that means that both of them are constant. Otherwise, that wouldn't work. And that means that we can do a one-dimensional equation for each one. So let's try the trial product solution. Again, we used capital X of X. Let's use capital theta of theta. It's just some function. We don't know what it is, but we don't care at this point. We use capital phi of phi and substitute it in. So we've got the Legendrean on this thing. And now we get a number and here epsilon, this number minus epsilon is just 2iE over h bar squared and i is the moment of inertia. So that just cleans it up so that we don't have to write a lot of extra terms. Now, if we substitute this in, here's what we find. We have 1 over sine squared, second derivative with respect to phi of this product. Plus 1 over sine theta, first derivative with respect to theta of sine theta, first derivative with respect to theta of the product is equal to minus epsilon times the product. I do the same thing. I divide both sides by the product. And first, I say, haha, I've got the product of two things. Second derivative of, with respect to phi of some function of theta and a function of phi. The function of theta is a constant so I can pull it out because it doesn't matter where it is. That's what the partial derivative means. In the second term where the derivative is with respect to theta, I can pull the function big phi out in front as a constant because it's just a bunch of derivatives with respect to theta and we treat phi as a constant. So let's pull it out just like the same way we pull out a constant in a derivative. And if we pull those out in front, that makes it much easier to see. Now we've got big theta out in front, sine theta squared, second derivative with respect to phi squared and then the rest that you can see is equal to minus epsilon times the product. And now if I divide by the product on both sides, the theta goes away and the term with phi and the phi goes away and the term with theta. But we still have this sine squared so we have to multiply through the whole equation by sine squared. And then if we do that, we finally find the following. We have 1 over phi times the second derivative with respect to phi. That's one term. Plus a bunch of gobbledygook but it doesn't matter what it is because it doesn't have phi. It's all theta and it has an epsilon sine squared theta. That's equal to zero. That's good enough because I have this thing over here which is just phi and this thing over here which is just theta and so I've got two equations. One of them is just phi, the other one is just theta and I'm in. And we can switch, once we've got it down to one variable, we can switch to the regular D instead of the funny D because it's the same thing and we can solve the differential equation. So the first term, as I said, is going to give us exactly what we had before in the particle on a ring because it's the same equation basically as the particle on a ring equation. It's what it's actually going to be will depend on what theta happens to be but that doesn't bother us. We have e to the i m phi and then whatever theta happens to be is going to be some other function that's going to be our business to solve in the second part of the equation which is a different equation to do. And so we can break this very complex problem up into this series of problems and solve them and it turns out for a real problem with an atom, what we're going to do is we're going to first break up the problem into the particle on a sphere and then we're just going to let R be the other variable and not surprisingly we're going to try a product, we're going to say G. I think that the wave function is a product of some function of R times some function of theta times some function of phi and see if that doesn't work and then that gives us the clue then of how to factorize the thing and see that it works. One will that fail? Well if it'll fail right away unfortunately if we have two electrons because if we have two electrons then where each of them happens to be and they're repelling each other and they're attracting to the nucleus, it gets too difficult. We can't separate the equation and so we run into problems. In that case what we do is we treat each electron separately, solve it and then take the electron, electron repulsion part that we couldn't handle and we couldn't separate and we treat that as a perturbation. How good that will be will depend a lot on how close the electrons are getting. You can imagine that if the electrons happen to get pretty close in space that the potential could get quite high and so the electrons will tend to avoid each other and that means that their motion is correlated. They're sort of like a cat chasing its tail. When one starts going this way the other may go that way and so forth and so the electrons may not be independently moving around and that kind of electron correlation is a very important aspect of multi-electron atoms and higher dimensional systems but we'll touch on that later on in the course and for now we'll leave it there. Please take time and look through these transformations and spend a little time in a quiet room with a pencil and a piece of paper and just methodically go through and take each step and at each step say what does it mean? What am I doing? Why can I do that and go through it and see if you can't figure out why these things have the structure they do. The d by d theta sine theta stuff with the one sine theta, one over sine theta out in front, that's going to be a little bit tricky to get but you can get that too if you work on it. And next time what we'll do is we'll pick up our solution. We can guess the solution in phi but we can't figure out the solution in theta yet because that's a different differential equation with sine theta in it and we haven't figured out anything to do with that equation yet. So that's a separate equation for us to solve and we'll have to figure out what kind of techniques we need to solve that equation and then figure out what these functions are and then hopefully we can get some idea of what these functions on a sphere actually look like. So we'll leave it there and then pick it up next time to figure out the actual wave functions for a particle on a sphere.
problem is. And then the problem will seem very well, how, what about the order of the x and y? classical mechanics. So the angular momentum, Lz, energy. This then is our final solution, e to the probability density is flat. And of course, it equation and the eigenvalues, then we know the real problem with an atom, what we're going to do of multi-electron atoms and higher dimensional