18. "Hydrogen" and its Discontents
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Adams, Allan
In this lecture, Prof. Adams continues the discussion on hydrogen atoms. Runge-Lenz symmetry and relativistic corrections are discussed. Zeeman effect and Pauli exclusion principle are also covered.
 The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. OK, so let me just quickly remind you of what we did last time. So unfortunately, today is going to be a little bit of what I wanted to do this time and a little bit of what I would have wanted to have been done in recitation yesterday because all recitations were canceled. So today is going to include a few steps along the way that I skipped over last time. OK, so to review, from last time we talked about central potentials, the energy operator had a radial derivative term, an angular momentum term, and a potential term. So this would be true regardless of whether we were central or not. But if we have a central potential where this only depends on the magnitude of the radius vector, then we can use separation of variables and write an energy eigenstate as a radial part, which we will sneakily pull out a factor of 1 over the radius. As we saw last time, that was particularly convenient. It took care of this r and made everything nice and simple. So a radial part times a spherical harmonic. The reason we use a spherical harmonic is that neither of these two terms depend on the angular coordinates. The angular momentum, we know what its eigenfunctions are. So if we use a spherical harmonic, this angular momentum becomes just a constant. And having separated in this form and pulling up that sinc factor of 1 over r, the energy eigenvalue equation reduces to a simple 1D effective energy eigenvalue equation with a simple dr squared and an effective potential where the effective potential is the original central potential plus an angular momentum term, often referred to as the angular momentum barrier, which is proportional to, as Professor Evans says in 804, one is forced to say h bar squared upon 2m repeatedly. And he pointed out that it would be much more efficient if I would just come up with a simple variable for saying it or a sound. So I will attempt to henceforth call it pfft. So the effective potential is l l plus 1 upon r squared plus the central part of the potential, props to Professor Evans. And this is true in general for a central potential. The last thing we notice of this is that since the eigenvalue equation has been reduced to this form, which depends on l but is independent of m, m appears nowhere here, the energy can't possibly depend on m, although it can depend on l, it's just e of l. And thus, for every different value of m, I will get the same energy, fixed l different m. And there are how many values of m for a fixed l? 12 plus 1. So the degeneracy of e l is 12 plus 1. That's general for any central potential that only requires spherical symmetry. So then we study the example of the Coulomb potential, where the potential is minus e squared upon r. And we quickly reviewed the solution of the energy eigenvalue equation. OK, let me write it that way. Where r of what I will call in place of e, I will simply write n, r n l of, sorry, is equal to e to the minus r over 2 r naught n times r over r naught to the l plus 1 times some simple function v sub nl of r over r naught, with the energy nlm being equal to e naught, or I should say e ridberg, in our notation of last time, upon n squared minus bound state. So let's talk about this quickly. This term, the fact that it falls exponentially large distances, came from the asymptotic analysis at r goes to infinity. And this is just saying it's a bound state. The probability density falls off exponentially as we go out to large radius. This term came from the asymptotic analysis near r goes to 0. And near r goes to 0, the only term that matters is, in v effective, this is 1 upon r. This is 1 over r squared, so this gets large more rapidly. This term is important, and that l l plus 1 gave us r to the l plus 1. We worked in dimensionless variables, rho, and here I replace rho with r over r naught, where r naught was the characteristic scale we computed last time. And finally, this function v sub nl satisfied, again, the energy eigenvalue equation, but having pulled out these factors such that asymptotically it's regular. So this function must be regular, smooth, no divergences, no poles, as r goes to infinity and to 0. And then we solved this by series expansion. And from termination of the series expansion, we got a relationship between the energy and l and m. And in particular, the relationship, this came from termination of the series just like it did for the harmonic oscillator. And the termination series gave us that the energy was equal to minus a constant over n squared. With one additional condition, which was that n had to be an integer greater than l, and l, of course, has to be greater than or equal to 0. So let me just quickly tell you how this arises, because this is an important fact about hydrogen, which I want you to understand. So the way that arises is we got the energy from a series expansion for this remaining function of r, v and l of r. And that series expansion started out by saying let v be of the form sum over j of a sub j r over r naught, the dimensionless variable rho, to the jth power. So it's a power series. And when we plug this whole form together with this series expansion into that differential equation, just like we did before for a harmonic oscillator, we get a set of relations between the various coefficients in that series. And we solve that by saying a sub j plus 1 by deriving a recursion relation is equal to a sub j times 2 square root of epsilon times, oh, I'm going to be careful, j plus l plus 1. j plus l plus 1, yes, good, minus 1 upon j plus 1. j plus 2l plus 2. So, OK, fine, we get this recursion relation. It looks absolutely awful. But it tells you that if you know a0, say, as a constant, one sec, if you know a0 as a constant, which is not 0, if it's 0, then they're all 0, then this function is identically 0. Because aj plus 1 is aj times some constant. So this had better not vanish, a sub 0. So if the first term, the constant term doesn't vanish, which is good, it tells us v doesn't vanish. So we've already learned something. If v doesn't vanish identically at the origin, then we can determine a1 from a0 by multiplying by this. j is 0 in this. And l is whatever number we fix from ylm. And so on, we can deduce all the higher a sub l's. It's a series expansion. This is the recursion relation for it. However, note the following thing. As j gets large, as we go up to higher and higher orders in the series, as we go to higher and higher order in the series, then j is much larger than whatever a fixed number l is, and much larger than whatever a fixed number 2 is. So this asymptotes to aj times roughly 2 epsilon j upon j plus 1j. At very large j, this just goes to 2 epsilon over j. Now let's think about what that means. If I have that aj plus 1 is roughly equal to aj times root epsilon times 2 over j plus 1, I guess, then this tells me that every time I increase j, I get a factor of j plus 1 and another factor of 2 to the epsilon. This tells me that a sub j is equal to a sub not times 2 root epsilon to the j over j factorial. But those are the expansion coefficients of an exponential. Quantity to the n over n factorial. But that's bad, because if this series is the exponential series with argument 2 root epsilon, then this series becomes e to the 2 root epsilon r over r not. But that's bad, because that would exactly swamp this term. That would defeat the asymptotic analysis we did at the beginning. If this term is v is growing so large twice as rapidly as this is going to 0, then the net is diverging at infinity. That's bad. So in order for this to describe, for the series expansion to describe a good function that really does vanish at infinity, we need that this doesn't happen. So it must be true that j doesn't get large enough that the exponential in v overwhelms the decaying exponential from the asymptotic analysis. Ever recall that? So it has to terminate. But if it terminates, let's look at the coefficient here. If it terminates, this says that 2 root epsilon times j plus, sorry, if it terminates, that says that there's some j maximum. Well, let me write it over here. If it terminates, that means there's some value of j, such that a j plus 1 is 0. But if a j plus 1 is 0, how can that possibly be? It must be that for that maximum value of j, the numerator vanishes, specifically for that maximum value of j. But what's the condition for that numerator to vanish for a specific value of j? I'll call j sub m, or j max, the maximum value such that a sub j max plus 1 vanishes. So j plus l plus 1 must be equal to 1. Or said differently, epsilon, which is the energy in units of E0, must be equal to 1 over 4 j max plus l plus 1 quantity squared. Cool? So that means that our solutions, our states, psi, are now labeled by three integers. They're labeled by l and by m from the spherical harmonic. But they're also labeled by the value of j max. And j max can be any number from 0 to infinity. Strictly, it shouldn't be infinite, but any finite number, any integer. Because eventually, the series terminates at j max. So it's labeled by these three numbers, and the energy is equal to of l m j max is equal to E0 minus over 4. And this E0 over 4 we already saw was E Rydberg, 13.60 v, times what value? Well, here's the energy in units of E0, j m plus l plus 1 quantity squared, where l is greater than or equal to 0. j sub m is greater than or equal to 0, and 1 is 1. But this is cumbersome. We can just as well call this quantity n, where n is greater than l as an integer. It can't be equal to l, but it can be 1 greater than l, or any greater integer. Is that cool? So instead of labeling by j m the point at which it terminates, the series terminates, we'll label it by n, and you can deduce what j m is by subtracting off l plus 1 from n. That tells you where the series terminates. Notice that this tells you something nice. This power series for v, if n is equal to l plus 1, has how many terms in it. So don't say it out loud. Just think about it for a second. If n is equal to l plus 1, how many terms are there that are non-vanishing in the power series? I won't call an Uber. Raise your hand when you have an answer. What's the value of j max if n is equal to l plus 1? So what term is the first one to vanish in the series? OK, so how many terms are non-vanishing? Yeah, just the one. Just the first term. So when l is as large as it can possibly be, when l is equal to n minus 1, v is a constant function. It's just constant. When l is one less, when l is n minus 1 or n is l plus 2, then v has two terms in it, a constant and a linear. So the greater the difference between l and n, the more terms there are in the series. OK? Cool. So anyway, this leads to this form of the energy. But as is usually the case, this sort of a brute force analysis doesn't give us a whole lot of insight into why we get the qualitative features we do. In particular, the qualitative feature that stands up most obviously to me, I don't know about you guys, is the fact that the degeneracy is now much, much larger. In particular, we know from basic principles of a central potential, as we just reviewed, that any time a potential is symmetrically invariant, symmetrically symmetric, spherically symmetric, symmetrically symmetric. And he said the potential is spherically symmetric, rotationally invariant. Then there must be a degeneracy of 2l plus 1 for every energy I can see. There must be. That l could be 0. But if there's any angular momentum, then that degeneracy must be 2l plus 1. But this is much more degenerate. This is much more degenerate. And in fact, how degenerate is it? Well, the total degeneracy is, well, for any given En, the degeneracy is equal to, well, what are the possible values of l? So we have the sum. We have a state for every different value of l. For l could be equal to 0. And it can go up to n minus 1. Because, again, n is some integer which could vanish l plus 1. So the smallest this can be is 0. So n can be l plus 1 or anything greater, i.e. l can go up to n minus 1. But it can't be any greater. Then for every value of l, there is a state m. There's a value of m, which could be minus l all the way up to l in integer steps. So how many states are there? Well, there's one state for every value of n, l, and m. So it's this sum. The sum on m from minus l to l is 2l plus 1. So this is the sum n minus 1, l equals 0, of 2l plus 1. And this is equal to kind of beautifully n squared. Just arithmetic series. So we have this huge degeneracy, which is much larger than 2l plus 1. It's a sum over 2l plus 1 from l is 0 to n minus 1. Where is this degeneracy coming from? Why? So classically, or quantum mechanically, we know the following. We've learned the following. And I hope it's under your fingernails at this point. That when you see a degeneracy, you should expect that degeneracy to follow from some symmetry. There's some symmetry of the system that is leading to a degeneracy by virtue of the fact that the generators of that symmetry, for example, for rotation, that's angular momentum, commute with the energy. And when you have an operator that commutes with the energy operator, as we've seen, you can generally construct new states given a single energy eigenstate by acting suitably with that operator. So we should expect there to be some new symmetry. But it's very hard to see what symmetry. I mean, hydrogen is the central potential. It's just some stupid central potential. What's so special about the hydrogen system, as opposed to, for example, the harmonic oscillator, which hopefully you will see in recitation at some point, the harmonic oscillator in three dimensions has degeneracy 2l plus 1. It's a central potential. It has to have degeneracy 2l plus 1. But not n squared. Where did that come from? What's so special about the Kepler problem? So there must be some symmetry, but it's not obvious what it is. But now let's use a note from classical mechanics. No other theorem tells us when we have a symmetry, we have a conserved quantity. We also saw this in the quantum mechanical version for expectation values. We have a conserved quantity. And in quantum mechanics, having a conserved quantity means that the energy commutes with some quantity. Because that controls the time evolution of the expectation value. So there must be some quantity, which I've written by a question mark, that commutes with the energy operator, specifically in the case of the harmonic oscillator. So what is that quantity? So actually, this quantity in classical mechanics was studied because the same thing is true in the Kepler case. In the Kepler case, the orbits close, and there's simple ellipses. And they have all sorts of nice properties that beg for an explanation in terms of symmetry. And it was pointed out by a number of people, Laplace, Runge, Lenz, various people, that this is often called the Runge, Lenz vector. But I think you could blame it on any number of people, that there's a vector that in classical mechanics, for the Kepler problem or for the Coulomb problem, is conserved. And that quantity is p cross L, momentum cross angular momentum, minus m squared, sorry, m e squared. And let me break m sub electron. It's the mass rather than the quantum number little m. E squared R vector. So it turns out this quantity is classically conserved. And if you make a quantum operator out of it, then e with a is equal to 0. Hat, hat. It was enormously non-obvious. If it's obvious to you, you're a freak. It's really, really not obvious. You can go through and do the calculation. And when I say really not obvious, I mean, well, I'll show you what I mean in just a second. So you can go through and do the calculation classically, is this quantity conserved in the Kepler problem using Newton's law? And the answer is yes. That's amazing. Quantum mechanically, you can do the same thing. Although I'm going to warn you for anyone who has the chutzpah to try, because I encourage you to, but it takes a little bit of brawn. Think about this operator for a second. Does this classical quantity have an obvious interpretation as a quantum operator? No, because l is r cross p. But now we have prp. There's an ordering ambiguity. So you have to decide which order do you put that p and that r and the other p and you. Ppr, rpr, prr. You can write out which components you mean here. For example, you mean for the z component of a, you're going to get px, ly, and ly, px. But within that, how do you order the p's and the r's? Because ly contains px. And in particular, it contains x. So there's an ordering ambiguity. So just be a little bit careful about this as a quantum mechanical operator if you do play with this. But if you do it and you're thoughtful about it, it's pretty easy to see that, well, it can be computed and checked, that in fact, the commutator vanishes. So at this point, it's tempting to say, aha, there's an extra concert quantity we declare a victory. But the problem is, this didn't give us a symmetry. This just told us there's a conserved quantity. What's the symmetry behind this? What's the symmetry that ensures that this is a conserved quantity? It's obviously not rotations. It's something else. And so there are many answers to this question. One answer involves an explicit expression for in phase space the change of variables dx. And this is a variation that depends on, I'll write this as f of x and p. So there's a little vector parameter. And the change of the physical coordinates depends on both the coordinates and the momenta. This is a slightly strange thing, because it's like a change of variables where the position gets mapped into some nonlinear function of the positions of the momenta. That's a little bit weird. But you can do this. But I don't find that a terribly satisfying answer. And here's the answer I find most satisfying. It's not terribly useful for present purposes. But you're going to run into this again in 805 when you do what's called the operator method for the Coulomb potential. And it turns out to be an enormously useful machine. We're not going to use an 804, but I want to advertise it for you. I want to give you a description, though, that isn't the description you're going to get in 805. The description is the following. 1935, a guy named Fock, crazy guy, amazing, amazing physicist, played a huge role in the development of quantum field theory. I'm not exactly sure why he was thinking about this problem. But he was. And this was a very Soviet thing to do at the time, I guess, to be hardcore and mathematical. So Offi was going computing things. And he observed the following fact, how he observed this again, crazy long story. He observed the following fact. If you take the Kepler problem, and you work in momentum space, so write everything in terms of the momenta, it turns out that the Kepler problem, or he referred to it as a Kepler problem, but all referred to the Coulomb problem, the Coulomb problem in 3D, three spatial dimensions. This is really crazy. Is exactly equivalent. There's a change of variables. It's kind of complicated, but it's just a projection, it turns out, is exactly equivalent to the problem of a free particle in, wait for it, four dimensions constrained to the surface of a sphere, to a three sphere in the four dimensions. So you take a three sphere in four dimensions, you take a marble, you kick it, but make sure that it's stuck to the sphere, it will move in perfect circles. Obviously, it's moving on a sphere, it's just going to move in great circles. OK, that's nice and simple. What's the symmetry group of that system? Well, the sphere is invariant under any rotation in four dimensions. Ah, SO4. There is a symmetry, SO4, special orthogonal transformation. Special means it doesn't change the length, orthogonal means it's a rotation. So it's orthogonal rotations in four dimensions. And so now you can, so this should sound familiar, because the angular momentum group, angular momentum in 3D, is SO3. It's rotations that don't change the length in three dimensions. How many elements does this have? How many conserved quantities go along with this? Well, how many conserved quantities go along with rotations in three dimensions? What's conserved by virtue of rotational invariance in three dimensions? Angular momentum. How many components of angular momentum are there? Three. Great, there are three. So it turns out that if you calculate in SO4, how many conserved quantities are there? There are six. And when you do this mapping back down from the free particle in 4D to the Coulomb problem in 3D, three of them are the L's. And the other three of them are this A vector. Comes out spot on. So what it's telling you is that the spectrum of hydrogen, or at least the spectrum I shouldn't say of hydrogen, the spectrum of the Coulomb system, has an enhanced symmetry. It's the symmetry group of rotations in four dimensions. Why? I don't know, it's just kind of awesome. But I think that's about the best that can be said. It's just true. So for certain very special potentials, we find that the sort of thing happens. And it happens in other systems too. You discover that there are accidental symmetries in your system. They shouldn't be referred to as accidental symmetries. There are enormously non-trivial functions of the rest of the system. And we're going to break the symmetry in a minute, and you'll see why we shouldn't refer to it as accidental. But in any case, the answer to the question, why do we have this giant degeneracy in the Coulomb problem? Why is it n squared rather than 12 plus 1? Is that there is an enhanced symmetry group. There's more symmetry than you would have thought. And there's a conserved quantity associated with it. And that conserved quantity corresponds quantum mechanically to an operator that commutes with the energy operator. OK? Yeah? I have a question. So in quantum mechanics, do you remember that there was only a set of like five, like two, like that, that was where it was? Indeed. Indeed. So how do you get like three certain components of a model in quantum mechanics? Excellent. What do I mean in classical mechanics for quantity to be conserved? I mean that the time rate of change of that quantity under the classical dynamics, under classical evolution, is preserved. So what's the quantum mechanical version of that? So the quantum mechanical version of that is, well, can we say that d dt of some operator, or of some observable a, is 0? Does this make sense? No, because this is, I mean, it turns out, in the language we've been using, in the Schrodinger evolution description that we've been using, this doesn't really make sense because it's an operator. That's not a quantity. The quantity is something you measure. So what do you measure? Do you measure the operator? In a state, you measure its expectation values. You might measure its eigenvalues, but the more general thing is its expectation values. So this is something that depends on time, because the wave function depends on time, psi of t. So the best we can do to say something is conserved is to say d dt of a is 0. But we've seen what the condition is for the expectation value to not change in time. In general, for any operator a, the time rate of change of its expectation value is given by the commutator of e with a. I can't do the symbols. Good. OK. So in particular, if this commutator vanishes, if the commutator vanishes, equals 0. This is equal to. Sorry? Where is the sum of the time varying? Well, the wave function can be time varying. And I'm going to assume for the moment that the energy operator and the a operator are not themselves explicitly time dependent. Yeah, that's an important assumption. So let me just assume for simplicity that neither the energy nor the operator a that I'm looking at are explicitly time dependent. Like, it shouldn't be like x plus tp. That would be annoying. So let me assume that that's not the case. You do have to deal with that sometimes. For example, if you are dealing with a system that changes in time, you can do that. If you are dealing with a system that changes in time, where there's a background magnetic field or something like that that changes in time. But modular, that sort of detail. And also, there's an ih bar. So where do the ih bars go? Do the units make sense here? A, a, good. E and d dt. So where do the ih bars go? Well, you go on either side. You have to tell me if they go up or down. So what is the Schrodinger equation? So there's, yeah, exactly, ih bar. Good, OK, good. So we need an ih bar. Ah, good. OK, man. OK, so the spring has got to us all. So now, in the case that E with A is 0, imagine we have an operator A that commutes, well, 0, then d dt of the expectation value of A in any state is equal to 0. And that's what I'll call conserved. So classically, to be a conserved quantity means that it's time-independent. Quantum mechanically, to be a conserved quantity means that it computes with the energy. Now, coming back to the question. The question was, look for angular momentum, there are only two numbers you can know at any moment in time. You can't know three. You can know the total angular momentum, little l. And you can know the angular momentum in a particular direction, which we'd conventionally call z, little n. But you can't know some other angular momentum, like the angular momentum in x direction, simultaneously. So what does it mean to say that there are three conserved quantities? I can't know three conserved quantity at the same time. They're not even well-defined at some moment in time. The answer is, there are three operators, lx and ly, and lz, all of which, which I will simply write as l vector, all of which commute with e. So while it's true that the systems don't have definite values of lx, ly, and lz simultaneously, nonetheless, lx, ly, and lz all commute with the energy operator, and as a consequence, their time expectation values are all 0 and essential potential. So the time derivatives of their expectation values are all 0 and essential potential. They're all constant and essential potential. Yeah? You mean 1 over r? You mean 1 over r? Well, for the force of order of charge, so you have 1 over r. Yeah, 1 over r. So the question is, look, if you don't have 1 over r, do you get extra special things that happen for other potentials that aren't 1 over r, but something else? It's not quite the unique case. So the question is, look, this is a very good question, though. So the question is this. This is a great question. Look, there's something peculiar about 1 over r classically. The orbit's closed. You get ellipses. You don't get possessing ellipses. That's what you get if you have an anharmonic, if you have an extra 1 over r to the sixth term or something, small correction, the 1 over r law, which we do in the real solar system. It's a many-body system instead of a two-body system, so our orbit doesn't close. It's closed. So that's special. And here we see that there's an enhanced symmetry. Are these things related to each other? So the answer is almost, in the following sense. For the harmonic oscillator, we also have it closed, but there isn't a runs-lens vector. However, the orbit's closed. So there's something special about that one, too. Is there a conserved quantity there? And it turns out that the answer to that is yes, there's an enhanced symmetry there as well. But it's not like this. It doesn't lead to this gigantic degeneracy. There's just extra structure that's nice about the harmonic oscillator. But for the harmonic oscillator, of course, it also closes. It's not this orbit, this. So there's a nice structure there, but it's not quite so simple. OK. Yeah, one last question. Yeah? I was just wondering if thinking about the symmetric symmetry, degeneracy, is the problem? Like, is it equal times an equal area? Yeah, it's not unrelated. I don't know how to make that connection sharp. But the thing I can say sharp, so the question is, look, is that conserved quantity somehow related to the fact that there's equal time in the equal area swept on equal time at all points in the orbit? Which is a true thing about the Kepler problem. And that's a good observation. I don't know how to make that connection, but it is true in the following sense that both of those facts are guaranteed by the fact that you have closed orbits with a particular ratio. And that is indeed can be blamed on the conservation of the Runge-Lenz vector. I don't know how to make that connection direct and explicit, but indeed it is. That's a fun question. I'll have to think about that. That's a good one. One of the totally awesome things about teaching U04 is that every single time someone asks a question I've never even thought of before. Forget never answered. Never even thought of it. I never thought of that. It's a great question. OK. So everyone cool with the degeneracy? We have this gigantic degeneracy. So knowing what the degeneracy is or where it's coming from, at least more or less, is progress. But if you really understand why something is conserved, then you can break that conservation. You can break that degeneracy. So here's my question. Is it possible, can we, or indeed must we, can we break or lift this degeneracy? So first let me tell you what those words mean. So we have this gigantic degeneracy of the, what's this guy? Reasted. OK, we have this gigantic degeneracy of the, oh there it is, up at the top, n squared of the Coulomb potential. Can we break that degeneracy or lift it? So what do I mean by lifted degeneracy? What do you mean by lifted degeneracy is the following. Suppose I have a system that's degenerate. So it's got two energy eigenstates with the same energy eigenvalue. So this is energy. Then if I can kick the system in somewhere, or modify it, or bend it, such that the energy levels are no longer degenerate, one state has been lifted above the other. And so we use the word to lift the degeneracy. Now you can see that there are two different states by measuring the energies. Is that cool? It's also said to split a degeneracy or to break a degeneracy. So those are just words. So can we break or lift this degeneracy? So sure, how do we break a degeneracy? How do we lift a degeneracy? We need to break the symmetry. So if we don't have the symmetry, we won't get the degeneracy. We've seen this before. So first off, let's think about what are the two symmetries that we want. Well, we got this degeneracy of 2L plus 1 from spherical symmetry. And we got this extra degeneracy of n squared from conservation of the Runge-Lenz vector. And conservation of the Runge-Lenz vector came from the fact that we had exactly the Coulomb system. So anything that isn't exactly the Coulomb system, but is still rotationally invariant, should still have the 2L plus 1 degeneracy, but not the n squared degeneracy. Cool? Like the harmonic oscillator, for example. So let's change the system. Let's find a small correction to the system that leaves it mostly the Coulomb problem, so it's some small correction. So what comes to mind? So one good way to think about a physical, natural way to break this degeneracy, to break the symmetry, is to think about a real system that's described by the Coulomb potential. So the system that we usually talk about is hydrogen. Hydrogen is a proton that's got a plus charge. It's got an electron, and it's attracted with a minus e squared upon R. So in hydrogen, is this the energy operator for hydrogen? Is this a good description of the energy operator for hydrogen? Well, it's a good start. But it's not exactly right for a bunch of reasons. And in fact, on your first problem set, when you computed the time it takes for hydrogen classically to decay from the classical bore radius, or from the classical, well, from the bore radius, one of the things you discovered is that the electron starts out being non-relativistic, but very rapidly becomes relativistic. Rapidly is like 10 to the minus 15 seconds. This is a non-relativistic p squared upon 2m kinetic energy. That's probably not a perfect approximation. So let's ask, so the first example, example one, is going to be relativistic corrections. Now, will relativistic corrections break rotational symmetry? They can't, right? Because relativity doesn't say this is the z direction. In fact, it does quite the opposite. Relativity, among other things, says, don't get too hung up on what's the z direction. But relativistic corrections will not break the symmetry. So if we include relativistic corrections to the kinetic energy, this will change E. Change the energy operator, but it will preserve the rotational symmetry, E with L is equal to 0. So we won't break the rotational symmetry, and so we'll preserve the 2L plus 1. So let's check to see that this is true. So to check to see that this is true, let's actually estimate out the correction to the energy due to the first non-trivial relativistic effect in the atom. The first non-trivial relativistic effect can be calculated in the following way. What's the kinetic energy in special relativity of a particle of mass m? So that's easy. The total energy is the square root of m squared c to the 4th plus p squared c squared minus mc squared. That's the rest energy. That's the total energy. So this is the kinetic energy. But this is equal to, well, if we pull out a factor of m squared c squared, this is m squared, sorry, mc squared times square root of 1 plus p squared over m squared c squared minus 1. And now if we Taylor expand this guy, because if the momentum is small compared to mc squared, or mc, I should say, if the momentum is small compared to mc, which is morally speaking at low velocities, the velocity is small compared to c, then this is 1 plus a small number. And we can Taylor expand this to get 1 plus 1 half a small number. And the ones will cancel, giving us only 1 half. So this is going to be equal to 1 half that small number, p squared over mc squared times m squared c squared times mc squared upstairs, giving us the first term, which is p squared upon 2m. That's the familiar classical kinetic energy. Rock on. What's the next correction? The next correction is the next term from this Taylor series, which is going to involve this quantity squared. So it can be p to the 4th over m to the 4th, c to the 4th times mc to the 4th. The coefficient is a minus 1 eighth, so we get minus p to the 4th over 8m cubed c squared. And that's down by a factor of m squared c squared from this guy and a p squared upstairs, which is what you'd get, p squared m squared over c squared. So the correction to the kinetic energy for giving momentum p is to correct it down by p to the 4th term. Ever cool with that? Yeah? Why do you think the substances p select our red and the p to the small of that? Because you, like Schrodinger, did problem set 1. And problem set 1 said, if you have an electron at the Bohr radius classically, its velocity is about a 400th of the speed of light. 200th? Something. It's small. It's a couple of order of magnitude small in the speed of light. So that number, that ratio there is 1 upon 200 or 400, I remember, squared. That's a very small number, so that's a reasonable approximation. Why is it the typical? Because what is the typical? That's an excellent question. So why am I talking about it as if the electron is sitting at that radius? And the answer is, well, look, what is the average momentum? What is the average value of p squared in, let's say, the ground state of the hydrogen atom? What's the average value of p? What must be the average value of p in the ground state of the hydrogen atom? 0 on two grounds. First off, p is a vector. And the ground state of hydrogen, the rotation of the invariant, doesn't break anything. Second, it's not going anywhere. It's a bound state, so it's just sitting there. It's got an average momentum 0. Curved momentum, it would be cruising. So it's got average momentum 0. OK, fine. But what's the average momentum squared? What's the average of p squared? Well, we can actually do out that calculation. You're going to have to do that calculation on the problem set. And it's a small number. And it's, in fact, almost exactly what you'd guess classically from the Bohr radius. So you have to know the h bar to compute the Bohr radius in the first place. So the answer is to the best of anyone's ability to define what's the momentum or the typical scale of momentum. That typical scale of momentum for the electron in the ground state of hydrogen is extraordinarily small compared to the speed of light by factors of 100 or 200, 400. Exactly. Does that answer your question? It's a very good question. So here's the kinetic energy. And another way to say, by the way, another way to say that this approximation must be good is that we already computed ignoring the relativistic correction. We computed the ground state energy. And it was 13.6EV, which is pretty good since the binding energy of hydrogen is about 13.6EV. So apparently, that was a good approximation. And you might be disgusted by answer analysis like that. But that's what physics is. You write down some stupid cockamamian model that you know is wrong. But if it does a good job of fitting the data, you declare triumph and you call staco. It depends on the problem you've just solved, approximately. So here's the first correction. So what this tells us is, therefore, the energy of hydrogen, or at least a better model of hydrogen, I'll call it hydrogen tilde, is equal to E coulomb minus p the fourth over 8. Whoops, that's an 8. m cubed c to the fourth. Oh, squared. Wow, that's a deformed 8. Let's try it again. Much better. OK, so this, however, we know is small compared to the kinetic energy because 13.6 is pretty close to 13.6. You'll notice I didn't include the rest of the significant digits. So this must be in the insignificant digits. So this must be small. So here's what I want to ask. We know that the energy of the coulomb problem, nlm, is equal to E naught squared minus E naught squared over n squared, independent of ln m. What's the energy in this toy model of hydrogen that includes the first relativistic correction, nlm? Well, to answer that question, we have to resolve the energy eigenvalue equation. We have to refine the energy eigenstates. We have to solve that. And this is now going to be a much worse problem because it's going to involve a p to the fourth term, which is going to involve four derivatives. That sounds horrible. Can anyone think of a better way to just approximate, estimate the magnitude and the value of the leading correction from this quantity? Yeah? Some variation, take an eigenvalue, plug it in, see what the error is, correct for that error, plug it in again. That's a very, there's an iterative method for doing that. That's a very good guess. But there's an even easier way to estimate it. But that's an excellent, that's something we will call perturbation theory when we do this in 805. And doing that systematically is exactly the right answer. However, I want to do the just leading part of that. Yeah? Oh, dimensional analysis would be awesome. That's fantastic. So how would you compute p? OK, so we didn't calculate p. So you could do a dimensional analysis and get a p. Because you have an e, you have an l, you can probably do it. Here's the problem with that, though. This is the correct first answer to any question of this kind. Correct first answer is plug in dimensional analysis and get an estimate for the order of magnitude. But we want more than the order of magnitude here. We're interested in splitting the degeneracy due to this interaction. So what that means is we want to see that different values of l lead to different energies. What we care about is the l dependence that comes out of this correction. And we're not going to be able to get the l dependence from dimensional analysis because l is an integer, it is dimensionless. Yeah? So that will give you the correct magnitude, and that is the first thing you do. What's the second thing you do? Yeah? We're going to stick in the expectation value of p to the fourth. Excellent. Let's just stick in the expectation value of p to the fourth. That's not exactly the right thing to do. But if this is small, that's probably pretty reasonable. And what you'll find when you do it systematically using perturbation theory, as was pointed out earlier, we can take the entire system, the entire energy eigenvalue equation, we can think of this 1 over m cubed c squared. We can think of that as a small number, and we can do perturbation theory in that small number. We can Taylor expand everything in that small number. And when you Taylor expand the exact equation, what you discover is the leading term in that Taylor expansion is the expectation value of p to the fourth over 8 m cubed c squared. So this is a fantastic guess. OK, good. So let's compute it. So e hydrogen is going to be equal to e nlm, which I'm just going to write as minus e nought squared over n squared. Oh, sorry. I don't know why I wrote squared. It's just e nought. If the energy were equal to the energy squared, at that point every respectable physicist goes out and asks themselves, OK, so what is the correct answer here? So it's this quantity minus, so this is making it more tightly balanced. Notice the sign, because p to the fourth is strictly positive. p is real. So minus e nought over m squared. And if you do this expectation value, which I really should have put on the problem set, oh, I still can. It hasn't been posted yet. So it'll be posted this afternoon. But I won't, because I won't. It's not that bad, actually. Matt, what do you think? Through. We'll discuss this afterwards. OK, so which may or may not appear on your problem set. So if you go out and you do this calculation, what you find is the correction is e n squared. So e nought squared over n to the fourth. This time it really is n to the fourth. And now you have to worry about dimensional analysis. This is two energies. We want one energy divided by, wait for it, mc squared, because it's a relativistic correction. That shouldn't be too surprising. Times the quantity 4n over, and here's where life gets awesome, l plus 1 half minus 3. See, you laugh, but I feel delighted by this for so many reasons. So one is, that's beautiful. What a crazy combination of symbols. But the second and more important one is, look, it has an l in it. And the l is downstairs. That's good. It has more leverage that way. So we've got this l appearing. So now the energy depends on l. Good, we've broken the degeneracy. Different values of l give different energies. So let's do that graphically. So in particular, what this tells us is that now, while the Coulomb energy commuted with the Runge-Lenz vector, it must be true that the energy of this model of hydrogen, which includes the first relativistic correction, does not commute with the Runge-Lenz vector. And indeed, if you check this, which is effortful, but not intellectually difficult, indeed, it doesn't vanish. And the failure of this conservation equation, the failure of this commutation relation to vanish, tells us that it's possible for the system to not be degenerate. And lo and behold, it is not degenerate. Did you say before that the actual first frame you count the relativity just includes substituting the non-relativistic value of 4? Yes. No, sorry. Yeah, I should have said that. There are other things that are important for relativity that we'll come to in a little bit. But if you just look at the kinetic energy, this is the first non-trivial correction. Yeah? We still have the 2L plus 1 degeneracy. We still have the 2L plus 1 degeneracy, exactly. Because does this correction violate rotational invariance? No, that's p squared. That's a scalar under rotation. It's just a number, not a vector. But e still commutes with L. And you see that because this is independent of little m. This is the mass of the electron. So this is independent of the quantum number little m. It only depends on the quantum number little l. So indeed, it is still degenerate. So let's go through this and see the degeneracy explicitly. And let me do that by drawing you the following diagram. So what happens if we have the e1, which is the ground state? It has some energy, which is negative, which is minus e0. And when we include this first correction, what does it do? Well, it decreases it by some amount. Ever recall that? It doesn't change the number. It doesn't split any degeneracy because there was only one ground state in the first place. Because L had to be 0 when n is 1. L has to be 1 less than n or smaller. And if n is 1, L has to be 0, or if a camping is smaller than 0, it's 0. There's just the one state with degeneracy 2, 0 plus 1, which is 1. So that's fine. It just changes the ground state by some amount. Let's consider the second excited level. So in the second excited level, we have the states. And let me name this slightly differently. So this has energy minus e0. The ground state had L equals 0 and m equals 0. Whoops. m equals 0. It had to. For the second excited state, n equals 2, how many states are there? 2 squared is 4. So there are four states, 2, 3, 4. And what are their values of L and m? Well, if n is 2, L can be either 0 or 1. If it's 0, then m has to be 0. If it's 1, m could be 1, or it could be 0, or it could be minus 1. Yeah? So this again is L, m, l, m. So those are the four states with energy e2 in the hydrogen atom. And now, when we ask what are the energies of these three states when we include this so-called fine structure, what is this going to do? Well, the ground state is going to get pushed down by some amount. I should draw this differently to make it more vivid. This gets pushed down by some amount with some value of L. L here is 0. But for these three states, which all have the same value of little l, 1, the denominator is larger. So the amount by which it gets pushed down is less. So these three states get pushed down less, but the same amount is each other. So that cool? So these three states, the generacy is 3, and this state degeneracy is 1. Now notice that 1 plus 3 is equal to 4 is equal to n squared. So these states split into multiplets, groups of states, which have to have the same energy by rotational invariance. If they have L equals 1, there must be three states with the same energy. There must be. And states with L equals 0, there's just the one state. Is that cool? And if we did the same thing for E3, let's just think about what would happen. So here we have, I'm just going to write the L. I'm not going to include the m. We have the L equals 0 state times 1. We have the L equals 1 state times 3. And we have the L equals 2 state. And how many states are there in the L equals 2 state? Times 5. Good. And again, this gets pushed down to some value. This gets pushed down to some value. This gets pushed down to some value, where this is 1, 3, and 5. And again, these add up to 9, which is 3 squared. So the degeneracy. So by turning on the relativistic corrections, we break this degeneracy. We lift the degeneracy. You break the symmetry, you lift the degeneracy. Cool? Now, unfortunately, this isn't one we can control by turning a dial in the lab. No experiment that I can do in the lab will change the value of the speed of light. Or the, well, actually, there are experiments that will change the value of the speed of light. If you work in medium, if you work in medium, the effective speed of light is different. So that's going to change relativistic effects. So that's interesting. That tells you you can actually vary this continuously if you can vary continuously the index of refraction of the ambient system. I invite you to think about how you might do that experimentally, because it's a fun question. However, what I'd really like is something where Professor Evan, in his lab, can actually turn a dial and change the spectrum in a way that explicitly breaks the symmetry. Meanwhile, I also want an example that breaks the rotational invariance. I want to break the 2L plus 1 states up into 2L plus 1 states with different energies. I want to make them depend on the angular momentum in some direction. What must I do in order to lift that degeneracy? Fantastic. If you add a magnetic field, why is that going to break the degeneracy? Yeah? Yeah? The difference in angular momentum is sort of on the different dimensions. Like different radii. Give me an answer that depends on, not on your knowledge of chemistry, but give me an answer that depends on symmetries. Why does turning on a magnetic field have the potential to split the degeneracy of the 2L plus 1 states? Thank you. Exactly. You have no rotational symmetry. Exactly. So that's pretty good. So let's break the 2L plus 1 by breaking rotational symmetry. And there are many ways we could break rotational symmetry. We could turn an electric field. We could break the magnetic field. We could put a shovel in the room. But anything that breaks a rotational symmetry will do it. But a very convenient way is to do with a magnetic field. Magnetic field is particularly beautiful. So this effect is now called the Zeeman effect. I believe it was Faraday. One of the classic electromagnetists of the 19th century attempted to look at the spectral lines coming off glowing gas and see if they changed by turning on an electric or magnetic field. It turns out to be a tricky experiment to do because you need to have a very well-controlled magnetic field. Otherwise, what you get is just a bunch of schmutz. So to get a clean spectrum, you have to have a very uniform magnetic field, which is not trivial. It has to be time-independent. So Zeeman, at the very end of the 19th century, in fact, this experiment was done, I believe, in 96. So I think the paper came out in 97. But the experiment was done in 96. Zeeman did this experiment. He said, look, if I turn on a magnetic field next to a hot glowing gas and look at the spectrum that comes off, anything happen? So why would you think that anything, well, why would you think that anything might happen? And the answer to this one is really quite nice. Look, if we compute the energy, the energy is going to be equal to the coulomb energy plus a correction, which is the following. If you think of the atom as an electron carrying some angular momentum in the presence of the proton, if you think of the electron as a thing carrying some angular momentum and with a charge, if it has a charge and an angular momentum, it's going to have a magnetic moment, mu. And the magnetic moment is going to be equal to the current of the charge times the area swept out divided by c, because we work with sensible units. And this is equal to minus, for the case of the atom, EL over, man, I really made this too small. Let me write it here, minus mu, the magnetic moment of the system dotted into the magnetic field that you turn on in the background. So this is a background magnetic field. I am turning on an electromagnet, which induces magnetic field in some known direction with some known magnitude. So my dial is with the magnitude of that B field. It could make it larger. It could make it small. And it's in a known direction. And the point is, our electron carries some, because it's orbiting, because it carries some angular momentum and it carries some charge, has some magnetic moment. So it behaves like a little bar magnet. And that bar magnet wants to anti-align with the background magnetic field that I've turned on. So there's an energy penalty to not being aligned. This is the usual dipole-dipole interaction from a dipole potential from electrostatics. And mu here is equal to minus E over 2mc angular momentum. So this can be computed from a simple model of the atom. Whatever. It's easy to compute. So this is slightly annoying, because B is in some arbitrary direction. Let's just simplify our life and say that it's equal to E coulomb minus mu B. And I want to turn on a magnetic field of known amplitude in a known direction. Let's call it the z direction. So this is mu z. But this mu z, that's just mu z is Lz times minus E over 2mc. So this is going to be plus E over 2mc Bz. And I'm going to put this all together, Bz Lz. Everyone cool with that? And so where this is coming from, again, is just the angular momentum is telling you the velocity. The velocity is telling you the current. And then the area is the radius. So that's what's giving you the radius and the angular momentum. OK. So this is our energy. What does this tell you about the energy eigenvalues? Yeah, they depend on M. And do we have to solve the energy eigenvalue equation again? No, because our energy eigenfunctions of Ec are proportional to Ylm. So they're already eigenfunctions of Lz. Everyone cool with that? So when I take this beast and I act on the weight functions I erased, but on the 1 upon little r capital R sub nl, Y sub lm. When I act on the Ylm, this is just going to give me h bar M. So this is just a constant when acting on the Coulomb energy eigenfunctions. So this is a much easier problem than when we don't even have to make any approximations. We can just say exactly energy Zeeman is equal to minus E0 upon n squared nlm plus Eh bar upon 2mc Bz, the backer magnetic field. And this is the M of electron mass times little m. And I'm going to write little m sub l just to distinguish little m sub l from m sub e. Cool? So when we turn on the prediction here, is when we turn on the magnetic field, all Zeeman, our energy levels will get split according to the different values of M. Now if instead of working with the pure Coulomb system, if we were in fact a little clever and we had already noticed that there was a fine structure, because they already knew that there was a fine structure at this point, they could see the different spectra in lines coming off of hot gas. So it's just an experimental fact. You can't turn off the fine structure. This is going to be equal to E nlm of our correction to the hydrogen atom with the fine structure corrected. So the l equals 1 states and the l equals 0 states are split. Let's focus, for example, on E2. So these guys have a different energy. The l equals 1 guys have a different energy than the l equals 0 guys. What does this predict is going to happen if I look at my spectral line? So what this predicts is the following. It says that look, if you have some spectrum that looks like this, so this is a function of wavelength. This is a function of wavelength. And I'm looking literally at what comes off the prism. I'm looking at spectral lines. And I'll have a line here. And these are going to be the l equals 0 states. This is n equals 2. l equals 0 states. And then there's the l equals 1 states. This is with 0 magnetic field. B is equal to Bz equals 0. If we turn on a magnetic field, Bz not equal to 0, what should we see? Well, what happens to this state? Nothing. It's the same thing because the m is strictly 0. So it's the same value. And this guy, what happens to it? The m equals 0 state is the same. But then it breaks up, as everyone said, it breaks up into three lines that are equally spaced. Yeah? It breaks up into three lines because one of them has m which is plus 1. One of them has m which is 0, and one is m is minus 1. So the energies will increase or decrease. Let's say this is frequency. So we get this splitting. And another way to draw this is, as a function of B, let's draw the energy levels. So there's the 1. There's the 0. And this stays constant. And these guys, 1 is constant, 1 increases, and 1 decreases. Everyone cool with that? When you start getting to strong magnetic fields, weird things happen, like these guys cross, and all sorts of subtle things can happen. But let's ignore the strong magnetic field case and just focus on this. So we see that these guys split. So this is what the Schrodinger theory, which came in 19, or the full story would predict. It came a little bit after this. But this experiment was done in 96. So there's actually kind of an entertaining story about people trying to explain this effect classically. And you say, well, look, could have some angular momentum, z direction, some angular momentum, and the x direction, some angular momentum, the y direction. The part that has angular momentum, the x direction, doesn't do anything. And then the z direction is basically part with the x and part with y. You can have different polars. There's a whole classical song and dance. It's crazy. It doesn't make any sense. But that's because they were trying to do classical mechanics to do quantum mechanics, which doesn't work. So I'm not even going to bother explaining what they were thinking. But anyway, so people came up with a sort of classical crutch to say, well, that's not totally insane. So this is what Zeeman got, for example, by looking at hydrogen. But he did more than just look at hydrogen. I can't remember exactly. I think he did it with sodium. Matt, do you remember? Sodium? I think he did sodium and iron. But anyway. But here's the funny thing that he got with sodium and iron. So this is a different experiment. Same, exactly the same setup, but he was looking at spectral lines in sodium. And the spectral lines in sodium did a very similar thing. There were some where one went to one. And there were others where one went to, why did we have three? Here, just as a quick reminder, why did we have three states here? Because we had three states for l equals 1. This one state went to 1, 2, 3, 4 states with none of them in the middle. And in fact, it's worse than that, because they didn't go to equally spaced states in general. In general, they did something slightly funny, and the beautiful plot in his paper looks like this. There are four states. And as you take the magnetic field to 0, these four states all coalesce exactly to this one state. So this one line is representing four different states. And this already included the fine structure. So we know that this is a single value of l. But that apparently has four states. That should worry you. What does that tell you about l? What does l have to be in order to have four states in your tower? Half integer, in particular, what value? Three halves. Crap. We already know that these wave functions are described by the YLMs. We solved that problem. We explicitly solved it. Here's the exact answer. It's the YLMs. But we've observed experimentally that l has me 3 halves. But if l has 3 halves, YLM is equal to 0, because it's equal to minus itself. This is bad. This is 0. So this effect is called the Zeeman effect. And this effect is called the anomalous Zeeman effect, which is strange for two reasons. The first reason it's strange is they showed up in the same paper. And the second thing is, OK, you call it the Zeeman effect, because there's some guy named Zeeman. Is there some guy named anomalous Zeeman? It's a very strange name. Anyway, so this was called the anomalous Zeeman effect, despite appearing in the same paper, because it's weird. It was deeply disconcerting to people. We now just call it the Zeeman effect, although we have the bad habit for entertainment value of referring to it still as the Zeeman effect. I'm not the only one responsible for that bad habit. I'm going to point out that at the end of this paper, it's totally awesome paper, by the way. It's very readable. It's short. He says, I was looking at it last night, Zeeman says, quote, possibly this observed, the observed phenomenon, will be regarded as nothing of any consequence. OK, so a few years later, Polly says the following. So Polly says this actually in a also, this is also totally lovely in science from 1946. Polly was in the US during the war. After the war, he was at the Institute for Advanced Study. At the Institute for Advanced Study, while he was there, he got the Nobel Prize. And he gave a little for the exclusion principle, which we're about to get to. And he gave a little spiel, and it's written up in this edition of Science. And he says, quote, a colleague who met me strolling rather aimlessly in the beautiful streets of Copenhagen said to me in a friendly manner, you look very unhappy to whereupon I responded fiercely, how can one look happy when he is thinking about the novel of Zeeman Effect? OK. So this troubled people. Yeah? Is there a scale of people who speak of a very non-sparely, or one-house, or any of those? Yeah, he found all sorts of crazy states. So this isn't the only example. It's just the only one I'm going to draw. But indeed, you can find five half states. You can find states with six. Why is it the airy quote one and airy quote one half of the messier? Oh, I didn't draw all the states. The spectrum is complicated and messy, because it's sodium. There's all sorts of crap. So I'm just not drawing. I'm just focusing on a specific set of spectral lines. Maybe the best way to draw that is. OK. So this was really troubling. But let's put this on pause for the moment. And we'll come back to it in just a second. So meanwhile, there's something else that's sort of annoying about this system. We've got the Coulomb potential. And the Coulomb potential has the following beautiful property. The degeneracy of energy energy state is equal to n squared. Meanwhile, various people were fond of observing the fact that the number of states, which I will in an abuse of language called the degeneracy of the level n in the periodic table, is equal to, well, the first one there's 2. And then there's 8, and then there's 18. So 2, 8, 8, 18, that's almost 2 n squared. 2, 4, 8, 9, 18. Cool. Maybe there's a relationship. And meanwhile, it's very tempting to think that there's a relationship, because imagine you have four electrons. If you have four electrons, what do you do? Well, you put an electron here, and then you could put an electron here, or you could put an electron and another electron here. You can also put another electron here and here, or you could put them in any combination. 2 here, you could put 2 here, 2 here, 1 1. You do all sorts of stuff. But if for the moment, you imagine that electrons have this funny property. Imagine for the moment that electrons have the property that can't be in the same state at once. Then if you have four electrons, I don't know why. I'm desperate. I'm thinking about the anomalous aim on effects, among other things. I have nothing better to do. So just for fun, let's imagine if I put an electron here, I can't put another one. So if I have a 2 electron atom, the second one goes in one of these, a 3 electron atom, a 4 electron atom. And these guys are all related to each other, because they live in the same energy level, but they have different L values. So they'll have slightly different properties. That suggests that the 1L equals 0 state should be different from the 3L equals 1 state chemically. But wait, there aren't groupings of 2 and 1 and 3. They're groupings of 2 and 6. Ah, 2. Well, so Pauli, staring at this, in 25, said the following. He said, look, I'm going to conjecture two things. The first is that no two electrons can live in the same state in habit, the same quantum state. That doesn't give us the periodic table, though. It gives us half the periodic table. There are twice as many states as you think in hydrogen as you think. And you might think that I'm being facetious, that he said something a little more sophisticated, but he really didn't. What he said is, I posit the existence of an additional quantum number, which the electron can have, which takes one of two values, which translated says, there are twice as many states as you think. So that's exactly what he said. So this is called the Pauli exclusion principle. And now this does a totally spectacular thing. And you've probably all seen this in high school chemistry or even college chemistry. Now, first electron goes here. Second electron goes here because there are twice as many states as you think. But now you can't put another state in there because there's as many as you can have. One can go here and you can put up to six in here. And that's it. So the E2 states, the n squared, which is four times two states, are somehow related to each other naturally. And this gives you the structure of the periodic table. And in fact, an awful lot of chemistry falls directly from this. So this is a Pauli exclusion principle. However, it's a little bit disappointing because while this is whatever some ridiculous rule, but we're doing quantum mechanics and all they're all ridiculous rules. On the other hand, this one is just stupid. This is just like, look, it wouldn't be nice. So in 25, a couple of guys, really interesting characters named Goschmitt and Ulinbeck. So Ulinbeck, wow, we're really low on time. OK, so I'll just tell you this and then I'll get to the last of it later. So Goschmitt and Ulinbeck, who were young, kind of naive, they said, look, we've got two effects here. One effect is the anomalous Zeeman effect. And there's this weird fact that states have the wrong total angular momentum. And we have the second thing that there are twice as many states for an electron as you think in the Coulomb potential. How can these two things fit together naturally? Well, here's a guess. Suppose that the angular momentum that we calculated when we did the estimate of the breaking due to the Zeeman effect. Suppose that angular momentum was not the right angular momentum. Maybe there's more angular momentum in the system. There's angular momentum in the system from the electron orbiting around. But maybe like a little tiny Earth, the electron itself can have some intrinsic angular momentum. It turns out Pauli had had this idea before. In fact, Cramer had suggested this to him, who was a very young guy at the time. And Pauli said, you're a blithering idiot. Because if you calculate how small an electron has to be in order to fit all the other things we know, and you figure out how fast it has to rotate to explain the anomalous Zeeman effect, surface would have moving faster than the speed of light. That's ridiculous. Leave my office. So then Ullinbeck and Goldschmidt, I don't know how to put it, Goldschmidt, G-O-U-D. Anyway, these two guys write a paper and say, aha, well, we can explain this. But let's just not assume that the super thing is rotating. Let's just say an electron has some intrinsic angular momentum. I don't know. Why not? We'll have some electrons have intrinsic angular momentum. And if you assume that electrons have an intrinsic angular momentum, which I'll call s for the moment. So then the total angular momentum, j, is equal to l. It's orbital angular momentum. Plus some intrinsic angular momentum, which I don't remember what symbol they used, we'll call it for the moment s, where this has the property that s squared for an electron is equal to, sorry, the principal quantum number, which I will call little s, is equal to 1 half. So m sub s is equal to plus or minus 1 half. So this is like the l equals 1 half state, but it's not l. It's something intrinsic. Has nothing to do with anything rotating. It's just a fact about electrons. Suppose electrons have a little bit of spin. Then what you discover is, if you have the l equals 1 state and the electron has spin 1 half, what's the total angular momentum? 3 halves. So what do you get? You get quadruplets like that. So this turns out to explain, if you include a small relativeistic effect, this turns out to explain the anomalous Zaman effect bang on. Meanwhile, there was an experiment done in 1922, which was a Stern-Gerlach experiment, in which people discovered that Nick, Stern-Gerlach, not people, discovered that nickel, when sent through a magnetic field gradient, bent into one of two different spots, never 3, never 0, always 2. How could that be? Electron spin. That wasn't realized until 1929, that the connection was there. So these guys came up with this ridiculous postulate. This was Ullinbeck and Goeschmet. And let me just quickly, U-H-L-E-N-B-E-C-K. So Ullinbeck, an amazing, amazing scientist, also the father of the mathematician Ullinbeck. And she was a total badass and has inspired an awful lot of physics. Karen Ullinbeck, who's at UT. So this is a pretty interesting and prolific eye. But he was also prolific in the following way. He ran a program at the Rad Lab at MIT during World War II. And Goeschmet worked with him. And when they were finishing up, when the war was ending, Goeschmet became the scientific advisor to something called Project Alsos. Project Alsos was the project where the military went to the conquered territories in Germany to catch the German atomic scientists and bring them to a place called Farm Hall in England, where they listened to them, eavesdropped on them, Heisenberg, all the good guys. Well, it depends on all the great physicists in the German territory at the time who were deeply complicated people. And they listened to them in something called the Farm Hall transcripts, or the transcripts of those recordings. They were written up in a book called the Epsilon Project, which is totally breathtakingly awesome. And Goeschmet wrote a book called Alsos about this process of hunting down the German scientists. So Goeschmet, U-D-S-M-I-T, I think. And he wrote a book called Alsos, which I heartily recommend to you, because it's like a combination adventure story and beautiful bit of physics history. So these guys both had it. And they were both at MIT during the World War. There's a nice connection here. So these guys, fascinating characters, and they came up with this idea of some intrinsic angular momentum. Polly then calls it Spin. He gives it the name. And he develops a mathematical theory of spin. And that mathematical theory of spin will lead to quantum field theory relativistic quantum mechanics. And it will eventually lead to quantum computation, which is going to be the topic of the last week of our course. So what we've done so far is we've explained the discreteness of atomic spectra. And we've explained the structure of the periodic table. What we haven't done is we haven't explained why atoms form molecules or solids. And we also haven't explained what spin is at all. Those are going to be the topics of the next three weeks. See you next time. Thank you.
 The following content is provided under a Creative  Commons license.  Your support will help MIT OpenCourseWare continue to  offer high quality educational resources for free.  To make a donation or to view additional materials from  hundreds of MIT courses, visit MIT OpenCourseWare at  ocw.mit.edu.  OK, so let me just quickly remind you of what we did  last time.  So unfortunately, today is going to be a little bit of  what I wanted to do this time and a little bit of what I  would have wanted to have been done in recitation yesterday  because all recitations were canceled.  So today is going to include a few steps along the way that  I skipped over last time.  OK, so to review, from last time we talked about central  potentials, the energy operator had a radial derivative  term, an angular momentum term, and a potential term.  So this would be true regardless of whether we were  central or not.  But if we have a central potential where this only  depends on the magnitude of the radius vector, then we can  use separation of variables and write an energy eigenstate  as a radial part, which we will sneakily pull out a factor  of 1 over the radius.  As we saw last time, that was particularly convenient.  It took care of this r and made everything nice and simple.  So a radial part times a spherical harmonic.  The reason we use a spherical harmonic is that neither of  these two terms depend on the angular coordinates.  The angular momentum, we know what its eigenfunctions are.  So if we use a spherical harmonic, this angular momentum  becomes just a constant.  And having separated in this form and pulling up that  sinc factor of 1 over r, the energy eigenvalue equation  reduces to a simple 1D effective energy eigenvalue  equation with a simple dr squared and an effective  potential where the effective potential is the original  central potential plus an angular momentum term, often  referred to as the angular momentum barrier, which is  proportional to, as Professor Evans says in 804, one is  forced to say h bar squared upon 2m repeatedly.  And he pointed out that it would be much more efficient if  I would just come up with a simple variable for saying it  or a sound.  So I will attempt to henceforth call it pfft.  So the effective potential is l l plus 1 upon r squared plus  the central part of the potential,  props to Professor Evans.  And this is true in general for a central potential.  The last thing we notice of this is that since the  eigenvalue equation has been reduced to this form, which  depends on l but is independent of m, m appears nowhere here,  the energy can't possibly depend on m, although it can  depend on l, it's just e of l.  And thus, for every different value of m, I will get the  same energy, fixed l different m.  And there are how many values of m for a fixed l?  12 plus 1.  So the degeneracy of e l is 12 plus 1.  That's general for any central potential that only  requires spherical symmetry.  So then we study the example of the Coulomb potential, where  the potential is minus e squared upon r.  And we quickly reviewed the solution of the energy eigenvalue  equation.  OK, let me write it that way.  Where r of what I will call in place of e, I will simply write  n, r n l of, sorry, is equal to e to the minus r over 2 r  naught n times r over r naught to the l plus 1 times some  simple function v sub nl of r over r naught, with the energy  nlm being equal to e naught, or I should say e ridberg, in  our notation of last time, upon n squared minus bound state.  So let's talk about this quickly.  This term, the fact that it falls exponentially large  distances, came from the asymptotic analysis at r  goes to infinity.  And this is just saying it's a bound state.  The probability density falls off exponentially as we go  out to large radius.  This term came from the asymptotic analysis near r  goes to 0.  And near r goes to 0, the only term that matters is, in v  effective, this is 1 upon r.  This is 1 over r squared, so this gets large more rapidly.  This term is important, and that l l plus 1 gave us r  to the l plus 1.  We worked in dimensionless variables, rho, and here I  replace rho with r over r naught, where r naught was the  characteristic scale we computed last time.  And finally, this function v sub nl satisfied, again, the  energy eigenvalue equation, but having pulled out these  factors such that asymptotically it's regular.  So this function must be regular, smooth, no  divergences, no poles, as r goes to infinity and to 0.  And then we solved this by series expansion.  And from termination of the series expansion, we got a  relationship between the energy and l and m.  And in particular, the relationship, this came from  termination of the series just like it did for the  harmonic oscillator.  And the termination series gave us that the energy was  equal to minus a constant over n squared.  With one additional condition, which was that n had to be  an integer greater than l, and l, of course, has to be  greater than or equal to 0.  So let me just quickly tell you how this arises, because this  is an important fact about hydrogen, which I want you  to understand.  So the way that arises is we got the energy from a series  expansion for this remaining function of r, v and l of r.  And that series expansion started out by saying let v be  of the form sum over j of a sub j r over r naught, the  dimensionless variable rho, to the jth power.  So it's a power series.  And when we plug this whole form together with this series  expansion into that differential equation, just like we  did before for a harmonic oscillator, we get a set of  relations between the various coefficients in that series.  And we solve that by saying a sub j plus 1 by deriving a  recursion relation is equal to a sub j times 2 square root of  epsilon times, oh, I'm going to be careful, j plus l plus 1.  j plus l plus 1, yes, good, minus 1 upon j plus 1.  j plus 2l plus 2.  So, OK, fine, we get this recursion relation.  It looks absolutely awful.  But it tells you that if you know a0, say, as a constant,  one sec, if you know a0 as a constant, which is not 0, if  it's 0, then they're all 0, then this function is  identically 0.  Because aj plus 1 is aj times some constant.  So this had better not vanish, a sub 0.  So if the first term, the constant term doesn't vanish,  which is good, it tells us v doesn't vanish.  So we've already learned something.  If v doesn't vanish identically at the origin, then we can  determine a1 from a0 by multiplying by this.  j is 0 in this.  And l is whatever number we fix from ylm.  And so on, we can deduce all the higher a sub l's.  It's a series expansion.  This is the recursion relation for it.  However, note the following thing.  As j gets large, as we go up to higher and higher  orders in the series, as we go to higher and higher order in  the series, then j is much larger than whatever a  fixed number l is, and much larger than whatever a fixed  number 2 is.  So this asymptotes to aj times roughly 2 epsilon j upon j  plus 1j.  At very large j, this just goes to 2 epsilon over j.  Now let's think about what that means.  If I have that aj plus 1 is roughly equal to aj times root  epsilon times 2 over j plus 1, I guess, then this tells me  that every time I increase j, I get a factor of j plus 1  and another factor of 2 to the epsilon.  This tells me that a sub j is equal to a sub not times 2  root epsilon to the j over j factorial.  But those are the expansion coefficients of an exponential.  Quantity to the n over n factorial.  But that's bad, because if this series is the exponential  series with argument 2 root epsilon, then this series  becomes e to the 2 root epsilon r over r not.  But that's bad, because that would exactly swamp this term.  That would defeat the asymptotic analysis we did at  the beginning.  If this term is v is growing so large twice as rapidly as  this is going to 0, then the net is diverging at infinity.  That's bad.  So in order for this to describe, for the series  expansion to describe a good function that really does  vanish at infinity, we need that this doesn't happen.  So it must be true that j doesn't get large enough that  the exponential in v overwhelms the decaying exponential  from the asymptotic analysis.  Ever recall that?  So it has to terminate.  But if it terminates, let's look at the coefficient here.  If it terminates, this says that 2 root epsilon times j  plus, sorry, if it terminates, that says that there's some  j maximum.  Well, let me write it over here.  If it terminates, that means there's some value of j,  such that a j plus 1 is 0.  But if a j plus 1 is 0, how can that possibly be?  It must be that for that maximum value of j, the  numerator vanishes, specifically for that  maximum value of j.  But what's the condition for that numerator to vanish for  a specific value of j?  I'll call j sub m, or j max, the maximum value such that a sub  j max plus 1 vanishes.  So j plus l plus 1 must be equal to 1.  Or said differently, epsilon, which is the energy in units  of E0, must be equal to 1 over 4 j max plus l plus 1  quantity squared.  Cool?  So that means that our solutions, our states, psi, are now  labeled by three integers.  They're labeled by l and by m from the spherical harmonic.  But they're also labeled by the value of j max.  And j max can be any number from 0 to infinity.  Strictly, it shouldn't be infinite, but any finite  number, any integer.  Because eventually, the series terminates at j max.  So it's labeled by these three numbers, and the energy is  equal to of l m j max is equal to E0 minus over 4.  And this E0 over 4 we already saw was E Rydberg, 13.60 v,  times what value?  Well, here's the energy in units of E0, j m plus l plus 1  quantity squared, where l is greater than or equal to 0.  j sub m is greater than or equal to 0, and 1 is 1.  But this is cumbersome.  We can just as well call this quantity n, where n is greater  than l as an integer.  It can't be equal to l, but it can be 1 greater than l, or  any greater integer.  Is that cool?  So instead of labeling by j m the point at which it  terminates, the series terminates, we'll label it by  n, and you can deduce what j m is by subtracting off l plus  1 from n.  That tells you where the series terminates.  Notice that this tells you something nice.  This power series for v, if n is equal to l plus 1, has how  many terms in it.  So don't say it out loud.  Just think about it for a second.  If n is equal to l plus 1, how many terms are there that  are non-vanishing in the power series?  I won't call an Uber.  Raise your hand when you have an answer.  What's the value of j max if n is equal to l plus 1?  So what term is the first one to vanish in the series?  OK, so how many terms are non-vanishing?  Yeah, just the one.  Just the first term.  So when l is as large as it can possibly be, when l is  equal to n minus 1, v is a constant function.  It's just constant.  When l is one less, when l is n minus 1 or n is l plus 2,  then v has two terms in it, a constant and a linear.  So the greater the difference between l and n, the more  terms there are in the series.  OK?  Cool.  So anyway, this leads to this form of the energy.  But as is usually the case, this sort of a brute force  analysis doesn't give us a whole lot of insight into why we  get the qualitative features we do.  In particular, the qualitative feature that stands up most  obviously to me, I don't know about you guys, is the fact  that the degeneracy is now much, much larger.  In particular, we know from basic principles  of a central potential, as we just reviewed,  that any time a potential is symmetrically  invariant, symmetrically symmetric,  spherically symmetric, symmetrically symmetric.  And he said the potential is spherically symmetric,  rotationally invariant.  Then there must be a degeneracy of 2l plus 1  for every energy I can see.  There must be.  That l could be 0.  But if there's any angular momentum,  then that degeneracy must be 2l plus 1.  But this is much more degenerate.  This is much more degenerate.  And in fact, how degenerate is it?  Well, the total degeneracy is, well, for any given En,  the degeneracy is equal to, well,  what are the possible values of l?  So we have the sum.  We have a state for every different value of l.  For l could be equal to 0.  And it can go up to n minus 1.  Because, again, n is some integer which could vanish l  plus 1.  So the smallest this can be is 0.  So n can be l plus 1 or anything greater,  i.e. l can go up to n minus 1.  But it can't be any greater.  Then for every value of l, there is a state m.  There's a value of m, which could be minus l all the way up  to l in integer steps.  So how many states are there?  Well, there's one state for every value of n, l, and m.  So it's this sum.  The sum on m from minus l to l is 2l plus 1.  So this is the sum n minus 1, l equals 0, of 2l plus 1.  And this is equal to kind of beautifully n squared.  Just arithmetic series.  So we have this huge degeneracy, which  is much larger than 2l plus 1.  It's a sum over 2l plus 1 from l is 0 to n minus 1.  Where is this degeneracy coming from?  Why?  So classically, or quantum mechanically,  we know the following.  We've learned the following.  And I hope it's under your fingernails at this point.  That when you see a degeneracy, you  should expect that degeneracy to follow from some symmetry.  There's some symmetry of the system  that is leading to a degeneracy by virtue of the fact  that the generators of that symmetry, for example,  for rotation, that's angular momentum,  commute with the energy.  And when you have an operator that  commutes with the energy operator, as we've seen,  you can generally construct new states given a single energy  eigenstate by acting suitably with that operator.  So we should expect there to be some new symmetry.  But it's very hard to see what symmetry.  I mean, hydrogen is the central potential.  It's just some stupid central potential.  What's so special about the hydrogen system, as opposed  to, for example, the harmonic oscillator, which hopefully  you will see in recitation at some point,  the harmonic oscillator in three dimensions  has degeneracy 2l plus 1.  It's a central potential.  It has to have degeneracy 2l plus 1.  But not n squared.  Where did that come from?  What's so special about the Kepler problem?  So there must be some symmetry, but it's not obvious what it is.  But now let's use a note from classical mechanics.  No other theorem tells us when we have a symmetry,  we have a conserved quantity.  We also saw this in the quantum mechanical version  for expectation values.  We have a conserved quantity.  And in quantum mechanics, having a conserved quantity  means that the energy commutes with some quantity.  Because that controls the time evolution of the expectation  value.  So there must be some quantity, which I've written by a question  mark, that commutes with the energy operator,  specifically in the case of the harmonic oscillator.  So what is that quantity?  So actually, this quantity in classical mechanics  was studied because the same thing is true in the Kepler case.  In the Kepler case, the orbits close,  and there's simple ellipses.  And they have all sorts of nice properties  that beg for an explanation in terms of symmetry.  And it was pointed out by a number of people, Laplace,  Runge, Lenz, various people, that this is often  called the Runge, Lenz vector.  But I think you could blame it on any number of people,  that there's a vector that in classical mechanics,  for the Kepler problem or for the Coulomb problem,  is conserved.  And that quantity is p cross L, momentum cross angular  momentum, minus m squared, sorry, m e squared.  And let me break m sub electron.  It's the mass rather than the quantum number little m.  E squared R vector.  So it turns out this quantity is classically conserved.  And if you make a quantum operator out of it,  then e with a is equal to 0.  Hat, hat.  It was enormously non-obvious.  If it's obvious to you, you're a freak.  It's really, really not obvious.  You can go through and do the calculation.  And when I say really not obvious,  I mean, well, I'll show you what I mean in just a second.  So you can go through and do the calculation classically,  is this quantity conserved in the Kepler problem using Newton's  law?  And the answer is yes.  That's amazing.  Quantum mechanically, you can do the same thing.  Although I'm going to warn you for anyone  who has the chutzpah to try, because I encourage you to,  but it takes a little bit of brawn.  Think about this operator for a second.  Does this classical quantity have an obvious interpretation  as a quantum operator?  No, because l is r cross p.  But now we have prp.  There's an ordering ambiguity.  So you have to decide which order do you put that p  and that r and the other p and you.  Ppr, rpr, prr.  You can write out which components you mean here.  For example, you mean for the z component of a,  you're going to get px, ly, and ly, px.  But within that, how do you order the p's and the r's?  Because ly contains px.  And in particular, it contains x.  So there's an ordering ambiguity.  So just be a little bit careful about this  as a quantum mechanical operator if you do play with this.  But if you do it and you're thoughtful about it,  it's pretty easy to see that, well, it  can be computed and checked, that in fact,  the commutator vanishes.  So at this point, it's tempting to say, aha,  there's an extra concert quantity we declare a victory.  But the problem is, this didn't give us a symmetry.  This just told us there's a conserved quantity.  What's the symmetry behind this?  What's the symmetry that ensures that this  is a conserved quantity?  It's obviously not rotations.  It's something else.  And so there are many answers to this question.  One answer involves an explicit expression  for in phase space the change of variables dx.  And this is a variation that depends on,  I'll write this as f of x and p.  So there's a little vector parameter.  And the change of the physical coordinates  depends on both the coordinates and the momenta.  This is a slightly strange thing,  because it's like a change of variables  where the position gets mapped into some nonlinear function  of the positions of the momenta.  That's a little bit weird.  But you can do this.  But I don't find that a terribly satisfying answer.  And here's the answer I find most satisfying.  It's not terribly useful for present purposes.  But you're going to run into this again in 805  when you do what's called the operator method  for the Coulomb potential.  And it turns out to be an enormously useful machine.  We're not going to use an 804, but I  want to advertise it for you.  I want to give you a description, though,  that isn't the description you're going to get in 805.  The description is the following.  1935, a guy named Fock, crazy guy, amazing, amazing physicist,  played a huge role in the development of quantum field  theory.  I'm not exactly sure why he was thinking about this problem.  But he was.  And this was a very Soviet thing to do at the time,  I guess, to be hardcore and mathematical.  So Offi was going computing things.  And he observed the following fact,  how he observed this again, crazy long story.  He observed the following fact.  If you take the Kepler problem, and you work in momentum space,  so write everything in terms of the momenta,  it turns out that the Kepler problem, or he referred to it  as a Kepler problem, but all referred  to the Coulomb problem, the Coulomb problem in 3D,  three spatial dimensions.  This is really crazy.  Is exactly equivalent.  There's a change of variables.  It's kind of complicated, but it's just a projection,  it turns out, is exactly equivalent to the problem  of a free particle in, wait for it, four dimensions constrained  to the surface of a sphere, to a three sphere  in the four dimensions.  So you take a three sphere in four dimensions,  you take a marble, you kick it, but make sure  that it's stuck to the sphere, it  will move in perfect circles.  Obviously, it's moving on a sphere,  it's just going to move in great circles.  OK, that's nice and simple.  What's the symmetry group of that system?  Well, the sphere is invariant under any rotation  in four dimensions.  Ah, SO4.  There is a symmetry, SO4, special orthogonal transformation.  Special means it doesn't change the length,  orthogonal means it's a rotation.  So it's orthogonal rotations in four dimensions.  And so now you can, so this should sound familiar,  because the angular momentum group, angular momentum in 3D,  is SO3.  It's rotations that don't change the length in three  dimensions.  How many elements does this have?  How many conserved quantities go along with this?  Well, how many conserved quantities  go along with rotations in three dimensions?  What's conserved by virtue of rotational invariance  in three dimensions?  Angular momentum.  How many components of angular momentum are there?  Three.  Great, there are three.  So it turns out that if you calculate in SO4,  how many conserved quantities are there?  There are six.  And when you do this mapping back down  from the free particle in 4D to the Coulomb problem in 3D,  three of them are the L's.  And the other three of them are this A vector.  Comes out spot on.  So what it's telling you is that the spectrum of hydrogen,  or at least the spectrum I shouldn't say of hydrogen,  the spectrum of the Coulomb system,  has an enhanced symmetry.  It's the symmetry group of rotations in four dimensions.  Why?  I don't know, it's just kind of awesome.  But I think that's about the best that can be said.  It's just true.  So for certain very special potentials,  we find that the sort of thing happens.  And it happens in other systems too.  You discover that there are accidental symmetries  in your system.  They shouldn't be referred to as accidental symmetries.  There are enormously non-trivial functions  of the rest of the system.  And we're going to break the symmetry in a minute,  and you'll see why we shouldn't refer to it as accidental.  But in any case, the answer to the question,  why do we have this giant degeneracy in the Coulomb  problem?  Why is it n squared rather than 12 plus 1?  Is that there is an enhanced symmetry group.  There's more symmetry than you would have thought.  And there's a conserved quantity associated with it.  And that conserved quantity corresponds  quantum mechanically to an operator that  commutes with the energy operator.  OK?  Yeah?  I have a question.  So in quantum mechanics, do you remember  that there was only a set of like five, like two,  like that, that was where it was?  Indeed.  Indeed.  So how do you get like three certain components of a model  in quantum mechanics?  Excellent.  What do I mean in classical mechanics for quantity  to be conserved?  I mean that the time rate of change  of that quantity under the classical dynamics,  under classical evolution, is preserved.  So what's the quantum mechanical version of that?  So the quantum mechanical version of that is, well,  can we say that d dt of some operator,  or of some observable a, is 0?  Does this make sense?  No, because this is, I mean, it turns out,  in the language we've been using, in the Schrodinger  evolution description that we've been using,  this doesn't really make sense because it's an operator.  That's not a quantity.  The quantity is something you measure.  So what do you measure?  Do you measure the operator?  In a state, you measure its expectation values.  You might measure its eigenvalues,  but the more general thing is its expectation values.  So this is something that depends on time,  because the wave function depends on time, psi of t.  So the best we can do to say something is conserved  is to say d dt of a is 0.  But we've seen what the condition is for the expectation  value to not change in time.  In general, for any operator a, the time rate  of change of its expectation value  is given by the commutator of e with a.  I can't do the symbols.  Good.  OK.  So in particular, if this commutator vanishes,  if the commutator vanishes, equals 0.  This is equal to.  Sorry?  Where is the sum of the time varying?  Well, the wave function can be time varying.  And I'm going to assume for the moment  that the energy operator and the a operator  are not themselves explicitly time dependent.  Yeah, that's an important assumption.  So let me just assume for simplicity  that neither the energy nor the operator a that I'm looking at  are explicitly time dependent.  Like, it shouldn't be like x plus tp.  That would be annoying.  So let me assume that that's not the case.  You do have to deal with that sometimes.  For example, if you are dealing with a system that  changes in time, you can do that.  If you are dealing with a system that changes in time,  where there's a background magnetic field or something  like that that changes in time.  But modular, that sort of detail.  And also, there's an ih bar.  So where do the ih bars go?  Do the units make sense here?  A, a, good.  E and d dt.  So where do the ih bars go?  Well, you go on either side.  You have to tell me if they go up or down.  So what is the Schrodinger equation?  So there's, yeah, exactly, ih bar.  Good, OK, good.  So we need an ih bar.  Ah, good.  OK, man.  OK, so the spring has got to us all.  So now, in the case that E with A is 0,  imagine we have an operator A that commutes, well, 0,  then d dt of the expectation value of A in any state  is equal to 0.  And that's what I'll call conserved.  So classically, to be a conserved quantity  means that it's time-independent.  Quantum mechanically, to be a conserved quantity  means that it computes with the energy.  Now, coming back to the question.  The question was, look for angular momentum,  there are only two numbers you can know at any moment in time.  You can't know three.  You can know the total angular momentum, little l.  And you can know the angular momentum  in a particular direction, which we'd conventionally  call z, little n.  But you can't know some other angular momentum,  like the angular momentum in x direction, simultaneously.  So what does it mean to say that there are three  conserved quantities?  I can't know three conserved quantity at the same time.  They're not even well-defined at some moment in time.  The answer is, there are three operators, lx and ly, and lz,  all of which, which I will simply write as l vector,  all of which commute with e.  So while it's true that the systems don't have definite  values of lx, ly, and lz simultaneously,  nonetheless, lx, ly, and lz all commute with the energy  operator, and as a consequence, their time expectation values  are all 0 and essential potential.  So the time derivatives of their expectation values  are all 0 and essential potential.  They're all constant and essential potential.  Yeah?  You mean 1 over r?  You mean 1 over r?  Well, for the force of order of charge,  so you have 1 over r.  Yeah, 1 over r.  So the question is, look, if you don't have 1 over r,  do you get extra special things that  happen for other potentials that aren't 1 over r,  but something else?  It's not quite the unique case.  So the question is, look, this is a very good question, though.  So the question is this.  This is a great question.  Look, there's something peculiar about 1 over r classically.  The orbit's closed.  You get ellipses.  You don't get possessing ellipses.  That's what you get if you have an anharmonic,  if you have an extra 1 over r to the sixth term or something,  small correction, the 1 over r law,  which we do in the real solar system.  It's a many-body system instead of a two-body system,  so our orbit doesn't close.  It's closed.  So that's special.  And here we see that there's an enhanced symmetry.  Are these things related to each other?  So the answer is almost, in the following sense.  For the harmonic oscillator, we also have it closed,  but there isn't a runs-lens vector.  However, the orbit's closed.  So there's something special about that one, too.  Is there a conserved quantity there?  And it turns out that the answer to that is yes,  there's an enhanced symmetry there as well.  But it's not like this.  It doesn't lead to this gigantic degeneracy.  There's just extra structure that's  nice about the harmonic oscillator.  But for the harmonic oscillator, of course, it also closes.  It's not this orbit, this.  So there's a nice structure there, but it's not quite so simple.  OK.  Yeah, one last question.  Yeah?  I was just wondering if thinking about the  symmetric symmetry, degeneracy, is the problem?  Like, is it equal times an equal area?  Yeah, it's not unrelated.  I don't know how to make that connection sharp.  But the thing I can say sharp, so the question is, look,  is that conserved quantity somehow related to the fact  that there's equal time in the equal area swept  on equal time at all points in the orbit?  Which is a true thing about the Kepler problem.  And that's a good observation.  I don't know how to make that connection,  but it is true in the following sense  that both of those facts are guaranteed by the fact  that you have closed orbits with a particular ratio.  And that is indeed can be blamed on the conservation  of the Runge-Lenz vector.  I don't know how to make that connection  direct and explicit, but indeed it is.  That's a fun question.  I'll have to think about that.  That's a good one.  One of the totally awesome things about teaching U04  is that every single time someone asks a question  I've never even thought of before.  Forget never answered.  Never even thought of it.  I never thought of that.  It's a great question.  OK.  So everyone cool with the degeneracy?  We have this gigantic degeneracy.  So knowing what the degeneracy is or where it's coming from,  at least more or less, is progress.  But if you really understand why something is conserved,  then you can break that conservation.  You can break that degeneracy.  So here's my question.  Is it possible, can we, or indeed must we,  can we break or lift this degeneracy?  So first let me tell you what those words mean.  So we have this gigantic degeneracy of the,  what's this guy?  Reasted.  OK, we have this gigantic degeneracy of the,  oh there it is, up at the top, n squared of the Coulomb  potential.  Can we break that degeneracy or lift it?  So what do I mean by lifted degeneracy?  What do you mean by lifted degeneracy is the following.  Suppose I have a system that's degenerate.  So it's got two energy eigenstates  with the same energy eigenvalue.  So this is energy.  Then if I can kick the system in somewhere,  or modify it, or bend it, such that the energy levels are  no longer degenerate, one state has been lifted above the other.  And so we use the word to lift the degeneracy.  Now you can see that there are two different states  by measuring the energies.  Is that cool?  It's also said to split a degeneracy  or to break a degeneracy.  So those are just words.  So can we break or lift this degeneracy?  So sure, how do we break a degeneracy?  How do we lift a degeneracy?  We need to break the symmetry.  So if we don't have the symmetry,  we won't get the degeneracy.  We've seen this before.  So first off, let's think about what  are the two symmetries that we want.  Well, we got this degeneracy of 2L plus 1  from spherical symmetry.  And we got this extra degeneracy of n squared  from conservation of the Runge-Lenz vector.  And conservation of the Runge-Lenz vector  came from the fact that we had exactly the Coulomb system.  So anything that isn't exactly the Coulomb system,  but is still rotationally invariant,  should still have the 2L plus 1 degeneracy,  but not the n squared degeneracy.  Cool?  Like the harmonic oscillator, for example.  So let's change the system.  Let's find a small correction to the system  that leaves it mostly the Coulomb problem,  so it's some small correction.  So what comes to mind?  So one good way to think about a physical, natural way  to break this degeneracy, to break the symmetry,  is to think about a real system that's  described by the Coulomb potential.  So the system that we usually talk about is hydrogen.  Hydrogen is a proton that's got a plus charge.  It's got an electron, and it's attracted with a minus  e squared upon R. So in hydrogen, is this the energy  operator for hydrogen?  Is this a good description of the energy operator for hydrogen?  Well, it's a good start.  But it's not exactly right for a bunch of reasons.  And in fact, on your first problem set,  when you computed the time it takes for hydrogen classically  to decay from the classical bore radius,  or from the classical, well, from the bore radius,  one of the things you discovered is  that the electron starts out being non-relativistic,  but very rapidly becomes relativistic.  Rapidly is like 10 to the minus 15 seconds.  This is a non-relativistic p squared  upon 2m kinetic energy.  That's probably not a perfect approximation.  So let's ask, so the first example, example one,  is going to be relativistic corrections.  Now, will relativistic corrections  break rotational symmetry?  They can't, right?  Because relativity doesn't say this is the z direction.  In fact, it does quite the opposite.  Relativity, among other things, says,  don't get too hung up on what's the z direction.  But relativistic corrections will not break the symmetry.  So if we include relativistic corrections  to the kinetic energy, this will change E. Change  the energy operator, but it will preserve  the rotational symmetry, E with L is equal to 0.  So we won't break the rotational symmetry,  and so we'll preserve the 2L plus 1.  So let's check to see that this is true.  So to check to see that this is true,  let's actually estimate out the correction to the energy  due to the first non-trivial relativistic effect  in the atom.  The first non-trivial relativistic effect  can be calculated in the following way.  What's the kinetic energy in special relativity  of a particle of mass m?  So that's easy.  The total energy is the square root of m squared c to the 4th  plus p squared c squared minus mc squared.  That's the rest energy.  That's the total energy.  So this is the kinetic energy.  But this is equal to, well, if we  pull out a factor of m squared c squared,  this is m squared, sorry, mc squared times square root of 1  plus p squared over m squared c squared minus 1.  And now if we Taylor expand this guy,  because if the momentum is small compared to mc squared,  or mc, I should say, if the momentum is small compared  to mc, which is morally speaking at low velocities,  the velocity is small compared to c,  then this is 1 plus a small number.  And we can Taylor expand this to get 1 plus 1  half a small number.  And the ones will cancel, giving us only 1 half.  So this is going to be equal to 1 half that small number,  p squared over mc squared times m squared c squared times mc  squared upstairs, giving us the first term, which  is p squared upon 2m.  That's the familiar classical kinetic energy.  Rock on.  What's the next correction?  The next correction is the next term  from this Taylor series, which is going  to involve this quantity squared.  So it can be p to the 4th over m to the 4th, c  to the 4th times mc to the 4th.  The coefficient is a minus 1 eighth,  so we get minus p to the 4th over 8m cubed c squared.  And that's down by a factor of m squared c squared  from this guy and a p squared upstairs,  which is what you'd get, p squared m squared over c squared.  So the correction to the kinetic energy for giving momentum  p is to correct it down by p to the 4th term.  Ever cool with that?  Yeah?  Why do you think the substances p  select our red and the p to the small of that?  Because you, like Schrodinger, did problem set 1.  And problem set 1 said, if you have an electron at the Bohr  radius classically, its velocity is about a 400th  of the speed of light.  200th?  Something.  It's small.  It's a couple of order of magnitude  small in the speed of light.  So that number, that ratio there is 1 upon 200 or 400,  I remember, squared.  That's a very small number, so that's  a reasonable approximation.  Why is it the typical?  Because what is the typical?  That's an excellent question.  So why am I talking about it as if the electron is sitting  at that radius?  And the answer is, well, look, what is the average momentum?  What is the average value of p squared  in, let's say, the ground state of the hydrogen atom?  What's the average value of p?  What must be the average value of p in the ground state  of the hydrogen atom?  0 on two grounds.  First off, p is a vector.  And the ground state of hydrogen,  the rotation of the invariant, doesn't break anything.  Second, it's not going anywhere.  It's a bound state, so it's just sitting there.  It's got an average momentum 0.  Curved momentum, it would be cruising.  So it's got average momentum 0.  OK, fine.  But what's the average momentum squared?  What's the average of p squared?  Well, we can actually do out that calculation.  You're going to have to do that calculation on the problem set.  And it's a small number.  And it's, in fact, almost exactly what you'd  guess classically from the Bohr radius.  So you have to know the h bar to compute the Bohr radius  in the first place.  So the answer is to the best of anyone's ability  to define what's the momentum or the typical scale of momentum.  That typical scale of momentum for the electron  in the ground state of hydrogen is extraordinarily small  compared to the speed of light by factors of 100 or 200, 400.  Exactly.  Does that answer your question?  It's a very good question.  So here's the kinetic energy.  And another way to say, by the way,  another way to say that this approximation must be good  is that we already computed ignoring  the relativistic correction.  We computed the ground state energy.  And it was 13.6EV, which is pretty good  since the binding energy of hydrogen is about 13.6EV.  So apparently, that was a good approximation.  And you might be disgusted by answer analysis like that.  But that's what physics is.  You write down some stupid cockamamian model  that you know is wrong.  But if it does a good job of fitting the data,  you declare triumph and you call staco.  It depends on the problem you've just solved, approximately.  So here's the first correction.  So what this tells us is, therefore,  the energy of hydrogen, or at least a better model of hydrogen,  I'll call it hydrogen tilde, is equal to E coulomb minus p  the fourth over 8.  Whoops, that's an 8.  m cubed c to the fourth.  Oh, squared.  Wow, that's a deformed 8.  Let's try it again.  Much better.  OK, so this, however, we know is small compared  to the kinetic energy because 13.6 is pretty close to 13.6.  You'll notice I didn't include the rest of the significant  digits.  So this must be in the insignificant digits.  So this must be small.  So here's what I want to ask.  We know that the energy of the coulomb problem, nlm,  is equal to E naught squared minus E naught squared  over n squared, independent of ln m.  What's the energy in this toy model of hydrogen  that includes the first relativistic correction, nlm?  Well, to answer that question, we  have to resolve the energy eigenvalue equation.  We have to refine the energy eigenstates.  We have to solve that.  And this is now going to be a much worse problem  because it's going to involve a p to the fourth term,  which is going to involve four derivatives.  That sounds horrible.  Can anyone think of a better way to just approximate,  estimate the magnitude and the value of the leading  correction from this quantity?  Yeah?  Some variation, take an eigenvalue, plug it in,  see what the error is, correct for that error, plug it in again.  That's a very, there's an iterative method for doing that.  That's a very good guess.  But there's an even easier way to estimate it.  But that's an excellent, that's something  we will call perturbation theory when we do this in 805.  And doing that systematically is exactly the right answer.  However, I want to do the just leading part of that.  Yeah?  Oh, dimensional analysis would be awesome.  That's fantastic.  So how would you compute p?  OK, so we didn't calculate p.  So you could do a dimensional analysis and get a p.  Because you have an e, you have an l, you can probably do it.  Here's the problem with that, though.  This is the correct first answer to any question of this kind.  Correct first answer is plug in dimensional analysis  and get an estimate for the order of magnitude.  But we want more than the order of magnitude here.  We're interested in splitting the degeneracy  due to this interaction.  So what that means is we want to see that different values of l  lead to different energies.  What we care about is the l dependence  that comes out of this correction.  And we're not going to be able to get the l dependence  from dimensional analysis because l is an integer,  it is dimensionless.  Yeah?  So that will give you the correct magnitude,  and that is the first thing you do.  What's the second thing you do?  Yeah?  We're going to stick in the expectation value of p to the fourth.  Excellent.  Let's just stick in the expectation value of p to the fourth.  That's not exactly the right thing to do.  But if this is small, that's probably pretty reasonable.  And what you'll find when you do it systematically  using perturbation theory, as was pointed out earlier,  we can take the entire system, the entire energy eigenvalue  equation, we can think of this 1 over m cubed c squared.  We can think of that as a small number,  and we can do perturbation theory in that small number.  We can Taylor expand everything in that small number.  And when you Taylor expand the exact equation,  what you discover is the leading term in that Taylor expansion  is the expectation value of p to the fourth over 8 m cubed c  squared.  So this is a fantastic guess.  OK, good.  So let's compute it.  So e hydrogen is going to be equal to e nlm, which I'm just  going to write as minus e nought squared over n squared.  Oh, sorry.  I don't know why I wrote squared.  It's just e nought.  If the energy were equal to the energy squared,  at that point every respectable physicist  goes out and asks themselves, OK, so what  is the correct answer here?  So it's this quantity minus, so this  is making it more tightly balanced.  Notice the sign, because p to the fourth is strictly positive.  p is real.  So minus e nought over m squared.  And if you do this expectation value,  which I really should have put on the problem set, oh,  I still can.  It hasn't been posted yet.  So it'll be posted this afternoon.  But I won't, because I won't.  It's not that bad, actually.  Matt, what do you think?  Through.  We'll discuss this afterwards.  OK, so which may or may not appear on your problem set.  So if you go out and you do this calculation,  what you find is the correction is e n squared.  So e nought squared over n to the fourth.  This time it really is n to the fourth.  And now you have to worry about dimensional analysis.  This is two energies.  We want one energy divided by, wait for it, mc squared,  because it's a relativistic correction.  That shouldn't be too surprising.  Times the quantity 4n over, and here's  where life gets awesome, l plus 1 half minus 3.  See, you laugh, but I feel delighted by this for so many  reasons.  So one is, that's beautiful.  What a crazy combination of symbols.  But the second and more important one is, look,  it has an l in it.  And the l is downstairs.  That's good.  It has more leverage that way.  So we've got this l appearing.  So now the energy depends on l.  Good, we've broken the degeneracy.  Different values of l give different energies.  So let's do that graphically.  So in particular, what this tells us is that now,  while the Coulomb energy commuted  with the Runge-Lenz vector, it must be true  that the energy of this model of hydrogen, which  includes the first relativistic correction,  does not commute with the Runge-Lenz vector.  And indeed, if you check this, which is effortful,  but not intellectually difficult,  indeed, it doesn't vanish.  And the failure of this conservation equation,  the failure of this commutation relation to vanish,  tells us that it's possible for the system to not be degenerate.  And lo and behold, it is not degenerate.  Did you say before that the actual first frame  you count the relativity just includes  substituting the non-relativistic value of 4?  Yes.  No, sorry.  Yeah, I should have said that.  There are other things that are important for relativity  that we'll come to in a little bit.  But if you just look at the kinetic energy,  this is the first non-trivial correction.  Yeah?  We still have the 2L plus 1 degeneracy.  We still have the 2L plus 1 degeneracy, exactly.  Because does this correction violate rotational invariance?  No, that's p squared.  That's a scalar under rotation.  It's just a number, not a vector.  But e still commutes with L. And you  see that because this is independent of little m.  This is the mass of the electron.  So this is independent of the quantum number little m.  It only depends on the quantum number little l.  So indeed, it is still degenerate.  So let's go through this and see the degeneracy explicitly.  And let me do that by drawing you the following diagram.  So what happens if we have the e1, which is the ground state?  It has some energy, which is negative, which is minus e0.  And when we include this first correction, what does it do?  Well, it decreases it by some amount.  Ever recall that?  It doesn't change the number.  It doesn't split any degeneracy because there was only  one ground state in the first place.  Because L had to be 0 when n is 1.  L has to be 1 less than n or smaller.  And if n is 1, L has to be 0, or if a camping is smaller than 0,  it's 0.  There's just the one state with degeneracy 2, 0 plus 1,  which is 1.  So that's fine.  It just changes the ground state by some amount.  Let's consider the second excited level.  So in the second excited level, we have the states.  And let me name this slightly differently.  So this has energy minus e0.  The ground state had L equals 0 and m equals 0.  Whoops.  m equals 0.  It had to.  For the second excited state, n equals 2,  how many states are there?  2 squared is 4.  So there are four states, 2, 3, 4.  And what are their values of L and m?  Well, if n is 2, L can be either 0 or 1.  If it's 0, then m has to be 0.  If it's 1, m could be 1, or it could be 0,  or it could be minus 1.  Yeah?  So this again is L, m, l, m.  So those are the four states with energy e2 in the hydrogen atom.  And now, when we ask what are the energies of these three  states when we include this so-called fine structure,  what is this going to do?  Well, the ground state is going to get pushed down  by some amount.  I should draw this differently to make it more vivid.  This gets pushed down by some amount with some value of L.  L here is 0.  But for these three states, which all  have the same value of little l, 1,  the denominator is larger.  So the amount by which it gets pushed down is less.  So these three states get pushed down less,  but the same amount is each other.  So that cool?  So these three states, the generacy is 3,  and this state degeneracy is 1.  Now notice that 1 plus 3 is equal to 4 is equal to n squared.  So these states split into multiplets, groups of states,  which have to have the same energy by rotational invariance.  If they have L equals 1, there must be three states  with the same energy.  There must be.  And states with L equals 0, there's just the one state.  Is that cool?  And if we did the same thing for E3,  let's just think about what would happen.  So here we have, I'm just going to write the L.  I'm not going to include the m.  We have the L equals 0 state times 1.  We have the L equals 1 state times 3.  And we have the L equals 2 state.  And how many states are there in the L equals 2 state?  Times 5.  Good.  And again, this gets pushed down to some value.  This gets pushed down to some value.  This gets pushed down to some value, where this is 1, 3,  and 5.  And again, these add up to 9, which is 3 squared.  So the degeneracy.  So by turning on the relativistic corrections,  we break this degeneracy.  We lift the degeneracy.  You break the symmetry, you lift the degeneracy.  Cool?  Now, unfortunately, this isn't one  we can control by turning a dial in the lab.  No experiment that I can do in the lab  will change the value of the speed of light.  Or the, well, actually, there are experiments  that will change the value of the speed of light.  If you work in medium, if you work in medium,  the effective speed of light is different.  So that's going to change relativistic effects.  So that's interesting.  That tells you you can actually vary this continuously  if you can vary continuously the index of refraction  of the ambient system.  I invite you to think about how you might do that experimentally,  because it's a fun question.  However, what I'd really like is something  where Professor Evan, in his lab,  can actually turn a dial and change the spectrum in a way  that explicitly breaks the symmetry.  Meanwhile, I also want an example  that breaks the rotational invariance.  I want to break the 2L plus 1 states up into 2L plus 1  states with different energies.  I want to make them depend on the angular momentum  in some direction.  What must I do in order to lift that degeneracy?  Fantastic.  If you add a magnetic field, why is that  going to break the degeneracy?  Yeah?  Yeah?  The difference in angular momentum is sort of  on the different dimensions.  Like different radii.  Give me an answer that depends on, not on your knowledge  of chemistry, but give me an answer  that depends on symmetries.  Why does turning on a magnetic field  have the potential to split the degeneracy of the 2L plus 1  states?  Thank you.  Exactly.  You have no rotational symmetry.  Exactly.  So that's pretty good.  So let's break the 2L plus 1 by breaking rotational symmetry.  And there are many ways we could break rotational symmetry.  We could turn an electric field.  We could break the magnetic field.  We could put a shovel in the room.  But anything that breaks a rotational symmetry will do it.  But a very convenient way is to do with a magnetic field.  Magnetic field is particularly beautiful.  So this effect is now called the Zeeman effect.  I believe it was Faraday.  One of the classic electromagnetists of the 19th century  attempted to look at the spectral lines coming off glowing gas  and see if they changed by turning on an electric  or magnetic field.  It turns out to be a tricky experiment  to do because you need to have a very well-controlled  magnetic field.  Otherwise, what you get is just a bunch of schmutz.  So to get a clean spectrum, you have  to have a very uniform magnetic field, which is not trivial.  It has to be time-independent.  So Zeeman, at the very end of the 19th century, in fact,  this experiment was done, I believe, in 96.  So I think the paper came out in 97.  But the experiment was done in 96.  Zeeman did this experiment.  He said, look, if I turn on a magnetic field  next to a hot glowing gas and look at the spectrum that  comes off, anything happen?  So why would you think that anything, well, why would you  think that anything might happen?  And the answer to this one is really quite nice.  Look, if we compute the energy, the energy  is going to be equal to the coulomb energy  plus a correction, which is the following.  If you think of the atom as an electron carrying  some angular momentum in the presence of the proton,  if you think of the electron as a thing carrying  some angular momentum and with a charge,  if it has a charge and an angular momentum,  it's going to have a magnetic moment, mu.  And the magnetic moment is going to be  equal to the current of the charge  times the area swept out divided by c,  because we work with sensible units.  And this is equal to minus, for the case of the atom,  EL over, man, I really made this too small.  Let me write it here, minus mu, the magnetic moment  of the system dotted into the magnetic field  that you turn on in the background.  So this is a background magnetic field.  I am turning on an electromagnet,  which induces magnetic field in some known direction  with some known magnitude.  So my dial is with the magnitude of that B field.  It could make it larger.  It could make it small.  And it's in a known direction.  And the point is, our electron carries some,  because it's orbiting, because it carries some angular momentum  and it carries some charge, has some magnetic moment.  So it behaves like a little bar magnet.  And that bar magnet wants to anti-align with the background  magnetic field that I've turned on.  So there's an energy penalty to not being aligned.  This is the usual dipole-dipole interaction  from a dipole potential from electrostatics.  And mu here is equal to minus E over 2mc angular momentum.  So this can be computed from a simple model of the atom.  Whatever.  It's easy to compute.  So this is slightly annoying, because B  is in some arbitrary direction.  Let's just simplify our life and say  that it's equal to E coulomb minus mu B.  And I want to turn on a magnetic field of known amplitude  in a known direction.  Let's call it the z direction.  So this is mu z.  But this mu z, that's just mu z is Lz times minus E over 2mc.  So this is going to be plus E over 2mc Bz.  And I'm going to put this all together, Bz Lz.  Everyone cool with that?  And so where this is coming from, again,  is just the angular momentum is telling you the velocity.  The velocity is telling you the current.  And then the area is the radius.  So that's what's giving you the radius and the angular momentum.  OK.  So this is our energy.  What does this tell you about the energy eigenvalues?  Yeah, they depend on M. And do we  have to solve the energy eigenvalue equation again?  No, because our energy eigenfunctions of Ec  are proportional to Ylm.  So they're already eigenfunctions of Lz.  Everyone cool with that?  So when I take this beast and I act on the weight functions  I erased, but on the 1 upon little r capital R sub nl,  Y sub lm.  When I act on the Ylm, this is just  going to give me h bar M. So this is just  a constant when acting on the Coulomb energy eigenfunctions.  So this is a much easier problem than when we don't even  have to make any approximations.  We can just say exactly energy Zeeman is equal to minus E0  upon n squared nlm plus Eh bar upon 2mc Bz,  the backer magnetic field.  And this is the M of electron mass times little m.  And I'm going to write little m sub l  just to distinguish little m sub l from m sub e.  Cool?  So when we turn on the prediction here,  is when we turn on the magnetic field,  all Zeeman, our energy levels will get split  according to the different values of M.  Now if instead of working with the pure Coulomb system,  if we were in fact a little clever  and we had already noticed that there was a fine structure,  because they already knew that there was a fine structure  at this point, they could see the different spectra in lines  coming off of hot gas.  So it's just an experimental fact.  You can't turn off the fine structure.  This is going to be equal to E nlm of our correction  to the hydrogen atom with the fine structure corrected.  So the l equals 1 states and the l equals 0 states are split.  Let's focus, for example, on E2.  So these guys have a different energy.  The l equals 1 guys have a different energy  than the l equals 0 guys.  What does this predict is going to happen  if I look at my spectral line?  So what this predicts is the following.  It says that look, if you have some spectrum that  looks like this, so this is a function of wavelength.  This is a function of wavelength.  And I'm looking literally at what comes off the prism.  I'm looking at spectral lines.  And I'll have a line here.  And these are going to be the l equals 0 states.  This is n equals 2.  l equals 0 states.  And then there's the l equals 1 states.  This is with 0 magnetic field.  B is equal to Bz equals 0.  If we turn on a magnetic field, Bz not equal to 0,  what should we see?  Well, what happens to this state?  Nothing.  It's the same thing because the m is strictly 0.  So it's the same value.  And this guy, what happens to it?  The m equals 0 state is the same.  But then it breaks up, as everyone said,  it breaks up into three lines that are equally spaced.  Yeah?  It breaks up into three lines because one of them  has m which is plus 1.  One of them has m which is 0, and one is m is minus 1.  So the energies will increase or decrease.  Let's say this is frequency.  So we get this splitting.  And another way to draw this is, as a function of B,  let's draw the energy levels.  So there's the 1.  There's the 0.  And this stays constant.  And these guys, 1 is constant, 1 increases, and 1 decreases.  Everyone cool with that?  When you start getting to strong magnetic fields,  weird things happen, like these guys cross,  and all sorts of subtle things can happen.  But let's ignore the strong magnetic field case  and just focus on this.  So we see that these guys split.  So this is what the Schrodinger theory, which came in 19,  or the full story would predict.  It came a little bit after this.  But this experiment was done in 96.  So there's actually kind of an entertaining story  about people trying to explain this effect classically.  And you say, well, look, could have some angular momentum,  z direction, some angular momentum,  and the x direction, some angular momentum, the y direction.  The part that has angular momentum, the x direction,  doesn't do anything.  And then the z direction is basically part with the x  and part with y.  You can have different polars.  There's a whole classical song and dance.  It's crazy.  It doesn't make any sense.  But that's because they were trying  to do classical mechanics to do quantum mechanics, which  doesn't work.  So I'm not even going to bother explaining  what they were thinking.  But anyway, so people came up with a sort of classical crutch  to say, well, that's not totally insane.  So this is what Zeeman got, for example,  by looking at hydrogen.  But he did more than just look at hydrogen.  I can't remember exactly.  I think he did it with sodium.  Matt, do you remember?  Sodium?  I think he did sodium and iron.  But anyway.  But here's the funny thing that he got with sodium and iron.  So this is a different experiment.  Same, exactly the same setup, but he  was looking at spectral lines in sodium.  And the spectral lines in sodium did a very similar thing.  There were some where one went to one.  And there were others where one went to,  why did we have three?  Here, just as a quick reminder, why  did we have three states here?  Because we had three states for l equals 1.  This one state went to 1, 2, 3, 4 states  with none of them in the middle.  And in fact, it's worse than that,  because they didn't go to equally spaced states in general.  In general, they did something slightly funny,  and the beautiful plot in his paper looks like this.  There are four states.  And as you take the magnetic field to 0,  these four states all coalesce exactly to this one state.  So this one line is representing four different states.  And this already included the fine structure.  So we know that this is a single value of l.  But that apparently has four states.  That should worry you.  What does that tell you about l?  What does l have to be in order to have four states in your tower?  Half integer, in particular, what value?  Three halves.  Crap.  We already know that these wave functions are  described by the YLMs.  We solved that problem.  We explicitly solved it.  Here's the exact answer.  It's the YLMs.  But we've observed experimentally that l has me 3 halves.  But if l has 3 halves, YLM is equal to 0,  because it's equal to minus itself.  This is bad.  This is 0.  So this effect is called the Zeeman effect.  And this effect is called the anomalous Zeeman effect, which  is strange for two reasons.  The first reason it's strange is they showed up in the same paper.  And the second thing is, OK, you call it the Zeeman effect,  because there's some guy named Zeeman.  Is there some guy named anomalous Zeeman?  It's a very strange name.  Anyway, so this was called the anomalous Zeeman effect,  despite appearing in the same paper, because it's weird.  It was deeply disconcerting to people.  We now just call it the Zeeman effect,  although we have the bad habit for entertainment value  of referring to it still as the Zeeman effect.  I'm not the only one responsible for that bad habit.  I'm going to point out that at the end of this paper,  it's totally awesome paper, by the way.  It's very readable.  It's short.  He says, I was looking at it last night, Zeeman says,  quote, possibly this observed, the observed phenomenon,  will be regarded as nothing of any consequence.  OK, so a few years later, Polly says the following.  So Polly says this actually in a also,  this is also totally lovely in science from 1946.  Polly was in the US during the war.  After the war, he was at the Institute for Advanced Study.  At the Institute for Advanced Study, while he was there,  he got the Nobel Prize.  And he gave a little for the exclusion principle, which  we're about to get to.  And he gave a little spiel, and it's  written up in this edition of Science.  And he says, quote, a colleague who  met me strolling rather aimlessly  in the beautiful streets of Copenhagen  said to me in a friendly manner,  you look very unhappy to whereupon I responded fiercely,  how can one look happy when he is thinking about the novel  of Zeeman Effect?  OK.  So this troubled people.  Yeah?  Is there a scale of people who speak of a very  non-sparely, or one-house, or any of those?  Yeah, he found all sorts of crazy states.  So this isn't the only example.  It's just the only one I'm going to draw.  But indeed, you can find five half states.  You can find states with six.  Why is it the airy quote one and airy quote one  half of the messier?  Oh, I didn't draw all the states.  The spectrum is complicated and messy, because it's sodium.  There's all sorts of crap.  So I'm just not drawing.  I'm just focusing on a specific set of spectral lines.  Maybe the best way to draw that is.  OK.  So this was really troubling.  But let's put this on pause for the moment.  And we'll come back to it in just a second.  So meanwhile, there's something else  that's sort of annoying about this system.  We've got the Coulomb potential.  And the Coulomb potential has the following beautiful property.  The degeneracy of energy energy state is equal to n squared.  Meanwhile, various people were fond of observing the fact  that the number of states, which I will in an abuse of language  called the degeneracy of the level n in the periodic table,  is equal to, well, the first one there's 2.  And then there's 8, and then there's 18.  So 2, 8, 8, 18, that's almost 2 n squared.  2, 4, 8, 9, 18.  Cool.  Maybe there's a relationship.  And meanwhile, it's very tempting to think  that there's a relationship, because imagine you  have four electrons.  If you have four electrons, what do you do?  Well, you put an electron here, and then  you could put an electron here, or you could put an electron  and another electron here.  You can also put another electron here and here,  or you could put them in any combination.  2 here, you could put 2 here, 2 here, 1 1.  You do all sorts of stuff.  But if for the moment, you imagine  that electrons have this funny property.  Imagine for the moment that electrons have the property that  can't be in the same state at once.  Then if you have four electrons, I don't know why.  I'm desperate.  I'm thinking about the anomalous aim on effects,  among other things.  I have nothing better to do.  So just for fun, let's imagine if I put an electron here,  I can't put another one.  So if I have a 2 electron atom, the second one  goes in one of these, a 3 electron atom, a 4 electron atom.  And these guys are all related to each other,  because they live in the same energy level,  but they have different L values.  So they'll have slightly different properties.  That suggests that the 1L equals 0 state  should be different from the 3L equals 1 state chemically.  But wait, there aren't groupings of 2 and 1 and 3.  They're groupings of 2 and 6.  Ah, 2.  Well, so Pauli, staring at this, in 25, said the following.  He said, look, I'm going to conjecture two things.  The first is that no two electrons  can live in the same state in habit, the same quantum state.  That doesn't give us the periodic table, though.  It gives us half the periodic table.  There are twice as many states as you think in hydrogen  as you think.  And you might think that I'm being facetious,  that he said something a little more sophisticated,  but he really didn't.  What he said is, I posit the existence  of an additional quantum number, which the electron can have,  which takes one of two values, which translated says,  there are twice as many states as you think.  So that's exactly what he said.  So this is called the Pauli exclusion principle.  And now this does a totally spectacular thing.  And you've probably all seen this in high school chemistry  or even college chemistry.  Now, first electron goes here.  Second electron goes here because there are twice as many states  as you think.  But now you can't put another state in there  because there's as many as you can have.  One can go here and you can put up to six in here.  And that's it.  So the E2 states, the n squared, which is four times two states,  are somehow related to each other naturally.  And this gives you the structure of the periodic table.  And in fact, an awful lot of chemistry  falls directly from this.  So this is a Pauli exclusion principle.  However, it's a little bit disappointing  because while this is whatever some ridiculous rule,  but we're doing quantum mechanics and all  they're all ridiculous rules.  On the other hand, this one is just stupid.  This is just like, look, it wouldn't be nice.  So in 25, a couple of guys, really interesting characters  named Goschmitt and Ulinbeck.  So Ulinbeck, wow, we're really low on time.  OK, so I'll just tell you this and then  I'll get to the last of it later.  So Goschmitt and Ulinbeck, who were young, kind of naive,  they said, look, we've got two effects here.  One effect is the anomalous Zeeman effect.  And there's this weird fact that states  have the wrong total angular momentum.  And we have the second thing that there are twice as many states  for an electron as you think in the Coulomb potential.  How can these two things fit together naturally?  Well, here's a guess.  Suppose that the angular momentum that we calculated  when we did the estimate of the breaking due to the Zeeman  effect.  Suppose that angular momentum was not  the right angular momentum.  Maybe there's more angular momentum in the system.  There's angular momentum in the system  from the electron orbiting around.  But maybe like a little tiny Earth,  the electron itself can have some intrinsic angular momentum.  It turns out Pauli had had this idea before.  In fact, Cramer had suggested this to him,  who was a very young guy at the time.  And Pauli said, you're a blithering idiot.  Because if you calculate how small an electron has  to be in order to fit all the other things we know,  and you figure out how fast it has to rotate to explain  the anomalous Zeeman effect, surface  would have moving faster than the speed of light.  That's ridiculous.  Leave my office.  So then Ullinbeck and Goldschmidt,  I don't know how to put it, Goldschmidt, G-O-U-D.  Anyway, these two guys write a paper and say, aha, well,  we can explain this.  But let's just not assume that the super thing is rotating.  Let's just say an electron has some intrinsic angular  momentum.  I don't know.  Why not?  We'll have some electrons have intrinsic angular momentum.  And if you assume that electrons have  an intrinsic angular momentum, which I'll call s for the moment.  So then the total angular momentum, j, is equal to l.  It's orbital angular momentum.  Plus some intrinsic angular momentum,  which I don't remember what symbol they used,  we'll call it for the moment s, where this has the property  that s squared for an electron is equal to, sorry,  the principal quantum number, which I will call little s,  is equal to 1 half.  So m sub s is equal to plus or minus 1 half.  So this is like the l equals 1 half state, but it's not l.  It's something intrinsic.  Has nothing to do with anything rotating.  It's just a fact about electrons.  Suppose electrons have a little bit of spin.  Then what you discover is, if you have the l equals 1 state  and the electron has spin 1 half, what's  the total angular momentum?  3 halves.  So what do you get?  You get quadruplets like that.  So this turns out to explain, if you include  a small relativeistic effect, this  turns out to explain the anomalous Zaman effect bang on.  Meanwhile, there was an experiment done in 1922,  which was a Stern-Gerlach experiment, in which people  discovered that Nick, Stern-Gerlach, not people,  discovered that nickel, when sent through a magnetic field  gradient, bent into one of two different spots, never 3,  never 0, always 2.  How could that be?  Electron spin.  That wasn't realized until 1929, that the connection  was there.  So these guys came up with this ridiculous postulate.  This was Ullinbeck and Goeschmet.  And let me just quickly, U-H-L-E-N-B-E-C-K.  So Ullinbeck, an amazing, amazing scientist,  also the father of the mathematician Ullinbeck.  And she was a total badass and has inspired  an awful lot of physics.  Karen Ullinbeck, who's at UT.  So this is a pretty interesting and prolific eye.  But he was also prolific in the following way.  He ran a program at the Rad Lab at MIT during World War II.  And Goeschmet worked with him.  And when they were finishing up, when the war was ending,  Goeschmet became the scientific advisor  to something called Project Alsos.  Project Alsos was the project where the military went  to the conquered territories in Germany  to catch the German atomic scientists  and bring them to a place called Farm Hall in England,  where they listened to them, eavesdropped on them,  Heisenberg, all the good guys.  Well, it depends on all the great physicists  in the German territory at the time  who were deeply complicated people.  And they listened to them in something  called the Farm Hall transcripts, or the transcripts  of those recordings.  They were written up in a book called the Epsilon Project,  which is totally breathtakingly awesome.  And Goeschmet wrote a book called  Alsos about this process of hunting down the German  scientists.  So Goeschmet, U-D-S-M-I-T, I think.  And he wrote a book called Alsos,  which I heartily recommend to you,  because it's like a combination adventure story  and beautiful bit of physics history.  So these guys both had it.  And they were both at MIT during the World War.  There's a nice connection here.  So these guys, fascinating characters,  and they came up with this idea of some intrinsic angular  momentum.  Polly then calls it Spin.  He gives it the name.  And he develops a mathematical theory of spin.  And that mathematical theory of spin  will lead to quantum field theory relativistic quantum  mechanics.  And it will eventually lead to quantum computation, which  is going to be the topic of the last week of our course.  So what we've done so far is we've  explained the discreteness of atomic spectra.  And we've explained the structure of the periodic table.  What we haven't done is we haven't  explained why atoms form molecules or solids.  And we also haven't explained what spin is at all.  Those are going to be the topics of the next three weeks.  See you next time.  Thank you.
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