OK, so we're going to do this thing of the hydrogen atom and the algebraic solution. And I think it's not that long stuff, so we can take it easy as we go along.

I want to remind you of a couple of facts that will play a role. One result that is very general about the addition of angular momentum that you should, of course, know is that if you have a J1 tensor J2, what does this mean?

You have some states of a first angular momentum J1. So you have a whole multiplet with J1

equal little j1, which means the states in that multiplet have J1 squared, giving you h squared little j1 times little j1 plus 1. That's having a J1 multiplet. You have a J2 multiplet.

And these are two independent commuting angular momentums acting on different degrees of freedom of the same particle or different particles. And what you've learned is that this can be written as a sum of representations,

as a direct sum of angular momenta, which goes from J1 plus J2 plus J1 plus J2 minus 1, all the way up to the representation

with J1 minus J2. And these are all representations or multiplets that live in the tensor product, but are multiplets of J equal J1 plus J2.

These states here can be reorganized into these multiplets. And that's our main result for the addition

of angular momentum. Mathematically, this formula summarizes it all. These states, when you write them as a basis here, you take a basis state here times a basis state here. These are called the uncoupled basis.

And then you reorganize. You form linear combinations that you've been playing with. And then they get reorganized into these states. And these are called the coupled basis, in which we're talking about states of the sum of angular momentum. So that's one fact we've learned about. Now, as far as hydrogen is concerned,

we're going to try today to understand the spectrum. And for that, let me remind you what the spectrum was. The way we organized it was with an l versus energy levels. And we would put an l equals 0 state here.

Well, maybe there's color. So why not using color? Let's see if it works. Yeah, it's OK. l equals 0.

And this was called n equals 1. There's an n equals 2 that has an l equals 0 state and an l equals 1 state. There's an n equals 3 set of states

that come with an l equals 0, an l equals 1, and an l equals 2. And it just so goes on and on with the energy levels En equal minus e squared over 2a0.

That combination is familiar for energy, Bohr radius charge of the electron with a 1 over n squared. And the fact is that for any level for each n,

l goes from 0, 1, 2, up to n minus 1. And for each n, there's a total of n squared states.

And you see it here. You have n equal 1, 1 state. n equal 2, you have l equals 0 is 1 state. l equals 1 is 3 state, so it's 4. Here we'll have 4 plus 5, so 9.

And you can do it. It's a famous thing. There's n squared states at every level. So this pattern that, of course, continues and it's a little difficult to do a nice diagram of the hydrogen atom in scale because it's all pushed towards the 0 energy with 1

over n squared. But that's how it goes. For n equal 4, you have 1, 2, 3, 4, for example. And this is what we want to understand.

So in order to do that, let's return to this Hamiltonian, which is p squared over 2m minus e squared over r, and to the Runge-Lenz vector

that we talked about in lecture and you've been playing with. So this Runge-Lenz vector r is defined to be 1 over 2m e squared p cross l minus l cross

p minus r over r. And it has no units. It's a vector that you've learned

has interpretation of a constant vector that points in this direction, r. And it just stays fixed. Wherever the particle is going, classically, this is a constant vector that points in the direction of the major axis of the ellipse.

With respect to this vector, this vector is Hermitian. And you may recall that when we did the classical vector, you had just p cross l and no 2 here. There are now two terms here, and they

are necessary because we want to have a Hermitian operator. And this is the simplest way to construct the Hermitian operator, r. And the way is that you add to this this term that if l and p commuted, as they do in classical mechanics,

that that term is identical to this. And you get back to the conventional thing that you have in classical mechanics. But in quantum mechanics, of course, they don't commute. So it's a little bit different. And moreover, this thing, r, is Hermitian. l and p are Hermitian.

But when you take the Hermitian conjugate, l goes to the other side of p. And since they don't commute, that's not the same thing. So actually, the Hermitian conjugate of this term is this. There's an extra minus sign in Hermiticity when you have a cross product.

So this, the Hermitian conjugate of this is this. The Hermitian conjugate of this second term is the first. And therefore, this is actually a Hermitian operator. And you can work with it. Moreover, in the case of classical mechanics, it was conserved.

In the case of quantum mechanics, the statement of conservation, quantum mechanics, is something that, in one of the exercises, you were asked to try to do these computations. And these computations are challenging.

They're not all that trivial and are good exercises. So this is one of them. This is practice. So OK, this is the vector r.

What about it? A few more things about it that are interesting. Because of the Hermiticity condition, or in a way, you can check this directly. In fact, was one of the exercises for you to do. Was p cross l.

You did it a long time ago, I think. Is equal to minus l cross p plus 2i h bar p. This is an identity.

And this identity helps you write this kind of term in a way in which you have just one order of products and a little extra term, rather than having two complicated terms.

So the r can be written as 1 over m e squared alone p cross l minus i h p minus r over r.

For example, by writing this term as another p cross l minus that thing gives you that expression for r.

You have an alternative expression in which you solve for the other one. So it's 1 over m e squared minus l cross p plus i h bar p.

Now, r, we need to understand r better. That's really the challenge of this whole derivation. So we have one thing that is conserved. Angular momentum is conserved. It commutes with the Hamiltonian.

We have another thing that is conserved, this r. But we have to understand better what it is. So one thing that you can ask is, well, r is conserved. So r squared is conserved as well. So r squared, if I can simplify it,

if I can do the algebra and simplify it, it should not be that complicated. So again, a practice problem was given to do that computation. And I think these forms are useful for that to work less. And the computation gives a very nice result.

r squared is equal to 1 plus 2 h over m e to the fourth l squared plus h bar squared. Kind of a strange result, if you think about it. People that want to do this classically first

would find that there's no h squared. And here, this h is that whole h that we have here. It's a complicated thing. So this right-hand side is quite substantial. You don't have to worry that h is in this side

whether it's on the other side, because h commutes with l. l is conserved. So h appears like that. And this, again, is the result of another computation.

So we've learned something. Oh, I'm sorry. This is r squared. Apologies. r is conserved. r squared must be conserved, because if h commutes with r, it commutes with r squared as well. And therefore, whatever you see on the right-hand side,

the whole thing must be conserved. And h is conserved, of course. And l squared is conserved. Now, we need one more property of relation. You have to do these things. Even if you probably don't have an inspiration at this

moment, how you're going to try to understand this. There are things that just curiosity should tell you you should do. We have l. We do l squared. It's an important operator. OK, we had r. We did r squared, which is an important operator. But one thing we can do is l dot r.

It's a good question what l dot r is. OK, so what is l dot r or r dot l?

What is it? Well, a few things that are important to note are that you did show before that you know that r dot l,

little r dot l, is 0, and little p dot l is 0. These are obvious classically, because l is perpendicular to both r and p.

But quantum mechanically, they take a little more work. They're not complicated. But you've shown those two. So if you have r dot l, you would have, for example, here r dot l.

You would have to do and think of this whole r and put an l on the right. Well, this little r dotted with the l on the right would be 0. Well, that p dotted with l on the right would be 0.

And we're almost there. But p cross l dot l, well, what is that? Let me talk about it here, p cross l dot l.

So this is part of the computation of this r dot l. We've already seen this term will give nothing. This term will give nothing. But this term could give something. So when you face something like that, maybe you say, well, I don't know any identities I should be using here.

So you just do it. Then you say, this is the i-th component of this vector times the i-th of that. So it's epsilon ijk pj lk li.

And then you say, look, this looks a little. You could say many things that are wrong and get the right answer. So you could say, oh, ki symmetric

and ki anti-symmetric. But that's wrong because this k and i are not symmetric, really, because these operators don't commute. So the answer will be 0, but for a more complicated reason. So what do you have in here, ki?

Let's move the i to the end of the epsilon. So jki pj lk li. And now you see this part is pj l cross lj

is the cross product of this. But what is l cross l? You probably remember this is i h bar l.

l cross l, that's the commutation relation of angular momentum, in case you kind of don't remember, was i h bar l, like that. So now p dot l is anyway 0, so this is 0.

So it's a little delicate to do these computations. So since that term is 0, this thing is 0.

Now you may ask, well, r dot l is 0. Is l dot r also 0? Quantum mechanics is not all that obvious. You can even do that. Well, in a second, we'll see that that's true as well. l dot r and r dot l, capital R, are 0.

OK, so let's remember. Let's continue. Let's see, I wanted to number some of these equations.

We're going to need them. So this will be equation one. This will be equation two. It's an important one.

Now, let me remind you of a notation we also had about vectors and their rotations. Vector under rotations.

So what was a vector vi under rotations? Was something that you had li vj was i h bar epsilon ijk vk.

So there's a way to write this with cross products that is useful in some cases. So I will do it. You probably have seen that in the notes,

but let me remind you. Consider this product, l cross v plus v cross l, and the i-th component of it. i-th component of this product. So this is epsilon ijk.

And you have lj vk plus vj lk. Now, in this term, you can do something nice.

If you think of it like expand it out, you have the second term has epsilon ijk vjk change j for k. If you change j for k, this would be vk lj. And this would have the opposite order.

But this order can be changed up to a cost of a minus sign. So I claim this is ijk. First term is the same, minus vk lj. So in the second term, for this term alone,

we've done, for this term multiplied with this, of course, we've done j and k relabeling. But this is nothing else than the commutator of l with v.

So this is epsilon ijk lj vk. That's epsilon ijk. And this is epsilon jk p, or l, vl.

Now, two epsilons with two common indices is something that is simple.

It's a Kronecker delta on the other indices. Now, it's better if they're sort of all aligned in the same way. But they kind of are, because this l, without paying a price, can be put as the first index. So you have jk as the second and third, and jk as the second and third, once l has

been moved to the first position. So this thing is 2 times delta il. And there's an h bar, ih bar, I forgot here, ih bar. 2 delta ik ih bar il ih bar vl.

So this is 2ih bar vi. So this whole thing, the i-th component of this thing using this commutation relation is this. So what we've learned is that l cross v plus v cross l

is equal to 2ih bar v. And that's a statement as a vector relation of the fact

that v is a vector and a rotation. So for v to be a vector and the rotations means this. And if you wish, it means this thing as well. It's just another thing of what it means.

Now, r is a vector and the rotations, this capital R. Why? You've shown that if you have a vector and the rotations, and you multiply it by another vector and the rotations under the cross product, it's still

a vector and the rotations. So this is a vector and the rotations, this is, and this is a vector and the rotations. r is a vector and the rotations. So this capital R is a vector and the rotations,

which means two things. It means that it satisfies this kind of equation. So l cross r plus r cross l is equal to ih bar r.

So r is a vector and the rotation is a fact beyond

doubt. And that means that we now know the commutation relations between l and r. So we're starting to put together this picture in which we get familiar with r and all the commutators that are possible.

So I can summarize it here. l dot r li rj is ih bar epsilon ij k rk.

That's the same statement as this one, but in components. And now you see why r dot l is equal to l dot r.

Because actually, if you put the same two indices here, i and i, you get 0. So when you have r dot l, you have r1l1 plus r2l2 plus r3l3. And each of these two commute when the two indices

are the same because of the epsilon. So r dot l is 0, and now you also appreciate that l dot r is also 0, too.

OK. Now comes, in a sense, the most difficult of all calculations, even if this seemed a little easy. But you can get quite far with it. So what do you do with l's? You computed l commutators, and you

got the algebra of angular momentum over here. This is the algebra of angular momentum. And this is kind of a nontrivial calculation. You did it by building results. You knew how r was a vector in the rotation,

how p was a vector in the rotation. You multiply the two of them, and it was not so difficult. But the calculation that you really need to do now is the calculation of the commutator, say, of ri with rj.

And that looks like a little bit of a nightmare. You have to commute this whole thing with itself. Lots of p's, l's, r's, 1 over r's. Those don't commute with p, you remember. So this calculation done by brute force,

you're talking a day, probably. I think so. And probably it becomes a mess, but you find a little trick how to organize it better. It's less of a mess, but still you don't get it. And try several times. So what we're going to do is try to think

of what the answer could be by some arguments. And then once we know what the answer can be, there's still one calculation to be done that I will probably put in the notes, but it's not a difficult one. And the answer just pops out.

So the question is, what is r cross r? r cross r is really what we have when we have this commutator. So we need to know what r cross r is, just like l cross l. Now r is not likely to be an angular momentum.

It's a vector, but it's not an angular momentum. It has nothing to do with it. It's more complicated. So what is r cross r, quantum mechanics? Classically, of course, would be 0.

So first thing is you think of what this should be. Be a vector, because it's a cross product of two vectors. Now I want to emphasize one other thing,

that it should be this thing, r cross r, is tantamount to this thing. What is this thing? It should be actually proportional to some conserved quantity.

And the reason is quite interesting. So this is a small aside here. If some operator is conserved, it commutes with the Hamiltonian. Say if s1 and s2 are symmetries,

that means that s1 with h is equal to s2 with h is equal to 0. Then the claim is that the commutator of this s1 and s2

claim s1 commutator with s2 is also a symmetry. So the reason is because commutator of s1, s2

commutator with h is equal actually to 0. And why would it be equal to 0? It's because of the so-called Jacobi identity for commutators. You remember, when you have three things like that,

this term is equal to 1. This term, plus 1 in which you cycle them, and plus another one where you cycle them again, is equal to 0. That's a Jacobi identity. And in those cyclings, you get an h with s2, for example,

that is 0, and then an h with s1, which is 0. So you use these things here, and you prove that. So I write here by Jacobi. So this is the great thing about conserved quantities.

If you have one conserved quantity, it's OK. But if you have two, you're in business. Because you can then take the commutator of this 2, and you get another conserved quantity, and then more commutators. And you keep taking commutators, and if you're lucky, you get all of the conserved quantities. So here, r cross r refers to this commutator.

So whatever is on the right should be a vector and should be conserved. And what are our conserved vectors?

Well, our conserved vectors, candidates here, r, l, r itself, and l cross r, that's pretty much it.

l and r are our only conserved things, so it better be that. Still, this is far too much. So there could be a term proportional to l,

a term proportional to r, a term proportional to l dot r. So this kind of analysis is based by something that Julian Schwinger did, this same guy that actually did quantum electrodynamics along with Feynman and Tomonaga.

And he's the one that also invented the trick of using three-dimensional angular momentum for the two-dimensional oscillator and had lots of bags of tricks. So actually, this whole discussion of the hydrogen atom, most books just say, well, these calculations are hopeless.

Let me give you the answers. Schwinger, on the other hand, in his book on quantum mechanics, which is kind of interesting but very idiosyncratic, finds a trick to do every calculation. So you never get into a big mess.

It's absolutely elegant, and it keeps pulling tricks from the bag. So this is one of those tricks. So basically, he goes through the following analysis now and says, look, suppose I have the vector r, and I do a parity transformation.

I change it for minus r. What happens under those circumstances? Well, the momentum is the rate of change of r, should also change sign. Quantum mechanically, this is consistent

because the commutation between r and p should give you h bar. And if r changes, p should change sign. But now, when you do this, l, which is r cross p, just goes to l.

And r, however, changes sign because l doesn't change sign, but p does and r does. So under this changes, so this is the originator, the troublemaker, and then everybody else follows.

r also changes sign. So this is extremely powerful because if you imagine this being equal to something, well, it should be consistent with the symmetries.

So as I change r to minus r, capital R changes sign, but the left-hand side doesn't change sign. Therefore, the right-hand side should not change sign. And r changes sign, and l cross r changes sign.

So computation kind of finished because the only thing you can get on the right is l. So this is the kind of thing that you do. And probably, if you were writing a paper on that, you would anyway do the calculation the silly way,

the direct way. But this is quite a save of time. So actually, what you have learned is that r cross r is equal to some scalar conserved

quantity, which is something that is conserved. It could be like an h, for example, here, but it's a scalar, and l. Well, once you know that much, it doesn't take much work to do this

and to calculate what it is. But I will skip that calculation. This is the sort of thoughtful part of it. And r cross r turns out to be ih bar minus 2h again. h shows up in several places, like here.

So it has a tendency to show up. m e to the fourth l. So this is our equation four.

And in a sense, all the hard work has been done, because now you have a complete understanding of these two vectors, l and r. You know what l squared is, what r squared, what l dot r is. And you know all the commutators.

You know the commutators of l with l, l with r, and r with r. You've done all the algebraic work. And the question is, how do we proceed from now to solve the hydrogen atom? So the way we proceed is kind of interesting.

We're going to try to build from this l that is an angular momentum and this r that is not an angular momentum, two sets of angular momenta. You have two vectors.

So somehow we're going to try to combine them in such a way that we can invent two angular momenta. Just like the angular momentum in the two-dimensional harmonic oscillator that was not directly true angular momentum but was mathematical angular momentum,

these two angular momenta we're going to build, one of them is going to be recognizable. The other one is going to be a little unfamiliar. But now I have to do something that may sound a little unusual, but it's

necessary to simplify our life. I want to say some words that will allow me to think of this h here as a number. And will allow me to think of this h as a number. So here's what we're going to say.

It's an assumption. It's no assumption. It sounds like an assumption, but it's no assumption whatsoever. We say the following. This hydrogen atom is going to have some states. So let's assume there is one state.

And it has some energy. If I have that state with some energy, well, that would be the end of the story. But in fact, the thing that I want to allow the possibility for is that at that energy, there are more states.

One state would be OK. Maybe sometimes it happens. But in general, there are more states at that energy. So I'm not making any physical assumption to state that there is a subspace of degenerate states. And in that subspace of degenerate states,

there may be just one state or two states or three states, but there are subspace of degenerate states that have some energy. And I'm going to work in that subspace. And all the operators that I have are going to be acting in that subspace.

And I'm going to analyze subspace by subspace of different energies. So we're going to work with one subspace of degenerate energies. And if I have, for example, the operator r squared acting on any state of that subspace, since h commutes with l squared, h can go here,

acts on the state, becomes a number. So you might as well put a number here. You might as well put a number here as well. So it has to be stated like that carefully. We're going to work in a degenerate subspace of some energy, but then we can treat the h as a number.

So let me say it here. We'll work in a degenerate subspace with eigenvalues of h equal to h prime, where h prime.

Now I want to write some numbers here to simplify my algebra. So without loss of generality, we put what is dimensionless.

This is dimensionless. I'm sorry, this is not dimensionless. This has units of energy. This is roughly the right energy, with this one would be the right energy

for the ground state. Now we don't know the energies, and this is going to give us the energies as well. So without solving the differential equation, we're going to get the energies. So if I say, well, that's the energies, you would say, come on, you're cheating. So I'll put 1 over nu squared, where nu can be anything.

Nu is real. And that's just a way to write things in order to simplify the algebra. I don't know what nu is. Yeah, you say, you don't know, but you have this in mind. It's going to be an integer. Sure.

That's what good notation is all about. You write things, and then it's nu. You don't call it n, because you don't know it's an integer. You call it nu, and now you proceed. So once you've called it nu, you see here that, well, that's what we call h, really.

This h prime is kind of not necessary. This is where h becomes in every formula. So from here, you get that minus 2h over me to the fourth is 1 over h squared nu squared.

I have a minus here, I'm sorry. 2h, the minus, me to the fourth down is h squared nu squared. So we can substitute that in our nice formulas, the hme to the fourth.

So our formulas 4 and 5 have become, I'm going to use this blackboard. Any blackboard where I don't have a formula boxed can be erased, so I will continue here.

And so what do we have? r cross r from that formula is, well, this thing is over there, minus 2h over me to the fourth.

You substitute it in here. So it's i over h bar 1 over nu squared l. Doesn't look that bad. And r squared is equal to 1 minus 1 over h bar nu squared

like this, l squared plus h squared. 2h, that's minus h squared nu squared. So these are nice formulas.

These are already quite clean. We'll call them five, equation five. I still want to rewrite them in a way that perhaps is a little more understandable or suggestive.

I will put an h bar nu together with each r. So h nu r cross h nu r is equal to i h bar l. Makes it look nice.

Then for this one, you'll put h squared nu squared r squared is equal to h squared nu squared minus 1 minus l squared. It's sort of trivial algebra.

You multiply by h squared nu squared, you get this. You get h squared nu squared minus l squared, because it's all multiplied, minus h squared. So these two equations, five, have become six.

So five and six are really the same equations. Nothing much has been done. And if you wish, in terms of commutators, this equation says that the commutator h nu ri with h nu

rj is equal to i h bar epsilon ijk lk. h cross this thing, h nu r, h nu r cross, equal ihl in components means this.

That is not totally obvious. It requires a small computation. Remember, it's the same computation that shows that this thing is really l i l j equal i h bar epsilon ijk lk, in which these l's have now become r's.

So we've cleaned up everything. We've made great progress, even though at this moment

it still looks like we haven't solved the problem at all. But we're very close. So are there any questions about what we've done so far? Have I lost you in the algebra or any goals here? Yes. AUDIENCE MEMBER 3 R cross R map definition.

Why should we expect that mapping definition? PROFESSOR SINGH NGUYEN In general, it's the same thing as here. l cross l is this. The commutator of two Hermitian operators is anti-Hermitian. So there's always an i over there.

Other questions? It's good. You should worry about those things. The units right, or the right number of i's on the right-hand side, that's a great way to catch mistakes. OK.

So we're there. And now it should really almost look reasonable to do what we're going to do. h nu r with h nu r gives you like l. So you have l with l form angular momentum.

l and r are vectors under angular momentum. Now r cross r is l. And with these units, h nu r and h nu r looks like it has the units of angular momentum. So h nu r can be added to angular momentum

to form more angular momentum. So that's exactly what we're going to do. So here it comes. Key step, J1. I'm going to define two angular momenta.

Well, we hope they are angular momenta. l plus h nu r. And J2, 1 half l minus h nu r. These are definitions.

It's just defining two operators. We hope something good happens with these operators. But at this moment, you don't know. It's a good suggestion because of the units match and all that stuff. So this is going to be our definition, 7.

And from this, of course, follows that l, the quantity we know, is J1 plus J2. And r, or h nu r, is J1 minus J2.

You solve them the other way. Now, my first claim is that J1 and J2 commute. Commute with each other.

So these are nice commuting angular momenta. Now, this computation has to be done. Let me, yeah, we can do it.

J1i with J2j is 1 half and 1 half gives you 1 quarter of li plus h nu ri with lj minus h nu rj.

Now, the question is, where do I? I think I can erase most of this blackboard.

I can leave this formula. It's kind of the only very much needed one. So I'll continue with this computation here. This gives me 1 quarter.

And we have a big parenthesis. i h bar epsilon ijk lk for the commutator of these two. And then you have the commutator of the cross terms.

So what do they look like? They look like minus h nu li with rj and minus h nu ri

with, no, yeah. So I have minus h nu li with rj. And now I have a plus of this term, but I will write this as a minus h nu of lj with ri.

Those are the two cross products. And then finally, we have this thing, the h nu with h nu ri jk. So I have minus h nu squared.

And you have then ri rj. No, I'll do it this way. I'm sorry.

You have minus over there. And I have this thing. So it's minus i h bar epsilon ijk lk from the last two commutators. So this one you use essentially equation six.

Now look, this thing and this thing cancels. And these two terms, they actually cancel as well. Because here you get an epsilon i jr.

And here is an epsilon ji something. So these two terms actually add up to 0. And this is 0. So indeed, j1i and j2i 2j is 0. And these are commuting things.

I wanted to say commuting angular momenta, but not quite yet. Haven't shown their angular momenta. So how do we show their angular momenta? We have to try it and see if they really

do form an algebra of angular momentum. So again, for saving room, I'm going to erase this formula. It will reappear in lecture notes. But now it should go.

So the next computation is something that I want to do, j1 cross j1 or the j2 cross j2 to see if they form angular momenta. And I want to do them simultaneously. So I will do 1 quarter of j1 cross j2

would be l plus minus h nu r cross l plus minus h nu r.

OK, that doesn't look bad at all, especially because we have all these formulas for products. So look, you have l cross l, which we know. Then you have l cross r plus r cross l. That is convenient.

here. And finally, you have r cross r, which is here. So it's all sort of done in a way that the competition should be easy. So indeed, 1 over 4, l cross l gives you an ih bar

l from l cross l. From these ones, you get plus minus with plus minus is always plus, but you get another ihl. So you get another ihl. And then you get plus minus l cross h nu r plus h nu r cross l.

So here, you get 1 quarter of 2 ihl.

And look at this formula. Just put an h nu here, an h nu here, and an h nu here. So you get plus minus 2 ih from here and an h nu r.

OK, so the 2's and the 4's and the ih's go out. And you get ih times 1 half times l plus minus h nu r, which is either j1 or j2.

So very nicely, we've shown that j1 cross j1 is ih bar j1 and j2 cross j2 is ih bar j2.

And now finally, you can say that you've discovered two independent angular momenta in the hydrogen atom. You did have an angular momentum and an r vector.

And all our work has gone into showing now that you have two angular momenta. So pretty much we're at the end of this because after we do one more little thing, we're there.

So let me do it here. We'll not need these equations anymore, except this one I will need.

So from l dot r is 0.

So from l dot r equals 0, this time you get j1 plus j2 is equal to times j1 minus j2 is equal to 0.

Now, j1 and j2 commute. So the cross terms vanish. j1 and j2 commute. So this implies that j1 squared is equal to j2 squared.

Now, this is a very surprising thing. These two angular momenta have the same length squared. Let's look a little more at the length square of it. So let's, for example, square j1.

Well, if I square j1, I have 1 fourth l squared plus h squared nu squared r squared. No l dot r term because l dot r is 0.

And h squared nu squared r squared is here. So this is good news. This is 1 fourth l squared plus h squared nu squared minus 1 minus l squared.

The l squared cancels. And you got that j1 equals to j2 squared. And it's equal to 1 fourth of h squared nu squared minus 1.

OK. Well, the problem has been solved, even if you don't notice at this moment. It's all solved. Why? You've been talking a degenerate subspace

with angular momentum with equal energies. And there's two angular momenta there. And they're squared equals to the same thing. So these two angular momenta, our squares are the same.

And the square is precisely what we call h squared j times j plus 1, where j is quantized. It can be 0, 1 half, 1, all of this. So here comes the quantization.

j squared being nu squared. We didn't know what nu squared is, but it's now equal to these things. So at this moment, things have been quantized. And let's look into a little more of detail

what has happened and confirm that we got everything we wanted. So let me write that equation again here.

j1 squared is equal to j2 squared is equal to 1 quarter h squared nu squared minus 1, which is h squared j times j plus 1. So cancel the h squares and solve for nu squared.

Nu squared would be 1 plus 4j times j plus 1, which is 4j squared plus 4j plus 1, which is 2j plus 1 squared.

That's pretty neat. Why is it so neat? Because as j is equal to 0, all the possible values of angular momentum, 3 halves, all these things, nu, which is 2j plus 1, will be equal to 1, 2, 3, 4,

all the integers. And what was nu was the values of the energies.

So actually, you've proven the spectrum. Nu has come out to be either 1, 2, 3. But you have all representations of angular momentum. You have the singlet, the spin 1 half. Where are the spins here?

Nowhere. There was an electron, a proton. We never put spin for the hydrogen atom. But it all shows up as these representations in which they come along. Even more is true, as we will see right away and confirm that everything really shows up the right way.

So what happened now? We have two independent equal angular momentum. So what is this degenerate subspace we were inventing is the space j, which is j1, and m1 tensor product

with j, which is j2, but has the same value because the squares are the same, m2. So this is an uncoupled basis of states

in the degenerate subspace. And now it's all a little surreal because this don't look like our states at all. But this is the way algebraically they show up. We choose a value of j.

We have then that nu is equal to this. And for that value of j, there are some values of m's. And therefore, this must be the degenerate subspace. So this is nothing but the tensor product of a j multiplet with a j multiplet,

where j is that integer here. And what is the tensor product of a j multiplet? The first j is for j1. The second j is for j2.

And what is? So at this moment, of course, we're calling this n for the quantum number. But what is this thing? This is 2j plus 2j minus 1 plus all the way up

to the singlet. But what are these representations of? Well, here we have j1 and here is j2. This must be the ones of the sum. But who is the sum?

l. So these are the l representations that you get. l is your angular momentum, l representations. And if 2j plus 1 is n, you got a representation with l equals n minus 1, because 2j plus 1 is n,

l equals n minus 2, all the way up to l equals 0. Therefore, you get precisely this whole structure.

So just in time, as we get to 2 o'clock, we finish the quantization of the hydrogen atom. We finished 805. I hope you enjoyed. I did a lot. Aaron and Will did, too.

And good luck, and we'll see you soon.