 Felly'n amser, mae'n amser yn gallu gynnal oherwydd yn dda i'n fawr o'r corfodau am hyn o'r logic o'r Aanglion Maenthau. Rwy'n rŵf yn dda i'r iawn o'r gyro yn y bobl. Felly'n dda i'r rhaid o'r rhaid o'r rhaid o'r gwbl yn dda i J1, o'r rhaid o'r rhaid o'r gwbl yn dda i J2, ac mae'n rhaid o'r gwbl eisiau'n byddol, ystafell o'r bobl yn dda i'r aanglion maentham, ac mae'n gwybod a'r yrhyw ystod, feddwl, i ddiddordeb hynny y bod yn ddiddordeb bod yma ystod yn ddiddordeb pethau. I ddiddordeb yma, ychydig o ffysigwyr yw'r cyfeirio. Yn y ddiddordeb yw'r cyfeirio a'r cyfeirio'r cyfeirio'r cyfeirio pethau, yna'r ddiddordeb yma yna'r ddiddordeb yw'r argynno, lle mae'n dechrau'r argynno i'r ddiddordeb yw'r cyfeirio, ac we would get a state with total angular momentum J1 plus J2 and apparently all of it down the z-axis. And then we used the J-minus operator, the reorientation operator, J-minus, to create this state in which we still had the same large amount of angular momentum because the two gyros are parallel to each other. But we didn't have it all parallel to the z-axis. And the algebra led us to this expression here that this state is a linear combination of the state in which the first gyro is offset from the z-axis, but the second gyro is on the axis and the state in which the first gyro is on axis and the second is offset from axis. And I was just saying as the lecture closed how to get this object here, this object here has to be a linear combination of the same two states. So this is a state of the box, this is a state of the box. Neither of these states of the box is a state of well-defined angular momentum of the box. This linear combination is, and there's another linear combination of these two which is a state of well-defined angular momentum. That's this state which has less total angular momentum of the box and this state that we're looking for has to be orthogonal to that and one good way of writing it is to say that j minus one, j minus one, so this is the state in which the gyros are not parallel to each other, quite, but all of the angular momentum available given that they're not parallel is along the z-axis that this is the linear combination orthogonal which you could write as j2 over j of j1, j1 minus one, j2, j2 minus root j1 over j of j1, j1, j2, j2 minus one. So in here we have the slightly strange thing. We have the state in which, so this is a state in which the two gyros are not quite parallel to each other, which is why the total angular momentum of the box is less than maximum. Here they are parallel to each other and yet when you look in here it looks as if they're not because one of them is aligned with the z-axis and the other isn't and here we have a linear combination of the same two states of the contents of the box but with different coefficients out front and a crucially a minus sign here and that has the physical interpretation of the two gyros not being parallel to each other. So let's try and clarify this strange situation, well get used to it I suppose is the state of affairs by doing a concrete example. What does it look like in the very important case that we say j1 is one and j2 is a half? That means obviously that j, the maximum angular momentum we can get is three halves and we're going to have a diagram now that looks like this. So that was all in general now we're going to be looking more, we can say more concretely what we're going to have. We're going to have three halves, three halves at the top here which is going to be the same as one, well we can just say one and plus. So you're using a shorthand notation here. So I've got that j1m is now going to be objects like 1, nothing and minus 1. Cos there's no need to write this down, I'm writing down the possible values for m, m is 1, m is nothing, m is minus 1 and j2m can be because j2 is a half, I can write this as plus and minus where I'm writing down the values of m in the sense of plus a half and minus a half, right? That's a shorthand notation that makes life easier. So this is just a different notation for that state and more compact notation for that state. If we would come down here what would we have? We would have three halves, one half, right? That's cos j is three halves and what would it be? It would be the square root of one over two, sorry, not at all true. It would be j1 which is one over j which is three halves. I'm in danger of running out of space, let's just shave that off. It would be one over three halves times nothing plus, and now I want this state which is going to be a half over three halves, the square root of a half over three halves of one and minus. So what does that, let's just clean that up a little bit. That's equal to root two thirds of nothing plus, plus this is going to be one third, root one third of one minus. Notice a nice thing about this is that the state, the linear combination of these states of what's in the box that we generate comes out beautifully normalised. This thing squared plus this thing squared, two thirds plus a third comes to one. It comes out normalised automatically and that provides a nice check on your algebra so it's good to check that it does cut, it is properly normalised because if it isn't, the algebra has gone wrong somewhere. We now have a physical prediction if you look at this state here and what we might be talking about now that j1 equals one, that might be the orbital anglement of an electron and that j equals that j2 equals a half, might be the spin anglement of the electron. So we might be talking about the total anglement of the electron due to both its spin and its orbital motion and if you would look inside the box, if you would examine the atom, the electron in detail when it was in this state you would find that there was a probability of this thing squared, i.e. two thirds, that the orbital anglement in the z direction would be nothing and the spin would be along the z direction and there would be a probability of one third that the orbital anglementum would be all parallel to the z axis or as parallel to the z axis as it can be and the electron spin would be pointing downwards. So that's the physical interpretation of this. So we have that p spin up equals two thirds and p spin down in this particular state is equal to one third. That's the physical meaning of these numbers here. If we would, okay, so now let's ask what's this state here? We would like to, there are some more states to find. This is the state in which the, we have a half less, sorry we have one unit less of anglementum than we have on the outer circle. So this is the state, a half, a half of the box. It's going to be a linear combination of these two things and it's going to be the linear combination which is orthogonal to these two things because it's an eigencat of the total anglementum squared operator for the box which it's an eigencat of that which has eigenvalue different from this. That must be orthogonal to this by the orthogonality of the eigencats of Hermitian operators. And what is it going to be? It's going to be root one third of this nothing plus minus one third, root two thirds of one minus. So now, so in this state, the odds when you look in the box of what you find are changed. This, in this state which has less anglementum in total, the probability of spin up is this thing squared, i.e. third, and the probability of spin down is two thirds. So that's the physical implication of this. It's an interesting exercise to apply the J minus operator here to generate this. So if we take this state and apply the J minus operator to it, on the left side we're going to get three halves minus a half which is this state here. Let's just have a new diagram because we're running out of space there. So here we have three halves, three halves minus a half and it's going to be obtained by using the J minus operators on the left and the right of that equation. I recommend that you do this. I'll just write down what the answer is for the moment. It's going to be one over, well it's actually obvious, one over three of minus one of minus one, sorry, plus two, three square root of nothing minus. Now it should be this because there should be symmetry between this state down here physically, let's write it in here. This state here physically it's evident that this has to be, this is obviously minus three, this is three halves and all of it anti-parallel to the axis, three halves minus three halves and physically it has to be that both of them are pointing down. So it has to be minus one minus orbital angular momentum down, spin down which is obviously the sort of negative of what we put at the top there. This thing similarly has to be, physically it should be that you can get this thing by changing one to minus one above minus to plus, naught stays alone and plus to minus and indeed it does, right? So it's an exercise that I recommend that you check that when you use J minus to go from here you do indeed arrive here which is where you expect to arrive by the symmetry between plus and minus and you do. So what do we, so what's happening here physically, let's see if we can form some kind of physical picture. We can only do this to a limited extent because of the big roll that quantum uncertainty plays with small spins. But the physical idea here is that, so let's take this three halves a half state. What do we have? We have some angular momentum vector which is in some sense three halves long and it's only got a half of it in the z direction and that is some superposition of the angular momentum being, the orbital angular momentum being more or less in the xy plane and the spin carrying you up. So you add this vector to this vector, you get this vector. That's sort of what the first term up that the root two thirds of nought plus symbolically indicates, spiritually indicates. We then also have another linear combination which is one over root three of one minus. Now how do we understand that? Well we have to draw a diagram that's something like this. So we're now combining the orbital angular momentum which is sort of vaguely along the z axis. Remember I stress that when you're dealing with small spin systems you can never get the angular momentum exactly parallel to the z axis. There's always a significant amount in the xy plane. So this is the direction of z, this is the xy plane. So this vector shouldn't be going straight up and I shouldn't have drawn this vector going straight up really either. That should have been at some funny angle. In fact let's improve the quality of the diagram a bit by making it not go straight up, let's make it go like that. And then I've got minus pointing down but again it's not pointing straight down because there's always, I stressed with spin a half, there's as much angular momentum parallel to each axis at all times. So this is the sort of diagrammatic representation of that expression up there and how do you think about this, how a possible way of thinking about this physically is to say to yourself, well the angular momentum of the orbital and the spin angular momentum are interacting with each other and as a result of it processing around this fixed vector, this is the total angular momentum of the box which by conservation of angular momentum must be a fixed thing. So you can imagine that these two vectors are processing around this vector here and here we see two snapshots of possible configurations. So if you imagine this thing moving around like that now we see this, sometime later we see this and then it'll process back to that. Now that is not really strictly speaking a legitimate proceeding because in doing all this stuff we never said anything what the Hamiltonian was. We never said anything about that, we just had these two gyros in a box and they weren't physically interacting in any way, consequently they have no means mechanically for exchanging angular momentum. And yet when the box is in a state of well-defined angular momentum we have these results up here and we have this state of the box is a superposition of these states of the contents of the box. So you'd be aware of this picture but there is a certain amount of intuitive satisfaction in this picture and it does at least give you a physical understanding of why it is that a state of well-defined angular momentum for the box is not a state of well-defined angular momentum of the contents of the box because already classically that would be the case. What's happened is by insisting that the box has a well-defined angular momentum we have forced the particles to be correlated because if the orbital angular momentum or the first gyro is doing this in order that the total angular momentum is this, the other thing has to do that. So we have forced a correlation between the two gyros between the spin and the orbital angular momentum and that correlation is reflected in the entanglement of these particles in the sense that we discussed when we talked about composite systems. And in real physical circumstances like an electron, if we do have an orbital angular momentum and spin angular momentum then there is a physical coupling between the two provided by the electromagnetic field and it is then legitimate to think about these things that are processing around. The other thing that I should probably say is that this diagram doesn't really work. If you try and make this diagram work with proper lengths, you give a proper length to this and this thing should equal this thing and this thing should equal this thing, you won't be able to make it happen. And the reason you won't be able to make it happen is because this is showing something in only two dimensions and what's really happening is in three dimensions. So you've got to imagine, so we don't know anything about what's happening in the XY plane. Those were the terms, that was the deal we did. We said we were going to have eigenfunctions of L squared or J squared and Jz and having chosen to know something about what Jz is doing, we've given up on, we've surrendered knowledge of what Jx and Jy are doing. So what's happening in the plane perpendicular? This is the XY plane, right? It's not just, it's not X and it's not Y, it's just things happening in that plane means that you can't really draw this as a two-dimensional diagram. So that's why you can't make it work vectorially. Well I think on that we should leave the addition of angular mentor and turn to our final topic, very important one, which is hydrogen. So obviously atoms are terribly important, we're made of them, that's most of what we see. Here and elsewhere and they're also played a crucial role in the development of quantum mechanics. Quantum mechanics was developed in order to build models of atoms. It is amazing that this enterprise was successful because even simple atoms like an oxygen atom is substantially less friendly dynamical system than say the solar system because it contains, right? An oxygen atom contains eight electrons and the nucleus. So it's sort of same order of the number of particles as the solar system but it is much more horrible dynamical problem than the solar system because the electrons attract each other much more strongly than the planets attract each other. So the approximation which is fundamental to understanding the solar system that the planets move around in the gravitational potential of the sun and we can neglect the gravitational potential of the other planets while we do that and make ourselves a model and then add in as a perturbation the action of Jupiter. The forces between the planets are not negligible, they play a crucial role in the structure of the solar system but you add them in later and they're a very small approximation very small matter relative to the electric attractions of the electrons which are really jolly large. Another problem about an oxygen atom is that the particles are moving with speed V which is on the order of 8 over 137 so several percent of the speed of light. You're talking about a system which is mildly relativistic. The contribution of relativity to motion in the solar system is very much smaller. 1 km a second, which is less than a thousandth of the speed of light. So relativistic corrections are much more important. Another very serious problem is that these particles which are moving around in an oxygen atom are all magnetised gyroscopes, right? They all have spin, a significant amount of spin because the Earth has spin but it spins very small, it's enormously small compared to its orbital angular momentum and the Earth isn't magnetised but the magnetic couplings between the Sun and planets and between planets and planets are completely derisory and negligible. And yet it took physicists, well, to get a pretty good understanding of the solar system was the work of the whole eighteenth and nineteenth centuries it was the work of Bessel, the classical structure of the solar system it was pretty much under control but as Poincaré who lived at the beginning of the twentieth century pointed out there was still an enormous gap and problem about the long-term life of the solar system and the long-term life of the solar system is still an active topic of discussion and it turns out to be a very interesting and finely balanced problem. So even though an oxygen atom is very much more complicated and more friendly a dynamical system than a solar system actually it's very much better under control quantum mechanics enables you to bring it under very much better control than even today we have brought the solar system. So it's an interesting point that these systems are in quantum mechanics actually rather easier to do than the corresponding classical system but they're nonetheless very complicated and we have to proceed by stages and what we're going to do is study, well hydrogen of course is very important it's a tremendously important atom but we're also going to use it as a building block for understanding atoms in general. So we're going to talk about, so what are we going to do? We're going to talk about the gross structure, what's called the gross structure, whoops, of hydrogen, whoops, hydrogen-like ion. So what do I mean by this? Gross structure, this means that we are going to, we're going to have no relativity, no spin, intimately related to relativity in fact. We're going to have, so we're going to be left with point spinless particles which interact electrostatics in non-relativistic mechanics and over here what are we going to do? We're going to say that the nucleus, the charge on the nucleus is going to be Z times the electron charge. So we're putting in here a number which in hydrogen will be one which we can make larger in order that we can discuss the motion of electrons on oxygen nuclei or other nuclei as a building block. No spin, electrostatics in other words, no magnetism. The key thing really is we're leaving out relativity because magnetism is a relativistic correction to electrostatics and spin arises naturally when you think about electrons in the context of relativity. I hope you'll appreciate next year. So we're leaving out the effects which are actually quite important but one has to proceed in steps. So now what we're going to do, what we're obviously trying to do is we're trying to solve, we're trying to find the stationary states of an atom, of a system which consists of one electron, one nucleus with that charge. So we want the stationary states because they provide the key to the dynamics, usual situation. So what's the Hamiltonian under these approximations? Well, it's going to be the nuclear kinetic energy, the kinetic energy of the nucleus, p-nucleus squared over the mass of a nucleus, twice the mass of a nucleus, plus the electrons, kinetic energy, plus the interaction energy between these two, which is going to be z e squared over 4 pi epsilon nought x e minus x n. Modules, right? So this is the sum of the energies, I mean of three distinct contributions to the energy, kinetic, kinetic potential, right? So we want to know what does this look like in the position representation, right? We're going to, we want to examine that equation concretely and the way to go is to put this into the position representation. So that's to say we bra-through by x e x n h. And then what we want is the position representation of this, which is going to be minus h bar squared del squared with respect to x n over 2 m nucleus, right? Because this is the kinetic energy operator, which we know that p is minus i h bar gradient, so p squared is minus i h bar, sorry, minus h bar squared nabla squared. So what does this mean? This sub x n means this involves derivatives with respect, this involves derivatives with respect to the components of the position of the nucleus, minus h bar squared over 2 mass of the electron plus h bar squared x electron minus z e squared over 4 pi epsilon nought. This is already in the position representation, if you like, x e minus x n. So all this stuff times psi, we'll do that for another space to put it in, times psi is going to equal e times psi, right? So this is the usual stuff that psi, which is a function of x e and x n. So it's a function of six variables is x n, whoops, x e e. It's a wave function of the stationary state of energy e. So we've got here now, so we've reduced our abstract equation to a very frightening partial differential equation in six independent variables, right? Because we've got the positions, the three components of x n and the three components of x e. So it's that we've got a PDE in six independent variables. The astonishing thing is that we can solve this exactly and without a huge amount of sweat. And as in the solution of any number of problems in physics, the key is to choose your coordinates correctly. That's quite generally the case that a problem which is very frightening in general with the clever choice of coordinates, you're all done. So we need new coordinates. We need to transform this equation to new coordinates. And the ones we take are big x, which is the center, is classically the center of mass coordinate. So that's going to be m e x e plus mass of the nucleus times the position of the nucleus over m e plus m n, right? So it's the center of mass coordinate. You may say, what authority have I got to use that in the context of quantum mechanics? And the answer is I make this strictly speaking, I made absolutely no claims as to the physical interpretation of this. It's just a suggestion of something we might use to simplify the algebra. But as rational beings, we know physically what that means. And then we'll have another, so that's three new variables, okay? Because it has three components, which are linear combinations of our old variables. And we're going to have another linear combination. And surprise, surprise, it's going to be x e minus x n, the separation. So all we do now is plodding mathematics in order to rip out of that differential equation x n and x e and insert the corresponding things with this. So let's see how this goes. Let's do d by dx e, right? Because that nabla squared e is sort of this operator dotted into itself. So let's see what is this. Well, the chain rule says that it's d by dx plus d by dr. So this is the chain rule. Mathematics, nothing to do with physics, but because it's mathematics, it's definitely true. And this dot implies a summation, right? Because this thing's got three components. So d by dx 1 by dx e, d by dx 1 plus dx 2 by dx e, d by dx 2, et cetera, et cetera, et cetera. Now fortunately these partial derivatives are nice friendly things because we just have linear combinations here. So I think we can easily see what this amounts to. This is going to be m e of m e plus m n. That's what this partial derivative comes to from up there is a partial derivative of d by dx plus, sorry, and this partial derivative is nice and simple. It's just 1, so we're just going to get plus d by dr. So I've still got, this is a shorthand for three equations because this is d by dx e 1 is equal to this thing times d by dx 1 plus d by dr 1, et cetera. We want this thing dotted into itself, so what we have to do is multiply this on itself with a dot between the two, and what do we get? We get that del squared x sub e is equal to, we get this thing squared, of course, so we have m e of m e plus m n squared d 2 by dx squared. Well, no, we can write that more handily as del squared with a big x, I think more clearly. And then we get this thing squared plus del squared with respect to del squared where we're talking about the components. Here's the usual expression for del squared but using the components of big x. Here's the usual expression for del squared but using the components of the separation vector r. And then, irritatingly, we get a mixed term because we're taking this operator and we're multiplying it on itself, so we get a mixed term of this operator multiplying the d by dr in the next bracket, and then we have this thing doing this, so we end up with plus 2 of m e of m e plus m n of d 2 by dx dr. So this is not very nice. Nobody wants this. This is excellent. We've found a relationship for this, which we want in terms of this and this, which you'll find, but this is definitely not required. It would kindly go away and we can make it go away easily by just working out what d by dx n is. That's going to be d big x by dx n, which is going to be mass n, mass of the nucleus of a mass of electron plus mass of the nucleus d by dx. And then it's going to be dr by dx n, which is going to give us a minus 1 instead of a plus 1, so this will be minus d by dr now. We square this up to work out what del squared of x n is. We get this thing squared, of course, m n of m e plus m n, squared del squared of big x, so that's this one squared, and then of course we get this one squared, and never mind the minus sign, because we're squaring up, so this becomes a plus del squared with respect to the separation variable components. And then we get this on this, where now the minus sign is manifest, we get minus twice this, which is m n of m e plus m n of d2 by dx dr. So we have expressions which are very similar, but include one has a plus sign, one has a minus sign. So what we want to do now is work out 1 over m e del squared with x e plus 1 over m n of del squared with respect to x n. So 1 over m e times that top equation plus 1 over m n times this, which is actually exactly what occurs in our Hamiltonian, if you go right up there, mercifully what we want is in fact the individual del squares weighted by 1 over the mass, and what's this equal to? This is equal to we get del squared x twice once from here and once from there. We used to have an m e squared over m e plus m n squared, but we divided through by m e, so we have an m e, and then similarly from here we have an m n over m e plus m n squared. That's the result of adding this with this weight to that. Similarly, what do we get here? We get a common factor del squared r and we have a 1 over m e plus a 1 over m n, and these terms go away, because by the time we divided through by m e, at the top we have just 2 over m e plus m n times that next derivative by the time we divided by m n we just have minus 2 over that, so they go away. So this that we want in the Hamiltonian is equal to this, which can be simplified and written as 1 over m e plus m n del squared x plus 1 over mu del squared of r, where mu is exactly equal to m e m n over m e plus m n, and goes by the name of the reduced mass, and you may already have met it in classical mechanics. I hope you have, but if you haven't, never mind. So, we can now write our Hamiltonian and I'll do it over here, I think, so we can keep in view the Hamiltonian at the top there. What do we have? We have that the Hamiltonian involves that. What we have is a minus h bar squared over 2, you know, right. So that's going to give us a minus h bar squared over 2 m n plus m e, or m e plus m n I've been writing, del squared with respect to big x. We will have minus h bar squared over 2 mu of del squared with respect to the separation vector, because this is the magic combination, minus h bar squared over 2 times this is what appears in the Hamiltonian, so we get minus h bar squared over 2 times this, which is what I hope I've written down, and then we have to add in the, we have to add in the potential energy term, which is a minus z e squared over 4 pi epsilon nought times the separation variable, where this is, if you understand me, the modulus of the separation variable. So that's what in the position representation with these new coordinates, our Hamiltonian is looking like this and this is really beautiful because we can define this to be h k, and we define the rest of it to be h sub r. And what do we have? We have that h sub k commutes with h sub r. It commutes because this is a function only of the x variables, the big x variables, right? It just involves derivatives with respect to big x. This, and those derivatives, these partial derivatives with respect to big x components are with all the other coordinates held constant and the other coordinates means the other components of big x and all the components of little r, because we made a legitimate change of coordinates. So this obviously commutes with this because partial derivatives always do and this commutes with this because this is being held constant while we're doing these partial derivatives. So h sub k commutes with h sub r correspondingly h sub k commutes with a Hamiltonian and also h sub r commutes with a Hamiltonian. That follows immediately from the first thing because obviously this thing is h k plus h r, h k obviously commutes with itself, every operator commutes with itself, only the established it commutes with r. So this vanishes and similarly this vanishes for the same argumentation. So we have three mutually commuting operators. That means there's a complete set of mutual eigenstates of h, hk and hr. So these we want eigenstates of this. That's what we set out to find and what we discovered is that if we find the eigenstates of this and this we will just, all we need to do is multiply them together and we will have eigenstates of this. So what does this imply? This implies that e is equal to some state of k times some state er or e sub k. Now I have to be a little bit careful. I think I've said something that's not strictly speaking true. What we know is that there is a complete set of eigenstates of this which is simultaneously eigenstates of this and this. It does not follow and indeed it's not true that every eigenstate of this is simultaneously an eigenstate of this and this. There are eigenstates of this which are not eigenstates of this and this but we will get a complete set of eigenstates of this. That's good enough for us which are simultaneous eigenstates of this and this. So we've reduced the problem of finding the eigenstates of this to the sub problems of finding the eigenstates of this and the eigenstates of this. So hk is just the kinetic energy is the kinetic energy operator of this particle. Physically it just describes the kinetic energy of the whole atom. The whole atom can cruise across your laboratory. It has kinetic energy. Are we interested in this? No. It's bloody obvious that the energy of an atom depends on how fast it's moving. We can deal with that. We've already studied the, we already last term studied the stationary states we know all about it boring, finished. So all we have to do is hammer away at this thing what we want to know is so this is basically trivial and of no interest. So we focus in on HR. Now we went to some considerable trouble to show this has to be studied. What we deal with is HRER is equal to ER ER and this is going to describe the internal energy of the atom a distinct from the translational kinetic energy that it has because it's moving. And we've been to some trouble we've to show del squared was equal to pr squared minus h over h bar squared. We showed a relationship between the between del squared and what we identified to be the radial momentum here squared and the orbital angular momentum operator. So this is the, basically we showed that L squared was r squared times the angular parts that we were familiar with inside the inside del squared the Laplacian. So this is the orbital, total orbital angular momentum total orbital angular momentum operator and pr, we write it down for you, pr we figured out what it had to be it turned out to be minus h bar of d by dr plus 1 over r, that's what it looked like in the position representation and it's the radial momentum operator. So the only so let's just say what we've got, let's write down hr now using this information. hr is minus h bar squared over 2 mu times nabla squared. So what's that? That's equal to pr squared over 2 mu plus l squared h bar squared, l squared over 2 mu r squared and then minus z e squared over 4 pi epsilon naught r. That's what this what this, so we focus, homed in on this operator our problem is solved once we find the eigen states of this operation and the eigen energies of this operator and this is what it looks like it involves radial momentum radial position radius and for the rest it involves the total orbital angular momentum operator. So the brilliant thing is that hr l squared equals naught. It's also true that hr comma z equals naught. Why is that? Because l squared, remember it can be represented as a partial differential operator in terms of dbd and dbd phi's so it clearly commutes with this and it commutes with this and it jolible commutes with itself. Similarly, lz turned out to be minus i h bar d by d phi and it has a muthal angle so it commutes and lz we know commutes with l squared and it clearly commutes with r and pr so what do we learn from this? There's a complete set set of mutual eigen states and l squared but we know all about the eigen states of l squared we've studied them at Norseam which means that we well so what we can do is we can say it's legitimate to say now it is not true that every eigen state of h is an eigen state of l squared well anyway it is not true that that's the case but it is true that there's a complete set of states of the form e and l squared sorry there's a complete set of states e and l which is such that l squared e l this is an eigen state that's what that's doing there denoting that it's an eigen state of this operator and of course has eigen value l plus 1 e l so I've learnt that it is possible it is legitimate to look for the to restrict the search for eigen states of this crucial internal energy Hamiltonian to states which are eigen states of the total angular momentum operator that will be tomorrow that then reduces the eigen value problem associated to HR to just a one-dimensional problem very closely analogous to the simple harmonic oscillator which we've already sorted and it will it will yield to the same line of attack that we used on the simple harmonic oscillator namely ladder operators