 OK, let's get going. So on Friday, we used the commutation relations that we had deduced for the angular momentum operators to find what the spectrum of j squared and any one of the j's, for example, jz could be. And now we can use this information to understand, to interpret the spectra, the infrared and the near infrared and terahertz spectra of diatomic molecules, which are very common, right? They're in the room here, O2 and N2, much of the mass of the baryons. The universe is contained in H2 and a very important, from practical point of view, a very important molecule, for reasons I'll explain, turns out to be carbon monoxide, CO. So a molecule like this consists, for our purposes, of two masses, being the nuclei, for example, of the oxygen and the carbon on a light spring. So we have two massive nuclei on a light spring formed by the electrons. And this is capable of being a simple harmonic oscillator that it does this, but also, at lower energies, it's capable just of rotating end of remed as a dumbbell and having rotational kinetic energy. So if you go to classical mechanics and ask yourself, what is the energy of this thing? The energy of this thing is given by the classical angular momentum around the x-axis squared over twice the moment of inertia around the x-axis plus the classical angular momentum squared around the y-axis over twice the moment of inertia around the y-axis plus the classical angular momentum squared over 2iz. So this is reminiscent of, this is a piece of classical physics, that the kinetic energy of a moving particle is p squared over 2m. So this is classical mechanics, classical mechanics. And those who did short option seven will, I hope, recognize this formula. So we have angular momentum x component here instead of x component of linear momentum divided by moment of inertia around the x-axis instead of divided by mass. So this is what the classical expression for the energy of a thing like this is. So what we say now, we conjecture that quantum mechanically we put, this is a conjecture, we put jx, the classical thing, goes to h-bar times jx, the operator. So we define these operators, many people define them to have dimensions of h-bar, but we would define them so they would dimension less. This thing has the dimensions of h-bar. So we make this transformation, we take this classical expression and turn it into h-bar times the quantum mechanical operator, et cetera, and then we infer that the Hamiltonian should be, or we guess, should be h-bar squared over 2 jx, the operator squared over twice, sorry, over ix, I've taken that out, plus gy squared the operator over iy plus jz squared over iz. So these are operators now. So you should think of this as a guess. It'll be confirmed by experiments to be shown soon. Now we say, for this diatomic molecule, we take the z-axis to be the symmetry axis of the molecule. That's the axis that runs through both nuclei. And then the moment of inertia around that axis, around that now z-axis, is negligible because the mass in this molecule is almost entirely contained in these nuclei, which are essentially point, they're very close to being point particles, and if you rotate it around this axis, those nuclei provide very little moment of inertia. The electrons could provide a certain amount of moment of inertia, so it's feasible for the electrons to have some angular momentum around this axis. And in some molecules they do, but in most molecules they don't. So the moment of inertia is, in any case, very small. So this implies that iz is very much less than iy than ix, which is the moment of inertia about an axis that sticks out here, where if you spin it around there, the two nuclei go around in a decent circle of radius half the length of that bonse, and for a diatomic molecule by symmetry, ix is going to equal iy. So given that ix is equal to iy, it's natural to rewrite this Hamiltonian as h is equal to h bar squared over 2 brackets j squared, which is gx squared plus jy squared plus jz squared, divided by ix, which is equal to iy. So those first two I've combined together into here. I've now in here also got a jz squared over ix, which I don't really want. So I simply get rid of this jz squared brackets 1 over iz minus 1 over ix. So this is just a repackaging of that expression to group together the terms which have a common factor 1 over ix. And now the beauty of this is that we know what the eigenvalues of this operator are and what the eigenvalues of this operator are. So we can immediately say what the eigenvalues of this operator are. So this says immediately what the spectrum of h is. Ejm, which is going to be h bar squared over 2 brackets jj plus 1. Because that, remember, we showed the eigenvalues of this with jj plus 1, where j is allowed to be an integer or a half integer, divided by ix plus m squared 1 over iz minus 1 over ix. So because the Hamiltonian for this rotating system is simply a function of the angular momentum operators, and because we have already found out what the spectrum of the angular momentum operators is, we can immediately say what the spectrum of the Hamiltonian is, i.e. we now know what the allowed energy levels are of these rotating molecules. There's almost no further investment of effort. We know that m lies in the range minus j less than or equal to m less than or equal to j. That was one of the things we showed on Friday. So this number is sort of generically of the same order as this number. But this 1 over iz, I've explained, is a very much bigger number. iz is a very much smaller number than ix. So this number is very much bigger than this number, which is the same as 1 over this number here. So the coefficient of m squared is huge. And that means that if m were anything other than 0, the energy would be enormous. So when you're dealing with molecules in the room or in interstellar space or somewhere where they only have moderate amounts of energy, so we can argue that this thing is going to be 0. So this number here is very large. I mean, it really is huge because iz is so very small for a diatomic molecule, which means that this is always 0. That's just saying that these molecules have their angular momentum around an axis, which is perpendicular to their symmetry axis. So that means that effectively the energy levels are simply ej h bar squared over 2 ix times j, j plus 1. So those are the energy levels. Now that's all been totally generic for diatomic molecules. And let's now talk about carbon monoxide. Carbon monoxide has a much more interesting spectrum, much more easily observed electromagnetic spectrum, because the carbon is slightly positive. Oxygen is slightly negative. I've lost this. Oxygen has a great affinity for electrons, so it borrows some of carbon's electrons and ends up being a little bit negative. This is a little bit positive. So carbon monoxide molecule, which is spinning end over end, is a rotating dipole, electric dipole. So CO, this is minus, will be a rotating electric dipole. The same will not be true of a hydrogen molecule, because obviously the two hydrogen atoms will have equal affinity for electrons, and so neither will be charged or an oxygen molecule or a nitrogen molecule. All of these molecules are not dipoles, so they can rotate without emitting electromagnetic radiation. But if CO is, if a carbon monoxide molecule is rotating, it will emit or potentially absorb electromagnetic radiation at some frequency, which we would imagine would be the frequency at which it rotated end of range. Classically that would make sense. A rotating dipole should emit or absorb radiation at the frequency that it rolls over. Let's see whether that's true, according to quantum mechanics. So it will emit or absorb em radiation. And the angular frequency, well, let's have the frequency. So the frequency is going to satisfy h nu is equal to E j minus E j minus 1. Sorry, another point. Photons carry one unit of angular momentum. It turns out, right? Not more, not less. So a photon can move a molecule which has a total angular momentum J to a state which has a total angular momentum J minus 1 or J plus 1, but it can't move it to J plus 2 or J plus 3 or it can only move it by one unit in J, because it's only got one unit of angular momentum. So by conservation of the angular momentum of the photon plus the molecule, right? So you can destroy the photon or create the photon. So you can add the angular momentum of the photon into the molecule or you can dump into a photon one unit of angular momentum and hence make some step like this. But only adjacent Js can be moved to. This is a key point by interaction with electromagnetic radiation. So we're going to have that h nu, the energy of the photon is equal to the difference in the energies of the molecule between the excited state it started and the state it finished. And what's that going to be? According to this formula here, that's going to be h bar squared over 2iz, ix, sorry, brackets J, J plus 1 minus the same expression with J put equal to J minus 1. So it becomes J minus 1 times J. And I think you can see that this cleans up very nicely to simply 2J. So we end up with h bar times J over ix. I think somebody must have a call. What does that tell us about nu? It tells us that nu, maybe this should get a subscript J to indicate that it's the frequency, which will be emitted as it goes from a state J to J minus 1. We can cancel one of the h bars. Well, this is an h, not an h bar, right? So this becomes equal to h bar J over 2 pi ix. So that's the coordinated quantum mechanics. That's the frequency of emitted light. Can we reconcile this with our classical picture? So how fast was the molecule going around? So classically, we can say that the angular momentum, Jx, is equal to ix times omega. This is the angular frequency of the tumbling, of the rotation. So this is a piece of classical mechanics again, the relationship between angular momentum and frequency. This is like the relationship between momentum. So this mirrors the classical statement that Px is equal to m times x dot. Omega is x dot, rate of change of angle. I mean, it's the analog of x dot. This is the rate of change of position. That's the rate of change of angle. I've already said that moment of inertia plays a role a bit like mass. And this momentum plays the role of, right? So that's where this thing comes from. It's just a piece of classical physics. So when you're in a state, what do we think that omega should be? Omega should be on the order of, well, it should be equal to, Jx over ix. That's classical. Doing it quantum mechanically, this is going to be h bar times the square root of J, J plus 1. So this is a heuristic sort of thing. We're saying that exposing its angular momentum is along the x-axis, right? Then Jx squared is going to be on the order of h bar squared J, J plus 1. Jx squared is going to be on the order of J squared, which will be this, divided through by ix. And we want to relate this to the frequency we had up there, a factor of 2 pi, the way I've done it wrong. Oh, it's because, sorry, this is the angular frequency. That's the actual frequency, right? So this is equal to 2 pi frequency of, this was angular frequency. This will be frequency of rotation. So we have the frequency of rotation. We expect to be h bar over 2 pi ix times the square root of J, J plus 1. So this is going to be, so what can we say? This is slightly bigger. This is greater than just a bit. Just if J is a larger number, it's only a smidgen bigger than h bar J over 2 pi ix. So in its upper state, when it had total angular momentum J, J plus 1, h bar squared, the rate at which it was tumbling was a bit larger. This is the frequency of emitted light, sorry. This is new J from above. So in its upper state, its rate of tumbling on this classical picture will be slightly larger than the frequency at which it emits the light. And you can check that in its lower state, we would have a minus sign here, a minus 1, so just a bit below the frequency emits the light. So the light is, in fact, emitted at just the average of the expected rotation tumbling rates at the upper and lower levels, which makes, I think, perfectly good physical sense. So in lower state, that's when J, well, EJ minus 1 tumbling rate is just a bit lower. Oh no, this thing has gone to sleep again. It's not my computer that's gone to sleep. It's the lecture room that's gone to sleep. It's too irritating. Let's get this to stop doing this. I really do want to show you something today. Is it coming back? OK, right. So this is the actual experimental spectrum of carbon monoxide. So this is in gigahertz. So this is a terahertz, 2 terahertz, 3 terahertz along there. And you have a line here. There's a line being drawn at the frequency measured. Associated with the transitions from J equals 1 to J equals nothing. Then you have, so you get photons out with this frequency. You get photons out at this frequency, which is almost but not quite twice the previous frequency. According to this mathematics that we've done so far, the next frequency up should be exactly twice. This frequency here should be exactly twice that frequency there. And you can't tell the difference actually on this plot and so on and so forth. So each line here is telling you the measured frequency of a line, a spectral line from these carbon monoxide molecules. And you can see that they do form a regular grid just like this says. So this is the transitions from 1 to nothing. This is the transitions from 1 to 2. Sorry, from 2 to 1. This is 3 to 2 and so on and so forth. And as we go up here, this is, I don't know, can't count, 10 to 9 or something, right? There are a couple of missing lines in here where I wasn't able to find what the, I mean, somebody hasn't published the measurement or something. The other interesting thing to note, so that even spacing is confirming this frequency, new J is proportional to J business, and that's the interpretation. There's another interesting thing to notice here, which is that as you go along here, the black lines get slightly to the left of the dotted lines. And the dotted lines are exactly at multiples of the frequency of this lowest transition, J equals 1 to nothing. So what's happening is that the measured frequencies almost conform to this rule up here, but not quite. They turn out to be slightly smaller than the numbers new J given up there. And the physical interpretation of that is interesting. It's that, OK, so new measured minus new J is slightly negative, if you like. And that's because new measured is equal to, or this is what you would think classically, is equal to h bar over 2 pi J over ix, which itself is a function of J. As you make this molecule, so you're firing in circularly polarized photons and making this molecule spin faster and faster end over end, obviously, the centrifugal force will stretch that spring out a bit. The spring is stiff, but not infinitely stiff. So when you're spinning it faster and faster, the centrifugal force pulls the spring longer, increases the distance between the nuclei, and in that way increases the moment of inertia. So this moment of inertia that's appearing on the bottom of this formula should really be itself a function of J, which increases. So some of this increase here is canceled by a slight increase in the bottom. And that's the interpretation anyway of these spectral lines falling behind the measured numbers, which are the black lines falling behind the perfectly evenly spaced grid of dotted lines. And so one of the problems on the problem set is to use this phenomenon to estimate the stretching, how stiff this spring is. Because you can calculate, using classical physics, you can calculate what this force is, the centrifugal force. You know how fast the molecule is going around, so you can calculate what v squared over r is, which is the force pulling the spring. From the change in the moment of inertia, I mean, from the change in the spectral line frequency, you can estimate this change in moment of inertia. So you have a given force and a given displacement, so you can work out the spring constant. And then you can check, then you make a prediction for what the frequency is at which the carbon and the oxygen would oscillate in and towards each other using the spring, which appears in a different part of the spectrum. And so that's what that problem is about. So I think that's all. There are other nice things you can do with carbon monoxide molecules, but I think we'll leave it at that because we have an important additional item to put on the agenda, which is orbital angular momentum. Maybe we should put it over here. OK, so classically, we know what angular momentum is for a particle. Classically, well, let's be careful. So the world, the Earth, has angular momentum about the center of the sun for two reasons. One is that it's moving around the sun once every year. And that motion contains a great deal of angular momentum. And the other is it's spinning on its own axis, which accounts for a slightly smaller amount of angular momentum. But it's still angular momentum. And orbital dynamics are very much involved with the interchange of angular momentum between orbital motion and spin motion and so on. It's the same quantity. It's exactly the same on atomic scales or whatever. And as I've said, electrons and protons and neutrons are all gyros that, like the Earth, they spin on their own axis. But of course, they also move around. And in moving around, they have angular momentum. So at the end of last term, we introduced these angular momentum operators by considering, by very general considerations, to do with what happened when we generated the operators that would generate new states that were like the set which gave us a system just like our old system except rotated through some angle. And that apparently has no connection with angular momentum as classically conceived. Or so in classical physics, we have a thing, orbital angular momentum, which is going to be given by L is equal to x cross p. So this is classical physics. This is classical orbital. So this describes the angular momentum of the Earth about the center of the sun by virtue of the motion of the Earth around the center of the sun where p is the momentum of the sun in a frame of reference in which the sun is stationary. Right. So we can define an operator. So we define, let's go down here. So in quantum mechanics, by analogy with this, it's natural to define an operator, L hat, which is equal to 1 over h bar x hat cross p hat. Let's write that out in components to make sure we know what we're doing. That's 1 over h bar, the sum epsilon ijkxj6i, sorry. No, which way around do I want to do this? It doesn't actually matter how I do do it. But for consistency, I should try and keep the same as we have here. Yeah, so that's consistent with what was on the book. So this is summed over r over j and k. So things to notice. One is we put a 1 over h bar in here to make this thing dimensionless. It's a close call whether you should make it dimensionless or not. Most people probably don't have it dimensionless. I think on balance, you're better off having dimensionless. And we introduced the angular momentum operators in such a way that they were dimensionless. They didn't have an h bar, which made our formula simpler. So for consistency, we need to make these dimensionless. So this has dimensions of this. And therefore, this ratio is dimensionless. Does this make sense? Do we have to worry about the order in which we write down x and p? In classical physics, we clearly don't. In quantum mechanics, you would think you would have to worry about the order. Does it matter whether this is xp or px? Well, it doesn't matter because in this sum, the only terms which occur are when the subscript on the x is different from the subscript on the p. Because this epsilon symbol, remember, vanishes if any two of its subscripts are the same. So for example, we have that lx is equal to 1 over h bar, the sum over j and k of epsilon ij epsilon xjk xjpk. So j sums goes over x, y, and z. But when it's x, you get nothing here. So there are any two places to consider when it's y and when it's z. And when it's y, we don't need to consider the possibilities that k are either x or y. Because this will vanish if k is either x or y. So this is 1 over h bar of xy, which is y hat pz minus, minus z hat py. So that's how it'll work out. And these two operators commute, and these two operators commute so it doesn't matter about the order. So order is not important. Lx is also going to be a Hermitian operator. Because if we take the dagger of this, the dagger of this equation, we will have the dagger of what? Well, let's do it. Lx dagger is going to be 1 over h bar. The rule is, when you take the Hermitian adjoint, you have to reverse the order of the operators. So it's going to be p dagger zy dagger minus py dagger z dagger. Well, these things are all operators. But each of these things is its own dagger, because the momenta and coordinates are Hermitian operators. And the order in which you write them down, we've already agreed, is unimportant. So this is, in fact, equal to Lx. So it's a Hermitian operator. So we expect it to be associated with an observable. And the observable is obviously going to be orbital angular momentum. That done, the next thing, oh, yes, one more thing. We introduce by analogy with angular momentum. We introduce a new operator, L squared, which by definition is Lx squared plus Ly squared plus Lz squared. And again, it will be Hermitian, because Lx is Hermitian. So Lx squared is Hermitian, and so on. So this is another Hermitian operator. The next thing to do is to work out some commutation relations. Now that we've defined these operators, let's find what commutation relations we have. And let's do Li xl. Well, might if I leave off these hats now, we're thoroughly stuck into quantum mechanical operators. So what's this commutator? Well, we should write that in what this is. This is epsilon i jk xj pk. This will be summed over j and k. So this is this commutator on xl. So this is the commutator of an operator with a product. So in principle, there are two terms. This thing stands idly by, whilst that commutes with that. And then there's this thing stands idly by in the back, where this commutes with this. But obviously, the x is all commutes, so forget that. So we only have to consider this commutator. So this is equal to the sum epsilon ij jk of xj standing idly by while we do the commutator pk xl. But this is minus i h bar delta kl. So we have minus i, sorry, I'm missing here, 1 on h bar. Am I not? Because Li is 1 upon h bar of this product. So we have a 1 on h bar here. Then this generates a minus i h bar. The h bars cancel. The minus i sticks around in this pane times epsilon ij k xj delta kl. This is still summed over j and k. When we sum over k, this becomes an l. So this becomes minus i summed over j of epsilon ij lxj. And we can get rid of this unattractive minus sign by swapping the order of these two. So we can write this as plus i is sum over j of epsilon i lj xl. So let me just write in the left-hand side again so we can see, appreciate the pattern that we're getting. The commutator of Li with xl is i times epsilon il. These two letters being those two letters, sorry, this was summed over j, jjjj, summed over j. So between here and here, I've merely reversed the order of the subscripts on the epsilon, picking up or dealing with the minus sign. Now this is exactly the same as a result we already had. So we need to recall at this point that ji comma xl is equal to i sum over j epsilon i lj xj. So the commutator of this orbital angular momentum operator with this position operator is the same as the commutator of this total angular momentum operator with the position operator. Similarly, just the same calculation, precisely analogous calculation, implies that Li comma pl. We just sit down and calculate this in exactly this way. We will find its i sum over j of epsilon i lj pj, which mirrors, and there's an analogous relationship between j and p. The next thing to calculate is, so what's Li comma lj? We've introduced a family of three operators. We should investigate what the commutation relations are between any two of them. So we know what the answer is in the case of the angular momentum operators, ji jj is i epsilon ijk jk. So this is setting us up for expecting what the answer is here, OK? But in order not to get it done, so it's an exercise that I would encourage you to do to work this out just as it stands, but I'm now going to do a simpler calculation, just one component. So I'm going to do, so there's a question mark associated with this. It's a good exercise to do that, but it's slightly complicated to do it in the general case. So let's do something that's slightly simpler. Let's work out what lx comma ly is. So what do we do? What we do is we replace one of these, shall we say, this one by its expansion in terms of x and p. So this is going to be 1 over h bar of lx comma ly. So what is ly? Well, ly must be, I think, zpx minus xpz. All right, so this product divided by h bar is that. If I've not got my signs wrong. And we know now how lx commutes with this and how lx commutes with that so we can work it all out. So this is going to be 1 on h bar, open a big bracket of lx commuted with z with px standing idly by plus z standing idly by lx commuting with px minus lx commuting with x and pz standing idly by minus x standing idly by, while lx works on pz. The easy terms here of this and this, because they're what? Zero, right? Because this would be epsilon xi epsilon xxk, that vanishes similarly here. So these two are nice and equal to zero. And these we have to work out using horrible cyclical everything. So this is going to be minus. So lx commuted with a component of x is going to produce i times the third component, which will be y. And I think we'll probably get a minus sign. So I think this will be 1 over h bar brackets, and there'll be an i, sorry, an i. Then I think we'll have a, well, maybe I should leave it in here, a minus i times y from here times px. And this was zero, this was zero. This is the other interesting thing, one. This is the other interesting one. It'll be the same thing. This commuted with this will be minus i p y. So we have two minuses, so we'll have a plus i x p y. So this is equal to i over h bar of x p y minus y p x, which this h bar and this stuff together make lz. So it's i lz. So in summary, this again mirrors a result we saw with the total angular momentum operators. Just so this is one component of, I'm going to now write down the answer for that calculation up there without proving it, but you can see that it's going to happen. Li comma lj is i summed over k of epsilon i j k lk. So let me repeat a result we've already got. Li xj is equal to i epsilon i j k x k. So l commuted with any component of a vector produces i times the third component of the vector. And that rule even works for l itself, which is itself a vector. So we regard this as analogous to this relationship here, because we also had, sorry, we also had that li comma pj is equal to i epsilon i j k pk. The point is that what goes in here for l to work on can be any vector, a component of any vector, and then you always get out i times the other component of the vector you put in here. So here also we get out i times the other component of the vector we put in here, which in this case was l itself. This is mirroring precisely. So these results exactly the same. If you replace all those l's with j's, everything remains true.