Okay. Okay. 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 JS, for example, Jay-Z 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 which are very common right there in the room. Hero to an end to much of the mass of the is a barrels.
The universe is contained in h two and a very important, from a practical point of view, a very important molecule for reasons I'll explain, turns out to be carbon monoxide. C0 So molecule like this consists for our purposes of two masses being the nuclei, for example, of the oxygen and the carbon. On a light springs we have two massive nuclei and a light spring formed by the electrons, and this is capable of being a simple harmonic oscillator in that it does this, but also at lower energies.
It's capable just of rotating and of rimmed 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 to I z. So this is reminiscent of this is a piece of classical physics that the energy of a chaotic energy moving particle is
p squared over two in this is so this is classical mechanics of classical mechanics and those who did short options. Seven will I hope recognise 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 and sort 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 a j x the classical thing goes to h bar times j x the operator. So we define these operators many people to find them to have dimensions of but we would define them so they would dimension less. This thing has the dimensions of h bar. So we make this transformation. We took, we take this classical expression and turn it into h by a times the quantum mechanical operator.
ET cetera. And then we infer that the Hamiltonian should be what we guess should be h bar squared over to j x the operator squared over twice. Sorry over taking that out plus q y squared the operator over i y plus jay z squared over it. So these are operators now. So this is this you should think of this as a guess. It'll be confirmed by experiments to be to be shown soon. Now we say if for this diatomic molecule.
We take. We take the X. 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 all 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 around this axis, those nuclei have this provide very little moment of inertia.
The electrons could provide a certain amount of moment of inertia. So it's possible for the electrons to have some. It's feasible for the electrons to have some momentum around this axis. And in some molecules they do. But in most molecules, they don't. So at the moment at the moment of inertia is, in any case, very small.
So this implies that Z is very much less than I Y with an eye X, which is the moment of inertia about an axis that sticks out here, where the if you spin it around there, the, the two nuclei go around in a decent circle of radius, half the length of that bond say.
And for a diatomic molecule by symmetry, x is going to equal a Y. So given that X is equal to a Y, it's natural to rewrite this Hamiltonian as H is equal to H bar squared over two brackets j squared which is g x squared plus two y squared plus jay z squared divided by x, which is equal to i y. So those first two I've combined together into here, I've now in here also got a jay z squared over x which I don't really want,
so I simply get rid of this jay z squared brackets one over A-Z minus one over i y ix. So this is just a repackaging of that expression to group together the terms which have a common factor one over i x and now we. 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 this says immediately what the spectrum.
Of H is e j m which is going to be h bar squared over two over two brackets. J j plus one. Because that remember, we showed the eigenvalues of this were j j plus one where j is allowed to be an integer or a half integer divided by x plus m squared one over i z minus one of i x. So we we have because we understand, because the Hamiltonian for this rotating system is simply a function of the angle meant by operators.
And because we have already found out what the spectrum of the incremental 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. Now we know that. We know that M lies in the range minus j less than or equal to where less than or equal to J.
That was one of the things we showed on Friday. So this number is sort of generically at the same order as this number. But this one over A-Z, I've explained is is a very much bigger number is it is very much smaller number than X. So this number is very much bigger than this number, which is the same as one over this number here. So the coefficient of m squared is huge. And that means that if you if m were anything other than zero, the energy would be enormous.
So when you are dealing with when you're dealing with molecules in the room or in interstellar space or somewhere where they only have moderate amounts of energy, we can we so we can argue that this thing is going to be is going to be zero. So this this number here is very large. I mean, it really is huge because of that. What because that is so very small for a diatomic molecule, which means that this is always zero zero.
Let's just say that these molecules have their angular momentum around an axis which is perpendicular to their symmetry axis. So that means that effectively. E the energy levels are simply e.j. H bar squared over two x times. J. J plus one. So those are the energy levels now. That's and that's that's all been totally generic for diatomic molecules.
Let's not talk about carbon monoxide. Carbon monoxide has a much more interesting spectrum, much more easily observed electromagnetic spectrum because the carbon is slightly positive and the effect on the oxygen is slightly negative. I've lost this. Right. Oxygen has a great affinity for electrons. It borrows some of carbon's electrons and ends up being a little bit negative.
This is a little bit positive. So a carbon monoxide molecule which is spinning and over end is a rotating dipole, electric dipole. So. So see, oh, this is plus. 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 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 and of range. Right. Classically, that would make sense. A rotating dipole should emit or absorb radiation at the frequency that it that it rolls over.
Let's see whether that's true according to quantum mechanics. So. So it will emit or absorb. Am radiation. And the and get a frequency whether it's have the frequency. So the frequency is going to satisfy h new is equal to e j minus e j minus one. Sorry, another point. Photons carry one unit of angular momentum, it turns out. Right? Not more, not less. So a photon is can move a molecule which has a total increment from J to a state which has a total element of J minus one or j plus one.
But it can't move it to J plus two or j plus three, or it can only move it by one unit in J because it's only got one use of angular momentum. So the angle and 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 the key point by interaction with electromagnetic radiation. So we're going to have that h new the energy. The photon is equal to the difference in the energies of the molecule between the the excited state he started in the state he finished.
And what's that going to be according to this formula here that's going to be h bar squared over to z i x sorry brackets j j plus one minus the same expression with j put equal to j minus one, so it becomes j minus one times j And I think you can see that this cleans up very nicely to simply two. J So we end up with h bar times j over. I think somebody must have a have a call. What does that tell us about new. It tells that new.
Maybe we 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 one. We can cancel one of the H bars. Well, this is an H on an H bar. Right? So this becomes equal to H bar J over to pi x. So that's the according to quantum mechanics. That's the frequency of emitted light. Can we reconcile this with with our classical picture? So how fast was the molecule? Was the molecule going around?
So classically. We can say that the angular momentum j x is equal to o x times omega. This is the angular frequency. Of 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 this mirrors the the classical statement that X is equal to m times x dot right. Omega is x dot rate of change of angle. I mean, it's the analogue of x dot.
This is the rate of change of position. That's the rate of change of angle. Well, I've already said that moment of inertia plays a role, a bit like mass, and this momentum plays the role of. Right, so this that's that's where this thing comes from. It's just a piece of classical physics. So when you so when you are in a state so what do we think that omega should be? Omega should be on the order of well, it should be equal to j x over i x that's classical doing it quantum mechanically.
This is going to be h bar times the square root of j j plus one. This is this is so this is a horrific sort of thing. We're saying that it's supposing it's angular momentum is along the x axis, right? Then J X squared is going to be on the order of h bar squared j j plus 1jx squared is going to be on the order of j squared, which will be this divide through by i x. And we want to relate this to the frequency we had up there, the factor of two pi.
I dead wrong. Oh, it's because this is this is the angular frequency. That's the actual frequency. Right? So this is equal to two pi frequency of this is this was angular frequency. This would be frequency of rotation. So we have the frequency of rotation. We expect to be h bar over two pi x times the square root of j j plus one. So this is going to be. So what do we 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 two pi by x so the in its upper state when it had angular momentum total increment m j j plus one 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 lowest state. We would have a minus sign here, a minus one. So we're just a bit below the frequency in which the light. So it. So the light is in fact emitted just the average of the expected rotation, tumbling rates at the upper and lower levels, which makes it, I think, perfectly good physical sense. So so in lower state. That's when j well. E J Minus one tumbling rate. It's just a bit lower. Oh, no.
This thing is going to sleep again. It's not my computer that's gone to sleep. It's the wretched lecture room that's gone to sleep. It's too irritating. We must get them to stop doing this. We do want to show you something today. Is it coming back? Okay. Right. Yeah. So this is this is the actual experimental spectrum of carbon monoxide. And so this is this is in gigahertz. So this is a terahertz two, terahertz three terahertz along there. And you you have a line here.
There's a line being drawn at the frequency, measured associated with the transitions from j equals one 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. Then you can't tell the difference actually on this plot and so on and so forth. So you get each line here is associate is telling you the measured frequency of a line, a spectral line from of 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 from one 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's a there are a couple of missing lines in here where people where I wasn't able to find what the mean. Somebody hasn't published a measurement or something. The other interesting thing to note, so so that's 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 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 one to nothing. So what's happening is that the measured frequencies almost conform to this, to this rule up here, but not quite. They, they, 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 okay so new measured minus new j is slightly negative if you like. And that's because new measured is equal to these what you would think classically is equal to edge bar over two pi j over i x, which itself is a function of j right? As these molecules you make this molecule show you firing in circular polarised photons and making this molecule spin faster and faster and 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 is sohc is cancelled by a slight increase in the bottom. And that's why 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 one of the so one of the problems on the problem set is to use this phenomenon to estimate the stretching and how stiff this spring is, because you can calculate using classical physics, you can calculate what the what this force is, the centrifugal force, you know, how fast the molecules are going around. So you can calculate what V squared overall is, which is the force pulling the spring from the change in the moment of inertia.
You can I mean, from this change in the spectral line frequency, you can estimate this change in a moment of inertia. So you have a given force, 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 to n and towards each other using the spring rate, 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 incremental. Maybe we should put it over here. Okay. 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 centre of the sun.
For two reasons. One is that it's moving around the sun once every year, and that contains 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 its langland mentum 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 we in the end of last term, we introduce 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. But and that apparently has no connection with angular momentum as classically conceived or so.
And in classical physics, we, we have a thing orbital angular momentum, which is going to be given by L is equal to X across P. So this is classical physics. This is all this is classical orbital. So this describes the angular momentum of the earth about the centre of the sun by virtue of the motion of the earth around the centre 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 quantum in quantum mechanics, by analogy with this, it's of it's natural to define an operator l hat which is equal to one over h bar x hat cross P hat. That's right. That out in components to make sure we know what we're doing. That's one over bar the some epsilon ijkxj6. Oh, sorry. No. Which way around do I want to do this? Doesn't actually matter how I do do it, but I for consistency, I should try and keep the same as we have to.
Yeah. So that's consistent with what was on the book. So this is some develop over J and K. So things to notice. One is we put a one 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 dimension as we introduce the angle momentum operators in such a way that they were dimensionless, didn't have an undesirable, and they didn't have an edge bar which made our formulae simpler. So for consistency we need to make these dimensionless, right? 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 X? Well, it doesn't matter, because in this sum, the only terms which occur when the subscript on the x is different from the subscript on on the P, because this epsilon symbol, remember, vanishes if any two of its subscripts are the same.
So for example, we have an x is equal to is equal to one over h bar the sum over j and k of epsilon i j f excellent x j k x j p k so j is sums goes over x, y and z, but when it's x you get nothing here. So there are only two classes to consider when it's when it's Y and when it's Z, and when it's why we don't need to consider the possibilities that K are either X or Y, because this will vanish.
If it's if they're either if K is either X or y. So this is one over x bar of x, sorry, x y. Which is why has p z minus minus z had p y. So that's how it'll work out. And these two operators commute. These two operators commute. So it doesn't matter about the order. The order is not important. Alex is also going to be is going to be a mission operator. Because if we take the dagger of this, right, the dagger of this equation, we will have the dagger of what?
Well, let's do it. Alex Dagger is going to be one over a bar. The rule is when you take the commissioner joint, you have to reverse the order of the operator. So it's going to be p dagger z y dagger minus p y dagger, z dagger. Well, these things are rule operators, but but each of these things is its own dagger because the momenta and coordinates, the permission operators and the order in which you write them down, we've already agreed, is unimportant.
So this is, in fact equal to Alex. So it's a commission operator. So we expect it to be associated with an observable. And the observable is obviously going to be orbital angle momentum that down 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 l squared plus l y squared plus L.Z. squared. Again, it will be emission because this it's Alex's emission.
So L x squared is emission and so on. Right? So this is another emission operator. The next thing to do is to work out some commutation relations. Now that we define these operators, find what computation relations we have and let's do l i x l well, mind if I leave off these hats now with thoroughly stuck into quantum mechanical operators? So. So watch this. COMMENTATOR Well, we should write in what this is.
This is epsilon i j k x j p k as it will be summed over j and k. So this is this commentator on x l So this is the commentator of an operator with a product. So in principle, there are two terms. There's is this thing stands idly by wells that competes with that. And then there's this thing stands idly by in the back where this computes with this. But obviously the X is all commute, so forget that. So we only have to consider this.
Commentator So this is equal to the sum epsilon i j oops j k of x j standing idly by while we do the commentator peak x l But this is minus h bar delta k l so we have minus I sorry I'm missing here. One on h bar am I not because l I is one upon bar of this product. So we have a one on h bar here. Then this generates a minus h bar.
The bars cancel the minus. I sticks around in this pain times epsilon i j k x j delta k l. This is still summed over j and k. When we sum over K, this becomes an L, so this becomes a minus. I summed over j of epsilon i. J l. X j. And we can get rid of this unattractive minus sign by swapping the order of these two. Right. So we can write this as plus i some over j of epsilon i l j excel. So let me just write in the left hand side again so we can see appreciate the pattern that we're getting.
The commentator l i with excel because i times epsilon i l these two letters being those two letters. Sorry. This was some of j j some of j. So between here and here, I've merely reversed the order of the subscripts on the epsilon picking up of of 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 j i comma x l is equal to i sum of a j epsilon i. L j x j. So the commentator of this orbital angle mentioned operator with this position operator is the same as the coming iteration of this total angular momentum operator with the position operator.
Similarly, just the same calculation imprecisely precisely analogous calculation implies that l i comma p l we just sit down and calculate this in exactly this way we will find it's i some over j of epsilon i l j p j which mirrors and there's an analogous relationship between j and P. The next thing to calculate is. So what's l i comma l j. 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 angle momentum operators g j j is i epsilon i j k j k. So this is setting us up for expecting what the answer is here. Okay. But in order not to get so so it's an exercise that I would encourage you to do to work this out just as it stands. But I'm not going to do a simple 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 slightly simpler. Let's work out what x, comma, l y is. So what do we do? What we do is re replace one of these, shall we say, this one by its expansion in terms of X and P. So this is going to be one over bar of l x, comma, l y. So what is l y are well l y must be I think z p x minus x p z.
All right. So this product divided by age bar is that if I'm if I've not got my signs wrong and we know now how Alex commits with this and how Alex commits with that so we can work it all out. So this is going to be one on edge bar open, a big bracket of X committed with Z with standing idly by plus Z standing idly by Alex commuting with x minus l x commuting with x and p, z standing idly by minus x standing idly by while Alex works on p z. The easy terms. Here are this and this.
Because they're what? Zero, right? Because this would be epsilon x ei, epsilon x x k that vanishes similarly here. So these two are nice and equal to zero. And these we have to work out using horrible cyclical things. So this is going to be minus. So Alex, communities with a component of X is going to produce high times. The third component which will be Y. And I think we probably get a minus sign. So I think this will be one over bar brackets and there'll be an I sorry.
And I then I think we'll have a well maybe I should give it a minus. I times Y from here, times X 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 computed with this will be minus i p y.
So we have two minuses. So we'll have plus i x p y. So this is equal to i overage bar of x y minus y he x which this h bar and this stuff together make l z so it's i l z. So in summary, what we what we this again mirrors a result we saw with the total agreement of operators. Just so this is one component of again. Now write down the answer for that calculation up there without proving it.
But you can see that it's going to happen. L i comma l j is I summed over k of epsilon i. J k. L k. So let me repeat a result we've already got l i x j is equal to i epsilon i j k x k so l computed with any components 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.
All right. So this is we regard this as analogous to this relationship here because we also had sorry, we also had that l i comma PJ is equal to i epsilon i j k p. Okay. The point is that what goes in here for l to work on can be any component, any vector, a component of any vector. And then you always get out times the other component of the vector you put in here.
So here also we get out times the other components of the vector we put in here, which in this case will itself and this is mirroring precisely. So these results exactly the same. If you replace all those L's with JS, everything remains true.