Okay. So yesterday. We slaved over a hot chain rule in order to recover this formula here. So what we're trying to do is find the. Is is find the wave functions that represent the states of well-defined orbital angular momentum. And I explain what the strategy was for doing that.
And that strategy involved knowing what these differential what these operators are as differential operators in the position representation and some rather tedious chain work was required in order to extract this formula here. And I finished with a triumphal acquisition of this formula here, and I made it stated that this was what you got if you pursued the line of argument, the find L minus. Okay, so we're now in a position to find these wave functions, let's call it Abassi.
L m this will be some function of our feature and fine because we're looking at this in spherical polar coordinates. So this of course is are Theta and Phi L m. Now, the radial dependence of this wave function is going to be completely unspecified because we're only going to require what we're going to require is that L.Z. on L m is equal to m l m an this thing is got to be an eigen function of this operator with the eigenvalue m which we now know to be an integer.
And we are going to require similarly that l squared on l m is equal to l l plus one of l m we're not going to direct and we will find we haven't yet calculated this operator this what L Squared looks like as a differential operator. We will get to that. But it will also turn out to be involved in derivatives, relate to theatre and PHI. So these operators, none of them involve anything about radius. And so this function is an arbitrary function of radius.
And all we're going to be able to discover is what its angular dependence is by imposing these requirements here. Okay, so what can we say? We can say, first of all, this equation is going to imply put into the position representation. It says that minus i dvt phi of Abassi is equal to m of a PSI, which we of course immediately recognise is telling us that if PSI at Theta and Phi is equal to e to the i m phi times of psi at r theta and nothing,
if you see what I mean. Nothing and nothing and nothing. So there's some kind of a constant here. Write this. This thing doesn't depend on PHI. If you differentiate this with respect to Phi, you bring down an I am the I's, make another minus sign that cancels this and you end up with them claims or whatever it is. So what we know is this should have it subscripts. I suppose so. What we know is that psi l m is equal to some function of our and theta times e to the I am fine.
I guess we kind of already knew that. Now we're going to. Oops, and we're going to. Well, yeah. We're now going to impose the condition that l plus on up ci l l is equal to nought because this this operator would create a state in which m this is we've put here the value of M equal to L, which is its largest value. This would try and make a value, make M even bigger than L and we know that's not possible. So we have this equals zero. So, uh, copying down what that is.
Turning this equation is the position representation. Each of the i fi sorry. Each of the yes, each of the i fi d by d seta plus i cotangent c to d by the fi operating on e to the i l fi times. This function k which depends on our theta especially. It depends on c to the our dependence we don't care about because we got no differential operators here with respect to R so this term looks only at that and bring this down and I. L Right, so, so what's that? I'm sorry, this has got to equal zero.
So we get two terms looking at this DVD C to term looks only at that so we discover the decay by D theatre minus cop theatre. Uh, sorry l cop theatre. We're bringing down an i l k equals nought there. Then there's a factor of e to the something or other phi which we can cancel the way. It's not interesting. So here we have this is a, this is a linear first order differential equation, the friendliest kind of differential equation. So we solve it with an integrating factor.
The integrating factor is e to the integral of this here each of the integral of minus l loops, minus l cut C to de seta. But I think that the integral of thi to dc to is log sine theta. So this e to this becomes e to the minus l log sine theta or e to the log of sine to the minus l feature or simply sine to the minus l theta. That is the integrating factor of this equation. In other words, the equation states the DVD theatre of the integrating factor, which is sine to the minus l feature times.
The function is equal to nought. In other words, this thing is equal to a constant. In other words, K is equal to a constant, which is obviously to be some kind of normalising constant times sine to the alpha beta. And we have discovered this constant in principle depends on all right, it's allowed to have any R dependence you like. So what we've discovered is that if sai l l is any function of r u like times sine to the l theta e to the i l fine.
This is an important kind of result. And now we're in a position to calculate anything else, because if we want to find what upside l, l minus one is, then it's equal to l minus divided by an appropriate normalisation factor, which happens to be l l plus one, minus l, l minus one. Remember, these letter operators come with square root normalising factors.
That was the case in harmonic oscillator. That's the case also with the anchor momentum operators operating on upside L, which we now know what it is. We now know that it's sine to the l theta e to the i l phi times the unspecified function of radius and this this l minus. Let's, as roughly speaking, put it in what it is. No, maybe we do it on the next board because we want to be able to see we want to be able to see those magic formulae.
Right. There they are. This tells me that upside l l minus one is equal to I think this is just a square root of two l So it's a function of radius over the square root of two. L Or being well times e to the minus ei phi times.
There's probably an overall minus sign coming from that formula at the top there d by de thetr minus I got theta d body phi working on the function we first thought of, which is signs of the l theta is the i l phi and what are we going to get this phi will again bring down an l etc. and then this exponential will take one off that. So we'll end up with something that goes like E to the high L minus one phi.
This will differentiate sine to the L and produce l signs the l minus one times a cosine this cotangent multiplying that because this is crossover sine will again produce me a cos time sine to the l minus one. So the whole thing is going to be minus unspecified function of radius of the square root of two l times is going to be everything's going to go like each of the minus I l minus one phi.
And then from here we're going to get an L from here differentiating that, we're going to get an L. Well there's going to be a factor sorry of sine to the L minus one theta cost two. And how many of them from here will have an L and from here we will have, we're going to bring down an L what a minus. Sorry. And I l they will cancel so I think it'll be plus another l of the same same stuff.
So you see that we have something like the square minus the square root of l over to whatever unspecified function of radius was e to the minus i l minus one five times sine l minus one cosine theta. And we could now apply EL Minus to this again and get and get the next in sequence, right? We're not going to do it because life gets very boring. L l minus two.
But it's just a matter of differentiating. Well, the thing to pick up is that when we do this next, when we when we differentiate this, this thing is going to become more complicated because we're going to be doing a derivative of this with respect to theta. So we will get a term that goes like sine to the L minus two times cos squared and then differentiating this will get our sine to the L back. So we'll get two different terms and then when we differentiate again to get a side.
So this is going to be an amount of this will, this will it's going to be amount of sine l minus two times cosine squared differentiating this will get l minus sign of the L minus two and then we get a cosine which goes on to that. And we will also have from differentiating this one plus an amount of call it B of sine to the alpha beta.
So there'll be two terms and it'll all be times E to the minus L minus two phi and when we differentiate this again in order to get up sine l l minus two, it'll get more Byzantine because this will generate me an l sign to the l minus three times cos cubed. This will get back what we had here and so on and so forth. You get more terms, you get a longer thing coming in front of the exponential. So what do these things actually turning out to be?
It turns out that what this is is is a normalising constant times p, l and of course the beta times e to the minus. So if you know. Well, let me make that clear. If we just keep going, this will turn out to be a normalising constant. At times, the associated illusions function PLM of costs 3 to 4 times easier than minus easily. Sorry. That's an iron or a minus. And it. Yeah. This should have been a plus e to the. I am fine. So this thing I think you may have met this.
Right. And Professor S Lewis lectures this is an associated nagendra function probably derived from solution in series. By using probenecid method. I'm not sure. Is that right? But fundamentally this is fundamentally I don't think this is very helpful knowing this is an associated genre function. I think it's much more helpful knowing how to do it this way. The the normalisation factors take care of themselves.
If we put in these square root animals and we start with this thing correctly normalised, how do we normalise this? Traditionally, what we do is we say a cycle is proportional, is equal to some function of radius to be discussed times while m of theatre where this thing the spherical harmonic. With this thing. This vertical harmony is is a multiple of plume times each of the I am fi normalised so that so that if you integrate d c to assign c to define over the sphere of y l m mod squared,
you get precisely one. So the y islands are correctly normalised. So if you mod square them, integrate them over the sphere, they come to one the LMS and have a daft normalisation. And that's why I don't think you should bother with plants. They're just stupid functions. They've been historically, they've been defined in a bad way, the way limbs, the things to go on. But the wireless is actually one of these functions.
Of course, theta times each of the I am phi so it has a very simple phi dependence this animal here and we need to undo it. So, so now let's all know something, let's just summarise what we have. So these things, y, l, m theatre and Phi are the wave functions. Essentially, they're the wave functions. Theta Phi L M They're the wave functions belonging to states of well-defined orbital angular momentum. That is to say if in the position representation you apply.
Well, yeah, L.Z. to y l m l and you get m times y l m which is a trivial result because this thing goes like e to the I am phi and if you apply l squared to y l m you get l l plus one of y 11. So if you have an electron, here's the nucleus. If you have an electron in orbit around the nucleus, it seems reasonable to say that. It's reasonable to ask, what does the orbit of what does this system look like? What of the wave function of the electron look like?
If the electron has well-defined orbital angular momentum, the answer is that its wave function is going to be a function of R, which will be we will see will tell you whether how much it's oscillating in radius as it goes round and round time is one of these while m things. So these while m things should give us we should be able to understand them in terms of in terms of orbits.
At some level. Right? So let's let's address ourselves to that. What can we understand about these mathematical functions while in terms of what we understand intuitively about how an electron should go in an orbit around its nucleus? So the place to start is not is when L is large, because when L is large is when we're approaching the classical regime for which we have some grip. And the pictures at the top here, a contour is these are contour maps of the real parts of y limb.
So while M is an inherently complex thing, right? While M consists essentially of plm, some real function of course theta times each of the I am fi. So by focusing on the real part of that complex function, we've got that p 11 times cos amphi and these ones at the top are all for L equals 15. This is for M equals 15. This is for some intermediate one, m equals seven and this is for m equals two. So what's the physical interpret? What what's the physical interpretation of these?
This thing, this, this. And why l m is a function on the sphere, right? It assigns a complex number to each point on the sphere. So this has been the real part is a real number on the sphere. And what's being plotted here are contours of constant value of this real number on the sphere. So you have to imagine that these are pictures of spheres. The so what do we see here is that around the equator we have.
So dotted contours mean negative values of the real part and and full contours mean positive values of the real parts, the large values of the real part or around the equator here. And that's apparent from this maths because we know that y this is y l l for l equals 15. So in fact, it's sign to the 15 seater e to the 15 of each of the 55.
Right. That's what this thing here is. And if sine theta is one on the equator and less than one everywhere else, if you take a number that's less than one and raise it to the 15th power, you have quite a small number. So you were expecting that the number get small quickly as we go away from the equator. That makes this exactly what we expect on physical grounds. Because why? Because the state L equals 15.
M equals 15 means you've got 15 units of angular momentum, broadly speaking, and they're all of them parallel to the Z axis. So this thing is an electron that's orbiting with its in a plane, classically orbiting in a plane, the equatorial plane that was perpendicular to the Z axis. So where do you expect to find the particle? You expect to find the particle in the equator and nowhere else. Where's the wave function? Peak in amplitude in the equator? Nowhere else. Why is it segmented like this?
Like an orange, right? It's we have sort of waves going around the equator here. It's big, small, big, small, big, small. That makes perfect sense. Because because the change in the because p p the momentum is minus ei h bar d by the DB position. Right. So if you have something with a large momentum, it, it's to do with a large gradient or a large rate of change of the wave function.
Now, the amplitude of the wave function does not change one iota as you go round the equator, because this thing has amplitude, which is sine to the 15th power of theatre. So it's completely constant one round the equator. But the phase of this wave function is changing like crazy as you go around the equator because it's e to the to the 15 EI phi and that is expressing the fact, according to this, that the momentum of the of the particle is around is directed tangentially around the equator.
It's rushing around the equator. In the equator, when it's rushing around the equator, what else would you expect? That's exactly what should be the case. So let's go now to this case, which is oops, I lost it and equals the the extreme right one m equals two. So we've still got sine sorry, we still got l equals 15 but we have m equals two. So we've got a particle which has 15 units of angle momentum, but only two of them are parallel to the z-axis.
So classically what this amounts to is that his US, his loss, his notional sphere. And you think that the orbital plane classically would be tilted like this one? Even more so sort of like this ish, very highly inclined, so that the the spin axis of the orbit was pointing almost in the X-Y plane. So what we're expecting is that the motion is mostly from the northern hemisphere down to the Southern Hemisphere and back up again.
So we expect that the contours on which the the phase of the wave function changes rapidly. Fiddlesticks this. So knowing the, the, the duration, which the phase varies should be from north to south. And lo and behold it is right. So the FE so now we have is having an orange peels plan. We have, we have sort of rings going around on which, almost on which the phase is so.
And if indeed we had, we put m equal to zero, we would have a wave function which had no variation as you went around the sphere. It would all be variations. You go from the Northern Hemisphere to the southern hemisphere, which corresponds to the fact that the particle is moving this way. Now, this particle has most of its angular momentum in the X-Y plane.
But the thing is, we don't because we know because we know the angular momentum in the Z direction and ls that does not compute with l x. We do not know how much angle momentum it has in the x direction. Most of his angle momentum is in the x and y directions, but we don't know whether it's positive or negative. So that means that we can, in this picture, see an orbital plane.
The probability of finding the of finding the particle is sort of large all the way down here and all the way down there. And and if M was zero and the angular momentum vector, where exactly in the X Y plane, we would we would have absolutely no variation in probability to find the particle as we went around and around the, uh, the sphere, when in fact even now we have no probability go round around the sphere.
The real part. So, so what this thing is, is a function of thetr times each of the i to fi so the phase is varying as we go around the sphere. But in fact the amplitude is not varying as we go around the sphere. The the amplitude to find the particle is constant as you go around the sphere on small circles. And that is associated with the fact that we do not know, we are not allowed to know. It is forbidden to it, to us to know which way this angular momentum vector is pointing.
But where are we most likely to find the particle? We likely to find the particle most likely in a given patch on the equator or most likely to find it on the pole. Well, this wave function is largest at the poles, the North Pole in the South Pole because it's going around this particle is going around all over the poles in a in a plane, which is of unknown orientation. So so we do we there's great uncertainty.
There are many places where it could cross the equator. But what we are sure of is it goes close to the pole. So that's why the probability there's this sort of crowding of the margin, a bunch of circles for a polar orbit going round the sphere, a different orientation as they all pass through the pole will be a great crowding of the circles near the pole,
and that generates the high amplitude to find the particle at the pole. A relatively low amplitude to find is at the equator, but not a vanishing amplitude to find the equator because it does cross the equator twice in each cycle. So this this amazingly, this sort of this is an intermediate case. M equals seven, L equals 15. This curious mass of squares in which the you can see the real part is alternately positive and negative.
The contours are dotted and full. This represents the situation where the orbital plane in classical physics, the orbital plane, which is moderately inclined at 45 degrees or 30 degrees or something to the equator. And there's absolutely no orbital plane visible there. And this is where we we come to a key point that if you want an orbital plane to be visible and after all, the orbital plane of the earth is entirely visible and and the earth presumably moves according to these principles, too.
We have to have how do we get an orbital plane to emerge? The way we get an orbital plane to emerge is by quantum interference between many states that look rather like this and have a patchwork of of pluses and minuses. If you have several of those patchworks of pluses and minuses, you can get the amplitude to cancel most places, except in some inclined orbital plane.
So it's uncertainty in the angular momentum which will generate for you, if you want it, some degree of certainty in the location of the orbit going around the sphere. It's the it's the old uncertainty principle over again. So those those are the classical. This is always the classical regime up here. Right. And of course, the as the earth goes round the sun, it's angular momentum is, who knows, ten to the 50 bar or something.
Right. It's simply I haven't worked it out. It's some staggering number. So you would have to imagine ten to the 50 little patches here of pluses and minuses or maybe it's tens of 50 squared. I think it probably is 10 to 100 patches of pluses and minuses. And then you can by taking a number of those, maybe you take ten to the 34 of those with ten to the 50 patches, you will be able to arrange for exquisite the pixels to cancel everywhere.
Except in some extremely narrowband, which is the client orbital plane of the earth. So atoms don't live in that regime up there of equals. 15 atoms live in this regime, this tiresome regime down here. This is. Where am I? This equals. I've lost it. This is l equals one. And this is. These are the three things. Four equals two. So this is. This is why one? One. So that means you've got one unit of angular momentum and well, it doesn't actually write because what does what does L equals one mean?
L equals one means that L squared has answer one, one plus one equals two. So the total angular momentum, the square root of L squared has answer root two, which is distinctly bigger than one. So we've got as much angular momentum along the Z axis is in this one one case as we can, which means the particle definitely is is going around the equator. So and you can see that it's going around the equator.
Well, I can't from this angle, but I hope you can in the sense that, that the, uh, the thing isn't constant. The wave function has gradient as you go around the, as you go around the equator, there's a gradient. On the other hand, the, there is not a very high probability of finding it in the, uh, the, this is only the real part of the wave function. If we would look at the imaginary part of the wave function. Well how is this one?
This one goes like sine theta, not like signs to the 53 to this function here is sine theta times e of the I phi. So as you as you in the equatorial, as you go away from the equatorial plane, the amplitude to find the particle falls but only falls like sine theta. So it's really quite likely not to be the equatorial plane and that's associated with the fact that we've done our best to get the angle momentum along the Z axis.
It isn't along the z-axis because its total angle mentum is is 1.4 something times each bar and only one of those units is along the z axis. So it's some sense, inevitably inclined and this is the case when we have no angular momentum r on the z axis.
So this is the case of polar orbits. The amplitude to find the particle is greatest at the two poles, smallest of the equator, etc., etc., etc. But the whole picture is less clear cut and I won't bore you by talking about these, but it's worth thinking about the equals and equals to case, to see to what extent you can make sense of these physical sense of these of these pictures here. Okay. So now we should address an important topic, which is the parity. This is practically an important topic.
The parity of why l and. So remember the parity operator p working on up sci makes a state whose amplitude to be at x is minus is the amplitude to be at minus x. If you were in the state of SCI as the definition of the parity operator and these states are well-defined angular momentum have well turns out have well-defined parity does what we're about to to show. And what's more the parity is minus one to the L so states of of different angular momentum have alternating parity.
Some are even parity. Some are old parity. Those what we want to show. Okay, so as as what we do now is so this is sort of imagined in Cartesian coordinates. We, we need since all Y's are all defined in terms of polar coordinates, we need to translate the operation of going from X to minus x it just spherical polar coordinates. So as x as we go from X to minus x, it's easy to check. But what happens is that theatre goes to Pi minus theatre and Phi goes to Phi plus Pi.
So this reflection action, you need a picture. Really? Well, we can just about show it. I suppose I hate three dimensional picture because the three dimensional picture. Okay, here's theta, the circle, polar coordinate theta. And what you do, what we have to do is take this point and move it down here. Right. And what we do is we, we, we move this point down to here.
That's the theatre goes to. So theatre this theatre goes to pi minus theatre and then having got it down here, we rotate it through out of the board and back into the board through Pi and Phi and that's how we get it down here. So these are the changes in polar coordinates that are associated with that. Now. Why? L l. Oh, yeah. Well, what else can we say? When? If you go some to pi minus theta, what does that have to say about sine theta?
Sine theta goes to sign pi minus theta and sine pi minus theta. It's easy to check from variety of arguments is actually equal to sine theta. So sine theta doesn't change and e to the i l phi what is what happens if you add pi to each of the I'll find where you're adding E to the R, you're getting an extra factor E to the l pi, which is minus one to the l times. E to the i l fi the right. Okay. Now, y l l is a constant, rather a yucky constant.
So I won't bore you with it. Times sine we've proved this sine to the l thetr e to the i l fi. So this thing does not change sign or it doesn't change at all. Right. So we can say now that y l l goes to this doesn't change sign. It doesn't change at all. And this one changes sign. So it goes to minus one to the l of y l l under x goes to minus x. So the this this means the parity of y l l is even if l is even odd. Otherwise, that's a very important result.
And moreover, it generalises because we have we have the y l l minus one is l minus over some square root. That's really boring. Well, it turned out to be two l. So when you put it in times y l l and what about this once l minus l minus is l x minus i l y in the position representation. What is this? This is minus h bar of y de by de z minus z de by t y plus minus who knows.
Rule of h bar. It doesn't much matter. The key thing is that we're going to have here is z d by x minus x divide is that when we change x to minus x, y to minus Y and z to minus said these things we get we get change of sign here and a change of sign here. Change the sign here. Change the sign there. So L minus. And also, as a matter of fact, L plus is is unchanged. By P. The strict mathematical statement is that the parity operator commutes with either of these animals.
Indeed, all the Anglo momentum operators commute with the parity operator basically because they contain products of position positions or if you like, ratios of positions, whatever, they don't change. So what that means is that what that means is that this is going to have the same parity as this, because if you apply the parity operators, this, you know, parity the parity operated this, those can be can swap in order. This turns to minus itself. The minus sign can be taken out.
And therefore we've shown that that leads to the conclusion that this thing has the same parity as this. Let me just write that argument down, perhaps. So we have the p l minus. Sorry, p on y l l minus one is equal to p l minus sy sorry not sy y l l over some square root for not interesting is equal to l minus p y l l of the square root, which is equal to minus one to the l times p times. Sorry. This thing produces y l l so we have l minus y l l over the square root, but this is y l l minus one.
So it's equal to minus one to the l of why l l minus one. So we conclude. But why? L l l. M has parity. Minus one to the L for all m. This is a very important fact because it enables you to set to zero all sorts of integrals, which would otherwise be very tough for them to work out. How we doing? Yeah. I realise that there's one other thing. So. Which we've unfortunately lost. Is it coming back? No. Well, what I wanted to do was show you the forms of the wire limbs, the first few.
You need to have some sense of how they go. Right. So why? Nothing. Nothing is one over the square root of four pi. Why? Which all they put in. So we want y one. Nothing is is basically cos theta that happens to be some factor of root three over four pi y11 is of course sine theta times some normalising factor e to the ei phi y one minus one would be the same thing with a minus sign here.
So the point is that the, the y ones go like cost data and sine theta and the y twos go like costly to sine theta. So y to nothing is equal to a normalising factor that happens to be five over 16 pi. That's not so interesting. Times three cost squared theta minus one and y21 is minus the square root of 15 over 32 pi, which is not so 32 pi, which is not so interesting.
What's important is it goes like sine to theta, which can also be written as sine theta cos beta and the other one goes like of course sine squared theta each, this one has e to the, to I sorry each of the I phi and this has e to the 2i5. So what do we need to remember? What we need to remember is that obviously y00 has no angular dependence. Yes. Okay, so think machine has finally come back to life and the correct formula here. That's what I wanted to show you. The Y ones have a cost or a sign.
The Y twos have a whiff. Well, that you can either think of. They have a strong whiff of cost to thetr and sign to feature. Right. Because cos squared theta is something like a half of one plus cost to thetr. So there's a whiff of this could be a rearrange to involve costs to theta.
Here we have a sign to feature and this sine squared has a whiff of cost to feature about it because we, because we know that sine squared feature is a half of one minus cos square of cost to feature something like that. Right. So we have these double angle formulae and so it would go on if we were looking at Y three we'd have a these things would have dependencies that look like cos cubed cos three theta and sine three theta.
Right. That's the pattern. But you don't need to know about the pattern beyond here, but these patterns here are expected or expected to be able to sense. So when you're given a function of theatre which is, which is made up, a linear combination of these things, you need to be able to unscramble it and write it as the right linear combination of those wise. All right. For next topic.
So in preparation for work on atoms, we need to get an important formula for how kinetic energy can be expressed in terms of L squared. And this finally obliges us to face up to the tedium of calculating what L squared is, what the what differential operator represents l squared in the position representation. Right. So we start by observing that L squared is it can be written as L which way round do I want to write it.
Yeah. Plus minus. Okay, I want to write. I can write either way, but I do it consistently like this. Well, let's see let's see what we're going to have to add to this to make l squared. This is x plus i l y l x minus i l y. What's that going to come to? That's going to come to l x squared plus l y squared plus well, minus i. L x comma l y. Commentator That's what this this thing multiplies up to. If we want to get l squared, we'd better. Here is a good start.
And L squared. Let's add L.Z. squared. But we need to get rid of this alex comer. L y is i l z. So we've got here. What would this minus sign? A plus. L.Z., we better taken L.Z. away in order to square the books. So that's what this should be. Sorry. This should be put equal to plus L.Z., squared minus L.Z. So that's that. So what we do now is we write down L plus L minus, which we have floating up there in the stratosphere.
So so we have an l squared is equal to e to the i phi d by d c to plus i cotangent c to d by define and that should operate on l minus which is minus. I'll take the minus inside the bracket e to the minus i phi divide phi d c to sorry. This minus sign was up there outside the bracket I think. Plus because I propagated the minus inside the bracket i cotangent c to d by defy.
So this disgusting mess is that product and then we have to add l z squared and take away L.Z. this thing is minus L.Z. is minus I DVT Phi. So with that minus sign, we get a plus. I debated Phi and this is going to be minus D to Buddy Phi squared. So the name of the game is to differentiate out this piggy mess and find out what it simplifies to. Some parts of it are easy, right?
We're going to have, for example, the end of the day, we will have terms where this is multiplying this and this is multiplying this. And these two exponentials have killed each other off. So we will have a term like D2 by dc2 squared. These eyes will generate sorry, they'll be a minus two by minus because one of these is got a minus sign. These two will create me a a minus called squared d2 by five squared. That's the easy part right now. The the mess.
There's going to be some mess because this differential operator is going to bang into that. Okay. And generate a minus I times what will kill this off so we'll have a minus I oops. Oh no. But then it's times this. So the minus sign that we're getting from here will meet this and generate a plus one. So we have cotangent theta that's this cotangent theta times this bracket. So that's the result of this differential operators seeing this.
When this differential operator sees this bracket, all we get, well, actually, we get a mixed derivative term. We get two terms. We get one term that we've already written down and we get a term two by DC to do by Defi. But that is going to be cancelled by by a term that comes from here when this differential operator looks at that. We'll deal with the differentiation of this in a moment. So I'm not going to write down those mixed derivative terms.
Otherwise, we have we've now so that almost on that understanding, we have dealt with the action of this on that. Now, what about this one? We've got the operation of this on this. We've got the operation of this on this. I've just said that. That's cancelled away. What we haven't got is this. When that differential operator meets this, we get the differential of what is squared, I think. So I think what we have is plus I kozak squared seta of d by define.
Now the sine should be checked at this point because because lines are are a pain, right? Well, I think it must be that the derivative is minus x squared. Votes can be taken afterwards. Right. So so that's that's the derivative of this on this. And then I claim that these brackets are dealt with. And all we have to do is write down the training terms here, which is a plus I d e phi and a minus due by the vice squad. Now we need to consolidate our various terms.
We have three terms one, two, three, which are just d by define terms and Gobi phrase. They all add up to nothing because we have a trig identity which is cot squared, minus x squared minus one. So we have the KOT squared, C2 minus Kozak squared, theta is minus one and here's our cot squared, there's our squared and I'm missing and this should have had an o i we have an eye problem, right? These have to be all. No, no, no. I'm not trying to mess with that one. I'm not trying to mislead.
Right. I'm going to have a cod squared here with with an associated attendant. I, I've got a squared with an I and I have here of a one. So let us by that that causes those all to add up to nothing. Then I also can use this identity to consolidate this double derivative and this double derivative. So we have a cot squared and a one and I can try that in four squared according to that formula of that.
Right. So we end up with minus D two by DC two squared it's going to be cot squared, minus cot squared. So it's going to be we've got a cot squared and a thing they both carry minus signs, which means I have to have them on the other side. So we get a minus six squared according to this. I'm slightly worried by this, so I'm going to end up with a Q6 squared D to Buddy five squared and I strongly suspect that sine is wrong,
but that's all I've honestly got. So that's this dealt with and the only this is so this has been dealt with this has been dealt with. This has been dealt with. I think we're all we're all tickety boo. We're not all we. What have I lost? To believe this is the easiest way to do it. It's hard to believe, isn't it? But it is. It is the single derivative that divided Peter. But this one. Yeah. Right. So. So that remains cop theatre DVD feature.
Thank you. Right. Yeah. So we now consolidate this all being well into one over minus one over sine theta d by d theta sine theta d body theta. And this should be, I think a minus that sine is wrong a one over sine squared C to this that's usually written D to buy D phi squared. So I've screwed up on the sign there somehow. So when you, when you differentiate this we get a cause which because of a sign is cut. So that's this term here we have the double derivative sine, etc. etc. etc.
What is this? This is all squared times the angular part. Of Del Squid. On that note, it's time to leave. We're not quite finished with the calculation, but that's the important bottom line that El Squid is is actually with a minus sign at minus L squared times the angular part of Del squared and we'll push that forward into the kinetic energy tomorrow. No, on Wednesday.