Okay. So we were talking yesterday about polling matrices and the way that they generalise for arbitrary spin. I and I was just reached this point. I was well, it's interesting to understand the connection between poly matrices and the slightly strange things that happened with Spin a Half. And then it's also good to study the case, spin one and there's a problem set problem on that.
I recommend to you. And then moving right along to the case, a very large spin so that we hope to recover classical mechanics and understand how bodies which have macroscopic have many, many balls worth of angular momentum end up pointing in some very well-defined direction. So. The procedure for generating the power matrices is completely general.
We're just working out the. We're just writing down a matrix, each entry of which is the value of of whatever operator here said, squeezed between states of a well-defined orientation between the states. So this is the matrix made up of s m primed as said s m and it's very straightforward to work out what these numbers are. They're perfectly trivial in the case of S said, because these are states of well-defined.
This is not even state of that operator with eigenvalue, m, etc. So we just have down the diagonal the possible allowed values of M, which range from S to minus S. So this is the bottom of that matrix. And in the case of x, z, sorry s x we replace s x with a half of s plus plus minus. And then we have nothing down the diagonal. But we have nonzero entries just on the diagonal that lies one above the main diagonal and one below and zero everywhere else.
So that's just the generalisation was apparently matrix in which only this part in the case of the party matrix only this part exists where these at this is a function alpha of this is this s minus one is playing the role of M so alpha of M sorry, this should be an M and this should be an m is what you get. Sorry. These were the racing operators. This should be in plus one. Right. So it's this. It's this object here. Just some square root.
So for example, you it's very straightforward to pick a large value of s in the this this diagram here which should be up there is for the case of s equals 40 and then to for this large value of s to have your computer find the eigenvalues. Sorry, not the eigenvalues. The We know the eigenvalues we're sorry. Out of these three matrices I can construct if I take n is equal to for example, nothing sine C to come across the term.
So this is a unit vector which lies in the this is theta, this is the e z direction, this is the e y direction. So this is the unit vector. And so if I take any unit vector whatsoever, it has some coordinates like this, then I get the matrix force spin down that direction, being an x, x plus and y, y plus NZ as said.
So choose some angle fit to take the appropriate linear combination of s, z and y. I haven't written down as y, but it's it's it's essentially the same as that with a one over two I and some minus signs and then have your computer calculate the state, calculate solve this problem that s n on a vector which will be a 40 component vector a1a2 up down to a 40 well, a s in general is equal to let us say sa1 through s then what are you doing?
You're finding the the state express the state in which you are guaranteed to get the value s and that was the maximum possible value for the angular momentum in the direction of N And you are expressing that state as a linear combination of states with different amounts of angular momentum down the Z axis. So this number, these numbers are the other the relevant linear combinations there.
So what we're saying is that that an s in the direction of N is going to be a one of S in the direction of Z plus a two of s in the direction of sorry of of of s minus one in the direction of Z plus and so on. And actually, this is of length, facets of length to S plus one because there are two plus one possible orientations. So if S is 40, this is going to be 81 and 81 component vector.
So have it do that. And what you find is what is shown in this picture up in this diagram here, this is for three different values, three different values of cost of cost of theta. So this is for cost. Theatre is 0.5. This is full cost theatre minus point five. This is some other value of cost. These are read it and so on.
Right so, so for this, if you, if you take this value of cost theta which opposes -60 degrees, then the then these A's, which of course the complex numbers have moduli that look like this. So there's they're non-zero in some interval around here, which is to say, so what does that mean physically? What are these numbers? This is the amplitude that if you would measure along the z-axis.
So first get your system into this state, your system being in that state, we would understand it to say that its spin is in the direction of theta. What then is the this is this becomes the probability to measure that it has s units of angle momentum along the z axis. This becomes the amplitude to find that you have this minus one units along the Z axis and so on.
So if the angular momentum, so if the an element and vector really were a direction theta, how much would we expect to find along the z-axis so classically in this state? Theatre n. We expect, as said, to return. S Cos theatre. S Cos theatre is the projection of a vector of length. S pointing in the direction of theatre. That is the projection of that down the x axis, down the z-axis.
Sorry. So what you're finding here is that these, these amplitudes peak around the place where classical physics would say this is the answer, and the quantum physics is saying, well, you you have a chance to get all these answers with probabilities which are given by the square of these numbers. So quite strongly, Pete. And as you change theatre, so you change the vector, you change the state, the input states, you change the direction of your spin.
And you you change the place where these amplitudes peak. So that's only four equals 40. And classical objects have as of ten to the 30 or whatever. And as you get more and more as as becomes bigger, there are more and more of these dots along this line here. There are here 81 dots. I suppose 81 numbers are being calculated, right, because there are 81 components in the vector.
By the time you've got two tens of the 31 dots, you'll find that they you know, that they're really completely peaked around here. So that's how we out of this quantum mechanical stuff we re recover at high spin the classical idea that things point in some definite direction and you can go on to show that the that the expectation value of Y which classically should be should be sine theta is indeed a sine theta.
And what's more, the uncertainty you can work out the RMBS, you can work out what the expectation value of s y squared is. Then you find that that's essentially the same as the expectation value of y itself squared. And there's no ones there's very little uncertainty at highest in what you will get for y so so these these and what's happening here is in quantum mechanics, we have to calculate a whole series of numbers, which is the components to this.
So to describe the, the, the spin state of something we have to construct in the case of spin a half, two numbers in the case of spin one, three numbers and so on, two X plus one numbers, we have to calculate being the amplitude to find the various possible answers on said if you would make the measurement asset. What we're doing essentially is recovering the probability distribution for the asset measurements, which in classical physics is a delta function glitch at s cost theta.
But in quantum mechanics we don't. Our probability distributions are not delta functions there some kind of spread out things and you're seeing what they are there. But as you go to higher and higher spin amounts, the probability distributions narrow down around the direction of spin, which classically so in classical physics we say the direction of this spin is given by the Euler angles, by the by the polar angles, Theta and Phi.
We just have some completely definite. Oh come on you stupid thing. We have some completely definite direction. And what? Whereas in quantum mechanics we need a whole load of numbers because we're defining a probability distribution in classical physics, it is strictly speaking a probability probability distribution, but it is a delta function. And all we have to do is specify the, the centre point of the, of the, of the delta function probability distribution.
And we do that with just two angles. And in quantum mechanics we need a load of different numbers to spell out the whole problem distribution properly. Now, the other thing I wanted to say on this topic of, you know, relating quantum mechanical levels of incremental and classic levels of angular momentum is is the importance of this. So we know that squared has the total incremental operator has values. S. S plus one, which is clearly greater than S squared.
And remember, as this thing came into the world, as the maximum value of the angle momentum around the given axis. So. And how much greater this is than this depends on the value of S. So when S is a half, we have s s plus one is clearly equal to three quarters, which is three times the quarter squared. Sorry, a half squared. Right. This is the maximum value and that's telling us that you can have you always have a third of your spin down each of the three axes.
If you have a spin half particle and the most you can never know is where the one component is pointing this way or that way. We never remotely get the spin properly aligned with one axis because there will always be two units of an element and somewhere other in the in the plane orthogonal to that chosen axis. So when we have, as is one, we have se plus one is equal to two, which obviously is two times one squared.
So now the amount of and mentioned we can have down one axis is twi is is a whole half here. It was only a third. Now it's become a half of the total angular momentum. And each each orthogonal each direction in the perpendicular plane has has less than in the direction that you've chosen to align the angular momentum with.
As you go down to large values of s, you have that SSE plus one is practically equal to squared because s because obviously squared is going to be by definition bigger than SSE. And that means that we can get essentially all of our angular momentum pointing down to give an axis. So the important message is from this that we're familiar with this regime where we can get something to point in a well-defined direction.
But the atomic world works in this regime where there's always loads of angular momentum in in the directions that you haven't been working on. Okay. So I now want to turn to a new topic, which is the addition of an element to. The last thing we have to do with angular momentum. So this is a very important topic for atomic physics because atoms contain I mean, the simplest atom, hydrogen already contains a proton that carries a half bar of spin.
An electron has the same amount of spin, and then the electron may have orbital angle mentum. It may have angular momentum by virtue of its orbit around the proton. So a generically hydrogen atom contains three units of angular momentum. And we want to know, so what are the states of the atom in which the atom has well-defined, angular momentum? So we're going to study and this is an application of the machinery that we introduced, I guess, early this term to discuss composite systems.
This is a classic. This is an application. Of the of our theory of composite systems. So if you feel unsure about the theory of composite systems, please go back and have a look, have another look at it, because this is what we're going to be applying. So all that stuff about Einstein but Wolski Rosen cetera. What underpins that? Because what we're going to do is we can we're going to consider. Two gyros. We're going to have we're going to have Jarrod one has j one.
So it has has has total. Jay Jay one. Jay one plus one. So it has m m lies between minus Jay one and Jay one. So that's the rate at which this gyro spins is fixed by some servo motor or something. Right. And it's spinning at this rate. And we're going to have gyro two, which obviously is going to have total angular momentum squared. Is this so? So this will be M1 and M2 is going to lie between minus J2 and J2.
So we've got these two gyros of there might be two objects belonging to a navigation system and we're going to stick them inside a box and they're not going to talk to each other. They're going to have there's going to be no Hamiltonian, as we know, coupling, physical coupling between these two gyros at all. But we are going to put them in a box, close the lid, and then say, hmm, so what are the states in which this box has well-defined, angular momentum?
And they will turn out to be. What we will find is that when the box has well-defined, angular momentum, if you open the lid and ask, What happens if I look at the angle momentum of Gyro one, I will get a variety of different answers. It will be uncertain what I'll find for Gyro one and Gyro two will have an angle momentum that will be correlated with Gyro one. So when the angle momentum of the box is well-defined, it has a different amount of angle mentum and it's pointing definitely in some.
Well, we know the amount parallels the z-axis is definite. When you look inside the box, you'll find it is uncertain what the angle momentum of the bits are. I will explain physically that it's a physical necessity. If that's the case, that's not mysterious. But but will, I hope make it make it evident that that's so. Right. But the moment we're going to address this kind of mathematical problem, we know that the states of the box and sorry.
So the states we have we have two sets of completes of complete sets of states. Complete sets of states are going to be j one. And there's a family like that and there's a family. J to m to three. This should have an in one shouldn't. And since we know what what the total momentum of the first zero is, the only thing to discuss is what its orientation is. And if I consider the set of states like this J one M1 with M1 ranging between minus J one and J one,
that's a complete set of states for the first Giro. This is a complete set of a states of the second gyro. In other words, we will be able to write any state of gyro gyro one as some, some uh, aam1, j one, m one, etc. Right. So one of the states wants a complete set of states of the box. It's the set of states J1, M-1 J to M2.
Right. We we discussed that, that if we have a system, a and a system be a complete set of states is obtained by taking a member of the complete set of a and maltreating it by a member of a of any other member of the complete set of be. If you take linear combinations of those, you get everything. In other words, the box can quite generally the state of the box can be written as some some of the M1 em to j one and one hops. It's meant to be a pointy bracket j to m2.
We want to find the status of the boss at any stage of the boss can be written like this, where these numbers could be up for suitable choices of these numbers, these amplitudes. What we want to do is find the states of the box, which are eigen functions of the boxes, angular momentum operators. And do you remember when we discussed these things, these composite systems, we had that.
That you added the operators of of systems of subsystems and it and you multiplied the and you multiplied their case. That's how it worked. So we want to consider now what the relevant operators are. Well, we're going to have for Gyro one, we have J1 squared, we have j1z and we have Jay one plus and Jay one minus the raising and lowering operators. Where this is equal to j1x plus or minus ij1y. And of course, we will have the same kit of operators for the for the second gyro.
That's for Jarrod, too. And then for the box we will have we will have J squared, which will be j one vector plus j two vector squared. And we will have Jay Z, which is equal to J one, Z plus J to Z, and we will have Jay Plus minus, which is equal to Jay one plus minus plus Jay two plus or minus. So we add the operators belonging to distinct systems here because it's a squared, you know, interval, right. And this thing should be the vector operator belonging to the box squared.
So we add the individual to the vector operators belonging to the individual boxes. So we have to do a bit of these. These are fairly straightforward. We have to be the footwork on this. So let's find out. Let's let's expand this. We have the j squared for the box is equal to j one plus j to toss it into j one plus g to. And this is not. I mean, this. There's nothing funny going on here because the operators belonging to distinct systems.
Another. Another thing we covered in the hole in the composite system discussion. Operators belonging to distinct systems always commute so we can multiply this out just as if they were ordinary weren't operators just ordinary boring vectors. And find that this comes to j one squared plus g two squared plus j one dot j two twice over. This is because j1i comma j to j commentator vanishes. Operators belong to distinct systems. Always commute. Well, this is fine.
This is our list of operations. But this is not in our list of operations. Right? J1 G2 is not up there. So we need to we need to we need to write this in terms of things that are up there. So we say G1. Well, okay, no, I want to get an expression for that in terms of the things already written up here.
And what I do is I say, let's consider J one plus times J two minus that is j1x plus ij1yj2x minus i j to y which is going to be j1x dot times g1 x times j2x plus this on this will give me aj1yj2y which these are two of the components of the elements that are buried inside the 1.2. But I get other stuff unfortunately, which is I get plus ij1yj2x minus j1xj2y. So this I want, this I don't want. But we can get rid of this by arguing that if I write down J one minus j two plus.
So reverse the plus and the minus. This will everything will carry across. The first two terms will will emerge. But what will happen here is that this will become this minus sign. We will migrate from here to here because I've changed where the minus sign happens here. So I'll get minus ij1yj to x plus sorry.
Minus j1xj to y. So when I add these, the left sides, these pesky terms that I don't want will go away and I will have the j one plus j two minus plus j one minus j two plus is equal to twice j one got j two minus j one. Z. J to z. Right. Because these two taken together make j one. Two, minus the z bits which are inside here. So now we have what we want, which is an expression I now go back to this j squared here and replace that with stuff to do with J plus and minus.
So I now write that j squared is equal to J one squared plus two squared. J And then I want, I want this. So I take this. I take plus. j1z. J to z plus. J one plus j two minus plus j one minus two plus. So this disgusting mess on the right express is j squared, the total element of operation of the whole box in terms of operators whose action upon the states of the box. I know that's the key thing.
What I've been doing here is getting an expression where I know what every one of these operators does on those states, those states of the box. J One and one. J two to right. I do not know what j x or j why does to those things. It makes a disgusting mess, but I know what every one of these operators does to those things. That's what I that's the purpose of this algebra. Okay, so now now a little physical argument.
Suppose you've got your first gyro pointing in the Z axis, sort of aligned with the Z axis, and you've got your second gyro pointed, aligned with the z-axis. Then you'd think that your total angular momentum would be will be the sum of the angle momentum of the two gyros because they were both parallel to the axis. You would argue they were parallel to each other and you'd have the total angle mentum.
In the Z direction. So what we do now is we investigate j1j1j to j to this physical argument suggests that this is the object. J one plus. J two comma. J one plus two. So this is the state of the box in which it has this much angling momentum and all of it pointing down the z-axis on the grounds that if you take two gyros, both pointing in the Z direction, surely you've got a box with sure they're meant to just add. So we want to show that this is the case. Physically, it seems reasonable.
Physically, is it true? We check that it is true by applying the relevant operators to both sides. Right. So if I if I do Jay Z on this, I'll just say Jay Z on the L on the left hand side, what do I get? I get j1z, j2z plus J2 Z. Right? Because Jay Total Z is the sum of the Z operators of the of the gyros operating on the right hand side, which is Jay one, jay two, jay one. J1 Jay two, Jay two.
So the way these composite operating system operators work is that this looks at this and we get Jay one because this is an eigen function of this operator with this eigenvalue times. Jay one. Jay one this stands by Jay two. Jay two. So that's that. And then I have plus this looks at that and produces a jay two Jay one, Jay one standing idly by Jay two Jay two produced as the I Can Cat. So indeed we get Jay one plus two times what we started with Jay one Jay one, Jay two Jay two.
So that confirms that this object is an eigen function of this operator for the box with the expected eigenvalue. Yep. Because the lines of both just way. One subject. Yeah, there probably is in that because this came across on to this side and we wanted to. Yes. Thank you very much. There is a factor and this was bound to be important, isn't it? There is a factor of two there because we wanted a two, one, g two from up there and we had twice this which came on to this side of the equation.
So. So that's. That's that. Now we check j squared. What does j squared do when it's applied? Well. It's going to be j squared on this. So j squared. I want to do j squared on on the right side and j squared we've discovered is j one squared plus j two squared plus 2j1zj2z plus j one plus j two minus plus j one minus two plus all that disgusting mess has to operate on. It operates on j one J1 j two.
J two. Well this operating this is an I kit of this operator with eigenvalue j1j1 plus one, and it will then return this and we'll find that this gets returned. So I'll just stick it in the back. I'll disappear as a common factor. Oh. Similarly, this one looks at that and produces. J two. J two plus one times itself. Then j1z looks at this and produces aj1 times this and j two looks at this and produces aj2 times this. So now we have a plus 2j1, j two.
And that's the action of this operator on this product. Then J one plus looks at this, tries to raise this trailing j one to j one plus one. But it can't because because we're already at the top, so it kills it. So the j plus operating on this kills it and it doesn't much matter. It does not matter what j two minus does to this because it's multiplied by nothing. Similarly, when this J two plus operates on this, it kills it, trying to raise that j 2 to 1 more so.
So the action of these two operators on this is to produce nothing. And I can close the bracket just that so so j squared actually this really should be on the right hand side. J squared on the right hand side. Produces this bracket times this card which shouldn't have been written so far to the right. And we can now rearrange this because we've got two j, one j twos. I can take one of those j1j twos and deal with it by putting it inside there so I can write. This is j1j1 plus j to plus one.
So I've to this bracket i ready to j1j2. That's one of those. And the other one I put inside this bracket by writing it is. J two times. J one plus two plus one. So this one, this, this j one produces aj1j2 which is the other one of those. So this is how much I've got of. J one. J one. J one. To you, too. And now I can immediately see that this is j j plus one of. J one. J one. J. One. J two. Due to. Where? Where, Jay? Is J one plus j too. So that proves the thing. It proves the conjecture.
We started with that. This object is an Oregon State of the box with the eigenvalue with a total incremental market value. J one. J two. So this this establishes. It establishes. So we've proved by hard work that j. J. Sorry. J. J. This being the state of the box is equal to. J one. J one. J two. J two. What Jay is Jay one plus. Jay two. That was rather hard work. The next bit easier because we can now apply the minus operator, the J minus operator to both sides of this equation.
And on the left side, we'll get some multiple of of of j j minus one. On the right side will get something more interesting. So now we apply J minus, which is equal to J one minus plus j two minus to both sides. J minus applied to j comma j produces. There's a square root here which turns out to be j. J plus one minus minus which should be m. M plus and minus one, but m is j so minus j j minus one times j j minus one.
So I've applied my, my the boxes tipping operator, lowering operator, whatever you want to call it, that tips its angular momentum away from the Z axis. And we get this multiple and appealing to the stuff we showed when we started on angular momentum for this for the square root times, the state tip, where it's where we got the tip to one unit away from the Z axis. So that's what we get on the left hand side. Is the US on the right? Yes, we have that J one, z plus J two.
No, no, not z minus minus two. Yeah, one minus J two minus. This sum is the same as this operating on j1j1j2j2 which we proved is the same as this. What does that give me? It gives me the square root. So this j one minus interrogates that it produces a square root. Well, let's evaluate this square root. So this square root simplifies because we have a j squared and a j square with a minus sign and a J and a J with the plus sign. So this becomes the square root, in fact, of 2jjj minus one.
So what we're going to get now is the same situation. We're going to have j one minus working on this is going to produce the square root of 2j1 operating on j sorry, and the output will be j1j1 minus one. This will stand idly by whilst that happens. J two. J two. And then we have two. And that's this plus sign here, the result of j two minus banging away. That was this stands idly by. We'll get a root to j2j1j1 standing idly by j to J two minus one being produced.
So we have that just as we can now equate the left and the right sides. Again, we can say that the State J, comma j minus one is the square root of j one over j of this Johnny one. It's not j1j1 minus one j to j two plus another square root, which is j two over j of J one J1. Cheetah. Cheetah minus one. So what do we show? We've shown that when the box has angular momentum, that's tipped a touch away from the Z axis.
If you open the box, you there are tooth and look and look at the individual jar is inside the box. There are two outcomes you might find. You might find that the first gyro is tipped away from the axis and the second is a little parallel to the axis. Or you might find that the first gyro is on the axis and the second is tipped away from the axis. And it's inevitable that that has to be the result.
One of these gyros has to be off the axis, but they can't both be off the axis because then we would have only we'd be two units of angry momentum on the axis short. So what we've got here is correlated states of gyros. The jurors have become entangled in this state in which the box has well-defined, angular momentum. The the results of measurements of the two individual of the Charas in the box are in or have become entangled.
Let's make a picture now to help organise these calculations, because we've just begun. Unfortunately, what in principle is a is an extensive exercise of calculating, but from now on, it's almost mechanical. The way to go is to is is to put your original state Jay Jay up here. It's why I'm putting it up there. Because the origin somewhere like here and this is J units up from the origin. And this is this is J units down. Right? So here we have a minus.
J. J units down. Here's the origin. We started with this state and established what it was that it was. j1j1 times. J two. J two. Then we used the lowering operator j minus to move around this semicircle to state here that we've just constructed, which is j j minus one, which turns out to be a linear combination of this and this. And we found out what the factors are that make the linear combination. And we can now apply.
We can now take this state and we can apply our lowering operator on this to generate state here, which will be j j minus two. Let's imagine doing that. We won't do that. But just imagine doing that. If we would apply the J minus to this, we would get some multiple some range, the square root times that target if then when we apply J one minus plus j two minus to this side, we get four terms because each of j one minus and j two minus works on this and this.
So each of these two things generates two terms. When j one minus works on this, we get j1j1 minus two times j to j to standing idly by. So let me just do this. We can say that j j minus two is an amount of and we could work out what this amount. I mean, it's straightforward to work out what this number is going to be, but we won't do it of j1j1 minus two times. J to j to. Then when J two minus works on this, it produces some amount of j1j1 minus one j2j2 minus one.
Right? Because because it lowers this to j two minus one whilst this stands idly by. Then when we use j one minus on this, we get some more of what we've already got. We get this gets lowered to J one minus one, and we get some more of this which we can absorb in this. B And then when J two minus works on this, that goes down to aj2 minus. So we get plus another amount of j1j1j2j2 minus two.
Physically, what does this say? It says that if your box has its angle momentum tip two units away from the Z axis, if you open the box three things you may find three things you will find one of three things. Either that the second gyro is still bang on the axis and the first is tipped to away, or that each of them is tipped a bit away from the axis, or that the first one is bang on axis, and the second one is tip two away.
So it's perfectly reasonable. What you see when you open the box is what in some sense, if you thought about it beforehand, you would have expected to see this apparatus will deliver you the numerical values of A, B and C, and that will tell you the probabilities of those three outcomes. So we have we're getting complete information of what we will see if we do open the box. And we can plod on like this until we're completely worn out.
The expressions will become. You can see it looks as if these expressions as we go around here are getting more and more horrific because if we apply J If we do if we do another lowering on this supply. J Minus to this and J one minus plus. J to minus two, this this will have a they'll this will give us the term j1j1 minus three whilst and so on.
So we'll have more terms. Mercifully, in the real world when you're dealing with small values of J. There comes a point at which this the lowering operator J one minus will simply kill this. Because, for example, if J one were the number one, this would be j one minus one. And when the when the lowering operator worked on that, it would try and lower this to a number more negative than that, and it would kill it. So the expressions get more and more complicated as we go down here.
And it turns out that when you go along here, they start to simplify because you get more and more of the lowering operators killing their targets and you get you get sorted out and you'll find that you arrive down here at J, comma minus J, you will find that this is simply what it has to be physically, but you will discover that it is. J one sorry. J one minus. J one times.
J To minus j to that is to say, you will find automatically that when the box is an element in the minus direction, open the box is any one thing you can find, which is that both gyros are pointing in the minus said direction. And it's worth doing that. Not in general. Jay But it's worth going all the way around. For example, for JAY For each of the Jay's are equal to one, one of the Jay's equal to one and one equal to a half.
So it's good to see that that happens. So in order now to complete the set up, there's one more thing we have to do, which is, well, strictly speaking, we should do some state counting, I suppose. Why don't we do some state counting? So the number of basis states. The number of basis states of the contents of the box is 2j1 plus one times 2j2 plus one. This is the number of ways we're allowed to to orient the angular momentum of the first gyro.
And for each such orientation of the first gyro. This is the number of ways you can orient the second gyro. So that's the number of possible states of what's in the box. So the number of states of the box. Should be the same because because whether it's our choice to either think about the whole box or to think about what's in the box. So there should be as many states of the box as there are. What's in the box and how many we got? So far we've got two. J one plus J two plus one.
Right, because this is J one plus two. And going around this circle, we get to j one plus two states and that's much less than this. If J1 and J to a big. So we haven't got enough states. And it's intuitively evident that if you have two gyros in a box, their angular momentum don't have to be parallel to each other. They can be inclined. They might, for example, be anti parallel, in which case you would have only, you think j one minus j two of angular momentum.
So what's the problem here is we've got all the states in which the two is a parallel to each other, despite despite what you might think by looking at these expressions here. Right? Remember, this is just remember these these these gyros have angular momentum other than what's appearing in the Z direction. They've got angle momentum in the X Y directions as well. So this may look as if the two gyros are not parallel to each other, but they are.
And there's a problem in the problem. In problem set five about hydrogen, which which illustrates that point. Okay. So these are all perils which are there. So what we need is the states which are not parallel to each other. And the way to go is to say, look, is, is to find what the what the expression is for the state. For this state, which is going to be. The State. J One while sorry. J Minus one.
J Minus one. Because the two gyros are not parallel, there is a bit of cancellation of their angular momentum. But all the engagement with the box is parallel to Z-axis. That's this state here. This state is a linear combination, which we've calculated a. It's a linear combination of where of this state and this state. And I argue on physical grounds that this state should be another linear combination.
And it must be orthogonal to this, because this is an eigen function and I can state of the total G squared operator with an eigenvalue different from. From this. So I now argue that. That J minus one. This is the state of the box. J Minus one is a linear combination of J one. J one minus one j2j2 and j one. J one. J to j to minus one. It must be a linear combination of these two. And and we have to choose A and B so that it's orthogonal to the state we've already got.
So comparing above, you can see by inspection. The condition that. J. J. J. Minus one. J. Minus one. J. Minus one equals nought. That equation implies that A is equal to minus the square root of J two over J and B is equal to the square root of J one over J. If you put in these choices for A and B, you you found a state which is orthogonal to that. And then we can. You can if you're if you're a sceptic, you you've you've got a well-defined state.
You can apply j squared to it and show that it produces you the expected eigenvalue. And it's trivial to see that this thing has an eigenvalue, one minus plus j, two minus one for the Jay Z. Having got this, we can apply the J minus operator mechanically to find this state and this state and so on all the way around down to here. And this is how we construct the states of the box. So we better talk a bit more about this. Its time is up, we better talk a bit more about this on Wednesday.
But we've got the main ideas.