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Okay. So we were talking yesterday about polling matrices and the way that they generalise for arbitrary spin.
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I and I was just reached this point.
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I was well, it's interesting to understand the connection between poly matrices and the slightly strange things that happened with Spin a Half.
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And then it's also good to study the case, spin one and there's a problem set problem on that.
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I recommend to you. And then moving right along to the case,
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a very large spin so that we hope to recover classical mechanics and understand how bodies which have macroscopic have many,
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many balls worth of angular momentum end up pointing in some very well-defined direction.
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So. The procedure for generating the power matrices is completely general.
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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,
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squeezed between states of a well-defined orientation between the states.
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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.
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They're perfectly trivial in the case of S said, because these are states of well-defined.
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This is not even state of that operator with eigenvalue, m,
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etc. So we just have down the diagonal the possible allowed values of M, which range from S to minus S.
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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.
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And then we have nothing down the diagonal.
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But we have nonzero entries just on the diagonal that lies one above the main diagonal and one below and zero everywhere else.
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So that's just the generalisation was apparently matrix in which only this part in the case of the party matrix only this
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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,
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this should be an M and this should be an m is what you get.
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Sorry. These were the racing operators. This should be in plus one. Right.
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So it's this. It's this object here. Just some square root.
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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
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is for the case of s equals 40 and then to for this large value of s to have your computer find the eigenvalues.
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Sorry, not the eigenvalues. The We know the eigenvalues we're sorry.
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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.
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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.
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So this is the unit vector. And so if I take any unit vector whatsoever, it has some coordinates like this,
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then I get the matrix force spin down that direction, being an x, x plus and y, y plus NZ as said.
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So choose some angle fit to take the appropriate linear combination of s, z and y.
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I haven't written down as y,
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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,
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calculate solve this problem that s n on a vector which will be a 40 component vector a1a2 up down to a 40 well,
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a s in general is equal to let us say sa1 through s then what are you doing?
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You're finding the the state express the state in which you are guaranteed to get
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the value s and that was the maximum possible value for the angular momentum
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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.
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So this number, these numbers are the other the relevant linear combinations there.
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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
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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.
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And actually, this is of length, facets of length to S plus one because there are two plus one possible orientations.
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So if S is 40, this is going to be 81 and 81 component vector.
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So have it do that. And what you find is what is shown in this picture up in this diagram here,
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this is for three different values, three different values of cost of cost of theta.
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So this is for cost. Theatre is 0.5. This is full cost theatre minus point five.
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This is some other value of cost. These are read it and so on.
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Right so, so for this, if you, if you take this value of cost theta which opposes -60 degrees,
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then the then these A's, which of course the complex numbers have moduli that look like this.
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So there's they're non-zero in some interval around here, which is to say, so what does that mean physically?
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What are these numbers? This is the amplitude that if you would measure along the z-axis.
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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.
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What then is the this is this becomes the probability to measure that it has s units of angle momentum along the z axis.
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This becomes the amplitude to find that you have this minus one units along the Z axis and so on.
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So if the angular momentum, so if the an element and vector really were a direction theta,
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how much would we expect to find along the z-axis so classically in this state?
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Theatre n. We expect, as said, to return.
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S Cos theatre. S Cos theatre is the projection of a vector of length.
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S pointing in the direction of theatre. That is the projection of that down the x axis, down the z-axis.
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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,
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and the quantum physics is saying, well,
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you you have a chance to get all these answers with probabilities which are given by the square of these numbers.
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So quite strongly, Pete.
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And as you change theatre, so you change the vector, you change the state, the input states, you change the direction of your spin.
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And you you change the place where these amplitudes peak.
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So that's only four equals 40.
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And classical objects have as of ten to the 30 or whatever.
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And as you get more and more as as becomes bigger, there are more and more of these dots along this line here.
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There are here 81 dots. I suppose 81 numbers are being calculated, right, because there are 81 components in the vector.
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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.
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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
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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.
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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.
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Then you find that that's essentially the same as the expectation value of y itself squared.
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And there's no ones there's very little uncertainty at highest in what you will
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get for y so so these these and what's happening here is in quantum mechanics,
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we have to calculate a whole series of numbers, which is the components to this.
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So to describe the, the, the spin state of something we have to construct in the case of spin a half,
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two numbers in the case of spin one, three numbers and so on,
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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.
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What we're doing essentially is recovering the probability distribution for the asset measurements,
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which in classical physics is a delta function glitch at s cost theta.
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But in quantum mechanics we don't.
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Our probability distributions are not delta functions there some kind of spread out things and you're seeing what they are there.
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But as you go to higher and higher spin amounts, the probability distributions narrow down around the direction of spin,
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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.
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We just have some completely definite. Oh come on you stupid thing.
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We have some completely definite direction.
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And what? Whereas in quantum mechanics we need a whole load of numbers because we're defining a probability distribution in classical physics,
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it is strictly speaking a probability probability distribution, but it is a delta function.
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And all we have to do is specify the, the centre point of the, of the, of the delta function probability distribution.
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And we do that with just two angles.
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And in quantum mechanics we need a load of different numbers to spell out the whole problem distribution properly.
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Now, the other thing I wanted to say on this topic of, you know,
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relating quantum mechanical levels of incremental and classic levels of angular momentum is is the importance of this.
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So we know that squared has the total incremental operator has values.
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S. S plus one, which is clearly greater than S squared.
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And remember, as this thing came into the world, as the maximum value of the angle momentum around the given axis.
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So. And how much greater this is than this depends on the value of S.
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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.
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Sorry, a half squared. Right.
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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.
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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.
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We never remotely get the spin properly aligned with one axis because there will always be two
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units of an element and somewhere other in the in the plane orthogonal to that chosen axis.
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So when we have, as is one, we have se plus one is equal to two, which obviously is two times one squared.
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So now the amount of and mentioned we can have down one axis is twi is is a whole half here.
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It was only a third. Now it's become a half of the total angular momentum.
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And each each orthogonal each direction in the perpendicular plane has has less
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than in the direction that you've chosen to align the angular momentum with.
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As you go down to large values of s,
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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.
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And that means that we can get essentially all of our angular momentum pointing down to give an axis.
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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.
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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.
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Okay. So I now want to turn to a new topic, which is the addition of an element to.
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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,
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the simplest atom, hydrogen already contains a proton that carries a half bar of spin.
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An electron has the same amount of spin, and then the electron may have orbital angle mentum.
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It may have angular momentum by virtue of its orbit around the proton. So a generically hydrogen atom contains three units of angular momentum.
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And we want to know, so what are the states of the atom in which the atom has well-defined, angular momentum?
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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.
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This is a classic. This is an application. Of the of our theory of composite systems.
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So if you feel unsure about the theory of composite systems, please go back and have a look,
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have another look at it, because this is what we're going to be applying.
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So all that stuff about Einstein but Wolski Rosen cetera.
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What underpins that? Because what we're going to do is we can we're going to consider.
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Two gyros. We're going to have we're going to have Jarrod one has j one.
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So it has has has total. Jay Jay one.
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Jay one plus one. So it has m m lies between minus Jay one and Jay one.
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So that's the rate at which this gyro spins is fixed by some servo motor or something.
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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.
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Is this so? So this will be M1 and M2 is going to lie between minus J2 and J2.
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So we've got these two gyros of there might be two objects belonging to a navigation system
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and we're going to stick them inside a box and they're not going to talk to each other.
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They're going to have there's going to be no Hamiltonian, as we know, coupling, physical coupling between these two gyros at all.
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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?
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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,
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What happens if I look at the angle momentum of Gyro one, I will get a variety of different answers.
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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.
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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.
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Well, we know the amount parallels the z-axis is definite.
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When you look inside the box, you'll find it is uncertain what the angle momentum of the bits are.
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I will explain physically that it's a physical necessity. If that's the case, that's not mysterious.
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But but will, I hope make it make it evident that that's so.
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Right. But the moment we're going to address this kind of mathematical problem, we know that the states of the box and sorry.
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So the states we have we have two sets of completes of complete sets of states.
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Complete sets of states are going to be j one.
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And there's a family like that and there's a family.
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J to m to three. This should have an in one shouldn't.
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And since we know what what the total momentum of the first zero is, the only thing to discuss is what its orientation is.
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And if I consider the set of states like this J one M1 with M1 ranging between minus J one and J one,
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that's a complete set of states for the first Giro. This is a complete set of a states of the second gyro.
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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.
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Right. So one of the states wants a complete set of states of the box.
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It's the set of states J1, M-1 J to M2.
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Right. We we discussed that, that if we have a system,
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a and a system be a complete set of states is obtained by taking a member of the complete
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set of a and maltreating it by a member of a of any other member of the complete set of be.
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If you take linear combinations of those, you get everything.
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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.
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It's meant to be a pointy bracket j to m2.
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We want to find the status of the boss at any stage of the boss can be written like this,
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where these numbers could be up for suitable choices of these numbers, these amplitudes.
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What we want to do is find the states of the box, which are eigen functions of the boxes, angular momentum operators.
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And do you remember when we discussed these things, these composite systems, we had that.
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That you added the operators of of systems of subsystems and it and you multiplied the and you multiplied their case.
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That's how it worked. So we want to consider now what the relevant operators are.
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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.
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Where this is equal to j1x plus or minus ij1y.
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And of course, we will have the same kit of operators for the for the second gyro.
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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.
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And we will have Jay Z, which is equal to J one, Z plus J to Z, and we will have Jay Plus minus,
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which is equal to Jay one plus minus plus Jay two plus or minus.
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So we add the operators belonging to distinct systems here because it's a squared, you know, interval, right.
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And this thing should be the vector operator belonging to the box squared.
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So we add the individual to the vector operators belonging to the individual boxes.
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So we have to do a bit of these. These are fairly straightforward. We have to be the footwork on this.
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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.
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And this is not. I mean, this. There's nothing funny going on here because the operators belonging to distinct systems.
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Another. Another thing we covered in the hole in the composite system discussion.
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Operators belonging to distinct systems always commute so we can multiply this out
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just as if they were ordinary weren't operators just ordinary boring vectors.
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And find that this comes to j one squared plus g two squared plus j one dot j two twice over.
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This is because j1i comma j to j commentator vanishes.
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Operators belong to distinct systems. Always commute. Well, this is fine.
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This is our list of operations. But this is not in our list of operations.
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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.
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So we say G1. Well, okay, no, I want to get an expression for that in terms of the things already written up here.
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And what I do is I say,
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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
191
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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.
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But I get other stuff unfortunately, which is I get plus ij1yj2x minus j1xj2y.
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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.
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So reverse the plus and the minus. This will everything will carry across.
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The first two terms will will emerge. But what will happen here is that this will become this minus sign.
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We will migrate from here to here because I've changed where the minus sign happens here.
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So I'll get minus ij1yj to x plus sorry.
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Minus j1xj to y.
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So when I add these, the left sides,
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these pesky terms that I don't want will go away and I will have the j one plus j
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two minus plus j one minus j two plus is equal to twice j one got j two minus j one.
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Z. J to z.
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Right. Because these two taken together make j one.
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Two, minus the z bits which are inside here.
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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.
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So I now write that j squared is equal to J one squared plus two squared.
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J And then I want, I want this. So I take this.
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I take plus. j1z. J to z plus.
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J one plus j two minus plus j one minus two plus.
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So this disgusting mess on the right express is j squared,
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the total element of operation of the whole box in terms of operators whose action upon the states of the box.
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I know that's the key thing.
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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.
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J One and one. J two to right. I do not know what j x or j why does to those things.
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It makes a disgusting mess, but I know what every one of these operators does to those things.
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That's what I that's the purpose of this algebra. Okay, so now now a little physical argument.
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Suppose you've got your first gyro pointing in the Z axis, sort of aligned with the Z axis,
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and you've got your second gyro pointed, aligned with the z-axis.
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Then you'd think that your total angular momentum would be will be the sum of the
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angle momentum of the two gyros because they were both parallel to the axis.
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You would argue they were parallel to each other and you'd have the total angle mentum.
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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.
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J one plus. J two comma.
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J one plus two. So this is the state of the box in which it has this much angling momentum and
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all of it pointing down the z-axis on the grounds that if you take two gyros,
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both pointing in the Z direction, surely you've got a box with sure they're meant to just add.
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00:28:33,040 --> 00:28:37,209
So we want to show that this is the case. Physically, it seems reasonable.
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Physically, is it true? We check that it is true by applying the relevant operators to both sides.
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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?
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I get j1z, j2z plus J2 Z.
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00:29:04,060 --> 00:29:13,500
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.
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J1 Jay two, Jay two.
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00:29:17,980 --> 00:29:23,680
So the way these composite operating system operators work is that this looks at this and we
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get Jay one because this is an eigen function of this operator with this eigenvalue times.
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Jay one. Jay one this stands by Jay two.
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Jay two. So that's that.
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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.
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00:29:48,010 --> 00:29:59,020
So indeed we get Jay one plus two times what we started with Jay one Jay one, Jay two Jay two.
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So that confirms that this object is an eigen function of this operator for the box with the expected eigenvalue.
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Yep. Because the lines of both just way.
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00:30:16,670 --> 00:30:25,700
One subject. Yeah, there probably is in that because this came across on to this side and we wanted to.
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00:30:25,780 --> 00:30:28,750
Yes. Thank you very much. There is a factor and this was bound to be important, isn't it?
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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.
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So. So that's. That's that. Now we check j squared.
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What does j squared do when it's applied?
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00:30:54,020 --> 00:31:02,040
Well. It's going to be j squared on this.
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00:31:02,040 --> 00:31:12,760
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
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plus 2j1zj2z plus j one plus j two minus plus j one minus two plus all that disgusting mess has to operate on.
249
00:31:31,030 --> 00:31:36,069
It operates on j one J1 j two.
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00:31:36,070 --> 00:31:49,540
J two. Well this operating this is an I kit of this operator with eigenvalue j1j1 plus one,
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00:31:51,250 --> 00:31:54,549
and it will then return this and we'll find that this gets returned.
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So I'll just stick it in the back. I'll disappear as a common factor.
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00:32:03,050 --> 00:32:09,830
Oh. Similarly, this one looks at that and produces.
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00:32:11,030 --> 00:32:16,190
J two. J two plus one times itself.
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00:32:18,440 --> 00:32:32,450
Then j1z looks at this and produces aj1 times this and j two looks at this and produces aj2 times this.
256
00:32:32,780 --> 00:32:36,830
So now we have a plus 2j1, j two.
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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.
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But it can't because because we're already at the top, so it kills it.
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00:32:52,760 --> 00:32:56,329
So the j plus operating on this kills it and it doesn't much matter.
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00:32:56,330 --> 00:33:00,590
It does not matter what j two minus does to this because it's multiplied by nothing.
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00:33:01,280 --> 00:33:07,490
Similarly, when this J two plus operates on this, it kills it, trying to raise that j 2 to 1 more so.
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So the action of these two operators on this is to produce nothing.
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00:33:11,330 --> 00:33:18,260
And I can close the bracket just that so so j squared actually this really should be on the right hand side.
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J squared on the right hand side. Produces this bracket times this card which shouldn't have been written so far to the right.
265
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And we can now rearrange this because we've got two j, one j twos.
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I can take one of those j1j twos and deal with it by putting it inside there so I can write.
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This is j1j1 plus j to plus one.
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So I've to this bracket i ready to j1j2.
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That's one of those. And the other one I put inside this bracket by writing it is.
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J two times. J one plus two plus one.
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So this one, this, this j one produces aj1j2 which is the other one of those.
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00:34:00,710 --> 00:34:04,520
So this is how much I've got of. J one. J one.
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J one. To you, too.
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00:34:12,670 --> 00:34:19,690
And now I can immediately see that this is j j plus one of.
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J one. J one. J.
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One. J two. Due to. Where?
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Where, Jay? Is J one plus j too.
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So that proves the thing. It proves the conjecture.
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We started with that. This object is an Oregon State of the box with the eigenvalue with a total incremental market value.
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J one. J two. So this this establishes.
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It establishes. So we've proved by hard work that j.
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J. Sorry. J. J. This being the state of the box is equal to.
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J one. J one. J two.
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00:35:13,840 --> 00:35:20,340
J two. What Jay is Jay one plus.
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Jay two. That was rather hard work.
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00:35:24,790 --> 00:35:33,430
The next bit easier because we can now apply the minus operator, the J minus operator to both sides of this equation.
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And on the left side, we'll get some multiple of of of j j minus one.
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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.
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J minus applied to j comma j produces.
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00:36:02,890 --> 00:36:06,040
There's a square root here which turns out to be j.
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00:36:06,490 --> 00:36:10,160
J plus one minus minus which should be m.
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M plus and minus one, but m is j so minus j j minus one times j j minus one.
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00:36:22,420 --> 00:36:26,920
So I've applied my, my the boxes tipping operator, lowering operator,
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whatever you want to call it, that tips its angular momentum away from the Z axis.
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00:36:30,730 --> 00:36:38,860
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,
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00:36:39,430 --> 00:36:45,760
the state tip, where it's where we got the tip to one unit away from the Z axis.
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00:36:46,690 --> 00:36:52,149
So that's what we get on the left hand side. Is the US on the right?
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00:36:52,150 --> 00:36:59,380
Yes, we have that J one, z plus J two.
299
00:36:59,740 --> 00:37:03,520
No, no, not z minus minus two. Yeah, one minus J two minus.
300
00:37:04,210 --> 00:37:13,780
This sum is the same as this operating on j1j1j2j2 which we proved is the same as this.
301
00:37:14,140 --> 00:37:22,000
What does that give me? It gives me the square root. So this j one minus interrogates that it produces a square root.
302
00:37:22,690 --> 00:37:24,069
Well, let's evaluate this square root.
303
00:37:24,070 --> 00:37:31,450
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.
304
00:37:31,450 --> 00:37:38,350
So this becomes the square root, in fact, of 2jjj minus one.
305
00:37:39,310 --> 00:37:42,580
So what we're going to get now is the same situation.
306
00:37:42,730 --> 00:37:50,490
We're going to have j one minus working on this is going to produce the square root of 2j1 operating on j sorry,
307
00:37:50,890 --> 00:37:56,920
and the output will be j1j1 minus one. This will stand idly by whilst that happens.
308
00:37:57,100 --> 00:38:00,639
J two. J two. And then we have two.
309
00:38:00,640 --> 00:38:05,260
And that's this plus sign here, the result of j two minus banging away.
310
00:38:05,350 --> 00:38:18,190
That was this stands idly by. We'll get a root to j2j1j1 standing idly by j to J two minus one being produced.
311
00:38:19,600 --> 00:38:24,219
So we have that just as we can now equate the left and the right sides.
312
00:38:24,220 --> 00:38:38,830
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.
313
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It's not j1j1 minus one j to j two plus another square root, which is j two over j of J one J1.
314
00:38:55,520 --> 00:39:01,960
Cheetah. Cheetah minus one. So what do we show?
315
00:39:02,520 --> 00:39:09,810
We've shown that when the box has angular momentum, that's tipped a touch away from the Z axis.
316
00:39:10,290 --> 00:39:14,900
If you open the box, you there are tooth and look and look at the individual jar is inside the box.
317
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There are two outcomes you might find.
318
00:39:17,250 --> 00:39:23,070
You might find that the first gyro is tipped away from the axis and the second is a little parallel to the axis.
319
00:39:24,210 --> 00:39:29,220
Or you might find that the first gyro is on the axis and the second is tipped away from the axis.
320
00:39:32,340 --> 00:39:35,830
And it's inevitable that that has to be the result.
321
00:39:35,850 --> 00:39:38,520
One of these gyros has to be off the axis,
322
00:39:38,520 --> 00:39:44,640
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.
323
00:39:45,330 --> 00:39:54,330
So what we've got here is correlated states of gyros.
324
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The jurors have become entangled in this state in which the box has well-defined, angular momentum.
325
00:40:00,720 --> 00:40:07,260
The the results of measurements of the two individual of the Charas in the box are in or have become entangled.
326
00:40:11,490 --> 00:40:15,840
Let's make a picture now to help organise these calculations, because we've just begun.
327
00:40:16,350 --> 00:40:22,740
Unfortunately, what in principle is a is an extensive exercise of calculating, but from now on, it's almost mechanical.
328
00:40:24,270 --> 00:40:35,070
The way to go is to is is to put your original state Jay Jay up here.
329
00:40:35,940 --> 00:40:44,580
It's why I'm putting it up there. Because the origin somewhere like here and this is J units up from the origin.
330
00:40:44,790 --> 00:40:48,569
And this is this is J units down. Right? So here we have a minus.
331
00:40:48,570 --> 00:40:57,260
J. J units down. Here's the origin. We started with this state and established what it was that it was.
332
00:40:57,260 --> 00:41:10,430
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,
333
00:41:10,430 --> 00:41:20,750
which is j j minus one, which turns out to be a linear combination of this and this.
334
00:41:20,760 --> 00:41:25,610
And we found out what the factors are that make the linear combination. And we can now apply.
335
00:41:25,610 --> 00:41:39,650
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.
336
00:41:41,840 --> 00:41:44,919
Let's imagine doing that. We won't do that. But just imagine doing that.
337
00:41:44,920 --> 00:41:49,360
If we would apply the J minus to this, we would get some multiple some range,
338
00:41:49,360 --> 00:41:57,849
the square root times that target if then when we apply J one minus plus j two minus to this side,
339
00:41:57,850 --> 00:42:05,170
we get four terms because each of j one minus and j two minus works on this and this.
340
00:42:07,180 --> 00:42:10,480
So each of these two things generates two terms.
341
00:42:10,900 --> 00:42:19,420
When j one minus works on this, we get j1j1 minus two times j to j to standing idly by.
342
00:42:19,420 --> 00:42:27,820
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.
343
00:42:27,940 --> 00:42:38,019
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.
344
00:42:38,020 --> 00:42:53,610
J to j to. Then when J two minus works on this, it produces some amount of j1j1 minus one j2j2 minus one.
345
00:42:55,440 --> 00:43:01,500
Right? Because because it lowers this to j two minus one whilst this stands idly by.
346
00:43:03,210 --> 00:43:08,880
Then when we use j one minus on this, we get some more of what we've already got.
347
00:43:08,880 --> 00:43:14,430
We get this gets lowered to J one minus one, and we get some more of this which we can absorb in this.
348
00:43:14,430 --> 00:43:19,589
B And then when J two minus works on this, that goes down to aj2 minus.
349
00:43:19,590 --> 00:43:26,850
So we get plus another amount of j1j1j2j2 minus two.
350
00:43:28,590 --> 00:43:34,890
Physically, what does this say? It says that if your box has its angle momentum tip two units away from the Z axis,
351
00:43:35,220 --> 00:43:39,570
if you open the box three things you may find three things you will find one of three things.
352
00:43:40,140 --> 00:43:45,810
Either that the second gyro is still bang on the axis and the first is tipped to away,
353
00:43:46,200 --> 00:43:53,940
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.
354
00:43:54,270 --> 00:43:58,770
So it's perfectly reasonable. What you see when you open the box is what in some sense, if you thought about it beforehand,
355
00:43:59,010 --> 00:44:04,350
you would have expected to see this apparatus will deliver you the numerical values of A,
356
00:44:04,590 --> 00:44:08,600
B and C, and that will tell you the probabilities of those three outcomes.
357
00:44:09,510 --> 00:44:13,020
So we have we're getting complete information of what we will see if we do open the box.
358
00:44:15,190 --> 00:44:20,740
And we can plod on like this until we're completely worn out.
359
00:44:21,730 --> 00:44:28,450
The expressions will become. You can see it looks as if these expressions as we go around here are getting more and
360
00:44:28,450 --> 00:44:34,389
more horrific because if we apply J If we do if we do another lowering on this supply.
361
00:44:34,390 --> 00:44:43,780
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.
362
00:44:44,020 --> 00:44:50,860
So we'll have more terms. Mercifully, in the real world when you're dealing with small values of J.
363
00:44:52,270 --> 00:44:59,170
There comes a point at which this the lowering operator J one minus will simply kill this.
364
00:44:59,440 --> 00:45:05,800
Because, for example, if J one were the number one, this would be j one minus one.
365
00:45:06,220 --> 00:45:13,180
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.
366
00:45:13,780 --> 00:45:17,620
So the expressions get more and more complicated as we go down here.
367
00:45:17,920 --> 00:45:24,280
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
368
00:45:24,280 --> 00:45:31,840
targets and you get you get sorted out and you'll find that you arrive down here at J,
369
00:45:31,870 --> 00:45:38,620
comma minus J, you will find that this is simply what it has to be physically, but you will discover that it is.
370
00:45:39,190 --> 00:45:43,810
J one sorry. J one minus. J one times.
371
00:45:44,020 --> 00:45:53,709
J To minus j to that is to say, you will find automatically that when the box is an element in the minus direction,
372
00:45:53,710 --> 00:45:59,560
open the box is any one thing you can find, which is that both gyros are pointing in the minus said direction.
373
00:46:01,720 --> 00:46:06,040
And it's worth doing that. Not in general. Jay But it's worth going all the way around.
374
00:46:06,040 --> 00:46:10,630
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.
375
00:46:10,840 --> 00:46:17,830
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,
376
00:46:18,520 --> 00:46:23,620
which is, well, strictly speaking, we should do some state counting, I suppose.
377
00:46:24,580 --> 00:46:28,840
Why don't we do some state counting? So the number of basis states.
378
00:46:35,410 --> 00:46:48,790
The number of basis states of the contents of the box is 2j1 plus one times 2j2 plus one.
379
00:46:49,630 --> 00:46:53,830
This is the number of ways we're allowed to to orient the angular momentum of the first gyro.
380
00:46:54,040 --> 00:46:58,390
And for each such orientation of the first gyro. This is the number of ways you can orient the second gyro.
381
00:46:58,900 --> 00:47:02,530
So that's the number of possible states of what's in the box.
382
00:47:04,030 --> 00:47:07,330
So the number of states of the box.
383
00:47:11,820 --> 00:47:18,390
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.
384
00:47:19,320 --> 00:47:24,720
So there should be as many states of the box as there are. What's in the box and how many we got?
385
00:47:25,110 --> 00:47:30,030
So far we've got two.
386
00:47:30,960 --> 00:47:34,950
J one plus J two plus one.
387
00:47:36,650 --> 00:47:39,980
Right, because this is J one plus two.
388
00:47:40,280 --> 00:47:45,649
And going around this circle, we get to j one plus two states and that's much less than this.
389
00:47:45,650 --> 00:47:49,910
If J1 and J to a big. So we haven't got enough states.
390
00:47:51,110 --> 00:47:58,550
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.
391
00:47:59,030 --> 00:48:07,100
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.
392
00:48:08,030 --> 00:48:13,100
So what's the problem here is we've got all the states in which the two is a parallel to each other,
393
00:48:13,310 --> 00:48:19,520
despite despite what you might think by looking at these expressions here.
394
00:48:19,520 --> 00:48:26,809
Right? Remember, this is just remember these these these gyros have angular momentum other than what's appearing in the Z direction.
395
00:48:26,810 --> 00:48:33,980
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.
396
00:48:34,610 --> 00:48:41,300
And there's a problem in the problem. In problem set five about hydrogen, which which illustrates that point.
397
00:48:42,770 --> 00:48:47,510
Okay. So these are all perils which are there. So what we need is the states which are not parallel to each other.
398
00:48:48,470 --> 00:48:58,040
And the way to go is to say, look, is, is to find what the what the expression is for the state.
399
00:48:58,670 --> 00:49:03,800
For this state, which is going to be. The State.
400
00:49:04,280 --> 00:49:07,909
J One while sorry. J Minus one.
401
00:49:07,910 --> 00:49:14,240
J Minus one. Because the two gyros are not parallel, there is a bit of cancellation of their angular momentum.
402
00:49:15,320 --> 00:49:19,280
But all the engagement with the box is parallel to Z-axis. That's this state here.
403
00:49:20,150 --> 00:49:24,290
This state is a linear combination, which we've calculated a.
404
00:49:25,010 --> 00:49:29,840
It's a linear combination of where of this state and this state.
405
00:49:30,440 --> 00:49:35,300
And I argue on physical grounds that this state should be another linear combination.
406
00:49:35,900 --> 00:49:37,790
And it must be orthogonal to this,
407
00:49:38,030 --> 00:49:45,800
because this is an eigen function and I can state of the total G squared operator with an eigenvalue different from.
408
00:49:45,920 --> 00:49:53,960
From this. So I now argue that. That J minus one.
409
00:49:54,470 --> 00:50:01,459
This is the state of the box. J Minus one is a linear combination of J one.
410
00:50:01,460 --> 00:50:09,200
J one minus one j2j2 and j one.
411
00:50:09,200 --> 00:50:15,330
J one. J to j to minus one.
412
00:50:15,630 --> 00:50:24,910
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.
413
00:50:25,390 --> 00:50:29,380
So comparing above, you can see by inspection.
414
00:50:36,390 --> 00:50:40,710
The condition that. J. J.
415
00:50:40,740 --> 00:50:44,970
J. Minus one. J. Minus one.
416
00:50:45,180 --> 00:50:47,650
J. Minus one equals nought.
417
00:50:47,670 --> 00:51:00,790
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.
418
00:51:00,810 --> 00:51:06,810
If you put in these choices for A and B, you you found a state which is orthogonal to that.
419
00:51:07,380 --> 00:51:13,709
And then we can. You can if you're if you're a sceptic, you you've you've got a well-defined state.
420
00:51:13,710 --> 00:51:19,230
You can apply j squared to it and show that it produces you the expected eigenvalue.
421
00:51:19,230 --> 00:51:27,390
And it's trivial to see that this thing has an eigenvalue, one minus plus j, two minus one for the Jay Z.
422
00:51:28,530 --> 00:51:38,070
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.
423
00:51:38,380 --> 00:51:43,650
And this is how we construct the states of the box. So we better talk a bit more about this.
424
00:51:43,650 --> 00:51:47,130
Its time is up, we better talk a bit more about this on Wednesday.
425
00:51:47,610 --> 00:51:48,840
But we've got the main ideas.