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Angular momentum is enormously important in physics, for example.
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It's central to all kinds of scattering experiments and scattering experiments or lie at the core of high energy physics.
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They play a very big role in condensed matter physics.
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Angry momentum plays a central role in the theory of the application of quantum mechanics to atoms to get atomic structure.
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So from the very beginning of the subject, it played a very big role. And people write whole books terrifically.
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They write whole books on angular momentum in quantum mechanics.
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So we are going to have to spend a few lectures on it, even though we hope there won't be quite as much physical content.
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We're building foundations for later work, is what I suppose I should be saying.
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But we we will on on Wednesday in the Lex lecture, we will at least be able to do something interesting and useful with Anglicanism.
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So the outlook is not entirely bleak, but I'm afraid today's lecture is a bit on the formal side.
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So you will recall, I hope at the end of last term we talked about we talked about operations that generated translations.
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They turned out to be momentum operators and we concluded that there must be operators that effect rotations.
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So so they must be a unitary operator.
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You have alpha which generates the state like the state you've already got that generates the state that is the same as the system you've already got,
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except rotated by an angle model for around the unit vector in the direction of alpha.
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Right. This there must be some unitary operator like this. This is a unitary operator depending on a continuous parameter.
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Right. You can either you can shrink the angle of your rotation down to nothing continuously.
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It's in that class of continuous sort of unitary operators.
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So it's generated we can we can write is the exponential of something or other by putting it in the this thing becomes the emission operator
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these JS So and because they're a because there are three components of this vector alpha which describes the rotation that you're planning.
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There must be three of these operators that generate these rotations and we're calling them, of course, G, X, Y and Z.
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I claimed I said that that they are the angular momentum operators, but we haven't really done a great deal.
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We didn't do a great deal of that time to justify this claim. Okay.
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So we have those three operators that the generators of rotations respectively run the X axis,
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the Y axis and the z-axis out of them because they're commission operators.
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We can construct another operator called J Squared is the sum of the squares of the operators and we have a set
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of four operators and we showed by considering what happens when you make rotations around different axes,
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we demonstrated that these operators must have the commutation relations that j squared commutes with every with all of them with all of j y and Z.
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And but these operators do not strangely commute with each other.
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They have the commutation relations that j x comma j why is it Jay Z and similar things which can be encapsulated in this
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way where Epsilon K is the object that keeps changing its sine and is zero if any two of its subscripts are identical.
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So what we want to do now, so, so that sort of show the existence of these things,
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what we have to do in the next section is find out more about these operators and the eigen status of these operators.
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We need to justify the claim that these operators really are the angle minimum operators we need to find crucially.
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Well, it will turn out that the orientation of something like an electron, well, indeed,
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the orientation of any quantum system is encoded in the amplitudes to find the possible results, the the possible eigenvalues.
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When you make a measurement of X, Jay, Y, or Z, you will there will be possible answers.
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It will be you will get a you'll get a number which is belongs to the spectrum of that.
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And the amplitude for that event strangely encodes the orientation of the object.
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Right. We need to understand about that. So what we want to know now really is what is the spectrum of these operations?
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You want to know what are the possible results of measuring J squared or G X squared or Jay Z squared or whatever?
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Right. So this is what the next spectrum is about. It's about the spectrum of J squared et al, right?
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These operators. So since the j the jigs, j, y and Z don't compete with each other,
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there isn't a complete set of mutual heads of J, X, Y, and z, but there is a complete mutual set of I can.
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It's because of that commutation relation, because J squared commutes with all of its subordinates.
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There is a complete set of mutual like in case of J squared and any one of any one of those.
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And it's conventional to study to to pick just at random that you we choose to have mutual aid and cats.
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J squared and Jay Z, which is just a convention that we choose Jay Z out of the three,
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the three things connected to the fact that Z is the is the singular access is the,
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is the special axis in systems of spherical polar coordinates, right?
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So in spherical coordinates, X and Y have pretty much the same role in life.
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But Z-axis is, is special and that's why we choose this one.
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Okay. So that's what we're going to do. So we're going to say, look, there must be some icon hits.
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We're going to label them by beta m this this label is going to tell us how the thing responds to this operator concretely.
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It's going to be this, right?
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So obviously we're labelling this cat by its eigenvalue with respect to J squared and Jay Z looks Jay Z on beta M is going to be M beta.
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So the second label in this thing tells you how it responds to Jay Z.
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This is by definition a member of the complete set of mutual liking cats of this operator and this operator,
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which the mathematicians have promised us exists. Okay, now we introduce some ladder operators.
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We're going to follow a line of reasoning that's very similar to how we got the eigenvalues of the Hamiltonian of a simple harmonic oscillator.
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We're going to introduce J plus as j x plus i j y.
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So this is a little bit analogous to when we introduce in the simple harmonic oscillator,
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the destruction operator, we said that A was equal to X plus AP similar game.
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So because of this I this is not a emission, it's not an observable it's a tool of the trade.
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And correspondingly, needless to say, we have j minus which is equal to j x minus i j y, and we also have that j plus dagger is equal to j j minus.
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All right. So this thing here is the commissioner joint of that thing there.
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Because, because if you take the dagger of this equation,
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this dagger goes into this because it's an observable that goes to minus II and this goes into this.
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So these are tools of the trade. Now we find what?
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Now we ask ourselves what happened? What are the computation relations?
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We have the j squared on j plus is nothing but j squared comma j x plus i j squared comma j y is nothing because this vanishes and this vanishes,
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right? So j squared communes with j plus and of course it competes with J minus as well.
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Right? So this is plus or minus, it vanishes. What does that tell us?
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That tells us that if you've got J, if you take J plus of beta M, you use this nontrivial operator on this state.
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You get some other states. What can we say about this other state? Well,
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one thing we can say is that J Squared applied to it because you can swap these two over is
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the same as J plus beta beta M so you swap these two over than J squared looks at this and
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says aha that's my I can cut out pops of beta this is a mere number can be popped over here
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is equal to beta j plus beta m so when you use J plus on this eigen state of J squared,
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you get another igen state of j squared for the same eigenvalue, right?
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Encouraged by that. The next thing to do is to have a look at Jay Z on J plus beta m.
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Now, when we swap these two over, we want to swap the two over, but of course we can't.
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So we do the usual business. J Plus Jay Z.
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I've swapped them over and then add in what we should have and take away what we're not entitled to.
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Jay Z Comma J Plus Commentator Brackets Brackets Beta M.
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Now this we we found what this thing was.
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We found that. Whoops. Sorry, we didn't.
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We didn't. Sorry. I'm getting ahead of myself.
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Bulges, right? So that's what we need to. Okay, well, we're going to find out what this is.
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We're going to find out what this is. That's the next thing we have to do. All right.
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So what is Jay Z, comma J plus?
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Well, it's Jay Z, comma Jay X plus I, Jay Z, comma, jay y.
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This is minus i j y from the rule giving way up there and this is minus i.
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So this is going to be I this i minus coming up another i j x a gain from the rule above.
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So this is going to be minus. Oh, sorry, this is going to be j plus because this is these two is going to make a minus sign.
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Cancel this. We're going to have j x at.
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It has to be. Plus. So what.
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What the hell's gone wrong here is I've goofed, presumably in that x y.
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Yes, I've goofed in that. Sorry. I'm always bad at this cyclical ordering so it is equal to J plus.
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So we take this this important result. We stuff it in there and we have that Jay Z on J plus B to M is equal to.
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So this is going to be J, this is going to be j plus and Jay Z working on that is going to produce an m times that.
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So we're going to have an M plus one times this.
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So what does that show?
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That shows that when you play J plus two this object, you get a new I can catch of this operator one which has this for an eigenvalue.
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So let's write that down. It says that J plus on beta m is equal to M plus one.
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Sorry is equal to some amount of which we will call alpha plus the state beta an m plus one.
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Okay. So the point is that what goes in here is the eigenvalue of this thing with respect to Jay Z.
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So this thing here, this thing here turns out to be a.
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This shows the reason I can catch of this operator with this eigenvalue m plus one.
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So that's what should go in there. And this is some this is some constant this is some normalising constant.
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So what have we achieved when we applied J plus to this state?
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JM We've made a new state with the same total amount of angular momentum, the same response to J squared.
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But the amount of this parallel to the z-axis has increased.
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So we have reoriented our system. Right.
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We have here are spinning the spinning top. Well, okay, some incremental along here and we've moved it a bit towards the Z axis.
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That's what J Plus does. Realigns the angle mentioned that you've got strictly speaking, it makes you a new state.
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And this new state has the same Anglicanism as the old state, but more of it's parallel to the z-axis.
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Okay, we could repeat all this stuff. I recommend that after the lecture you do repeat all this stuff and using j
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minus and you will find that j minus on beta m is going to equal some amount.
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Not to be determined. Not yet. Not known yet. Of beta and minus one.
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It does the reverse trick. It moves it away from the z axis.
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Or, if you like, towards the minus set axis.
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So it's showing this is is precise repeat of what was done up here except every plus sign gets turned into a minus sign.
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Okay. Now we have that the expectation value.
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All. For example, j x squared is equal to j x aci.
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Sorry for any state of sci. So take any state of SCI and work out this expectation value of x squared.
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It's jags of sci mod square, right?
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Because if you take the if you take the mod square of this, what you're doing is taking the emission out joint of this,
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which is that the commissioner joint of J X which is J itself and multiplying it into
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this so you end up with this and this is clearly this is the length squared of a vector.
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So it's greater than or equal to nothing for all of psi.
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So let's ask ourselves about J. J squared j.
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M That's clearly equal to beta because j squared onto sorry m of beta m.
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Beta m. So J squared on this produces better times.
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This this is correctly normalised. So we get pizza.
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But this can also be looked at as beta m j x squared b to m so it's equal to this plus beta m j y squared plus beta m jay z squared.
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But this this last one here is clearly m squared because j one of these JS looks at this and produces an
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m times beta m and the other one then looks at that and reduces another m times peach rim and we end up,
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we just m squared. So what have we got. We've got that beta is equal to what should we call this.
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We'll call this a and we'll call this B is equal to a plus B plus M squared where these numbers are greater than or equal to nought.
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In other words, we concluded that beta is greater than equal to m squared.
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So there's a problem. We've got an operator j plus which can make us a new state with M increased by one.
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Which has. But. But but has.
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But this new state has the same value of beta. So apparently we can make states with bigger and bigger m for the same beta.
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And we just shown mathematically that that's absurd.
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Physically, it's absurd because I'm saying that I've got a fixed amount of angry momentum and J plus just moves it towards the z-axis well,
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eventually you'll have it parallel to the z-axis and be able to it will be able to increase that many more.
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So what truncates this? It's something something has to has to give.
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And what? What? It's just like the harmonic oscillator. What gives is that eventually.
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So series of states bigger m truncated at beta and max for maximum value of M such that.
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How does this happen? It happens because when we use J plus on this state.
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We get exactly nothing. So what does this mean?
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This implies that alpha plus equals nought.
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In this particular case, that's the only way we can be stopped for making states a bigger and bigger ram.
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And it's clear we have to be stopped. So we all stopped in this way.
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So what we have to do now is look at the mod square of this of this state and show that it's zero.
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So we have that nought is equal to mod.
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So the mod square of this is going to be this emission emission at joint ID times J plus sorry J plus dagger,
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which is J minus times J plus times, beta and max.
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All right, so this thing here, this is J plus dagger, which is a pill appearing here.
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And I pointed out earlier on that J plus dagger is J minus. So let's have a look.
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See what we've got here by staring inside. So this is going to be.
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I don't need the mod square that's already taken care of.
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So this is beta and max j x minus i j y j x plus i j y close brackets, beta and max.
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So we multiply this stuff out and we get jake squared plus j y squared.
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And then we get we have a minus i j y x the plus i j x y.
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So we have plus i. Commentator j x.
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Comment j. Y. Well, when we've got this much of J squared, you might as well have the whole of J squared.
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So we write this as beta and max j squared minus jay z squared.
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Right. So we J we added we add a jay z squared and take it away again.
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And this, of course, is a Jay Z.
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So along with that, we get minus Jay, Z, Beta and Max,
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and now we're in clover because we know what every single one of these operations produces when it bangs into that.
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So we can evaluate this. This, of course, produces a beta.
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So this is going to be is going to be this is going to produce a beta times this thing.
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Then this thing will meet this thing and produce one.
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So I only need to write down beta this Jay Z will produce and max times this thing which will burn back into this thing, produce a one.
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So I have a minus and Max and this one is going to produce a max squared also with the minus sign.
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So in fact let me write this as never. Nevermind max squared.
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So this is more conveniently. Well, right.
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So what do we have? We have that nothing is equal to this stuff from which it follows that beta we've discovered
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now what beta is in terms of max it's equals when max brackets and max plus one.
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So if we apply j minus to be to m, I claimed that this was alpha minus beta m minus one.
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So M will start. Let's imagine M starts off positive as we take units from it.
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It's going to get smaller and if we keep going, presuming it'll become negative and M will start to be growing a negative number of growing magnitude.
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But we still have this condition that m squared is got to be less than m squared has got to be less than beta.
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So this series of operations has got to terminate as well. So series of cats with with with ever smaller.
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And has to stop. So there must be a minimum value of M, which we, as we imagine, will be negative.
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So we're going to have that beta and min times j.
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Well, I should write it differently, I should say. But nothing has to equal the mod square of J minus applied to beta and min.
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And when we expand that out will there'll be other things happen here and let me.
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So. In other words, nothing is going to be better.
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I mean, j plus J minus beta and min.
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That's awfully similar to what we had here, where we had J minus J plus.
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So you can see that it's going to produce the same stuff, except that the sign of the comet is going to be different.
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Otherwise everything will be the same. So this is going to be nothing is going to be beta and min, uh, j squared minus jay z squared plus jay z,
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which is going to lead to the conclusion that beta nothing is going to be beta minus a min squared plus a min.
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In other words, a beta is also equal to men and min minus one.
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So we have a relationship here between beta and the largest value the Dem can take and between beta and the smallest value that M can take.
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And we could. Well, we can. We can from these two.
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Between these two equations we can eliminate beta and learn that min squared minus min,
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which is this is equal to beta or minus beta equals nought, but minus beta is the same as minus and max and max plus one.
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So we have this equation and this can be thought of as a quadratic equation for a min in terms of max, right?
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So this is a quadratic equation and it tells me that our min is equal to minus B, well, b is minus one.
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So that's one plus or minus the square root of B squared minus for a C as one CS minus the stuff.
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So plus four and max brackets and max plus one all over two looks ugly.
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But actually it's very beautiful because this is going to be a half of one plus or minus the square root of this is this.
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Well, let me write down what it is and you'll can tell you tell me whether you agree with it.
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It's m max plus one squared.
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If you square this stuff up, you get four and max squared, you also get four and max from the cross from the cross terms two times to max four.
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So that's that and that. And you also get a one that's that. So we can extract the square root, right?
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Because we've got the square root of a square. So we have plus or minus this.
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And Min is obviously smaller than a max.
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So the plus root can be ignored because that would that would tell me that Min was bigger than Max.
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So only the minus the minus root is wanted and you soon find that that is equal to minus and max.
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So. So there's a biggest value the team can take and there's a smallest value that I can take.
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And we've shown that that's minus the the biggest value. In other words, we've got a picture of like this, we have a biggest value here.
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Then we have a next value. Then we have an X value. Then we have a next value.
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And suppose that this is this is the end, then zero lies.
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So this is a plot with M going up here. So here would be zero say.
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And in this case this would be a half.
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This will be three halves. This would be minus a half and this would be minus three halves.
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Or it might work out like this that we'd start.
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But the key thing is we could start we could start slightly higher up.
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And then we would have this one. And this one and this one and this one.
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So if we started at two, we could have one. Nothing, minus one, minus two.
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This is these are the possibilities.
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But the key thing is that I know that in an integer a number of steps here, three steps I can go from the biggest value to the smallest value.
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Here there are four steps. One, two, three, four.
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So the key thing is that twice a max is an is an integer.
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Now we could carry on talking about beta and Max, but it's extremely boring and nobody does that.
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What they do is they use a new notation. They say that the new mutation is that Jay is what you mean by max.
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The biggest value of M is called J little J.
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And what have we got? We've got the beta is equal to Max and Max plus one that's on the board just here.
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Is therefore equal to J. J plus one.
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And we know that two j is an integer. In other words, J is a half integer.
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It may be an even number of half integers, in which case it's an integer itself, or maybe an odd number of half integers.
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So in this left hand column, J is a half integer.
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All the values of J. J is a half integer. Consequently, all the values of m a half integers in the right column.
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J happens to be an integer, and therefore all the values of m are integers.
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Therefore this beta number is sometimes an integer.
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So if J is an integer, this is an integer.
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For example, if J what a possibility is that J comes out being nought, in which case pieces nought or J might come out being one,
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in which case Peter would be two or J might come out being two, in which case Peter would come out being six.
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We have a sort of funny selection of integers, but worse than that, when beta is sorry,
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when when J is a half integer, the values of beta are really quite weird.
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So we don't use beta as a label. So we we relabel.
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Beat em to j em instead of using as the label in here that tells you how this state responds to J squared.
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Instead of using the actual eigenvalue,
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you use this this number which is either an integer or half integer for which you can work out this eigenvalue because this eigenvalue is JJ plus one.
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That is to say we have j squared on j m is equal to j j plus 1jf and we have the Jay Z on j m is equal to M of j m.
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This is the new notation universally used.
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So the only we've changed notation only because we've discovered that the number beat
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the numbers beta are themselves rather unpleasant and don't make for handy labels,
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but they are related through this equation to something that's very simple, which we ensure a half integer.
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A moreover tells us immediately what the largest value of M is that you were allowed to have so four.
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So we have if j equals equals two there are five states.
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There is to commit to two comma, one to comma, nothing to comma minus one and two, comma minus two.
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And what does that mean? Well, statement is being made physically.
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It's being said that if my pen has two units of angular momentum, one has j equals two, which means,
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as I've said, it has three speaking j squared gives you has an eigenvalue of six, right?
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But if we call that tune Spangler mentum, it has five possible orientations.
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All right, this one, this one, this one, this one, this one, and this one.
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Only five. This is what they call Space Quantisation.
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When Stirling Stern and Gerlach discovered this experimentally, I think it's a terrible, terrible term, right?
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It's not. I wouldn't call it. I think it's no, I think it's a very bad it's got the space quantisation.
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But I just tell you historically that's what they called it. But this is the this is the bizarre conclusion that we have a discrete set of
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orientations anyway being possible for a pen with that amount of angular momentum.
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If J is a half, then what do we have?
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We have a half and a half and a half and minus a half.
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And that's it. Only two states. So that's why we've been talking about electrons and things.
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Two objects with angular momentum total with spin a half, half of units of spinning momentum like electrons, protons,
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positrons, etc. as the archetypal two state system, because there are there are only two possible orientations.
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Now, this is very misleading. Right. But I've already given two health warnings on this.
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But this the naive interpretation is that you're spin a half particle, you'll spin off zero, has two orientations.
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This one and this one. Nothing in between left.
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Okay. So that is a grossly oversimplified picture, which leads to misunderstandings, but it's it's gives us a bit of orientation.
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And people often do think in those terms. In the three halves case, we would have three halves, three halves, three halves,
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one half, three halves, minus one half and three halves, minus three halves.
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We would have four possible orientations. We'd have this, this, this and this, never pointing horizontally, etc., etc., etc.
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Okay. Almost done.
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Let's have a look at the effect of rotation.
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Around the z-axis. Okay.
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So upside goes to PSI primed, which is U of alpha of obsidian.
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So when these hanging momentum operators came in as the things you put in an exponential in order to generate a rotation,
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the unity matrix that makes you a new system, which is the old system rotated.
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So we want to see what we get now. So let's see what happens when we rotate a state of well-defined one of these eigen states here.
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Right? So let's do E to the minus.
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So we go about the Z axis. Then Alpha only has a component in the Z direction and this becomes and it has a magnitude five.
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So this becomes e to the minus I phi is the rotation angle.
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Jay Z And let's use that on one of these j m states.
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Well, this is a function of an operator used on and this is the function of an operator.
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So by the definition of a function of the operator, it has the same IK and states as the operator whose function it is.
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So this thing is an Oregon state of this operator and the eigenvalue is the function on the eigenvalue.
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So this is e to the minus. I am.
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JM So one of these states off of one of these Oregon states here,
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when you make when you rotate it using the this you rotation operator produces you the same state multiplied by this phase factor.
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Okay. So if we rotate through to PI, if we rotate the thing completely around.
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So if we put 5 to 2 pi, we are looking at what are we going to call this?
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We're going to call this a PSI prime to say, right.
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PSI primed is going to be e to the minus to pi.
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I m well, maybe we should say to m pi times what we first thought of.
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If M is an integer then so this is eight, then this is going to be a number one.
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So this is equal to J and if M is an integer, but it is equal to minus J.
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And if M is a half integer as we know it can be.
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So we have the surprising result that if you rotate a system with half integer angular momentum completely around,
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complete through an entire rotation, it state doesn't return to its original state.
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It returns to minus its original state. And this seems strange to us because we don't have any we don't have any concrete experience.
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We have no we have no experience of this kind of thing for the following reason
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that particles which have even the particles which have half integer j well,
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particles are described by fields of particles that have half integer j are described by fields whose whose value never becomes.
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This is a result of quantum field theory,
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whose value never becomes large compared to the quantum fluctuations in the field, the quantum uncertainty in the field.
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So so the values of these fields never become significant.
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And we have they these fields never enter classical physics.
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So the direct field whose exhortations are electrons and positrons are is not something that's part of classical physics.
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It's a part of the vacuum, just the same as electromagnetic field of the gravitational field.
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But it's never excited at a at a macroscopic level.
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So it doesn't enter classical physics.
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So we have no experience as classical beings within classical physics of the fields associated with these half integer values of M And therefore,
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we're unaware of this fact that if you if you turn the thing completely around, it changes its sign.
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And the fields we do have experience of the electromagnetic field and the gravitational field
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belong to integer values of m the electromagnetic field as well as j rather equal to one,
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and the gravitational field has j equal to two. And therefore these fields don't manifest this, this, this strange behaviour.
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Well, I think that is the right place to stop, even though it's a little early.
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And we will look at the rotating rotating molecules as a physical application on Wednesday.