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