Okay. So we finished last time.
We're just about pushed through the calculation of what else squared is the differential operator,
which we did, if you recall, by multiplying L the latter operators, L minus Snell plus together.
And it was rather a tedious calculation, but at the end of the day, with luck,
we ended up with this and we should recognise that this l squared is minus this combination of partial derivatives,
vector theatre and PHI is the it's minus r squared times the angular part of of del
squared the la plaza and when looked in when put into spherical polar coordinates,
if you take if you take this thing and put it here,
this minus sign cancels that minus sign and we get a one over all squared sine theta db peter sine teacher etc. which I hope you recognise as,
as del squared. So, so you might ask yourself so physically what's happening.
We have the kinetic energy operator i.e. so we got this is H sub k which means p squared over 2my,
this is squared plus p y squared plus p z squared is also minus h squared over two times del squared.
So in the position representation, this operator becomes this right because p is minus h bar times gradient.
And classically. We have that l is equal to v tangential times the radius.
So L squared is v tangential squared times radius squared l squared over our squared is equal to v tangential squared.
So we have, we have that HK is equal to.
Well sorry that's suggestive something isn't it, that this, this narrative squared can be written in terms of some radial derivative plus.
So we could say that. Yeah, it was alright. We can say that HK is equal to some radial minus h bar squared over to m one overall squared D by d
r squared D by the R and then we're going to have plus h bar squared L squared over all squared.
I think just by substituting into there over to m sorry, over to him.
And what's this going to be? We've defined the angular momentum operator l squared to be dimensionless,
so putting an edge bar in front of it h bar L is the classical animal, right?
So H bar operator is the analogue, I should say, is equal to it's the analogue of classical angular momentum, total angular momentum.
So this that you have here has the dimensions of total incremental squared.
It's the classically understood thing. So this term here is looks is looking awfully like v tangential squared.
Sorry, I need a mass here. Right.
The. So.
The classical angular momentum is m v tangential r So the square is m squared v tangential squared r squared.
Move the r squared down here. And this is the classical relationship that L squared overall squared is m squared V tangential squared.
So this is looking like this in the back here is looking like half am v tangential squared.
That's what this suggests to us, the quantum mechanical formula, which is correct.
But it's suggesting to us that it's this sort of natural translation of classical physics is this and this is clearly the tangential kinetic energy.
So this is the the k the tangential, the kinetic energy associated with tangential motion.
Which suggests that this here should be the kinetic energy associated with radial motion.
And that's what we want now to put on a firm intellectual footing.
So we're going to show that this thing is minus H by squared over two M times PR squared, where PR is the radial momentum.
So the question we want to address now is what is the radial momentum?
Operator. We found the tangential operator.
We found it in some sense, the tangential momentum operator in the sense Al.
Now we want to find the radial one. So classically momentum is a vector and we can say that the radial momentum is simply our dot p overall number
is the unit vector all dotted into P must surely be radial momentum momentum in the regular direction.
But there's a problem with this from the perspective of quantum mechanics, because this operator doesn't compute with this operator.
So it's well, what does that mean, this thing.
Oh sorry in QM oh dot P overall is not emission me prove that to you but it's easier to prove that in general than in particular.
Okay, so let me have two emission operators. A dagger is A and B, dagger is B, then let's look at a B dagger.
If I multiply these together, what is I guess an operator ab is this operator mission?
Find out that's b dagger. A dagger.
Because the rule for taking her emission that joint is you reverse the order and dagger the individual bits, but B's B dagger and A's a dagger.
So this is equal to be. So is this equal to AB?
Well, clearly it is if and only if b in a commute.
So. So this is only if woops a comma b equals nought in words.
The product of two emission operators is itself emission only if those two operators commute if and only if those two operators commute.
So this are dog P, which is shorthand of course, for x plus y,
if y plus there p z would be emission could be emission only if X and commuted Y and p y commuted.
Well, they don't. Therefore, this is not a mission. Therefore, this is not an observable.
So it can't be what we're looking for. We're looking for something, the momentum in the radical direction, which is which is observable.
All right. Well, there's a fix to this problem. There's a general fix, and we're going to use it.
But if we do a half of AB plus B.A., this is a mission.
Right. Because if you take the dagger of this. This one, we just proved the dagger of this one is that.
And the dagger of that one is this. So this thing, the dagger of this bracket is itself.
So when you've got two non-commissioned operators sorry, you've got two mission operators that don't commute and you want to make the product.
The way to go is to is to take the average of them.
You know, it's really naive thing to do. So let's do that. So let's let's so so we try.
Let's have a look at the emission operator PR,
which we're going to define to be a half of our dot p overall where it's
important that that all this R here is in front of the PR plus P dot R overall.
So this thing here will be emission and I'm going to show that it is what we require.
So in the position representation so we we you can do this calculation in the abstract,
not in the position representation, but it's easiest in position representation. So that's how we'll do it.
So PR is equal to. So this P gets replaced by minus H by a grad, right?
So this is going to be minus I h bar over to common factor.
We're going to have our dot grad overall.
But this of course is the scalar plus the divergence of our overall well.
Now the issue is this. When what?
This isn't quite the divergence of this. What this means is, is this is remember an operator.
It's waiting for a wave function to come and stand in front, get operated on.
Right. So this differential operator operates when everything to its right.
It operates on this and it operates on these two. This here, this differential operator operates everything to its right, which is only the ABC.
So we have to when we, when we expand this out,
we're going to get three terms because we're going to get this thing operating on this that that is standing idly by,
this thing operating on this with this and this standing idly by and this thing operating on this with these two standing idly by,
which is the same as that. So this is going to be a minus I HPR bar dot grad overall, right?
So I'm taking this one that I've got and the one that I'm promised at the end of all this reduction here.
So that's where the two went to. So that's, that's those two.
And now I've got these two bits minus my brackets.
We're going to have over two. Sorry, that's this factor here.
Then I'm going to have this thing operating on this, the divergence of RS three, and then I've got this thing operating on that.
It's going to be all dotted into the gradient of one overall.
So that's going to be minus oh dotted into the gradient of one overall.
The gradient of one overall has to be a well, it is the vector R overall squared.
This minus sign comes from the differentiating of the one overall. Right. Because I'm reminding you of prelims maths.
Now the gradient of R itself is the is the vector are divided by all.
This is dimensionless animal because that has dimensions of one of a length that has dimensions of length.
So it's the vector is the unit vector R And that's what we've been using.
Looks at this. Excuse me. So I made a mistake. This should be r cubed.
Maybe we need to do this. Sorry. So what? What?
Just to fill in here.
So the gradient of one overall is equal by the ordinary rules of differentiation, minus one overall squared times the gradient of R,
but the gradient of R I've just said is the vector are divided by R, hence the R cubed.
These two dot together make an R squared which cancel most of those.
So, so this minus sign, these two can be combined to a two overall.
The twos go away and guess what we end up with is minus h bar grade overall plus one overall.
Okay. So that's what this this stuff reduces to what we next want to know is so what is R dot gradient.
Well, I want to know what this is in spherical polar coordinates.
Well, the thing to do is just to write down let's write down r d by the R, it's easy to see that that is going to be x deep.
Well, let's, let's do it o then we're going to have by the chain rule the X by the r d by the x plus d.
Y. By the r. Debate Y plus d z by the R debate, you said that's just the chain rule.
But but what is the x by the r x is equal to our sine theta cos phi so d x by the r is equal to
x overall and for the same reason do y by the r is equal to Y over r and so on and so forth.
So this is equal to x d by the x plus y, DPD, y plus z, dvt z because this is x overall,
but this all cancels the R on the bottom y overall all cancels what's on the bottom and what's this?
This is a vector product of x, y, common z with nebula DVD, X, Y, or Z.
In other words, this is this is the animal we're interested in our dog.
Great. So I have now that PR is equal to minus I h bar r dot grad we've just agreed is deep idr.
So this is go away and we have D by the R plus one overall.
So we have an interesting result.
We have the momentum associated with the radius is not simply d by d radius like the momentum associated with x is D Barry X.
There's also this additional term in here. But just to convince you this really is the momentum associated with RADIUS.
Let's for fun calculate r comma pr.
So what is that? That is minus h bar of r deep 80 plus one which is r times pr minus D by the R of r minus one.
Yeah, right. Yes.
Sorry. I mean, we're getting zero on time. That's the trouble. I want to get minus.
I want to get bar out of this. What the [INAUDIBLE] did I do wrong?
DVD all PR minus DVD are plus one overall on working on.
Ah. Yeah. Yes.
Yes. Yes. Yes. Sorry. Sorry, sorry. Yes, yes. Yes. Yes. Yes. It's perfectly correct.
Okay, so the ones go away. This is this is this this sort of thing is confusing right now, as I say.
What does this mean? This means DVD are of everything to it's right.
And there's a phantom wave function here waiting to be operated on. So this is the derivative of ah upside.
When we take the derivative of the R, we get one times of psi and then the R stands idly by and we do the gradient of upside.
The second term cancels on this because our times, the gradient of upside is occurring here with a plus sign,
and there it will be occurring with a minus sign. So what we're left with is the D by the R times of PSI, the R, which makes one.
So this is equal to plus because it's a minus sign coming here h bar so that it's these two operators satisfy the canonical commutation relations.
Right. Canonical commutation relation.
So so PR really is the momentum associated with.
Ah. Okay. So what, what are we really trying to do here.
We're trying to show that that one overall square or square video is essentially PR squared.
So let's calculate PR squared. PR squared is going to be minus H bar squared because they'll be two, they'll be two minus bars and then it's D by t,
r plus one overall bracket D by the R plus one overall, which is equal to minus H bar squared.
Obviously this on this is D two by the R squared we will get this differential
operator will differentiate that and would use a minus one over squared.
We will otherwise get a one overall D by the R and also a one overall D by the R.
So we'll get to of one overall D by the R.
And sorry and I haven't finished. And I also get this thing on this thing is a plus one overall squared.
So these two terms, these two terms cancel and we're left staring at this.
Uh, I should have had two of these terms.
I think I said I was going to get two terms because I have a one over audio and I have a one over DVD after this.
Operator when this operator works on this, it reduces that, but also it works on the phantom wave function sitting over here without standing idly by.
So we get two of these. I think I said that, but I didn't write it.
I'm not sure. So we have a minus h bar squared D two by the R squared plus two overall D by the R,
which can also be written as minus h bar squared over all squared d, buddy r of r squared d, buddy r.
Right. Because if you differentiate out this product, you get R squared on R squared times D two by the R squared,
which is this term and you also get a2r overall squared two overall times D video.
So yeah, this, this term here, we've now shown the Del's,
we've now shown that HK the momentum operator which is minus H squared over two M Del squared is also minus h bar squared over two M of sorry.
Yeah. Of.
Well, let's leave that outside. Let's take the ball square into the bracket.
We're going to have a one over all square debris, all that stuff, which we've just shown is is PR squared.
And then oops. But there was a minus sign.
So that soaks up this minus sign, a PR squared. And then similarly there's plus H bar squared, L squared, overall squared.
This is a very important formula that we will need when doing hydrogen and therefore fundamental to.
So it's expressing your kinetic energy in terms of your radial kinetic energy and your tangential kinetic energy.
And that's one of the reasons why the total orbital increment operator is important,
because it encodes your sort of energy of going around and around.
So with that we are now finished with we can massively finish with orbital and lumentum and we can get on to spin.
And this is somewhat more interesting in the sense that it's quantum mechanics has more remarkable things to say and
it's less tedious because all that stuff with as part of the French operator's DVD theatre and stuff is not much fun.
It has to be set right. So we have identified two types of of generations of rotations, the angle, the total and element of operators.
And they generate we introduce them in order to generate complete road to complete rotations.
So you have alpha being e to the minus I alpha not j rotates system as on turntable.
So it moves. It moves your system around the origin.
If you put you you put your system.
It's as if you put your system on a turntable centred at the origin, the axis of the origin, and you turn the turntable around.
Your system moves through space and it rotates simultaneously.
Whatever internal structure it has. But we also have shown that ally the orbital angle momentum.
Moves system on circles.
So it moves it around physically. It translates it around the origin, but it is not rotated at the same time.
It leaves its orientation fixed. And we have some.
Well, okay. So we found the commutation relations here.
We found that we found that j i comma j j is equal to i some have a k excellent i j k j k.
And we found that it was also true that l i l j was equal to i sum David k
epsilon i j k l k They had the same commutation relations amongst themselves,
these operators, which is why we could use the work we did demonstrating what the eigenvalues of these could be.
Also down here, this this implied that and that that that j squared has a values.
J j plus one for j is nothing.
A half one three halves, etc.
And. From these commutation relations, we infer those are possible values for the eigen values of these operators.
But we also had the principle that if we wrote if we translated something completely around the origin,
we proved that that that was the identity transformation.
So we concluded that l l plus one had to be l equals l equals nought one to integers only allowed.
In this case, what we now going to do is introduce s i is by definition j i minus l i.
It's the difference between these two. What does that mean physically?
It means that s II is going to be the generator of rotations of a thing about its own axis.
So we're not going to be this rotated on a turntable.
So it rotates it and moves it. This simply moves it around a circle.
So this is going to only rotate it on its own axis.
It's not going to move it. It's only going to rotate it. That's what we expect to happen.
But we'll have to be guided to some extent by the mathematics of all these what is going to happen.
Right. So what about.
So what we want in having these new fangled operators, it's important to figure out what the computation relations are going to be.
Now, see, comma SJ is going to be j i minus l i comma j i minus sorry j j minus lj.
All right, we're going to get this commuting with this.
So we're going to get I some David K Excellent.
I j k this commuting with this will produce a j k this commuting with this will produce an elk this commuting with this.
Now, we didn't write that down, but I commuting with elk.
This is a vector operator and therefore this thing when you when commutes with this always produces the the missing component of this vector.
So this is going to be minus elk.
And similarly, this thing on this thing is going to produce swap them over and you're going to.
Well, there are several signs here that we could found down,
but this thing is we're looking fundamentally at the same thing as the commentator of this on this we're looking at.
Looking at minus l i comma jj is equal to is equal to obviously jj l i is equal to
i epsilon j i k l k is equal to minus i epsilon summed over k This is a sum of a
k of epsilon i j k l k So all right j i j i k And but I would like to have this in
the order k So I swap those two over and introduced a minus sign to compensate.
And then at the end and then you can see this thing, including that minus sign is the same thing as this thing, including that minus sign.
So we have a minus another L Okay. So this is the justification for that last term there.
So what do we end up with at the end of the day? These three L K's collapsed into just one.
Okay, it's going to be j k minus L. In other words, this is going to be I someday have a k epsilon i j k as k.
So these spin operators, so they say that they're going to be this we call them the spin operators.
They have exactly the same commutation relations as the J. Therefore, we know what their eigenvalues are.
So this implies that the eigenvalues.
Of X squared, which is of course s which is x squared plus x y squared plus z squared, r s s plus one where s is equal to a half.
Sorry, nothing. A half one three halves.
Blah blah blah. Okay. Because these these results follow ideally from the fact of having the commutation relations.
Asi como se j is I excellent? I ask all the half integer values allowed.
Answer you will have a half integer.
When Jay does. Because.
L does not. All right.
Why is that? That's because Se Z is equal to Jay Z minus L.Z. and Jay Z comma S.Z. equals nothing, which is also the same as L.Z.,
comma, Jay Z, etc. All these three operators commute with each other, so there's a complete set of mutual aid and states.
And if so, we can now see that if this is half integer, this is using half integer eigenvalues.
So the eigenvalues of this are going to be a going to be the difference between the eigenvalues of this and the eigenvalues of this.
So if this has half integer eigenvalues, therefore this will have to have half integer eigenvalues because this one has integer eigenvalues.
So. So if J has half integer eigenvalues then does correspondingly.
If J doesn't s doesn't. It just takes along behind J.
And indeed, that's how we tend to think about it.
We tend to think that the integer amounts of angular momentum come from orbital motion,
L.Z. and the half integer values, if present, come from S said.
And that's why J has half integer values. That's how we tend to think about it.
Okay. Now.
I think I've claimed a few times that spin is something to do with spin is something to do with the orientation of our system.
And now it's time to to make good this claim that.
That the eigenvalues of the spin operator or your response to the spin operators encodes how a particle is oriented.
And this is a strange area of very quantum mechanical area.
Okay. So in general, the internal configuration of a system could be written that we could write if PSI is equal to the sum of s m upside.
So we got a complete set of mutual we've got a complete set of mutual aid and states of S squared and s z.
Right? So this is the so we're saying that s squared on s m is equal to S s plus one whoops one of s m and we're
saying that s said on s m is equal to m s m and there should be a complete set of eigen states of this,
of these mutual sets, of these operators.
So I should be able to expand any, any, any state as a linear combination of these of these states.
And what are we going to have to sum over?
We're going to have to sum of S is equal to nought half and we're going to have to some of em is in modulus less than or equal to s.
This. This should be a generally valid statement,
but for the wave functions or the states of view of real occurring systems in the laboratory, the good news is you don't.
You only have to. There's only one value of s will occur in this sum.
So. So this is this this would be generally the case. But, uh, for the systems in the lab of microscopic systems, this wouldn't be true.
So if we were to embed a cricket ball, we would need to do a sum over all of these things up to ten to the 40 or something.
But if we're dealing with an electron or an atom or whatever, we don't have to do some of all these things.
We only only one value of s has non vanishing amplitude.
So this is going to vanish. For real systems. Except when s takes one particular value.
So for real. From microscopic. Microscopic.
Only one value of s occurs. Normal vanishing.
As any sign of this. We can say at the outset that the amplitude to find a value of s other than a value that's peculiar to the system is zero.
So we just don't need to consider it. So we're able to so we're able to write that up.
PSI is equal to the sum from m equals minus s2s of s and.
CI said. That means.
What does that mean? We're encoding? Well, the state of the system is described by these numbers here.
There will be two plus one complex numbers.
These these numbers here. And they're telling us they're encoding somehow the way the system is oriented.
So this is two plus one. Complex numbers, encoding, encoding orientation.
And what we want to do now is get some feel for how this encoding works.
In simple cases, how would you encode the orientation of a macroscopic body?
Well, the traditional procedure is you use only two angles. You write down three angles which describe how you get some access in the body,
you know which way it points with respect to the z-axis of some fixed fixed coordinates and then how it rotates around that.
So you usually encode the orientation of a body in three Euler angles for a classical system,
for quantum mechanical system, you encode it in a certain number of complex amplitudes.
The amplitudes defined find it in various orientations. It's carrying the same information.
Right, in a very funny way. Okay, so let's let's consider the various cases.
The case equals nought. In other words, if the only occurring thing here is nought, there's basically no information.
There's only one, one amplitude.
Nought. Nought. State of your system.
And also, if you if you if you you rotate your system, you try and rotate system.
With you, you alpha, which is of the minus I alpha dot s this is the so s is the generator of rotations of the system around its own axis.
Right? So I've written down the unitary rotation that makes me a new system which is rotated around the Axis Alpha by the magnitude of alpha.
Then if I use this you alpha on upside, what do I get?
I get using that expansion. I get nothing.
Nothing psi times you alpha or nothing.
Nothing. Well let me replace this by e to the minus I alpha dot s when when this operator sees this cat, it thinks you know nothing doing zero, right?
You simply get as all of these think, that that becomes nothing.
So we have easier than nothing, which is one. So this return, this goes back to itself.
Which is the same thing as upside.
In other words, when Earth is one, you can't tell the difference between the system before you rotated and after you rotated it.
It's like an absolutely immaculate and perfect sphere. If you rotate it, it stays the same.
So a particle that that does this that has x equals nought.
So no spin implies same after rotation.
So if you like, for a classical animal, we would say this is strictly symmetric.
And there's only one amplitude we need to bother with.
So we say it's a scalar. So the simplest particles are spin zero particles.
They there's just one amplitude, which is you say to say what the energy is or to say what the location is,
because there's no issue of how is it oriented. It's a silly question.
How is it oriented? It's it doesn't have an orientation. You can't tell it.
If it doesn't know it doesn't have an orientation. That's the best thing to say. Nothing changes if you try and reorient it.
And because of the one amplitude business, you say it's a scalar particle.
Right. So that first equals. Next in the hierarchy of.
And for an example, a pile is an elementary particle which is a scalar particle.
Right. Let's do X equals a half. So unfortunately, there are not many particles.
There's not much in physics that's a scalar particle or scalar field.
This is a rare case. The the really important cases are X equals one and as equals a half and this equals one.
We're all built out of X equals or half particles. Right. So this is electrons.
Protons. Quarks.
Therefore neutrons. Wide range of things as he made out of an odd number of quarks will be in as equals a half particle.
What will prove that later on? Okay. So this is a really big class we have that the state of the system can now be written as.
There's going to be a half a half of CI plus a half.
Minus a half. I'm sorry. I need two times a half.
Half plus upside times the half minus a half.
So any stage of the system of this particle can be written as a linear combination of this state, which is to say that M is,
you know, you're guaranteed to get the answer half if you measure a said and a linear and a linear combination of this.
So we now have a non-trivial linear combination with two possibilities. This is a very cumbersome notation.
So people don't use it. They either write that this is plus cy plus plus minus, abassi minus.
That's a handy notation, right?
Because no, there's really no point in writing down this half because we know the first half we know for certain will always be there.
And instead of writing a half, we write plus shorthand for plus a half and minus shorthand for minus a half.
Or since this is only a boring, complex number, we often write A plus plus, B minus.
So any state is a linear combination of these two basis states where we're guaranteed to get plus a half or so,
minus a half rest, and there are amplitudes. So this is the this is the amplitude.
It's Mod Square gives you the probability. If you would measure the Z components of spin of this particle, you would get a half.
And B is obviously the amplitude. What square of that would be?
The probability that you got said it was minus a half.
So let's do some stuff with. They would spend a half particles.
So if I take an arbitrary spin operator.
Well, so let that be the this is the the the spin component.
Along the unit vector. And what is it?
Mathematically, it's an dot sw. In other words, it's an x as x plus and y as y plus NZ said.
Wouldn't this that I've just written down would apply for any spin, not just spin a half.
And I would like to to do calculations like if I do an on ABC, what do I get?
I get some new state fi say, all right, this is an operator, I usual state, I get a new state.
I want to be able to to do calculations like this.
We will we'll find this is crucial.
Now, we know that this everything every one of these states can be written as a linear combination of plus and minus.
So this can be written a C plus plus.
D minus. His essay in.
And this can be written as a linear combination of A-plus plus B minus.
And the name of the game is given the numbers A and B, which characterise that to calculate the numbers C and D,
which characterise that we need an apparatus that does that for us and that's easily obtained.
What we do to find C of course is we draw through with plus.
And then on the left we get C plus nothing. So we find that C is equal to plus s n plus times, a plus plus and minus times B and growing through.
By minus we get an equation which tells us the value of D,
which is equal to minus S and plus of a plus minus s, n, minus B, and there's a handy a way of writing this.
We write this is C, D is equal to a matrix, but this is just a boring, complex number plus s and plus that's a boring complex number minus s n plus.
Plus S, N minus, minus S and minus.
And that's operating on the column vector a, b. So here we have a concrete apparatus.
This is a matrix, a two by two matrix of complex numbers, working on the given complex numbers that characterise ASI,
which gives us the two complex numbers that characterise FY. So for example, suppose we take.
Just got time to do this. Suppose we take an is equal to is equal to nothing come and nothing come of one.
Then this becomes s z. This becomes s.
I mean, all these s ends become assets. And then said on plus is a half of plus.
So this number evaluates to a half. Then see the is equal to a half.
As I said, on minus is minus a half times minus, but a minus is orthogonal to this.
So we get a nothing. Nothing. And as I said on this is minus, minus.
So this one minus half of minus. So this thing evaluates to minus a half.
So in this particular case, it evaluates to this which people write as a half of Sigma Z, A, B, a column B by Sigma.
That is the Pauli matrix. One minus one.
Nothing. Nothing. Apparently matrix.
So these matrixes we really want are these matrices, but it's handy to take the half outside of the matrix and write them in terms of these.
Well, this is and this is the first of them. And if we had time, which we haven't now, we'll have to do it tomorrow.
We will derive what Sigma X's and Sigma. Why are the three apparently matrices that enable us then to write the matrix
belonging to any one of these spin operators and then do calculations on these?
So it's it's time to stop.