Okay. So let's let's get going.
So yesterday we introduced these orbital angular momentum operators, the three of them,
the three components of X, cross, P and divided by H bar to make it dimensionless.
And we can make the total increment to operate two L squared by squaring the individual components.
Adding We establish these results that the commentator of for example, x with y is i times z.
So you if you.
And similarly, if you do a computation with ally with ally with Che having written this down here, you find EI times, peak times excellence, right?
So, so for example, times p computed with p y would be item's p z.
Basically what we're learning here is that L computed with the component of a vector operator gives you the third vector operator in that set.
And from that you can go on very easily to show it. The the demonstration is exactly the same as what we already did with the
angular momentum of the total angle minimum operator that any of these angle,
these orbital angle minimum operators commutes with scales like X squared or p
squared or x dot p a also L squared commutes with with any of these things,
any of the component, any of its components. So we have a situation so far which is precisely analogous.
To the commutation relations, to J comma, to the total Anglo mentum total.
But what we have so far called the angular momentum.
Operators. So, for example j i comma x j commentator is i some of a k epsilon i j k x k, etc.,
etc. So when we have that j i comma j j is equal to I some of the k epsilon i j
k j k which mirrors one which I didn't write down up here and should have done,
which is that l i comma l j commentator is equal to I some of the k epsilon i j k l k which is really a generalisation of this rule.
I mean application of this rule here to the vector l j being put in here as well.
K appears over there. She just says that L is in fact a vector operator.
And similarly, similarly, this is a generalisation. So we have this exact analogy and the question obviously arises are they the same operator?
Is L the operator which are the operators X, Y and Z?
The operators that generate rotations that we inferred had to exist just by thinking in the abstract about rotations.
At the end of last term. And so the answer to that question is, is no.
And here is and here is the demonstration of it.
Let's calculate j squared comma ally.
In other words, let's work out the commentator between the total angular momentum operator and one of these components.
Then this is going to be this is a combination of a product and I need to write this.
This is a sum of squares. So I need to I need to make that explicit.
So I write that as the sum of a over. I say of j I squared a comma.
Oops, no, no, I won't do because I've got I busy j ally.
And now each one of these is just a simple product. So we use the rule for taking a commutation of a product.
So this is equal to the sum of a j still of a sorry.
That's the j what we're summing over all I need to do.
I leave one of these standing idly by whilst the L operator commutes with the other one,
and then I have the L operator commuting with the other one, the first one, while the other one stands idly by in his position.
Now we know what the result of this is,
because this is a vector and we know we established a last term what the commentator of J with a component of a vector is.
It's going to be the third member so this I can replace by an epsilon j I say k so this
is going to be a sum over J And then there is going to be a sum over K coming up of J.
J There will be an I from the commentator.
So JJ that's this one here. Make sure that's clear.
This is going to be epsilon j i k l k So, so that's the, that's the fundamental rule.
We establish that the commentator of this with any vector gave you the third component, and then here we will have the same thing.
Epsilon j i k times l k times j j in the back so we can clean this up into the high times the sum of both J and K going from over X,
Y, and z of epsilon j i k times j j l.k. plus l k jj and.
So this is the sort of thing which often vanishes because this is anti symmetric.
If you swap those two over Jane K, you get a change of sign.
And if this were symmetric, we would get a change.
Oh, sorry. And if this was symmetric in J. In Canada, as if you swap them over, you've got no change of sign.
Then you just got the same thing back then we would have.
Then we would have. Zero.
But this is not symmetric.
If you interchange if you interchange the order of these, if you swap the indices.
So under a swap, what does this go to? It goes to J k, l j plus l j j k, which is not the same as this.
So this thing is not equal to zero. But we know that J squared does compute compute with all of its components.
So that sort of suggests that these are not the same and not the same thing.
So so let's try and understand what these operators,
these these orbital incremental angular momentum operators and really understand their relationship to the things that generate rotations.
So now we need to talk a little bit in abstract about rotation around a sorry motion around a path.
So if we just have a translation, we've already started translations last term through, through one displacement.
Hey that's, that's generated by. The unitary operator, which we called U of A, which is e to the minus i a dot P on each bar.
Right. That is the operator which generates out of a state the state that we would have
if our system was shoved along the bench by always was at the different location,
a distance or a vector displacement away.
So let's let's consider the following. Let's make a series of displacements.
Here's a one, here's a two, here's a three and so on.
Right? We're going to make a path by doing a series of displacements.
Then we have the you total is going to be the product of the this of this displacement and this displacement, this displacement, this displacement.
So it's going to be you a four operating on the result of using you a three operating on the result of using you a to you a one.
These each each of these things is an exponential.
So this is e to the minus i a for dot p over h bar is the minus ia3 dot p over h bar,
etc. and the operate is occurring in these exponentials all compute with each
other because all momentum approaches compute with all other momentum operators.
They compete with themselves obviously in any and they can mix with P, y,
etc. So when we multiply these exponentials together, we have the usual magic of exponential functions.
We don't have to worry that those are operators. So these are operators, but we don't have to worry about that because everything commutes.
So this can be written as e to the minus I for some of these vectors summed over I dotted into p over each ball.
So the operator that that generates you is the state that you get as a result of all of these displacements is given by this just a single exponential
it's just it's just one of these it's a it's an operator of exactly the original form where the displacement is simply the the sum of them.
So the result taking it this way and this way and this way and this way we've just shown is the same as this,
as the result of just taking it in a straight line over the sum of the I.
So that's that's that's the general argument.
And now we see a special case. So in particular particular for a closed path.
So if the path comes, if the path carries you all the way around, what does that mean?
It means the sum of the ai0.
All the vectors add up to nothing. And that implies immediately that you total because it's e to the nothing is the identity transformation.
Now, that might sound obvious, but you'll see in a minute that that's anything but an obvious result.
So. So now let's, let's specialise in our paths.
Let our path be made up of a series of of so let this be x let this angle in here be delta alpha,
left and right and be the unit vector point out of board.
Then this vector here, oops, this displacement vector, this is going to be a equals delta alpha and cross x, right?
That's just ordinary classical geometry that if I go like this and is out of the board, then that's the displacement that we generate.
So what does that do? That that means that the op,
the unitary operator that that moves my system from here to here you Delta Alpha is
going to be a to the minus I a which is that delta alpha and cross x dot p on bar.
So that's just applying the standard stuff with the displacement vector of this form.
This is Delta A or whatever. But what it what is that this scalar triple product can be reordered sort of standard
vector algebra tells us that we can reorder this thing into n dot x cross p.
So this is equal by vector algebra to I delta alpha and dot x cross p punch bar.
So that's the operator. Okay. And there's the bar that's here.
But this is what we define to be.
L So this is equal to E To the minus I Delta Alpha and dot.
L So that's what we've now discovered. Essentially what L does, l is the generator of of rotations, of movements around circles.
So we conclude. That L is the generator of translations around circles.
We can also get something interesting, very important by combining these two these two things.
So for a complete circle so far, we can multiply these things together.
So you for a circle is going to be the sum of all of this, for all sorts of of delta alphas.
Okay. Uh, these are the minus I Delta Alpha and L, which is going to be e to the minus I, alpha and dot l where this is where this is the sum.
Sorry, it's going to be the product of this, right?
If we make a series of transformations, each one by Delta Alpha,
then that product of these exponentials can be written as the exponential of the sum of the arguments, the sum of the arguments.
So it always has to end. Dot l is the operator and the sum of the alphas of the Delta Alphas we're going to call alpha.
So for a circle for circle, this is going to be two pi.
So we're going to come to the conclusion that e to the minus two pi i n dot l is equal to the identity
transformation because we've shown that you going all the way around and any close path has to be the identity.
So this thing has to be has to be an identity for any vector, any unit vector.
N and I will need that result shortly.
And what? Yeah. What? So what do we. Come on.
So what is the general. The general picture here that we can now understand what the distinction is between orbital angle mentum and angular momentum.
So here we have a sort of look at this L And then there's this arrow here if we.
So we're moving the we're moving the point at the base of the arrow around the circle.
We're just translate translating it. Let's just go back.
So we're we're just moving it round a circle, but we're not doing anything to its internal structure.
We're just translating it. So if it has a little arrow, it has an orientation as, for example, the direction of its spin.
That's that's not going to change its direction. It'll just be carried around.
Now, ask what Jay does.
What Jay does is it makes you the same system you would have had if you took what you've got and you rotated it on a turntable around the origin.
If you rotated it on a turntable around the origin, this is what happens.
The orientation of the particle changes as well as its location.
So J is changing locations. L so l is only changing locations.
Jay is doing a complete job by putting the particle on a turntable and rotating it at the same time is translating it.
That's the difference. Right.
So we've got these operators and we want to know obviously when you have operators,
you need to know what their spectra are, what the allowed values of their eigenvalues are.
And we can now immediately say what these are going to be.
So we know that we can sit, we can find simultaneous a complete set of simultaneous eigen states of l squared and any one of its components z.
That's the same as when we were dealing with the angular momentum operators and we obtained the eigenvalues of the Anglo momentum
operators at J Squared and Jay Z by using by exploiting the commutation relationships between the different components of J.
That was the only thing we used. Right. So we, we got that the e values of j squared Jay Z were they were j j plus one.
This is for J Square. J squared had eigenvalues of j j plus one where J was nought a half, one three half cases, three halves, etc.
Right. And we had that Jay Z had eigenvalues M which lay between minus J and plus J.
And we got these results only using that j comma.
J j equals i epsilon.
I. J k. J k. If you look back at what we did, you'll find that that's the case, that nothing else went into this.
But these computation relationships here for the L operators, we have the same commutation relations.
Right. So the L satisfy identical computation relations.
So this so the argument we had for J could be repeated line by line.
There's no point in repeating it literally, but virtually.
We now repeat that argument line by line with every J replaced by an L,
and we conclude that the e values can be l squared has numbers like l l plus one where l
could equal nought of half one level blah and ls that has m line between L and minus l.
But then we think. But so these are the kind of no other numbers are possible than those numbers.
Right. That's what that argument shows. Are all those numbers possible?
No, for the following reason. But E to the to Pi, L.Z. has to be the identity operator.
We've just shown that. That's a that's that's this statement with an put equal to the unit vector in the in the Z direction.
Okay. So I should I need a minus sign.
So we have that this thing has to be the identity operator.
Consequently, if we use this operator on one of our states, which are going to be called l m right.
This is the mutual eigen state of l squared and l.z. with eigenvalue l l plus one for l squared and m l said precisely by analogy with the.
With what we did with Jay. Then this has to be simply Elm, because this is the identity operator.
But what actually is this? If we have. So we're doing an exponential of this operator.
This operator looks at this and says, that's my eigen function and therefore it replaces itself with the eigenvalue.
So this is e to the minus two m pi i times l m comparing this side with this
side we have the e to the two pi and is equal to or is equal to the number one.
And that implies that m is an integer because if it were a half integer, this will be e to a certain number of I PI's, which should be minus one.
But we've shown it cannot be minus one, so it's an integer.
So this is the, this is where there is a difference between what happens, between what L does and what J does and why we have to keep track.
So L is looking very like J, but it is not J because it is.
Well, we've seen physically that it's not because what it what it merely translates you around a circle.
It doesn't rotate you round a circle. And that has the consequence that m has to be an integer and therefore L has to be integer.
So the eigenvalues of l squared are l l plus one where l equals nought one,
two, three, four, etc. and therefore M is also going to be stuck on integers.
So what's happening here? Is that what we what we what we learn from this in some sense is that.
Let's just do this again. So. It's when we have translated our system all the way around the circle.
It's come back to where it was, same orientation, everything the same right.
And and the result is that an identity operator is applied as described by applying an identity operator to our state.
If, on the other hand, we do we translate Jill way around the system.
You would think it came back to where? Well, it comes back to the same place, undoubtedly.
And the little arrow comes back to the same orientation.
And you think you were back in the same place. But quantum mechanics is telling us something.
Well, it's really experimental. Physics tells us that it's not. How do we know?
So that arrow I'm I'm having to describe the orientation of the spin of a particle using an arrow.
And I've already said this is a hazardous enterprise, because because quantum mechanics in quantum mechanics particles are gyros,
and they in some sense have a spin that point in a direction. But you do get into trouble.
Is the Einstein Podolsky Rosen experiment shows if you take this idea of pointing in a direction too seriously.
And here we have another example of the risks of taking too seriously the idea
that you can describe the spin of a particle by pointing in a direction,
because when it is gone all the way around the the state vector has changed sine in the event that
this is a half integer and four and and the and the Stern Gerlach experiment and other experiments,
zillions of experiments show that electrons and protons and so on do have half integer values.
Their angular momentum comes in half integer amounts. So when you take when you when you take a real electron on one of these terms around the origin,
its state is somehow different from the state it started from.
It's difficult for us to understand that. But but that's what that's what the mathematics combined with the experimental physics tells us.
Okay. So the next item on the agenda is, is the eigen functions of El Squared and El CID.
So we already know what the eigenvalues are.
What we now would like to know is what do the what do these states look like, these elm states look like in the position representation.
So what we're trying to find.
So what we want is our thetr phi l m we'd like to find expressions for we'd like wave functions which describe these things here and our strategy.
Is basically that we're going to.
So so the strategy is this the detail is rather is rather tedious, but the strategy is this we are going to apply l plus to l l and get nothing.
We're going to express this abstract. So these l x plus.
This is all right. I think we talked about this yesterday, but I want to see sun l x plus i l y.
So this is the operator which will try and raise the m entry here to one larger.
But that's not possible because it already is largest, largest value. So it'll kill it.
So this is this is an operator equation.
We will look at this operator equation in the position representation and then it will become a first order differential equation.
We will solve it. It'll turn out to be dead simple to solve.
So that will that will lead us to let me leave off the R because I'm not really interested enough at the moment.
It will lead us to this l l and then we will be able to say that Theta Phi l l minus one is equal to a seat of Phi.
So there's a horrible square root now l l plus one minus m m minus one times l minus two.
Sorry, this needs to the minus one times l minus.
Uh, so we will.
Having obtained this by this wave function,
we will apply l minus to it in the position representation that will give us essentially the next one down.
Lower this by one.
And by repeatedly doing this, we'll be able to generate all of the wave functions associated with a particular total incremental quantum number.
L That's the strategy to carry this out.
What we need is expressions. We need expressions in the position representation for these operators.
So this. Let's start by writing down what else that is.
What is L.Z.? It's one over a bar of s p y minus y x in the position representation.
What is that? P y and the position representation is minus h, bar d by d y.
So this becomes minus i x d by d y, minus y d by the x.
Now, to find out what that is, we could just we would we would like to express everything in terms of three,
four and five, because we know we know an element is to do with rotations.
That's why we want to use these revised spherical polar coordinates.
And we could just use the chain rule to turn this into derivatives in theatre or five,
but it's much easier to go to use the chain rule not going from this to to D by DC to
DVD five but to go to write to find out what is DVD PHI According to the chain rule,
it's the X by defy D by the x plus d, y by Defy D by d, y plus d z by defy divide you z.
Right. So that's just a chain rule from calculus. Now we put in what these X, Y, and Z are.
We know that x in polar coordinates is r sine theta cos phi we know that y is our sine theta sine phi and we know that Z is of course theta.
So this implies that. So take a derivative of this.
With respect to Phi we have the de x by de phi is going to be minus r sine theta
sine phi which is the same as minus y minus is going to be simply minus y.
We have that similarly d y by 85 that becomes a cosine and therefore this becomes x and we have the desired by defy is nothing at all.
So we take these results and stuff them back in here and that tells me that divided phi is equal to that's a y sorry.
That's a minus y. So let's write down minus y d by the x plus x divided y.
What is the relationship to what we have up there? That is okay.
So this is precisely what's in that bracket for L.Z. So this implies that L.Z. is minus I d by define.
Of course, this is no surprise because L.Z. is the thing that rotates you around the Z axis and everybody
knows that the spherical polar coordinate is defined as the angle around the z-axis.
So it's kind of obvious that this has got to be the case, but it's nice to see that the chain rule delivers the goods.
Now, the next bit is distinctly more tiresome.
What we now have to do is what we want to do is, as I said, is express L plus an L minus X plus l, you know, plus I l y.
In terms of DVD features and DVD phys, you could go with this just by brute force,
but the algebra would be heavy, so the result would be that wonderful.
Now here is I think this is the way to do it, right?
So let's just calculate what DVD theatre is using the chain rule.
It's obviously the x body theatre DBA X plus d y by d theatre d d y plus d z by d c to debate.
Is that okay? We have expressions up there.
So the X Body Theatre is going to produce this in our costs, theatre, cost fi, etc. this is going to produce this an r cost theatre.
Sine fi. So I got a common factor of our cost to open a bracket and then we will be able to write down sine.
Sorry. Cos fi d by the x.
Cost plus cost fi. Sorry, plus sign five dvt y.
And then finally we have d z by d thingy, which is going to be minus R because we are differentiating cosine theta d by d z.
Now let's take our expression. Full court.
Sorry. Four D by defy and multiply it by court theatre.
So we're going to write down Court Theatre Deep 85 just for the fun of it.
So. Right. So.
This has to be we're going to be doing it. We're going to replace well, let's take this one first.
It's going to be oh that is all sine theta cos phi when we multiply it by cost which is cause it was sine the, the sine is going to go into a cause,
so this is going to be o cause theta brackets sine phi excuse me cos phi dy by d y so this,
this is this here is our sine theta cos phi which is x times calls over sine which is not.
And then that why is going to be our sine theta sine phi and the Scott multiplication
will turn the sine teacher into a cost that I take out and we will have a sign phi here.
TBD, TBD. How much X? That's it.
Right. So what we now do is just again, for the fun of it, we take DVD theatre and add on it times this.
So we look at DVD theatre plus eight times court theatre, DVD Phi.
What happens then? We have our cost theatre as common as common factors.
Oops. Quick. I should have.
Yeah. Excuse me. Yes, that's right. That's fine. I'm looking at the wrong coefficients.
So what am I going to be doing? I'm going to be adding this and this.
So we're going to add eight times this to that.
So with a common factor of X, we're going to have cost Phi minus.
I signed Phi, which is well known to be written as E to the minus I fi that's how much of db x we're going
to have and then we're going to have sine fi plus ICOs Phi and that is going to be plus,
uh, high times each of the minus sci fi debate.
You y we haven't quite finished have we.
All right. So if you expand this out, it's going to give you an AI cost fi, which we want.
That's this one here times I and it will contain a minus I sign five times an I which makes it a plus,
just plain sign five, which is what we want there. Then finally we've got this in the back, so we will from here.
So we have minus sign three to debate.
You said. Yeah.
So supposing we. Oh, and this thing could be written.
Our sine theta is this.
No, no, no, no. Leave it alone. So what we do now is we multiply this by each of the five, so we have each of the if I brackets divided feature.
Plus I cut thetr d by 85, which will turn out to be our bottom line is equal to the.
This this is going to be cancelled because I'm multiplying through by each of the I fi this will be cancelled.
So we will have our costs theta brackets d by the x plus i d by d y,
and then we will have minus o oh sine theta times e to the i phi which I choose now to write is cost phi plus I
signed Phi and now our sine theta cos phi is x and this in the back is going to be y our sine theta sine phi.
So we'll be able to write the right hand side as our cost theta d by the x plus i d b d y minus.
So this is going to be x plus i y. Excuse me, this was times dpd z two which was where did that get lost.
It was minus R sine theta D by de z. Yes.
It just got lost in this line. In this line. Okay.
Phew. So I want to now establish that that is actually l plus.
So in order to do this, I write down so what is l plus l plus is l x plus i l y don't need those brackets, which is one upon h bar.
Okay, once x it's y, p, z minus z, p y plus i times l y is z, p x minus x, p z.
So now I want to turn this into x's. So this will be, uh.
So this is minus i h bar bag, he said.
So we're going to have know. It's just the sorry.
Before we do that, let's gather things together because we've got a p z and a p z.
So let's just for the moment, right. This is one over h bar we're going to have.
We're going to have a common factor of Z. We're going to have a i z, p x plus i, p y.
I think, right, this I and that's how I make the minus sign that we have there.
Z is the common factor. All right.
And then what about the Z factors? The P Z factors?
Well, we have we have a plus Y minus I, x, p, z.
Now we replace the P's by minus H, by D, by D, x, etcetera.
So you will get a minus H because we're going to go through out from here, we'll have a minus I times and I which will give us a plus.
So we'll have a z d by the X here.
We will have, we have two A's making a minus sign.
We have another minus sign coming in from the minus II.
So we end up with a minus i d by d y and the of by the sign there choice mean minus nine.
Now the two, the two made a minus and I soaked up. Sorry, these two i's made a minus.
I brought in another minus from minus h body by D the minus is cancelled, leaving me only with the I.
And this is going to bring in a minus I.
A minus I. Well, so we have a few propagate this minus sign side here.
We get a minus sign there. So we get a minus x and this is going to be plus i y because there's a minus I here and
there's the minus sign and here's the I by Z and that I hope agrees with what we have written.
It does because all costs theta is also known as Z.
So we have said X plus I did y and here we have a minus x, plus y, y, z.
So we have established a very important result, but l plus in the position representation is the differential operator in each of the EFI
brackets deep ADC to its pluses and it plus i cotangent vector d by fi close brackets.
There is an analogous calculation which I'm not going to do,
which tells us that L minus is equal to minus E to the minus sci fi d by dc to is it minus yeah it is minus i v to deep 85.
And if you try and do the two calculations together with plus and minus signs, good luck to you.
It's fantastically confusing, right? So. So at least I got one of them.
Right. So those will be us. Well, we let, let us know.
Well, now let's leave it at that that those are going to be our starting points for tomorrow,
deriving the eigen functions of these very important operations.