So I think where we finished on Friday was not quite at the end of the logic of ending angular momentum.
Remember, we had these two gyros in a box, the total, the rate at which one span was J1, the rate at which the other was spinning was J2.
And we were trying to understand what the States of the Box were,
which had well-defined angular momentum and what predictions we would get for if we opened the box and measured the individual gyros.
And we had shown that what you would expect on physical grounds was the case that the if you orient the first gyro with the z-axis in the second row,
both with the Z-axis also then the two angular momentum would add because that was if
they were parallel to each other and we would get a state with total angular momentum,
j one plus J two and apparently all of it down the Z axis.
And then we used the J minus operator, the reorientation operator,
j minus to create this state in which we still had the same large amount of angular momentum because the two gyros are parallel to each other.
And but we didn't have it all parallel to the Z axis.
And the algebra led us to this expression here that this state is a linear combination of the state in which the first gyro is offset from the Z axis.
For the second gyro is on the axis and the state in which the first zero is on axis and the second is offset from axis.
And I was just saying, is the lecture closed?
How to get this object here? This object here has to be a linear combination of the same two states.
So this is the state of the box. This is a state of the box.
Neither of these states of the box is a state of well-defined, angular momentum of the box.
This linear combination is and there's another linear combination of these two, which is a state of well-defined, angular momentum.
It's that's this state which has less total angle momentum of the box.
And it has this this state that we're looking for has to be orthogonal to that.
And one one good way of raising it is to say that.
J Minus one. J Minus one.
So this is the state in which the gyros are not parallel to each other.
Quite, but all of the angular momentum available, given that they're not parallel,
is along the Z axis, that this is the linear combination orthogonal,
which you could write is j two over j of j1j1 minus one j2j2 minus root j one over j of j1j1j to j two minus one.
So. So.
So here we have this slightly strange thing we have that the state in which so so
so this is a state in which the two genres are not quite parallel to each other,
which is why the total Anglicanism of the box is less than maximum.
Here they are, parallel to each other.
And yet when you look in here, it turns out it looks as if they're not, because one of them's aligned with the z-axis and the other isn't.
And here we have a linear combination of the same two states of the contents of the box,
but with a different coefficients out front and a and a crucially a minus sign here.
And that has the physical interpretation of the two gyros not being parallel to each other.
So let's let's try and clarify this strange situation.
Well, get used to it, I suppose, is the state of affairs by doing a concrete example.
What does it look like in the very important case that we say j one is one and J two is a half.
That means obviously that J The maximum angular momentum we can get is three halves, and we're going to have a diagram now that looks like this.
So that was in general. Now we're going to be looking more we can say more concretely what we're going to have.
We're going to have three halves, three halves at the top here, which is going to be the same as one.
Well, we can just say one and plus.
So so you're missing a shorthand notion here.
Shorthand notation here. Right. So I've got that j1m is now going to be objects like one nothing and minus one.
Right? Because there's no need to write this down. I'm writing down the values, the possible values for M and this one name is nothing.
M is minus one and J to M can be because j two is a half.
I can write this is plus and minus where I'm writing down the values of M in the sense of plus a half and minus a half.
Right? That's a shorthand notation that makes life easier. So this is just that.
This is just a different notation for that state, a more compact notation for that state.
If we would come down here, what would we have? We would have three halves, one half.
Right. That's because J is three halves. And what would it be?
It would be the square root of one over two.
Sorry, not at all true. It would be j one which is one over J which is three halves.
Oops. I'm in danger of running out of space. Let's just shaves that shave it off.
It'll be one over three halves. Times a times nothing.
Plus plus. And now I want this state, which is going to be a half over three halves, the square root of a half over three halves of one and minus.
So what does that let's just clean that up a little bit. That's equal to two thirds of nothing.
Plus plus, this is going to be one third.
One third. Of one minus note is a nice thing about this is that the state, the linear combination of these states,
of what's in the box that we generate comes out beautifully normalised to this thing squared plus this thing squared two
thirds plus a third comes to one comes out normalised automatically and that provides a nice check on your algebra.
So it's good to check that it does. It is probably normalised because if it isn't, the algebra is going wrong somewhere.
We now have a physical a physical prediction if you look at this state here and what we might be talking about now that that j one equals one,
that might be the orbital an element of an electron in that jake was that J two equals a half might be the spin angle mentioned the electron.
So we might be talking about the total and dimension of the electron due to both its spin and its orbital motion.
And if you would look inside the box, if you would examine the, the, the atom, the electron in detail when it was in this state,
you would find that there was a probability of this thing squared a two thirds,
that the orbital angular momentum in the Z direction would be nothing and the spin would be along the Z direction,
and there would be probability of one third that the orbital an element and would be all parallel to the z axis parallels this x as it can be,
and the spin and the electron spin be pointing downwards. All right.
So that's the physical interpretation of this.
So we have the P spin up equals two thirds and p spin down in this particular state is equal to one third.
That's the physical meaning of these numbers here, if we would.
Okay. So now let's ask, what's this state here?
We would like to there are some more states to find.
This is the state in which the we have a half less or we have one unit less of angular momentum than we have on the outer circle.
So this is the state, a half a half of the box.
It's going to be a linear combination of these two things, and it's going to be the linear combination,
which is orthogonal to these two things because it's an I can catch of the total angular momentum squared operator for the box which it's an I can
catch of that which has eigenvalue different from this that must be orthogonal to this by the orthogonal C of the I in case of commission operators.
And what is it going to be? It's going to be root one third of this nothing plus minus one third root what root?
Two thirds. Of one minus.
So now. So in this state, the odds when you look in the box of what you find a changed this in this state which
has less angular momentum in total the probabilities of probability of spin up.
Is this is this thing squared a third and the probability of spin down.
It's two thirds. That's the physical implication of this.
It's an interesting exercise to apply the J minus operator here to generate this.
So if we if we take this state and apply the J minus operator to it, on the left side,
we're going to get three halves, minus half, which is the state here.
I mean, that's let's just have a new diagram because we're running out of space.
The. So here we have three halves, three halves minus a half,
and it's going to be obtained by using the J minus operators on the left and the right of that equation.
And it's I recommend that you do this. I just write down what the answer is for the moment.
It's going to be one of the obvious one. Over. Over one.
Over three of minus. One of minus.
Minus one. Sorry. Plus +23 square root of nothing minus that should be this because there should be symmetry between between the plus.
This state down here physically. That's right in this right in here.
This state here physically it's evident that this has to be this is obviously minus three.
This is three halves and all of it. And parallel to the axis, three halves, minus three halves.
And physically, it has to be that both of them are pointing down.
Right. So it has to be minus one, minus orbital and lumentum down spin down, which is obviously the sort of negative of what we put at the top there.
This thing similarly has to be physically.
It should be that you can get this thing by changing one, two minus one above minus two plus nought and stays alone and plus two minus.
And indeed it does. Right. So it's it's an exercise that I recommend that you check that when you use J minus to go from here,
you do indeed arrive here, which is where you expect to arrive by the the symmetry between plus and minus.
And. And you do? So.
So what are we? So what's happening here physically?
Let's see if we can. We can form some kind of physical picture.
We can only do this to a limited extent because of the big role that quantum uncertainty plays with small with small spins.
But the physical idea here is that.
So let's take this three halves, a half state.
What do we have? We have some angular momentum vector, which is in some sense three halves long and it's only got a half of it in the Z direction.
And that is some superposition of the angular momentum being the orbital angle momentum being more or less in the.
X Y plane and the spin carrying you up.
So you add this vector to this vector. You get this vector.
That's sort of what the first that sort of what the first term up the root.
Two thirds of nought plus symbolically indicates.
Spiritually indicates. We then also have another linear combination, which is which is a one over root, three of one minus.
And how do we understand that? Well, we have to draw a diagram, something.
Something like this. So we're now combining the orbital angular momentum, which is sort of vaguely along the Z axis.
Remember, I stress that when you're dealing with small spin systems, you can never get the angular momentum exactly parallel.
And Z-axis is always a significant amount in the x y plane. Right? So this is this is the direction of Z, this is the x y plane.
So this vector shouldn't be going straight up and I shouldn't have drawn this vector going straight up really either.
That should have been at some funny angle. In fact, let's improve the quality of the diagram a bit by making it not go straight up.
Let's make it go like that. And then I've got minus pointing down.
But again, it's not pointing straight down because there's always I stressed with spin a half.
There is there are you know, there's this much angular momentum parallel to each axis at all times.
So so these this is the sort of diagrammatic representation of that expression up there.
And how do you think about this? How a possible way of thinking about this physically is to say to yourself, well.
I'm. The the angular momentum of the the orbital and the spin and the momenta are interacting with each other.
And as a result of it processing around this in this fixed vector, this is the total incrementing of the box,
which by cons of conservation of activities must be a fixed thing.
So you can imagine that these two vectors of processing around this vector here and here we see two snapshots of possible configurations, right?
So if you imagine this thing moving around like that, now we see this, sometime later we see this and then it'll process back to that.
Now that is not really, strictly speaking, a legitimate proceeding because in doing all this stuff,
we never said anything what the Hamiltonian was, we never said anything about that.
We just had these two gyros in a box and they weren't physically interacting in any way.
Consequently, they have no means mechanically for exchanging angular momentum.
And yet, when the box is in a state of well-defined angular momentum,
we have these results up here and we have this state of the box is a superposition of these states, of the contents of the box.
So beware of this picture. But there is a certain amount of there's a certain amount of intuitive satisfaction in this picture.
And it does at least give you a physical understanding of why it is the state of well-defined,
angular momentum for the box is not a state of well-defined, angular momentum of the contents of the box.
Because already classically that would be the case.
And what's happened is by insisting that the box has a well-defined, angular momentum we have, we have forced the particles to be correlated,
because if the angular momentum of the orbital angular momentum or the first gyro is doing this in order that the total angular momentum is this,
the other thing has to do that. So we have forced a correlation between the two.
The two gyros are between the spin and the orbital angular momentum.
And that correlation is reflected in the entanglement of these of these particles
in the sense that we we discussed when we talked about composite systems.
Okay. And in real physical circumstances, like an electron, if we do have an orbital angular momentum and spin angular momentum,
then there is a physical coupling between the two provided by the electromagnetic field,
and it is then legitimate to think about these things as processing around.
The other thing that I should probably say is that this diagram doesn't really work.
You won't if you if you try and make this diagram work with proper lengths, you know,
you get you give a proper length to this and that and this thing should equal this thing.
And this thing should equals this thing. You won't be able to make it happen. Right.
And the reason you won't be able to make it happen is because this is showing something in only two dimensions.
And what's really happening is in three dimensions. So you've got to imagine.
So we don't know anything about what's happening in the X Y plane. Those those were the those were the terms.
That was the deal we did. We said we were going to have eigen functions of L squared or J squared and Jay
Z and having chosen to know something about what Jay Z is up to is is doing,
we've given up on we've surrendered knowledge of what j, x and j we're doing.
So what's happening in the plane? Perpendicular. This is the x y plane, right?
It's not just it's not X and it's not Y. It's just things happening in that plane means that you can't really draw.
This is a two dimensional diagram. So that's why you can't make it work victoriously.
Well I think we on that we should leave the edition of England Centre and turn to our final topic, very important one which is hydrogen.
So obviously atoms are terribly important. We're made of them. That's most of what we see here and elsewhere.
And they're also played a crucial role in the development of quantum mechanics.
Quantum mechanics was developed in order to build models of atoms.
It is amazing that this enterprise was successful because even simple atoms, like an oxygen atom,
is substantially more is a substantially less friendly dynamical system than,
say, the solar system, because it contains an oxygen atom, contains eight electrons and the nucleus.
So it's sort of the same order of the number of particles as the solar system,
but it is much more horrible dynamical problem than the solar system because the
electrons attract each other much more strongly than the planets attract each other.
So the approximation, which is fundamental to understanding the solar system,
that's the way that the planets move around in the in the gravitational potential of the sun.
And we can neglect the gravitational potential of the other planets while we do that and make ourselves a model and then add in as a perturbation,
the the action of Jupiter, the forces between the planets are not negligible.
They play a crucial role in structure in the solar system, but you add them in later and they are a very small approximation relative,
very small matter relative to the electric attractions of the electrons which are really, truly large.
Another problem about the oxygen atom is that the particles are moving with speeds,
speed V, which is on the order of where it's on the order of eight over 137.
So several percent of the speed of light, you're talking about a system which is mildly relativistic.
The the contribution of relativity to motion in the solar system is very much smaller.
We're moving at 30 kilometres a second, which is less than a thousandth of the speed of light.
So relativistic corrections are much more important.
Another very serious problem is that these particles which are moving around in an oxygen atom are all magnetised gyroscopes, right?
They all have spin. Significant amount of spin because the earth has spin, but it spins very small.
Is enormously small compared to the orbital angle momentum.
And the earth isn't. And the earth is magnetised.
But the magnetic couplings between between the sun and planets and between planets and planets are completely derisory and negligible.
And yet it took it took physicists well to get together to get a pretty good
understanding of the solar system was the work of the whole 18th and 19th centuries.
It was the work of Bessel. The classical structure of the solar system was pretty much under control.
But when Carey, who, who lived at the beginning of the 20th century,
pointed out there was still there was still an enormous gap and problem about the long the long term life
of the solar system and the long term life of the solar system is still an active topic of discussion,
and it turns out to be a very interesting and finely balanced problem.
So. So even though an oxygen atom is very much more complicated and unfriendly at the nominal system,
in the solar system, actually, it's very much better under control.
Quantum mechanics enables you to bring it under very much better control than even today we have brought the solar system.
So it's it's an interesting point that these systems are in quantum mechanics,
actually rather easier to do than the corresponding classical system, but are nonetheless very complicated.
And we have to proceed by stages. And what we're going to do is study well.
Hydrogen, of course, is very important. It's nice that it's a it's a tremendously important atom.
But we're also going to use it as a building block for understanding atoms in general.
So we're going to talk about so what are we going to do? We're going to talk about the growth structure.
What's the growth structure? What's. Of hydrogen.
Oops. Hydrogen like iron.
So what do I mean by this growth structure?
This means that we are going to. We're going to have no relativity.
No spin intimately related to relativity.
In fact, we're going to have oh yeah.
So we're going to be left with points, spineless particles which interact electrostatic.
Right. In non-res specific mechanics.
And over here, what are we going to do? We're going to say that.
That the the nucleus, the charge on the nucleus is going to be Z times the electron charge.
So we're putting in here a number which is which in hydrogen will be one by which we can make larger in
order that we can discuss the motion of electrons around oxygen nuclei or other nuclei as a building block.
No spin. Oh yeah. Electrostatic is always no magnetism.
And these the key thing really is we're leaving out relativity because magnetism is relativistic correction
to electric statics and spin arises naturally when you think about electrons in the context of relativity,
as I hope you'll appreciate next year. So we're leaving out the facts, which are actually quite important.
But, you know, one has to proceed in steps. So now what we're going to do,
what we're obviously trying to do is we're trying to solve we're trying to find
the stationary states of of an atom of a system which consists of one electron,
one nucleus with that charge. So we want the stationary states because they provide the key to the dynamics usual, usual situation.
So what's the Hamiltonian under these approximations? Well, it's going to be the nuclear.
Kinetic energy, the kinetic energy of the nucleus. Nucleus squared over the mass of a nucleus, twice the mass nucleus plus the electrons,
kinetic energy plus the interaction energy between these two, which is going to be z e squared over four pi.
Epsilon nought x, e minus x.
And we'll do this right. So this is the sum of the energies, I mean, of three distinct contributions to the energy, kinetic, kinetic potential.
Right. So we want to know, what does this look like in the position representation?
Right. We're going to. We want to we want to examine that equation concretely.
And the way to go is to put this into the position representation. So that's to say we bra through by x, e, x and H.
Yeah. And then what we want is the position representation of this,
which is going to be minus h bar squared, del squared with respect to X and over to Nucleus.
Right? Because this is the kinetic energy operator which we and we know that P is minus H by a gradient.
So P squared is minus H bar is right is minus edge bar squared never squared.
So what does this mean? This this sub x end means this involves derivatives with respect.
This involves derivatives with respect to the components of the position of the nucleus minus h bar squared
over two mass of the electron del squared x electron minus z e squared over four pi epsilon nought.
This is already in the position representation if you like xy minus x and.
So all this stuff. Time's up, Cy.
We'll do it from outer space. To put it in terms of Cy is going to equal E.
Time's up, Cy. Right. So this is the usual stuff that have cy, which is a function of x, e and x, and so it's a function of six variables.
Is, is x and whoops, x, e, e.
It's the wave function of a stationary stage of energy. So we've got here now.
So we've reduced our abstract equation to a very frightening partial differential equation in six independent variables.
Right? Because because we've got the positions, the three components of X and and the three components of X E.
So it's that we've got a PD in six independent variables.
An astonishing thing is that we can solve this exactly.
And without a huge amount of sweat. And as in the solution of any number of problems in physics, the key is to choose your coordinates correctly.
That's quite generally the case that a problem which is very frightening in general with a clever choice of coordinates.
You're all done. So we need new coordinates. We need to transform this equation to new coordinates and the ones we take.
A big x, which is the centre is classically the centre of mass coordinate.
So that's going to be any XY plus mass of the nucleus times the position of the nucleus over me plus an.
All right, so that's the centre of mass coordinate. You may say, what authority have I got to use that in the context of quantum mechanics?
And the answer is I make this. Strictly speaking, I made absolutely no claims as to the physical interpretation of this.
It's just a it's just a suggestion of something we might use to simplify the algebra.
But as rational beings, we know physically what that means, and then we'll have another.
So that's three new variables, okay? Because it has three components which are linear combinations of our old variables.
And we going to have another linear combination and surprise, surprise, it's going to be Z minus X and the separation.
So what we do now is plodding mathematics in order to rip out of that differential equation X and An and insert the corresponding things with this.
So let's see how this goes. Let's do D by the XY, right?
Because that nebulous squared E is sort of this operator dotted into itself.
So let's see, what is what is this? Well, the chain rule says that it's D by the it's D by the x, d by the x plus d, by d r.
So this is the chain rule. Mathematics. Nothing to do with physics, but because it's mathematics is definitely true.
And this dot implies a summation, right? Because this thing has got three components.
So this is the x one by the x, i.e. D by the x one plus the x two by the xy debris x two, etc., etc., etc.
Now, fortunately, these partial derivatives are nice, friendly things because we just have linear combinations here.
So I think we can easily see the what this amounts to.
This is going to be m e of e plus m and that's what this partial derivative comes to from up there.
Of of. Of D by the x plus.
Sorry. And what's this? This partial derivative is nice and simple.
It's just one. So we're just going to get a plus DVD or.
So I've still got these a shorthand for three equations because this is DVD XY one is equal to this thing times DVD X one plus two,
liberty all one, etc. We want this.
We want this thing dotted into itself. So what we have to do is, is multiply this on itself with a dot between the two.
And what do we get? We get that bell squared x sub e is equal to we get this thing squared, of course.
So we have m e of m e plus M and squared d two by the x squared.
Well, now we can write that more handily. Estelle squared with a big x I think more clearly.
And then we get this thing squared plus del squared with respect to del squared where we're talking about the components.
Here's, here's the usual expression for del squared. But using components the big x, here's the usual component,
the usual expression for del squared, but using the components of the separation vector all.
And then irritatingly we get a mixed term because we get this because we're taking this operator and we're multiplying it on itself.
So we get a mixed term of this operator of multiplying the body are in the next bracket and then we have this thing doing this.
So we end up with plus two of m e of m e plus m and of D to the x.
They are. So this is not very nice.
Nobody wants this. This is excellent.
We've got a relationship.
We've found a relationship for this, which we want in terms of this and this, which is fine, but this is definitely not required.
She would kindly go away and we could make it go away easily by just working out what D by the end is.
That's going to be the big x by the x and which is going to be mass and mass of the nucleus of a
mass of electron mass of the nucleus divided big x and then it's going to be the R by the x n,
which is going to give us a minus one instead of a plus one. So this will be minus D by oh now.
And when we square this up to work out what Del squared a vaccine is, we get this thing squared, of course, and of m e plus m n squared del squared.
Whoops, big x. So that's this one squared.
And then of course we get this one squared and never mind the minus sign because we're squaring up.
So this becomes a plus del squared with respect to the separation variable components.
And then we get this on this where now the minus sign is manifest, we get minus twice this, which is an end of me plus and of D to by the x.
To O. So. So we have expressions which are very similar, but include one has a plus sign, one has a minus sign.
So what we want to do now is work out one over an e del squared with x e plus one of n of del squared with respect to x.
And so one over m times that type of equation plus one over many times this, which is actually exactly what occurs in a Hamiltonian.
If you go, yeah, if you go right up there, mercifully, what we want is in fact the Del Square,
the individual Del Squares weighted by one over the mass.
And what's is equal to this is equal to we get we get del squared x twice, once from here and once from there we are dividing.
We used to have an m e squared over me plus m n squared,
but we divide it through by me so we have an m e and then similarly from here we have an m and over m e plus and.
Squared that's a result of adding this with with this weight to that.
Similarly, what do we get here?
We get we get a common factor of del squared o and we have a one over m e plus a one over m and and these terms go away.
Right. Because they yeah.
They by the time we divide through by Amy at the top, we have just two over me plus many times that makes a derivative by the time we divide it by.
And then we just have minus two over that so they go away.
So this that we want in the Hamiltonian is equal to this,
which can be simplified and written as one over and e plus m in del squared x plus one over mu del
squared o where mu is exactly equal to me and of e plus and then goes by the name of the reduced mass.
And you may already have met it in classical mechanics. I hope you have, but if you haven't, never mind.
So we can now, right? Our Hamiltonian and I'll do it over here.
I think so. As not to so we can keep in view the Hamiltonian at the top.
There we have what we have. We have the Hamiltonian. Involves that what we have is a minus squared over two, you know.
Right. So that's going to give us a minus H bar squared over two and plus me or me plus I may not be writing del squared with respect to big x.
We will have minus h bar squared over two mu of dl squared with respect to the separation
vector because this is the magic combination minus h plus minus h bar squared over two times.
This is what appears in the Hamiltonian. So we get minus HP squared over two times this, which should I hope I've written down.
And then we have to add in the we have to add in the potential energy term,
which is a minus Z E squared over four pi epsilon nought times the separation
variable where this is if you understand the modulus of the separation variable.
So that's what in the position representation are with these new coordinates.
Hamiltonian is looking like this. And this is really useful because we can define this to be h k and we define the rest of it to be h sub are.
And what do we have? We have that h sub K commutes with h sub off.
Why? It commutes because this is a function only of the X variables, the big X variables, right?
It just involves derivatives with respect to big x.
This and those derivatives, these partial derivatives by two big X components with all the other coordinates held constant.
And the other coordinates means the other components of big x and all these and all the components of little are.
Because we we made a legitimate change of coordinates. So this part of this obviously competes with this because partial derivatives always do.
And this competes with this because this is being held constant while we're doing these partial derivatives.
So edge sub K commutes with H sub are correspondingly h h sub K commutes with the Hamiltonian and also H sub R commutes with the Hamiltonian.
That follows immediately from the first thing because obviously because this thing is HK plus h,
r, h, k obviously commutes with itself, every operator commutes with itself.
And we've established that commutes with with R so so this vanishes and similarly this vanishes in the same argumentation.
So we have three mutually commuting operators.
That means there's a complete set of mutual agent states.
Of H, h, k and H.
R. So these we want I can states of this this what we set out to find.
And what we've discovered is that if we find the iGen states of this and this,
we will just we all we need to do is multiply them together and we will have iGen status of this.
So what does this imply? This implies that E is equal to some state of k times.
Some state e r or e soc.
You have to be a little bit careful. I think I've said something that's not, strictly speaking, true.
What we know is that there is a complete set of I can cats of this, which is simultaneously I can catch of this and this it does not follow.
And in and it is not true that every eigen state of this is simultaneously and I state of
this and this there are eigen states of this which are not eigen states of this in this.
But we will get a complete set of eigen states of this that's good enough for us, which are simultaneously eigen states of this and this.
So we've reduced the problem of finding the eigen states of this to the sub problems, of finding the eigen states of this and the ego states of this.
So HK is just the kinetic energy.
Is the kinetic energy operator of a free particle.
Physically just describes the kinetic energy of the whole atom.
The whole atom can cruise across to the laboratory. It has kinetic energy.
All we interested in this? No. It's bloody obvious that that the energy of an atom depends on how fast it's moving.
We can deal with that.
We've already studied the we already last term studied the the stationary states and the spectrum and everything else of free particles.
We know all about it. Boring. Finish. So all we have to do is hammer away at this thing.
What we want to know is. So this is so these basically trivial.
And of no interest. So we focus in on H.R.
Now we went to some considerable trouble to show.
So so this has to be studied. What we want to deal with is h r e r is equal to e r e.
And this is going to describe the internal. Energy.
Of the atom as distinct from the translational kinetic energy that it has because it's it's moving.
And we've been to some trouble. We've. To show. Del squared was equal to PR squared minus h overage password.
We showed a relationship between the, uh, between Del Squared.
And what we identified to be the radial momentum.
He is squared and the orbital angular momentum operator.
All right. So this is the basically we showed that L Squared was.
All squared times the angular parts that we were familiar with inside the inside del squared that are placed in.
So this is the orbital total. Orbital and momentum total.
Orbital angular momentum operator and PR may write it down for you.
PR We figured out what it had to be.
It turned out to be minus H bar of D by the R plus one overall.
That's what it looked like in the position representation. And it's the radial momentum operator.
So. So the only.
So let's just say what we've got. Let's write down on our using this information.
H R is minus h squared over two mu times.
Natalie squared. Okay. So what's that. That's equal to p r squared over two mu.
Plus L squared H squared L squared over to you all squared and then minus z e squared over four pi epsilon nought r.
That's what this what this. So we've focussed homed in on this operator.
A problem is solved once we find the iGen status of this operation.
The I can energies of this operator and this is what it looks like.
It involves radial momentum, radial position radius and for the rest it involves the total orbital angle momentum operates.
Operator So the brilliant thing is that h r l squared.
Equals nought. It's also true that your comma, L.Z., equals nought.
Why is that? Because L Squared, remember, is could be represented as a partial differential operator in terms of DVD theatres and DVD files.
So it clearly commutes with this and it commutes with this and enjoyable commutes with itself.
Similarly, L.Z. turned out to be minus D by D Phi the azimuthal angle so it commute.
And L.Z. we know commutes with L squared and it clearly commutes with R and PR.
So what do we learn from this? There's a complete set ups, set of mutual aid and states.
Of H. And they'll squirt.
But we know all about the Asian states of elsewhere.
We've studied them ad nauseum. Right. Which means that we.
Well, so we.
What we can do is we can we can say it's legitimate to say now, it is not true that every Oregon state of age is an Oregon State of L squared.
Curiously that well, anyway, it is not true that that that that is the case.
But it is true that there's a complete set of states of the form e an L squared.
Sorry. Yeah, there's a complete set of states.
E and l, which is such that l squared on e l this is an igen state.
That's what that's doing there, denoting that it's an Oregon state of this operator and of course has eigenvalue l l plus 1el.
So I've learned that it is possible, it is legitimate to look for it to restrict the search for eigen states of this crucial internal energy.
Hamiltonian two states, which are eigen states of the total angular momentum operator.
And as we'll see tomorrow, that then reduces the eigenvalue problem associated to h r to just a one dimensional problem very closely,
analogous to the simple harmonic oscillator, which we've already sorted.
And it will it will it will yield to the same line of attack that we used on the simple harmonic oscillator, namely ladder operators.