So that's a summary of the important formulae we obtained yesterday.
We'd, uh, Wednesday we had reduced hydrogen's problem to a one dimensional Hamiltonian one h
sub l one for every particular value of the total angular momentum quantum number.
L And we found that this thing at the top here was a ladder operator in the sense that it's out of a state of a certain amount of energy,
in a certain amount of angular momentum. It constructed a state with the same energy and more angular momentum.
And using that and the idea that the sequence of states of more angry momentum with the same energy had to stop.
We concluded that the energy was given by a certain constant 13.6 V divided by end squared,
where n is one more than the maximum angular momentum that you can afford at that energy.
Put another way, the Anglo momentum quantum number is less than or equal to one minus.
This number n controls the energy and is called the principle quantum number.
So what we want to do now is move forward to get the energy eigen functions so as to get the.
These are of interest from the perspective of hydrogen.
If you want to do any detailed calculations, like how does hydrogen interact with the electromagnetic field?
What happens if you scatter electrons of hydrogen and that kind of stuff?
You'll need to know what these eigen functions, these wave functions are,
but they're also the building blocks for atomic for studies of atomic structure generally.
So there are a complete set of states which you can expand any state.
For example, the stationary state of an oxygen atom. You can expand in these states.
And that's what people do when they do atomic physics calculations.
For the most part, what they do. So these these play a very big role in atomic physics.
Okay. So we modify what they are. Now, from this from the fact that l squared on e l is equal to l l plus 1el from the fact that this thing,
this state here, well defined in a stationary state, is an eigen function of the total angular momentum operator we know all about the so the wave
function in question is this right the amplitude to find at X the the atom when it's in this state.
The reduced particles, strictly speaking, when it's in this state. We know all about the dependence, the anger, the dependence of this.
We know that this is takes the form of some function, which presumably depends on the energy and on the angular momentum.
Times are our times. Why l am a feature in PHI?
Because we know that these the spherical harmonics are the unique angular eigen functions of this operator with this eigenvalue.
So this thing, which we know is such an eigen function, its angular dependence must be given by, strictly speaking, a sum.
Potentially, it's a sum of these four different values of M and the same value of L,
but it's sensible to look for them one by one, assuming a particular value.
And so the angular dependence, you know, out of out of that statement up there is this.
And all we're looking for right now is the radial dependence on the radio dependence.
Uh, as I say, you have to expect it to depend on the energy.
I mean, it has to depend on what's in here, which is the energy and the.
The angle momentum, a quantum number. So that's what we're that's what we're looking for.
And we're going to get it just as we got the wave functions for the harmonic oscillator
stationary states by studying the equation that the the equation that this operator,
your letter operator kills the thing off in the appropriate circumstances.
So what we do is we look at the equation which says that a l well a l for the maximum value of l so we're not allowed to make.
The maximum value of this is N minus one. So we use a N minus one on E and minus one.
We get nothing. It's the end of the line. So we look at this equation in the position representation.
And so. We can forget about that a zero over two, because it's just a constant.
So what we want is we're going to be looking at IPR over each bar minus nine, putting L equal to and minus one.
So that L plus one, I suppose is going to be an end.
Overall. Plus he said, over.
And a zero that's operating on you and and minus one must give you nothing.
Now we we we know that PR is in the position representation.
We figured it out. It was minus h bar with deep idea plus one overall.
So we put that fact into this equation.
We have the I and the minus sign make a one the bars cancel.
So we're looking at D by the plus one overall minus this.
So we're going to have minus N minus one overall plus Z over and a zero.
You and n minus one is nothing.
Now, this is a premiums equation, right?
It's a first order linear differential equation.
So it has an integrating factor. E to the integral of with respect to all of the set of this stuff, right.
This just from prelims maths. This, of course on integrating just gives me an R on top.
This gives me a logo. And so we're looking at E to the minus some multiple of log R.
So that means that this is going to become R to the minus n minus one from the E to the minus n,
minus one log r times the exponential that we get from the E to the z r over and a zero.
So now we know pretty much what's going on because the original differential equation,
remember, says that D by the R the integrating factor times u and minus one equals nought.
In other words, this thing is equal to a constant. In other words, the wave function that we're seeking is is a constant over the integrating factor.
So we have the U and and minus one is some constant which must be determined by
the normalisation times R to the end minus one E to the minus Z and a zero.
So that's wonderfully simple expression. So let's let's ask ourselves a few things.
Let's have a look at the ground state of hydrogen.
So this is the case and equals one. How do we know that?
Because Elle has to be less than or equal to N minus one.
L Is the numbers the possible candidates for Ellen nought, one, two, blah, etc.
So when has to start at one and then go up to three, etc.
So the ground state, the state with the least energy is n equals one.
So what is the wave function? So you won.
And of course there should be a zero. Here is going to be a constant E to the minus in hydrogens.
That is one. So this is this is just going to be r over a zero.
So the ground state wave function is this mere exponential, a beautifully simple result.
What else was I going to say about this? Yeah.
One interesting given that that beautiful exponential one thing you notice is that this thing is never zero.
It it's the ground state wave function has nonzero modulus all the way to our equals infinity.
Although the particle is classically forbidden to go beyond a certain radius.
And in fact, so what this does was graph up here plots is the probability of finding the
reduced particle at a radius are measured in units of a zero of a over of a z.
So a radius bigger than this. And the classically forbidden region stops at that number two.
And it turns out there's a 20 there's a 24% probability that you'll find the reduced particle in the region that's classically forbidden,
where the kinetic energy as well would be negative. Right.
So if you go beyond R is to the potential energy is more than the total energy of the particle.
So there's less than nothing left the kinetic energy and there's a very significant probability of finding the particle that far out.
So so that's, I think, an entertaining result. It says that there's the p forbidden.
Looks forbidden. It's about 24%.
Let's get the normalisation of this thing sorted out so we can work out expectation.
A few expectation values working out in the normalisation is is fundamentally very straightforward.
What we require of course. So we require that the integral d cubed x over all space of the complete wave
function which consists of u and n minus one of radius y l m should equal one.
This thing we right of course, is the r r squared and then the integral sine theta dct defined to omega over the sphere.
Oops. Sorry should mod squirted this right.
That's our requirement when we integrate of angles over the sphere where we're integrating while m mod squared over the sphere and that comes to one.
So we're left staring at an equation which says that the r times r squared times times u this thing squared.
Now what did we say that was? That was c squared.
Well, we don't need to moduli make it a modulus. We can declare it to be real times r to the two and minus one e to the minus two.
Z over and a zero.
So this should be one. So the only thing to notice here is that there's an additional fact ask where you don't just take the the radial wave function,
square it and integrate the R to get one.
You have an additional factor, R-squared, because fundamentally this is the normalisation condition we wish to impose and the Y lens and normalised.
So integrating every sphere we get ones we've used here that the integral de to omega y l m squared is one that probably normalised.
Now this integral is is nice and easy.
So this can be written a C squared.
And what we need to do is to bring an integral like this under control is to declare that this is is ro.
So we introduce a new variable, ROE, which is to Z to Z of and a zero.
And I want to make these. So what I have here is an R to the to N, and I want to make all of those R's into rows,
which means I have to multiply by a load of factors and a zero of A to Z I've got or to the two N from these two and another one from there.
So it's two N plus one and then I can turn all these into rows d row, row two, the two in.
Well, that's the definition o missing the E to the minus row.
Now, this is a famous integral in mathematics.
Which is it? So it's often called gamma of two N plus one.
I believe Euler's responsible for that absurdity. But what it what it should be thought of as is two N factorial.
So this integral if if is simply the this exponent up here factorial.
Is the proof and, and you want to be able to recognise that because one often encounters that pattern and you want to just be able to say,
Aha, that's two factorial.
So this tells me what C is, C is equal to two Z over and a zero to the end plus a half, one over the square root of two and factorial,
which enables me to write down the the relevant way function u and minus one of radius.
I need this factor here and I'm going to write it as follows.
I'm going to say this is two Z over and a zero to the three halves power.
So I'm borrowing from that three halves power one over the square root of two and factorial.
That's too. I have to be clear.
So I need to bracket that to make sure it's the whole to end the gets factorial ized and then for the rest these other factors.
So. The rest of the factors in here can be put together with those odds to make this row to the end minus one each of the minus row.
Over to. That's the row is defined as to Z.
R. Etc. So. So. The factors.
Left over from here. Just what we need to bring make that which was an artsy and minus one into a rose at the end.
Minus one. So what does this.
Physically. What is this? We are looking at the states with the highest angular momentum for a given.
For given energy. So these are the quantum mechanical analogues of circular orbits, not eccentric orbits, but circular orbits.
So what do we expect qualitatively?
Well, we expect classically, if it was a circular orbit, our probability would be a delta function of the radius of a circular orbit.
And we know in quantum mechanics everything's a bit blurry because steep gradients of the wave function are associated with large kinetic energies.
So we're expecting it to be sort of like this ish.
So what's how is that how does that arise from this formula?
Well, when R is nought, row nought, this is going to be zero and then it's going to shift itself off zero.
Slower and slower, the bigger N is. So if end is ten to the 30 or whatever it would be for a classical classical particle,
then this would rise ever so slowly from zero and we'd have it would hug the origin for a long time.
It would then rise and then when it when it when Rho became on the order of one this
exponential which previously been harmless being ease the mind is something small,
would become a vicious cutting off thing. And that's how we get cut off on this side here.
So here we're looking at row two, the end minus one growth.
And here we're looking at E to the minus row over to.
Well, if this is the probability, then we need to multiply by we need to square up.
Right. And this is an exponential decline. So so precisely what it looks like is given in the next diagram.
So the top picture there shows just the first three.
So the pure exponential is end equals zero.
The ground state, then the one that rises steeply at the origin and falls off after an early peak is n equals two and n equals three.
Is, is, is the next one. And what you can see is that the characteristic radius is moving outwards.
Quickly. So let's calculate some let's calculate some expectation values, because that's now easy to do.
Let's work out the expectation value of the radius, right? So if we want to make a connection back to classical physics,
we should be thinking about expectation values because classical physics is the physics of expectation values.
So this is easy to work out. It's going to be the integral d r r squared times r times you and minus one.
Squared, right? That's what it should be. And now that we've got this normalisation, everything sorted, we can evaluate this.
So we're going to have. Yeah.
Well, actually. Let's just go back to which is the best way to do this.
All right. Let's turn this all into Rose. Let's turn these all into Rose now.
So this we've already got more or less as a function of Roe. So what we need to do is to.
Is to deal with these other ones. So there are four powers of all there are, and I need to turn those into rose,
which means I need I need an and a nought over two z race to the fourth power.
Then I need to write down this thing mod squared which is which is c squared which is two z over and a zero raise to the uh,
so that's two and +11 over two and factorial as the rest of that is the rest of C squared, then we need the integral D row row cubed.
So these three and then here we have row and I need to square that.
So it's going to be n to n minus two.
Easy. Linus Rowe. So what's this going to be?
This is going to be two and this will be row two, the two and plus one.
Each of the minus row. So this integral is going to be two and plus one factorial.
And on the bottom I've got two n factorial. So this on the top and that on the bottom gives me simply a2n plus one.
Everything else cancels in the editorial. And here something has gone wrong in that I've got far too many powers.
And what have I done wrong? What are we doing wrong? Uh.
Sorry. I got confused. Which one I was doing. Excuse me? I was using this formula here, which meant that the powers that I needed here.
Right. I was using this formula for you.
These require this. These powers. Now I'm using this formula.
So it's two the three halves power to the three because I've squared it up.
Exactly. So we end up with these three of these cancel.
So at the end of the day, I'm going to have an eight nought over two Z.
Just one of them. And we're going to have what we said was this was two and plus one.
In other words, we're going to have four. If I put that two inside there, we're going to have n n plus a half of a nought over Z.
So the expectation value of the radius is going sort of like N squared and it's going like the scale radius we defined for hydrogen divided by Z,
which tells you that if you increase the nuclear charge, the size of the orbit's going to shrink like one over the nuclear charge.
So the interesting fact here is that the expectation value of all is sort of like going like an
squared and that's exactly what we expect because e remember goes like minus one over and squared.
So therefore it's going like minus one over expectation value of R.
But where? But we have a particle moving in a Coulomb field, so the potential energy goes like one overall.
And from the visual theorem, we're expecting the potential energy to be minus twice the kinetic energy.
So the total energy should be sort of a half, minus a half of the potential energy.
So this is exactly what we're expecting. So that's that's a recovery of sort of classical ish stuff.
Interesting fact here is because this grows like this, it means the volume occupied by the atom is going like.
Which obviously goes like the expectation value of all cubed, which goes like end of the six power is growing very rapidly within.
So states so so this so this grows very rapidly.
This means that states in which you excite the electron to a large value of n cannot be seen.
You won't be able to observe these, to measure these unless you're in an incredibly high vacuum.
So. So. E.g. if ns a hundred the volume is going to be ten to the 12 times a regular atomic volume.
And in interstellar space you can see hydrogen atoms transitioning from any is 100 to
an equals 99 and stuff like that rate by making measurements of radio wavelengths,
centimetre wavelengths. But you can't do that kind of thing laboratory because you can't get high enough vacuum.
So in the brochure on earth we are restricted to relatively small values of n n less than ten typically.
Right. What else can we say? What about? What about? It's interesting to work at the expectation value of R squared.
It's essentially identical performance to what we've just done.
I mean, all we have is an extra R and that integral at the top there.
Right. So what are we going to have if we come down here, we will have an N,
a nought over two Z raised to the fifth power this time because we're going to have an extra power of oh before the use starts.
Then we will have two Z over and a nought to the third power coming from the you.
Then we will have to n plus one sorry to n factorial coming from the c and then
we will have to do the integral d row and we'll have an extra power of row.
So it'll be row two the fourth times that row to the to n minus two.
So we'll end up with row two, the two N plus two, each of the minus row.
In other words, this is going to be two and plus two factorial.
Right? So now we're taking two N plus two factorial, two N plus one factorial dividing by two N factorial.
So this is and we're going to get an extra power here.
So this is going to be n squared coming from here because because this fifth power we reduce to the second power by when we multiply this one.
So we left with n squared a zero over two Z squared and then we will have two N plus two, two N plus one.
It's interesting to express that as a multiple of as a multiple of of the expectation value of R,
which we've already derived as being an N plus a half a0 of a Z.
So A0 of Z is essentially so expectation value of R.
So this is going to be this these twos I can take out, there are two twos in here.
I take them out and use them to clean that up. So this is going to be n squared, n plus one n plus a half of a zero over z squared,
which itself is the expectation value of R squared over N squared and plus a half squared.
So we can cancel many things and we find that that's n plus one over n plus a half of the expectation value of R squared.
So what does that mean? That means that the uncertainty. Well, so, so what's the what's the arms variation.
And Oh, now you think this thing would go to zero, right?
Because what we're doing is we're looking at the quantum mechanical analogue of a circular orbit.
Circular orbit. The particle does not move in and out. So we would expect that this arms variation in all went to zero,
as in went to two large values and we would have thought we would recover classical physics.
We'll see that that's not the case because what is this are mass variation where it's all squared expectation value minus R expectation squared.
Take the square root of that. Okay, so here is the expectation value hit.
Let me write it down again. R squared expectation. So what I want to do is from this I want to take R squared and then take the square root.
So this is equal to the square root of n plus one over n plus a half minus one expectation value of R.
So you can easily see that this is going to come to something like the square root of a half over n plus a half of the expectation value of all.
So what's happening is that the arms variation in the radius is becoming small
with respect to the radius relative to the expectation value of the radius.
But. But jolly slowly. Right. That's on the order of expectation.
Value of are divided by root root n.
So it's becoming small relative to the radius itself, but only slowly.
But it's absolutely large, right? This because this thing is growing like an squared.
This is looking like ends of the three halves power. And I think you can just about see that in those pictures up there that as you as you well,
I've only shown the first three, but you can't see the peak becoming narrow.
It doesn't become narrow. So that's a remarkable result. Okay.
So, so those those are the wave functions for the essentially circular orbits.
What about the non-sequitur orbits? As we see, they're not very circular, but, you know, that's the best we can do.
So how do we expect to get these way functions for non circular orbits?
Well, in the case of the simple harmonic oscillator, we found the ground state wave function by solving a on ground state wave function equals zero.
And that's essentially what we've just done. And then we found the excited state wave functions by,
by taking that wave function we first found and multiplying it by a dagger an appropriate number of times.
And every time we multiply by a dagger, we got a more complicated wave function, right?
So the ground state wave function, the harmonic oscillator was a Gaussian, the ground state wave function here was a well.
It's a slightly more complicated problem, this more complicated problem because we have all these different values of the angular momentum.
So here the starting point is R to the L is R to the N minus one times an exponential as in the
same sense that our starting point in the case of a harmonic oscillator was just a Gaussian.
But that's but that's the strategy. And what we would hope is that a l dagger does the business right.
A l increased our angular momentum at fixed energy and drove us up against the equation that we solved to find the circular orbit
wave function and l dagger we would hope would move us from the circular wave function back down to more eccentric orbits.
But this has to be done in a slightly but in a slightly subtle in a slightly subtle matter manner.
Okay. So let's look at this for at a formula that we have here somewhere.
Let's look at this formula here. Ale, comma, ale.
Dagger is equal to the difference of h's. So let me write that down with l reduce by 1al minus one comma l minus one dagger.
This is just a relabelling operation, right? Is equal to can I remember which way up it is?
No h0 squared of amu is zero squared mu over h bar squared of h l minus h l minus one.
Now. Let me take the dagger. Sorry.
Let me commute this entire equation. Both sides of it, with respect to a L minus one dagger.
You'll see why we're doing this when we. When we've done it.
So we're going to say that this is a L minus one comma, a L minus one dagger comma, a L minus one dagger.
So that's the left side of the equation, commuted with a L minus one dagger.
And that's going to be boring, constant.
Times h l comma a l minus one dagger, which is what I want, minus h l minus one comma a l minus one.
Oops. Minus one dagger. Why am I doing this?
I'm doing this because I want to calculate this, which we haven't so far calculated.
We could calculate by going back to first principles and stuff, but working out these commentators is quite wearisome.
So it's best this is this is a reasonably slick route.
But what I want to do is calculate the commutation of this with with h sub l and
what I know at the moment is only the commentator of this with h l minus one K.
So I'm going to rearrange this equation now because this is my target as h l comma, a l minus one dagger.
That's what I want to find the value of, because it'll turn out to be the key.
Is equal to. Each password over a zero squared.
Mu open a big bracket. Then let's, let's write this out.
Turning this into its products. I'm going to expand this in a commentator.
As a general rule, I hate to expand commentators, but you'll see in a moment that it's an expedient thing to do.
So this is going to be expanded and it's going to be a L minus one times, a L minus one dagger.
Still, the outer commentator in place so computed with a L minus one dagger.
Close that bracket minus is that room?
Yes, minus the commentator a l minus one dagger a L minus one comma a L minus one close brackets dagger.
Sorry, that one's dagger writes this one here. So all I've done is taken the contents.
I've taken this in a commentator and expanded this into its two bits and then I should bring this onto this side of the equation.
Step one step to replace this h with the corresponding expression in terms of a dagger, a dagger, a right.
So a dagger it says up there is, is, is a multiple of l plus a constant.
I'm interested in h l inside a comitato so I don't need to care about that constant
because it will commute with everybody and produce the vanishing comitato. So for my purposes I need to replace h l by a l dagger a l times.
That constant. That constant is already present and correct.
So this term here becomes minus a l.
Gosh, which way round is it? It's dagger. On the left is a L minus one dagger a L minus one.
So that's this to get together with this. That's this insider commentator.
Comma. Now this a dagger.
L minus one close commentator. Close break bracket.
Now, we should find that two of these terms. Cancel.
Ow, ow, ow, ow. They don't know this.
And this is a minor catastrophe. Oh, because I brought it on to the other side of the question.
Thank you. Yep. People. So. So I brought this across here.
No, but. I thought I though, she says, then arrives.
It arrives with a plus sign. It rises. A plus sign, of course, is brilliant.
So we can kill to these two, fight each other, and leave a blank piece of board.
So. So what do we need to do now?
What we need to do is concretely evaluate this. COMMENTATOR Right.
Because so. So these two kill each other. And we're left looking at this.
This is the usual computation. Rubbish. So what we have is.
So I'll write it down explicitly because it is a bit of a mess. A zero squared mu, a l minus one comma, a l minus one dagger.
Commentator. Right. He works on this one while he stands idly by.
And the other term vanishes because it's because everybody commits with themselves.
So we don't need the big bracket because that's the end of the discussion.
And all we have to do now is plug in what this is, because it turned out to be the difference of the two.
Right. It's the difference of the two. Hamiltonians. So this is is this is h bar squared over a nought squared mu of h l minus h l minus one.
I have to do a little translation because his l minus one times a l minus one.
Okay. So now with that expression, we can, we can go to business because we can go to I'm sorry, should have gone away from the content out the front.
Should have gone away. Exactly. It was what I needed to.
Yes. There was that constant the other way up there. Thank you very much.
Get rid of that. So we have an unbelievably simple result after slightly scary computation.
So what do we do with that?
What we do with that, of course, is we go and say that a of a l minus one dagger e l is equal to a l minus one dagger h l e l right.
So h l l is e e l multiply both sides of the equation by this baby, and we have what we've got on the board.
Now, of course, we're going to swap these two over. So we right this is h l a l minus one dagger plus.
So we're not allowed to write that down. So we put in what should sort it out, which is a commentator, a l minus one dagger comma h l close brackets.
Brackets, brackets, e l.
This is what we've just laboriously worked out and found is is h l minus h l minus one.
Right. So this is going to give me an h l minus h l.
Minus one. Time's a crucially times h l minus one subscript times a l minus one dagger.
That's that's. That's right. So that this expression goes in where that commentator is, we can see.
Oh. The commentators the other way around.
Yes, the commentators are up there is h a not we want h so this becomes a minus.
Exactly.
So the undesired terms which are this on this cancels with this on this and we are left with minus minus is plus h l minus 1al minus one dagger e.
L. Which is the equation that we require, because it says that this object is an eigen function of this operator with the same old icon value.
Right. So this is establishes that e l minus one is actually equal to a is some multiple of to be discussed a l minus one dagger.
E. L. And that enables this by mere differentiation,
by just using more and more of these to work away from the circular orbit wave
function down to the wave function associated with no angular momentum at all.
Let's just begin to see how that how that pans out. So let's ask ourselves about.
So you kn and minus two you should get by using this a l dagger stuff on on you kn minus one which we already know what it is.
So that's proportional to I'm not going to chase down these proportionality constants now.
Any time it's proportional to.
Okay. Now we have to think what to put in here. This is the angular momentum with which you arrive and I'm interested in arriving with and minus two.
So this is going to be in minus two dagger operating on on the wave function associated with with where we were before,
which was U and N minus one of radius.
Let's write that out in the position where it's more or less in the position representation.
Let's be more concrete about it. So let's find our expression for a l we we're not interested in the constants in
front because we're just we're going to normalise this when we're all sorted.
So I want the I want the emission adjoin to if I want the dagger of that thing at the top there.
So this is equal to minus I pr on h bar.
Minus. Now, what's l? We put al equal to and minus two.
No, no, no, no. Uh, yes, I, uh.
This is. This is l it's been put equal to n minus two, and I'm supposed to have a L plus one there, so I'm supposed to have one more than this.
So that's n minus one overall plus z over.
Is it n minus one. Yep.
A zero. Close the bracket.
What's that working on? It's working on this which is ah to the end, minus one e to the minus, uh, z ease of the minus Z.
Over and. A zero.
If I did that right. And so now we have to we have to gain to replace PR with minus body or plus one overall.
So we we're going to have too many minus signs.
Now we'll get a minus D by the R minus one overall coming from here minus.
And minus one overall plus z over N minus one a zero plus brackets O to the n minus one E to the minus set of and a zero.
So we could write this, taking out an overall minus sign to make life a bit less negative.
This is an overall minus Z over N minus one, a zero or N minus one.
Okay. So what's going to happen?
One thing is I want to say what's going to happen? This is this is a this is a particular value.
This is this is a. This is a specimen of the L dagger sort of thing.
Right. If we change and L, we'll be changing this number here.
But the main idea is what's crucial is that this is going to contain every one of these is going to contain a derivative operator.
When the derivative operator meets this, it will produce a term which goes like R to the end minus two.
It will produce a term which has less R dependence. Also, there is this something over our term does the same thing.
So we get an amount of R to the end minus two.
But we also get from this or from differentiating the exponential, we get an amount of out of the end plus one.
And minus one. Sorry. So in other words and of course, the exponential will live on in the way that exponentials do and differentiated.
So what does the wave function look like? You and.
And minus two is now a linear function of radius.
It's going to be A plus B oh.
And actually that will be negative. Let me write it. I mean, concrete and it's going to be negative.
Uh, times are to the end, minus two.
So I've taken this out e to the minus said R over in a zero.
So since this is a linear function of R and moreover B and a a positive, we get one node, right?
So that is to say, you have some particular radius vanishes when we go to get you and minus two and sorry,
minus three and we'll use another of these differential operators and we will find that this is some it'll be sort of a primed minus,
B prime, double plus C prime to R squared.
It will be a quadratic expression E to the minus stuff.
And we will have two nodes because it'll turn out this quadratic has real roots and so on and so forth.
Right? So every time we, we use an a dagger, we get one more note.
Why? Why is that?
Physically, what we're doing is we're taking kinetic energy out of the out of the tangential motion, remember, and stuffing it into the radial motion.
And kinetic energy motion in quantum mechanics is associated with oscillating wave function.
So we're getting more and more wavelengths radially and therefore more and more nodes radially.
And here we are. And here is a picture of what a few of these things look like.
So, oops. These are these are the wave function, the radial wave functions.
Well, this is a picture. Sorry, this is a picture of the meridional plain. So this is radius here and this is z direction here.
Okay, so we're discussing the wave function in question is fitting a volume.
And it begins. Right. So so that's kind of how you should think about it.
This is the circular orbit case. Blackness means high probability of finding the electron or the reduced particle.
Whiteness means not much. So this is the circular orbit.
It's zero on the axis. Because, you know, we saw it with zero on the axis before.
It's reasonable. It's got a lot of tangential motion.
It can't get to the axis.
But then otherwise the the amplitude of finding it rises quite quickly, peaks and then very slowly falls away as you go off to infinity.
But it's rather boring, right? It's sort of everywhere. If we are using an A dagger and a N minus two dagger on this, we get this wave function.
This is for the case. N equals three. By the way, we get this situation where we have one, we have one radius.
This is the node. So it rises at the origin, it peaks sooner, so it reaches a high peak at a smaller radius than this,
which is associated on a plunging a more plunging orbit. Right.
It has a pair of centre, but then quantum mechanically it has this node where you won't find it.
You have zero probability of finding it around that circle and then you have a property of finding it further out.
Use another of these, a dagger things and you'll come over here where you have two nodes here you see them.
And now in this case, because N is three, you've run out of energy.
In this case, there's no orbital angular momentum. There's no tangential motion.
This is what a plunging orbit looks like in quantum mechanics.
Right? That particle in classical physics, we just diving at the nucleus,
slipping around it and coming back out again and in an arbitrary, elongated ellipse.
So it doesn't look at all like that. Well, that is clearly the moment to stop with that review of the hideaway functions.