 A dyna'r ffodol cyfnod o'r ffawr mewn gwahodau gwahodau ymlaen i ni yw. Yn ymlaen, roeddwn ni'n ymlaen o'r problemau hydrogens ar y 1-dymian i amlwntau perthynu 1h subl, 1 ar y cwm yng Ngwyloedd fath o'r FFM. ac we found that this thing at the top here was a ladder operator in the sense that it 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 and more angular momentum but the same energy had to stop we concluded that the energy was given by a certain constant 13.6EV divided by n squared where n is one more than the maximum angular momentum that you can afford at that energy put another way the angular momentum quantum number is less than or equal to one minus this number n that controls the energy and is called the principal quantum number. So what we want to do now is move forward to get the energy eigenfunctions so to get the these are of interest from the perspective of hydrogen and if you want to do any detailed calculations like how does hydrogen interact with electromagnetic field what happens if you scatter electrons of hydrogen that kind of stuff you'll need to know what these eigenfunctions these wavefunctions 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 a 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 play a very big role in atomic physics. Okay so you want to find what they are. Now from this from the fact that L squared on E and L is equal to L L plus one EL from the fact that this thing this state here of well-defined energy stationary state is an eigenfunction of the total angular momentum operator. We know all about the so the wavefunction in question is this right the aptitude 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 angular 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 R times Y L M of theta and phi because we know that these the spherical harmonics are the unique angular there the eigenfunctions of this operator with this eigenvalue. So this thing which we know is such an eigenfunction its angular dependence must be given by strictly speaking a sum potentially it's a sum of these for 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 of M and L. 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 radial dependence. As I say you have to expect to depend on the energy I mean it has to depend on what's in here which is the energy and the angular momentum 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 way functions for the harmonic oscillator station restates by studying the equation that the equation that this operator your ladder operator kills the thing off in the appropriate circumstances. So what we do is we look at the equation which says that Al well Al for the maximum value of L so we're not allowed to make the maximum value of this is n minus one. So if we use a n minus one on E n 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 root two because it's just a constant. So what we want is we're going to have to be looking at IPR over H bar minus nine putting L equal to n minus one. So that L plus one I suppose is going to be an n over r plus z over n a zero that operating on you and n 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 IH bar d by dr plus one over r. So we put that fact into this equation we have the I and the minus I make a one the H bars cancel. So we're looking at d by dr plus one over r minus this so we're going to have minus n minus one over r plus z over n a zero u n n minus one is nothing. Now this is a prelims equation right it's a first order linear differential equation so it has an integrating factor e to the integral of with respect to r of z of this stuff right. Just from prelims maths this of course on integrating just gives me an r on top this gives me a log r 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 there e to the z r. Over n a zero so now we know pretty much what's going on because the original differential equation remember says that d by dr the integrating factor times u n n minus one equals not in other words this thing is equal to a constant. In other words the way function that we're seeking is is a constant over the integrating factor so we have that u n n minus one is some constant which must be determined by the normalization times r to the n minus one e to the minus z r of n 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 n equals one how do we know that because L has to be less than or equal to n minus one L is the numbers the possible candidates for L a nought one to blar et cetera. So n has to start at one and then go up to three et cetera. So the ground state the state with the least energy is n equals one so what's the wave function so u one and of course there should be a zero here is going to be a constant e to the minus in hydrogen z is one so this is going to 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. Okay given that that beautiful exponential one thing you notice is that this thing is never zero it's the ground state wave function has non zero modulus all the way to r equals infinity. Although the particle is classically forbidden to go beyond a certain radius and in fact so what this graph up here plots is the probability of finding the reduced particle at a radius r measured in units of a zero over z there. So a radius bigger than this and the classically forbidden region stops at that number two and it turns out there's a 24% probability that you'll find the reduced particle in the region that's classically forbidden. Where the kinetic energy as it were would be negative right so if you go beyond ours to the potential energy is more than the total energy of the particle so there's less than nothing left for kinetic energy. And there's a very significant probability of finding the particle that far out. So so that's I think in an entertaining result. It says that there's a P forbidden. What's forbidden is about 24%. Let's get the normalisation of this thing sorted out so we can work out expectation a few expectation values. Working out the normalisation is fundamentally very straightforward. What we require of course so we require that the integral d cubed x overall 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 write of course as dr r squared and then the integral sin theta d theta d phi d two omega over the sphere. Oops sorry I should have mod square did this right. That's our requirement. When we integrate of angles over the sphere we're integrating y l m mod squared over the sphere and that comes to one. So we're left staring at an equation which says that dr 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 make it a modulus we can declare it to be real. Times r to the two n minus one e to the minus two z r over n a zero. So this should be one. So the only thing to notice here is that there's an additional factor r squared. You don't just take the radial wave function square it and integrate dr to get one. You have an additional factor r squared because fundamentally this is the normalisation condition we wish to impose. And the y l m's are normalised so integrating over a sphere we get one. So we've used here that the integral d two omega y l m mod squared is one. They're properly normalised. Now this integral is nice and easy. So this can be written as c squared. And what we need to do is to bring an integral like this under control is to declare that this is rho. So we introduce a new variable rho which is two z r of n a zero. And I want to make these. So what I have here is an r to the two n and I want to make all of those r's into rows which means I have to multiply by a load of factors n a zero over two z. I've got r 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, the rho, rho to the two n where that's the definition. Oh I'm missing the e to the minus rho. Now this is a famous integral in mathematics which is so it's often called gamma of two n plus one. I believe Euler is responsible for that absurdity but what it should be thought of as is two n factorial. So this integral is simply this exponent up here factorial. Easy to prove. 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 n factorial. So this tells me what c is. C is equal to two z over n a zero to the n plus a half one over the square root of two n factorial which enables me to write down the relevant wave function u n 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 n a zero to the three halves power. So I'm borrowing from that three halves power one over the square root of two n factorial. That's two. I have to be careful. Sorry I need a bracket there to make sure it's the whole two n that gets factorialised. And then for the rest these are the factors. So the rest of the factors in here can be put together with those rs to make this row to the n minus one e to the minus row over two. That's so row is defined as two z r etc. So the factors left over from here are just what we need to make that which was an r to the n minus one into a row to the n minus one. So what does this, physically what is this? We are looking at the states with the highest angular momentum for a given energy. So these are the quantum mechanical analogs 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 at the radius of a circular orbit. And we know in quantum mechanics everything is a bit blurry because steep gradients of the wave function are associated with large kinetic energies. So we are expecting it to be sort of like this ish. So how does that arise from this formula? Well when r is 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 n is 10 to the 30 or whatever it would be for a classical particle then this would rise ever so slowly from zero and it would hug the origin for a long time. It would then rise and then when rho became on the order of one this exponential which previously had been harmless being e to the minus something small would become a vicious cutting off thing and that's how we get cut off on this side here. So here we are looking at rho to the n minus one growth and here we are looking at e to the minus rho over two well. If this is the probability then we need to multiply by we need to square up right. This is an exponential decline 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 n 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 the next one and what you can see is that the characteristic radius is moving outwards quickly. So 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 you want to make a connection back to classical physics we should be thinking about expectation values because the classical physics is the physics of expectation values. So this is easy to work out it's going to be the integral dr r squared times r times u n n 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 rows let's turn these all into rows now so this we've already got more or less as a function of row so what we need to do is to deal with these other ones. So there are four powers of r there and I need to turn those into rows which means I need an n a naught over two z raised 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 n a zero raised to the so that's two n plus one one over two n factorial that's the rest of that's the rest of c squared. Then we need the integral d row row cubed that's these three and then here we have row n I need to square that so it's going to be n two n minus two ease of minus row. So what's this going to be this is going to be two n this will be row to the two n plus one ease of the minus row. So this integral is going to be two n 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 a two n plus one. Everything else cancels in the factorial and here something has gone wrong in that I've got far too many powers of n what have I done wrong what have I done wrong 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. I was using this formula for you these required this these powers I'm now using this formula so it's to 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 n a naught over two z just one of them. And we're going to have what we said was this was two n plus one. In other words, we're going to have if I put that two inside there we're going to have n n plus a half of a naught 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 orbits going to shrink like one over the nuclear charge. So the interesting fact here is that the expectation value of R is sort of like going like n squared and that's exactly what we expect because E remember goes like minus one over n squared. So therefore it's going like minus one over expectation value of R but we're but we have a particle moving in a Coulomb field so the potential energy goes like one over R. And from the virial 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 R cubed which goes like n to the six power is growing very rapidly within. 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 e.g. if n is 100 the volume is going to be 10 to the 12 times a regular atomic volume and in interstellar space you can see hydrogen atoms transitioning from n is 100 to n equals 99 and stuff like that. By making measurements at radio wavelengths, centimetre wavelengths, but you can't do that kind of thing in a laboratory because you can't get a high enough vacuum. So in the laboratory on earth we're restricted to relatively small values of n. N less than 10 typically. Right. What else can we say? 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 in 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 naught over 2z raised to the fifth power this time because we're going to have an extra power of R before the u starts. Then we will have 2z over n a naught to the third power coming from the u. Then we will have our 2n factorial coming from the c. Then we will have to do the integral d rho and we'll have an extra power of rho so it will be rho to the fourth times that rho to the 2n minus 2. So we'll end up with rho to the 2n plus 2 e to the minus rho. In other words this is going to be 2n plus 2 factorial. So now we're taking 2n plus 2 factorial not 2n plus 1 factorial dividing by 2n factorial. We're going to get an extra power here so this is going to be n squared coming from here because this fifth power will be reduced to the second power when we multiply this on. So we're left with n squared a0 over 2z squared and then we will have 2n plus 2 2n plus 1. It's interesting to express that as a multiple of the expectation value of R which we've already derived as being nn plus a half a0 over z. So a0 over z is essentially expectation value of R so this is going to be 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 1 n plus a half of a0 over z squared which itself is the expectation value of R. R squared over n squared n plus a half squared. So we can cancel many things and we find that that's n plus 1 over n plus a half of the expectation value of R squared. So what does that mean? That means that the uncertainty, well so what's the RMS variation R? Now you'd think this thing would go to zero right because what we're doing is looking at the quantum mechanical analog of a circular orbit. Circular orbit the particle does not move in and out so we would expect that this RMS variation in R went to zero as n went to 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 RMS variation? Well it's R squared expectation value minus R expectation squared. Take the square root of that. So here is the expectation value. Let me write it in again. R squared expectation. So all 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 1 over n plus a half minus 1 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 R. So what's happening is that the RMS variation in the radius is becoming small with respect to the radius relative to the expectation value of the radius. But jolly slowly right that's on the order of expectation value of R divided by 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 n squared. This is looking like n to 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 circular orbits? As we see they're not very circular but that's the best we can do. So how how do we expect to get these wave 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 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 this is a slightly more complicated problem. This is a more complicated problem because we have all these different values of the angular momentum. So here the our 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 this 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 a 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 form at a formula that we have here somewhere. Let's look at this formula here a l comma a l dagger is equal to the difference of h is. So let me write that down with l reduced by one a l minus one comma a l minus one dagger. This is just a relabeling operation right is equal to. Can I remember which way up it is? No a zero squared of a mu a 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've 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 commutators is quite worrisome. So this is a reasonably slick route. But what I want to do is calculate the commutator of this with h sub l and what I know at the moment is only the commutator of this with h l minus one. OK, 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 will turn out to be the key is equal to h bar squared over a zero squared mu open a big bracket. Then let's let's write this out turning this into its product. So I'm going to expand this in a commutator as a general rule. I hate to expand commutators, but you'll see in a moment that that'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 commutator in place. So commuted with a l minus one dagger close that bracket minus is that room? Yes, minus the commutator a l minus one dagger a l minus one comma a l minus one close brackets dagger. Sorry, that one's dagger right this one here. So all I've done is taken the contents I've taken this inner commutator and expanded it into its two bits. And then I should bring this onto this side of the equation, step one, step two, replace this h with the corresponding expression in terms of a dagger a dagger a. So a dagger a says up there is is a is a multiple of a gel plus a constant. I'm interested in a gel inside a commutator so I don't need to care about that constant because it will commute with everybody and produce a vanishing commutator. So for my purposes I nearly can merely replace a gel 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 around is it? It's dagger on the left. Yes, a l minus one dagger a l minus one. So that's this to get together with this that's this inside a commutator comma. Now this a dagger l minus one close commutator close break bracket. Now we should find that two of these terms cancel. Oh, oh, oh, oh. They don't. And this is a minor catastrophe. Oh, because I brought it onto the other side of the equation. Thank you. The people fight so so I brought this across here. No, but I thought I had a lot. It arrives with a plus sign. It arrives with a plus sign. Of course it arrives with a plus sign. Brilliant. So we can kill so 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 commutator, right? Because so these two kill each other and we're left looking at this. This is the usual commutator of a product rubbish. So what we have is so I'll write it down explicitly because it is a bit of a mess. A0 squared mu a l minus one comma a l minus one dagger commutator, right? He works on this one while he stands idly by. And then other term vanishes because it's because everybody commutes 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 this is h bar squared over a naught squared mu of h l minus h l minus one. I have to do a little translation because this is l minus one times a l minus one dagger. So now with that expression we can we can go to business because we can sorry should have gone away. The constant 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 e 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 write this as 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 commutator 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 times a crucially times h l minus one subscript times a l minus one dagger. That's that's that. Right. So that this expression goes in where that commutator is. We can see the commentator is the other way around. Yes. The commentator up there is h a not we want a 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 one a l 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 eigen value. Right. So this this 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 us 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 u n n minus two. You should get by using this a l dagger stuff on on u n n 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. Ain't time. It's proportional to a. Now we have to think what to put in here. This is the angular momentum with which we arrive and I'm interested in arriving with n minus two. So this is going to be n minus two dagger operating on on the wave function associated with with where we were before, which was u n n minus one 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 were not interested in the constants in front because we're just we're going to normalize this when we're all sorted. So I want the I want the Hermitian adjoint of 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 L equal to n minus two. No, no, no, no. Yes, I this is this is L. It's been put equal to n minus two and I'm supposed to have L plus one there. So I'm supposed to have one more than this. So that's n minus one over r plus z over is it n minus one? Yep, a zero. Close a bracket. What's that working on? It's working on this, which is r to the n minus one e to the minus z e to the minus z r. Over n a zero. Hope I did that right. So now we have to we have to gain to replace PR with minus I H bar d by d r plus one over r. So we're going to have too many minus signs now. We'll get a minus d by d r minus one over r coming from here. This n minus one over r plus z over n minus one a zero plays brackets r to the n minus one e to the minus z r of n a zero. So we could write this taking out an overall minus sign to make life a bit less negative. This is n over r minus z over n minus one a zero r n minus one. Okay, so what's going to happen? The important thing is to understand what's going to happen. This is a this is a this is a particular value. This is this is a this is a specimen of the al dagger sort of thing, right? If we change nl will be changing this number here. But the main idea is what's crucial is that this is going to contain every one of these. It can contain a derivative operator. When the derivative operator meets this, it will produce a term which goes like r to the n 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 n minus two. But we also get from this or from differentiating the exponential. We get an amount of r to the n plus one in minus one. So in other words, and of course the exponential will live on in the way that exponentials do when differentiated. So what does the wave function look like? U n n minus two is now a linear function of radius. It's going to be a plus b r and actually that will be negative. Let me write it. I mean concretely it's going to be negative times r to the n minus two. So I've taken this out e to the minus z r over n a zero. So since this is a linear function of r and moreover b and a are positive, we get one node. So that is to say u of some particular radius vanishes. When we go to get u n minus two n sorry minus three n, we'll use another of these differential operators. And we will find that this is some it will be sort of a primed minus b primed r plus c primed r squared. It will be a quadratic expression e to the minus stuff. And we will have two nodes because it will turn out this quadratic has real roots and so on and so forth. So every time we use an a dagger we get one more node. 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 functions. So we're getting more and more wavelength radially and therefore more and more nodes radially. And here is a picture of what a few of these things look like. So these are the wave function, the radial wave functions. This is a picture of the meridional plane. So this is radius here and this is z direction here. So we're discussing the wave function in question is filling a volume. And it can... right. 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 we saw it was 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 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... using an a dagger, an a... n minus 2 dagger on this, we get this wave function. This is for the case n equals 3, by the way. We get this situation where 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 with the fact that it's on a plunging... a more plunging orbit, right? It has a pericenter. 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 probability 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 3, 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 will be just diving at the nucleus, slipping round it and coming back out again 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 hydrogen wave functions.