 So, where we got to yesterday was that we had reduced the horrible partial differential equation in six variables, which is posed by the time-independent Schrodinger equation for a hydrogen-like ion to this equation here. Sorry, not being premature to this. So, the internal motion had been separated from the translational motion of the atom as a whole, and we had exploited it by using a change of variables for central mass coordinates and the separation variable, and we had used the work that we did with the angular momentum operator earlier on to show that we could express the internal Hamiltonian in this format here. So, this was called R, I think, that's right, for the internal motion. Right, and then I pointed out that this Hamiltonian commuted with the total, with the angular momentum operators, because the only mention of the angles, theta and phi, sits inside this total angular momentum operator, and therefore we can seek stationary states. There's going to be a complete set of stationary states, which are simultaneous eigenfunctions of E and L. Right, so we will also have the statement that L squared on E and L is equal to L, L plus 1, E and L, and then when we're using these stationary states, the action of this operating will be replaced by this eigenvalue, and now we do end up with what I wrote originally, which is that we're going to have a Hamiltonian, which has a subscript L, meaning it's the one that's valid for stationary states, which have total angular momentum quantum number L, which is going to be PR squared over 2 mu plus L, L plus 1 h bar squared over 2 mu R squared minus z E squared over 4 pi epsilon naught R. And this is fundamentally a Hamiltonian analogous, this is a Hamiltonian analogous to that for the harmonic oscillator in the sense that we have now only one surviving coordinate, of our six coordinates, we're down to one, that's this here, the modulus of the separation vector, and it's conjugate momentum PR. So we've made an enormous simplification, made tremendous progress, and we're going to knock this into submission using the same approach as we did with the harmonic oscillator, we're going to define what will turn out to be a ladder operator AL, which is going to be defined to be dimensionless, it's going to be A0 over root 2, I better write this down from notes rather than my memory, because these factors do matter, I PR over h bar minus L plus 1 over R plus z over L plus 1, A0, where A0 is the following not very helpful item, 4 pi epsilon naught h bar squared over mu E squared. So this is an object I'll make it convincing, well let's make it convincing now. What is this object? This is the so-called Bohr radius, so it was introduced by Niels Bohr before there was proper quantum mechanics using what turns out to be a fallacious model of hydrogen, the Bohr atom, and the way to see what these dimensions are are actually to to write to bring this up onto this side and say look this is going to be E squared over 4 pi epsilon naught A naught, now on this side A naught is on the top, not the bottom, so therefore I have to say this is equal to h bar over A naught squared over mu, so out of this equation by putting that E on the top and this 4 pi epsilon naught on the bottom, dividing through by A naught squared to get this on the bottom, so this is something we understand to be, this is obviously electrostatic energy, and what's this that we see on the right side? Well, h bar k is momentum, so this is more or less p squared over mass, this is the reduced mass, recall this is the reduced mass, more or less the mass of an electron. So what we have on the right hand side is p squared, well it is p squared for k is equal to 1 over A naught, so if you have a wave which has a wavelength which is comparable to 1 over the this scale radius here, sorry a wavelength comparable to this scale radius here, not worrying about 2 pi's at the moment, then what you have on the right here is, so this side is 2 times kinetic energy associated with the particle which has a wavelength which is on the order of A zero, and what we have on the left side here is the electric energy, so the scale that's being set, this dimensional quantity, the Bohr radius, is the natural scale at which the kinetic energy associated with the uncertainty principle, the zero point in it, what with the harmonic oscillator we would have called the zero point motion, is on the order of the electrostatic energy, that's dimensionally where this number comes from, that number comes from. So with the thing to notice now is that Al is dimensionless, why is it dimensionless? Well, this cancels the dimensions of this, obviously this cancels this, this is all dimensionless, this factor apart from the A zero, and here we're looking at, again this is PR divided by, with this put on the bottom, H bar over A zero, therefore H bar K, therefore something with the dimensions of P. So this thing here, this operator here is dimensionless, same as the ladder operators in the harmonic oscillator were dimensionless. Okay, so what do we do with that? What we do is we calculate what A, and this thing of course carries the subscript little l because little l, the orbital quantum number is appearing in its definition, and what we do is we now calculate Al dagger Al. Right, so what does this give us? We have obviously an A zero squared over two out front because we're going to have two A zeroes and root twos, and then we're going to have PR, we showed was a Hermitian, well we engineered that it was a Hermitian operator, so when I take the Hermitian adjoint of what I have up there, I get a minus I PR, the minus sign is from the I over H bar, and then the rest of course is the same because it's Hermitian, well it's just boring numbers, plus Z, well actually this is an operator strictly speaking, but we're in the position representation, so it looks like a number, and that has to be multiplied onto I PR over H bar, so no minus sign because this is Al I'm writing down now, minus l plus one over R plus Z over l plus one A zero, so we have to multiply this stuff out, and the way we do it is we regard this in the back here as one factor, and that in the front is another factor, so this is looking like the product of a number minus some number, a number plus some number, right, the usual pattern, just so this is mirroring very closely what we did with the harmonic oscillator, so we get A zero squared over two, these two obviously multiply together, and we get PR squared over H bar squared, and then these I have to multiply together, so what do we get? We get the square basically of this number here, so we have plus l plus one over R squared plus Z over l plus one A zero squared minus the cross term here which is going to be, well there'll be two minus twice, two Z, those are going to cancel, and we will have A zero R, so that's that's the the two easy parts, right, because it's the front thing squared plus the back thing squared, and now we have to think about the cross terms which would vanish because this is A plus B into A minus B spiritually, these cross terms would vanish if we weren't dealing with operators, and they failed to vanish only because we are dealing with operators, so that there's a failure of commutation, in other words the PR, well okay, so PR commutes with this, so for the cross terms we do, we don't get anything from this thing on this, but we do get something from this thing on this, namely we get the relevant commutator, so the extra term that arises because we're working with operators is going to be I over H bar, that's that, so this minus sign and this minus sign cancel, we have that l plus one and then I have a PR comma one over R, plays big bracket, that's what I'm going to get from this term on this term not canceling on this term, on this term, so that's going, so what is this going to come to? Well let's rearrange things, A0 squared over 2 PR squared over H bar squared plus l plus one squared over R squared minus, I'm going to put this one down next, 2z over A0R, then this term plus z squared over l plus one squared A0 squared, and when I have to do this, what I remember is something we handled already last term, that when I had to do, it arose when I was doing, when we were doing the commutator of P and V, the potential function of X, that turned out to be the rate of dv by dx times the commutator of P and X, this is exactly the situation we have here because this is the operator canonically conjugate to R, so what we have here is the derivative of one over R, which is so minus i l plus one, l plus one over H bar times minus one over R squared because that's the derivative of one over R, times PR comma R, but this is a piece of a canonical commutation relation, this is equal to minus i H bar, right? We show that R comma PR is equal to i H bar, so in this order it's minus i H bar, so we have rather a load of minus signs, let me see, I think we have one, two, three minus signs and another minus sign coming from here, so I think in total we have a plus sign, so this stuff here I believe comes to l plus one over H bar R squared, sorry the H bar cancels, this H bar is on top, that's on the bottom, so we get an l plus one over R squared and this can now be combined with this, except I've got the wrong blinking sign, let me just double check that, yep I'm looking for minus, is it minus? Well anyway I want it to be minus so let's declare it to be minus and I'm sure it is minus, I'm sure it is minus, let's not spend time chasing down some rigid sign because now what we're going to do is combine this, so let's just a little side calculation here, we're going to have l plus one over R squared brackets l plus one from up here, that's the l plus one squared minus one times this stuff, so you can see we're going to put this and this is going to give me an l l plus one, which is exactly what I want, so this is going to be a naught squared over two P R squared over H bar squared plus l l plus one over R squared minus two z over a zero R closed brackets and then I'm going to take this and join it on that so we get plus garbage and the garbage term is z squared over two l plus one squared, this should all be dimensionless, I think it probably is dimensionless on the grounds that it's the product of two dimensionless operators, now what's the point of this ridiculous exercise, the point is that we should see the original Hamiltonians peeking out here, we should basically have our original Hamiltonian plus garbage, so in order to get our original Hamiltonian we need to have a mu under here and a mu under here so why don't we multiply by a mu on the top and the bottom, so this is a naught squared mu and we want to take this H bar outside and then that won't be under there as we want, we can allow that two to wander inside and then this bracket becomes HL, this stuff here becomes HL and we still got unwanted garbage in the back but that's exactly how it worked with the harmonic oscillator, remember a dagger a was equal to H, the Hamiltonian over H bar omega minus a half, there was garbage in the back which in that case was a half, so this is obviously some constant with the dimensions of energy, I better make sure that I've done that right, it's the H bar should be squared, yes of course it should because it was the H bar I took out from there otherwise it is correct and that's just the business, so we've expressed H, let's write this down in the other way, we've said that H is equal to H bar squared over a naught squared mu which must provide the dimensions of energy, a dagger L al minus z squared over 2 L plus 1 squared, another way of writing it, we've expressed H basically as dimensionless constant times as of a dagger a, so this is the harmonic oscillator trick and it's all just looking a little bit messier but this is only a boring number, what's the difference between a half and this, it's just a number, they're both just numbers, rational numbers, okay so what do we do next, what do we do in the harmonic oscillator, we calculated a dagger comma a, we found the commutator, that was the next thing that we did and that's what we do just right now, make sure I'm doing it the right way, no it's more convenient to work it out the other way around, later use, all right so what is that, we have to write this horrible thing down again, so we're going to have an a naught squared over 2, sorry open a square bracket because we're talking commutators now, write down I PR on H bar because I'm writing down a now which is just there, I need to write down minus L plus 1 over R, I don't need to write down the boring constant in the back, the z over L plus 1 a0, that will commute with everything in sight and that will make no contribution to the commutator, then I have to write down the corresponding parts of a dagger which is minus I PR over H bar minus L plus 1 over R and now I can rest easy. So this is what the commutator that I have to do, obviously so PR commutes with itself, nothing doing there, PR on the other hand does not commute with this, so we are going to get a0 squared over 2, I L plus 1, that's that L plus 1 over H bar, that's that H bar PR comma 1 over R commutator, I get that from that, or at least live in hope that I do, what did I do with the minus sign, I put it in the bin, I shouldn't have done, there should be a leading minus sign, then I have the same term actually because here I'm going to get plus, but everything plus this except I'm going to have a 1 over R comma PR, which of course is minus this, so I'm going to get this all over again, so why don't you just rub out the 2 and then it's right. What's this commutator, we've already discussed that problem, it's going to be the rate of change is going to be d by dr of this times the commutator, so this is equal to minus a0 squared I L plus 1 over H bar, and then I'm going to have a minus 1 over R squared for the derivative of this times PR comma R commutator, because that's how these things work, but this once again is minus I H bar, so the 2 minus is here cancel, the 2 I's make another minus sign which kills this minus sign, all being well this is equal to a0 squared L plus 1, the H bars cancel over R squared, check sign, yeah, should have a, no, that's correct, that's correct, good, all right, so we want to express this, remember the commutator doing the harmonic oscillator case, what was this commutator, this commutator was actually a 1, so the bad news is, it's not looking very promising, at this point you think oh no, it's not good, because our commutator is some damn function of R, but we've seen that damn function of R somewhere before up in the Hamiltonian basically, I've lost the Hamiltonian, there it is, so we've got L L plus 1 H bar squared over 2 mu R squared is appearing in the Hamiltonian, so supposing I would take H L plus 1 and from it I would take H L, then everything in the Hamiltonians would cancel except that middle term which has the right form, namely it contains a 1 over R squared and what would we have, we'd have H bar squared over 2 mu I think, R squared brackets L plus 1, L plus 2 minus L L plus 1, did I do that right, living home, okay, so obviously there's going to be a factor of L plus 1 common and then we're looking at the difference between L plus 2 and L in other words 2, the 2 is going to cancel this and this is going to equal L plus 1 H bar squared over mu R squared, so we can express with this little side calculation we can go back up the board and express this as an appropriate multiple of the differences in the Hamiltonians, there's going to be a naught squared, that's this a naught squared, then we will want to multiply by mu divided by H bar squared and then we'll be able to say these H L plus 1 minus H L, check that we haven't got any horrible factors in this, all right, what was the next thing we did in the harmonic oscillator, having got the commutator of the A and the A dagger and having expressed H as a product of A and A dagger, the next thing we did was calculate the commutator, use these results to calculate the commutator of A with H, so that's what we do now, I will do it here so that we can see our results, so I want to calculate the commutator A L comma H L and I can do that by expressing this H L, you see H L is basically a product of the A's, now I've only got to locate the retrofit product, it's at the top there isn't it, it's so hard from this position to see what you need to see, this product, et cetera, et cetera, there's a statement, this is a statement I'm looking for, I want that statement, all right, so this H L can be traded in for that product, so we have H bar squared over A zero squared mu times the commutator of A L and A L, A L dagger A L and I can rest easy there, I no need to put this stuff inside a commutator because it commutes with everything inside so it can't contribute to the commutator, so this is the commutator I have to evaluate and this is easy because this is a commutator of product, some A with B and C which in principle is the commutator of A with B C standing idly by and then the commutator of A with C B standing idly by but A of course commutes with itself so there's only one term which is an A L, A L commutator with A L dagger, so this is equal to H bar squared over A zero squared mu A L comma A L dagger and we just work that one out and the answer was, and the answer was, it was here, sorry I did something wrong, yes it's very important, I need an extra factor A L, thank you very much in the back, this is standing idly by while A L works on his companion, all right, so I now need to plug this in for this commutator and you can see that all these factors are going to cancel, this factor is going to cancel on this factor and so we're simply going to get H L plus 1 minus H L times the A L I've been very helpfully told to include so let me just, right, so we're now in wonderful shape, so we've completed all three steps of the simple amount of oscillator calculation and now we just need to go for the point of the exercise which is, so we're given that we always were given that H L on E and L is equal to E of E and L, right, what we want to do is make ourselves a new, so we've got one stationary state, we want to make ourselves a new stationary state by multiplying by A L, obviously both sides of the equation, so let me write down the right side of the equation first, this implies that E, which is a boring number, times A L, E and L is equal to A L, H L of E and L, usual business, swap these over, H L, A L which I'm not entitled to plus the commutator that restores order on E and L, we just worked that out and found that it was the difference of two H's times A L, so this becomes H L plus 1 minus H L with an A L in back, guess what? We have an H L, A L here with a minus sign and H L, A L here with a plus sign, so the whole thing is equal to H L plus 1, A L, E and L, so we have achieved what we wanted to achieve, that is to say we have shown that this state is an eigenstate, not of H L, but of H L plus 1 and for the same energy, so the map's looking a little different now from harmonic oscillator, but nonetheless we have a very powerful result, we have generated ourselves from a state which had energy E in angular momentum quantum number L, we've made ourselves a state which can only be characterized as E L plus 1, we've made ourselves a state of the same energy but more angular momentum, so what have we done? This operator, this is some normalizing constant, right? We had just this kind of thing in the case of the harmonic oscillator, so physically what have we done? Well, classically what have we done? We've taken an orbit that might look like this and we've turned it into an orbit, I'm trying to make an orbit that has about the same semi-major axis and is rounder, you know about Kepler orbits, so with the same supply of energy, we've increased the angular momentum so we've made the orbit less eccentric. So in the several-line harmonic oscillator case, what did we do? We made ourselves an orbit with less energy and then we argued that the energy we were able to show that the energy could never be negative, so we said to ourselves so there must be some, so given this state now, we could apply a l plus one, a sub l plus one to this and make ourselves e comma l plus two, state with even more angular momentum, so like in the harmonic oscillator case, we said we could make ourselves an orbit with even less energy. Is this possible with a given supply of energy with a bound orbit to have more and more and more angular momentum? No, it's not. At some point, you've got the maximum angular momentum you can have for that given energy, which in classical physics is what we would call a circular orbit. We've completely destroyed the radial motion. See, what we've been doing here is we've been shifting kinetic energy. We've shifted ke from pr squared over two mu to l squared h bar squared over two mu r squared. This was the tangential kinetic energy. This was the radial kinetic energy. We've shifted energy from here to here. When we've got no energy left in there or as little as the theory as quantum mechanics allows, which won't be zero, but will be some amount, then we won't be able to shift any more. So there must be a maximum angular momentum for a given energy. We'll call this curly l. This is the maximum angular momentum and it is a function of energy, but we won't write that it's a function of energy. But we're gonna find out what function of energy it is. So what does that mean? That means if we take the, if we take the, what are we gonna call this, the circularization operator, AL, belonging to this maximum angular momentum and we use it on the state which has the maximum angular momentum for the given energy, what do we get? Nothing. That's the only way we can be prevented from getting a state which has even more angular momentum for the same energy is if this operator simply kills this state. So we've used this argument twice before, once in the harmonic oscillator case and also in the case of the angular momentum operators. What do we mean by nothing? What we mean is the mod square of this is nothing. What does that map to? That maps to e maximum angular momentum, AL dagger AL, e curly thingy is not. Where have we seen a dagger a before? I think we must have seen it in the Hamiltonian. We need to replace that by the Hamiltonian times some horrible factors. Yeah. All right, so we, well, we already have it here. So a dagger a comes right down to this line here. So this line here can replace the a dagger a in here. So we get to have that e curly L onto a naught squared mu over h bar squared h curly L plus z squared over two curly L plus one squared closer bracket e curly L, ain't much. But this thing, this is an eigenfunction of this operator with eigenvalue e. This is a boring number. So it stands by whilst this bangs into that and makes a one. This gives me e times this and this is left over and it bangs into this and makes a one. So this implies that a zero squared mu over h bar squared e plus z squared over two L plus one squared is nothing or perhaps I should write this as equals minus. So what have we done? We've got a relationship between the energy and the maximum allowed angular momentum. More than that, we know that these angular momentum quantum numbers, because this is orbital angular momentum we're talking about, we prove that those had to be integers. So n being defined to be is an integer, integer. What integer? We know that curly L is allowed to be nothing one, two, three, four. So n is equal to the numbers it's allowed to be is one, two, three, four, blah, blah. Nothing not included in the list because of that plus one. So we have shown that e, the energy, has to be of the form minus z squared h bar squared over a naught mu, we've done that right, a naught squared mu one over, no we need a two here, one over n squared. So we have found the possible energies of a hydrogen atom. Well, in fact, for a hydrogen like ion, because z remembers this integer, which controls the number of charge units on the nucleus, we have found this with the possible values, right? It's given by this constant, which we know what it is. We'll simplify it in a moment, we know what it is times one over n squared, where n is one, two, three, four. So this gives the energy levels. We write this as minus z squared times r, over n squared, where curly r is whatever you see it to be, which is h bar squared over two a naught squared mu, which is not very intuitive. The way to make this intuitive is to take those a zeros, there are two of them and turn one of them back into its h bars and things. Now, where did we define a zero for heaven's sake? It was right over here somewhere, right? There it is. So one of those two I'm going to replace by that garbage there, all right? So this is going to become, on the bottom we're gonna have an eight pi epsilon naught a zero, all right? That's the four pi epsilon naught. The h bar squared will cancel top and bottom, so that goes away. The mu and the e squared, well the mu will go away with this and the e squared will sit on the top. So the Rydberg is what? E squared over four pi epsilon naught a zero would be the potential energy of two charges of charge, you know, two electric charges that were separated by a zero. So this is half of the potential energy at a separation of a zero. And so this is the fundamental energy scale of atoms. And what is it equal to? 13.6 electron volts plus, you know, 13.6 to three significant figures. So the energy range at which we work, you know, the battery that you stick into your, you stick into your camera or something has 1.5 volts basically because of that 13.6 EVs. It's all of, all of condensed matter physics is a mere reflection of the energy. Condensed matter physics is a mere reflection of that number. Rather, you know, this is why we live at one EV not at one MEV or one millievery or whatever. So what do we need to do next? Yeah, jargon. This is called the principle, principle, pal, AL, quantum number. So in these hydrogen like ions we've discovered that there are a whole series of different, of distinct states which have the same energy and different angular momenta. So let's talk a bit about degeneracy. Okay, so for principle quantum number n equals one, we have that L, which is equal to n minus one is equal to zero. And almost the largest angular momentum you can have is nothing. And in the ground state of hydrogen, there's one electron, it sits in the state with the lowest energy which is going to be associated with n equals one and it has no angular momentum. It's on a totally radial orbit in classical physics, right? Not going round and round at all. It just goes in and out, in and out. I mean in classical physics, quantum mechanics. But it doesn't have any angular momentum. So that's a surprising result. For n equals two, L is equal, the maximum angular momentum is equal to one. That means that L can be naught if you like and L can be one, right? This is the maximum angular momentum. So there's a slightly funny thing going on here. n was introduced as one plus the maximum angular momentum. But now I'm saying it's better to what we one standardly thinks about it. One thinks about what's the value of n from it? One takes as n minus one the maximum angular momentum. So, sorry. And in this sense, we have one state. It'll be two states, well, there's one. Here, basically we have, this is for a spinless particle, right? It'll turn out to be two states when we include the spin of the electron. But remember, we were doing the growth structure which means we said we were gonna forget about the spin of the electron. Here, we would have one state and here we would have three states, right? Because for L equals one, we got total angular momentum one which means we got three possible orientations of it. m can be one, nothing or minus one. So we have three quantum states here, one here. So we got four states all with the same energy for n equals two, one for n equals one and so it goes down the line. So the number of states is increasing rapidly because there'll be five states for n equals three. The maximum angular momentum will be two. For two units of angular momentum, you've got five possible orientations and then you've still got three of these and one of those, so that's nine states, et cetera. So the structure we've derived is extremely degenerate. What does this have to say about experiments? So stick some hydrogen atoms in a vessel and pass an electric current through and get the electrons knocked out of their comfortable, out of their comfort zone and you will get photons coming out at discrete frequencies. Nu is going to be the difference in the energies over H which is going to be z squared, the Rydberg constant over H, over H null H bar, one over n prime squared minus one over n squared. This is for n goes to n prime. So if you were in one of these higher states, for example, n equals two, you will have less, your energy will be a smaller negative number, right? You'll have one over two squared, you'll have a quarter here and this could then fall down to the state n prime equals one, in which case this will be one, so this bracket will be, say, three quarters and you will get three quarters of this number coming out. So that gives you some frequency and what we have is series of lines. So the way we think about this is that we have a series of lines each for fixed n primed, i.e. bottom level. So if we fix n primed at one, we can have transitions from n is two to one or n is three to one or n is four to one and these are the successive lines of the Lyman series. So here we have, so here is the energy of n equals one, here's n equals two, here's n equals three and Lyman alpha is the name used for the spectral line associated with an electron tumbling from n equals two down to n equals one and Lyman beta is associated with from n equals three down to n equals one, which is further to four so it emits more energy so the line appears at higher frequency, longer, shorter wavelength. So the Lyman series is for n primed equals one, if n equals two we're looking at Lyman alpha, that's what it's conventionally called, if n equals three it's Lyman beta and this has I think it's 112 nanometers, is that right? 121, sorry. And as you go down to n equals infinity, in other words, if you fall all away from not being bound into the bottom of the atom, then this is the beginning of the Lyman continuum and that's what is it, 92 nanometers roughly speaking, I've got a more accurate number here, 91.2. So these lines are all in the, these are all vacuum ultraviolet lines, they all carry, so this one is carrying 13.6 EV of energy and these are carrying, this is carrying three quarters of 13.6 EV of energy, so they're carrying enough energy to kick electrons out of the air molecules so these photons are heavily, are absorbed by all kinds of things, they're very, these are very easily absorbed photons because they carry enough energy to lift electrons out of most atoms. And then we have the next is the Balmer series, which is where the whole story started, which is so n primed is two, n equals three, if you go from three to two, that's called Balmer alpha, but it's written as H alpha because that stands for hydrogen alpha, so Balmer was a Swiss school teacher who empirically fitted the formula and we've derived, I've lost it, there it is, he fitted that formula empirically to measured frequencies of lines that he identified as being the Balmer series lines, well, A series of lines, he called the hydrogen series, so this is called H alpha and it's a pink photon, it's 600 and something nanometers, 656. So it's pink light. So many astronomical objects are pink because they are shining in, sorry, in H alpha, in Balmer alpha, this is H beta, et cetera. If you, then you go to passion, that's for n primed is three and obviously n can be four or five, et cetera, et cetera, et cetera. So these start off as pink and they get bluer, so as you go down this list, the wavelengths get shorter as you go to infinity where the series limit, I did write it down here, 364.6 nanometers. So they go from pink light right through the optical, the rest of the optical spectrum to the ultraviolet region and the passion series starts at 1875, I think. So n equals four, you are looking at 18, yep. So that's already, these are sort of more or less optical and by now we're in the near infrared, et cetera, et cetera, et cetera. So that's pretty much the right place to stop, I think. What we should do, just one other thing I would point out is that, so you can apply these formally to the inner electrons of, to the innermost electrons of other atoms, atoms that have more than one electron, you can't apply them to the outer electrons with any useful way because we've done all this right, remember with no other electrons present, we got one nucleus and one electron, but there's one very important thing to take home, which is that this energy scale goes like Z squared. So the energies, the characteristic energies of the innermost electrons are going up like Z squared by the time you get to uranium, which has 92 units of charge, so Z is 92, you're almost a factor of 10 to the four, you're almost a factor of 10,000 higher in energy, which means that these electrons are moving essentially relativistically. So that's just the thing to bear in mind. Okay, and we'll look at the wave functions that go with this lot tomorrow.