 We were at work on the harmonic oscillator that is the single most important dynamical system in physics. A model for a huge number of physical applications. We had this Hamiltonian which is basically P squared plus X squared. The crucial thing is that it's obviously quadratic in P. But especially it's also quadratic in X. That's its peculiar characteristic of its potential energy. We went the plan for what we're trying to do is find the stationary states, in other words, we're trying to find the states here, which are eigenfunctions of the Hamiltonian because they will enable us to follow the dynamics of the system, their crucial tools. We went after them by defining this thing A, the annihilation operator, which is a dimensionless operator made up of X and P, and we found that it, we evaluated, well, we found two, this thing had two properties. First, A dagger A is very nearly the Hamiltonian, A dagger A plus a half h bar omega is the Hamiltonian. First, important property. Second important property that its commutator with its Hermitian adjoint is minus one when done this way round. So using those properties, we said, suppose we have a state, a stationary state of energy E, if we multiply this equation through, this defining equation of that state through by A dagger and do some swapping of the order of operators using the results above, we were able to show, this is where we finished, that H on the state that you get by using A dagger on E is equal to E plus a h bar omega plus h bar omega times that self same state. In other words, this state here, this state here is essentially a state, well it is a state, a stationary state of energy increased by h bar omega. And because we could repeat this, it follows that the energy is, if there is an energy E, then there must be energies E plus h bar omega E plus two h bar omega, and far as we can see E plus any amount, any number of h bar omegers, any energy number of h bar omegers. And what remains at this point is to find out what the number E is that we first thought of. And we do that by applying not a dagger to that equation, but A to that equation. So then we have, if we take, if we start with now E E is equal to h E and we multiply both sides of the equation by A, not a dagger this time, then we swap the order here just as we did before into H, HA, and then we have to add in the commutator A comma H, all operating on E still. We yesterday, this is absolutely repeated what we did yesterday, this is HA. Now we replace that H as advertised up there by A dagger A, et cetera. We observe, so this should become the, this is going to be replaced by A dagger A plus a half H bar omeger. We can forget about the half because we're inside a commutator and a half commutes with everything. We can take the H bar omeger outside the commutator. And then what we're looking at is plus A comma A dagger A H bar omeger close brackets. E, we use our usual rules for taking the commutator of product, which is to say it should be the commutator of this with this, with that standing idly by, and then in principle the commutator of this with this with this standing idly by, but the second commutator vanishes. And this commutator by the result up there is equal to plus one, okay, because the turned around there A dagger A is minus one. So this commutator A comma A dagger must be plus one. So this is saying this becomes HA plus A H bar omeger E. So we have an AE with no operator in front on this side of the equation, but we have the same thing on that side of the equation. So we take these together and onto this side of the equation and we then have that E minus this H bar omeger goes to the other side and becomes a minus H bar omeger is equal to HAE, which establishes on the face of it what we might have hoped, which is that the states you get that you have after you used A on this stationary state E is another stationary state with an energy lower than E by H bar omeger. So A dagger raises your energy, it increases the excitation of the harmonic oscillator and A evidently lowers the energy of the harmonic oscillator. Now you have to this point start to worry because by the previous rhetoric it would follow that if there's E then there is also E minus H bar omeger and so on down through E minus N H bar omeger and it looks as if you can find states of lower and lower energy without limit. And you should be worried about that because thinking about the Hamiltonian up there because classically it looks manifestly positive because p squared should be positive and x squared should be positive. You can't trust such things in quantum mechanics, but what we can do is we can work out, we can say look E is the expectation value of the Hamiltonian in the state E obviously because H on E is equal to EE comes outside and this on this is one. So that's a self-evident equation. Replace what's inside here, this becomes over two M of E p p E plus M squared omega squared E x x E and because these are observables and therefore Hermitian I can put a dagger there if I want to. Doesn't make any difference because P dagger is P and I can observe that this is now manifestly P E mod squared. This thing is the mod square of that of the catch you get by using P on E and this is M squared omega squared of x E mod squared. Just deliver two M which is boring and this is manifestly greater than or equal to zero. Even in quantum mechanics there can't be any argybargy this has to be greater than or equal to zero. So all the energies, all the allowed energies, the entire spectrum of the Hamiltonian has to be positive. But we've apparently established that by using A we can get states which by successively using A and A and A and A we can get states of lower and lower energy because we take H bar omega off every time we use E. Sorry we use A. So there's a problem and the problem is the assumption. So let's just look carefully at what we've established here. What we've established is this equation here. This equation is absolutely copper bottoms. It's beyond criticism. It's true. It was obtained by totally legitimate operations and it establishes that this thing is an eigenfunction with this lower energy provided this thing is non-zero. But if at any stage in this chain of applying A's this thing here would vanish. So if when you get to a certain energy, lowest energy and you apply A, A simply kills it, it produces a ket of no length at all then we won't have a state of even lower energy. And since it's clear that this chain of operations, it's not, it's logically impossible to go on creating states of lower and lower and lower energy. This chain of results of applying extra factors of A has to stop somewhere and the only way it can stop since this equation is true is by this thing becoming zero. Then because H on zero is equal to any number you like on zero, so if this is zero this equation does not establish that this is an eigenvalue of H, right? But that's the only circumstance in which this equation would not establish that this was an eigenvalue of H. So for the lowest energy, the ground state energy, it has to be that A kills it. So that's the ground state. We must have that A on E zero equals naught, right? So I'm using this symbol now to indicate the lowest energy. So what does this equation mean to give more precise meaning to it? So what we say is that this ket that you get here has no length squared. So let's just evaluate the mod length squared. So we're saying that A E zero mod squared which is equal to E A dagger A E is nothing. But this thing, we have it somewhere up there, it's just in range. H is equal to A dagger A plus a half H bar. So this thing is, this is the expectation value of H over H bar omega minus a half. So these ones have zeros on it because we're talking about the ground state energy, not any old energy now. We're just talking about that one special lowest energy, the ground state energy that this equation is valid for. And so what we're discovering is that E zero over H bar omega is equal to a half. In other words, the ground state energy E zero is a half H bar omega. And now we know what the general energy is because we know that we can make states of higher energy by applying A dagger to the ground state moving up by H bar omega each time. The nth energy must be n plus a half H bar omega. So we have found the allowed energies and the next item on the agenda is to find the corresponding, find the wave functions of the corresponding eigenstates, stationary states. Let's move over here. So let's ask about wave functions of the stationary states. Now we've found the allowed energies. Actually, before I do that, it's probably good to generalise this calculation here. So what we've established is, so let's just talk about normalisation as a sort of preliminary. I think it's better to do it just now, normalisation. What we've established is that A dagger, and I'm going to introduce a new notation. We're going to say what previously I'd called E is going to go to n. So this is the state with E equal n plus a half H bar omega. We could write E n in here, but everybody writes just n. It just saves energy. It works well. Now we've discovered what the energies are and that they're labelled by an integer n 01234. So this is n equals 012 and so on. It makes sense to have a nice compact notation and to call our stationary states the state 0, the ground state the state 1, the first excited state the state 2, the second excited state. What we've established is that A on n is some constant, I'll call it k, times n plus 1. Because we've discovered that when we used A on the state E, we got a state which was an eigen ket of H with eigen value E plus H bar omega, which would be in the new notation the state n plus 1. If this E was n plus a half H bar omega, this will be n plus 3 halves H bar omega. And the issue arises, so what value does this thing have? So just to be clear, we've had relationships like this. We had a corresponding one up there. Yep, there we go. Yesterday we derived this equation and that equation establishes that this thing is an eigen ket of H. It doesn't establish that it's a properly normalized eigen ket of H. I want it to be properly normalized. And so I'm asking about what's the normalization constant I have to use after I have applied an A dagger. It's easily found because we just take the mod square of this side of the equation and the mod square of that side of the equation. So taking the mod square, so taking mod square of both sides, we get n A A dagger n is equal to k mod squared times 1 because that's by definition going to be correctly normalized. We can take k to be a real number, we can take k to be a complex number if we determine to but why don't we just agree to take it to be a real number and this just becomes k squared. We're just trying to get the thing normalized. We don't care about the phase. Let's ask ourselves what this is. If I swap this over, I want to swap this over because then I can relate it to the Hamiltonian. Then I have to add on A, A dagger. Oops, commutator. So that's that rewritten. This is the Hamiltonian minus, it's the Hamiltonian over H bar omega. Where is it? Yeah, it's up there. It's the Hamiltonian minus, it's the Hamiltonian over H bar omega minus a half. So this is H over H bar omega minus a half. That's this thing. And what's this? This is plus one I think because the commutator the other way around was minus one. So this is plus one. So that's that. H on n is equal to n plus a half H bar omega by definition. So this on this produces n plus a half H bar omega divided by the H bar omega and we have n plus a half. Take away a half, we have n, add one, we have n plus one. So this is in fact equal to n plus one. So what follows this? So comparing this with this, we find that k squared is equal to n plus one. Going back to up there, we have that n plus one is equal to one over the square root. Sorry, sorry. That's right, square root of n plus one of a dagger n. This is a very important equation. If we would go through this rigmarole, the same logic using a on n being some other constant times n minus one, we would find that n minus one is equal to one over the square root of n of a operating on n. We know that a operating on n is going to produce n minus one. It depletes the energy by H bar omega and therefore reduces n by one. The normalisation constant turns out to be one over the square root of n. I would recommend you check that after the lecture. It's a precise repeat of this logic here. It's worth remembering these rules because this is such an important relationship. They're very easy rules to remember. The normalisation constant is one over the square root of the biggest number that appears in the equation. So in this equation, the biggest number appearing is n plus one, that's what you use. These are two very important equations which physicists remember. What we have to do now, we're trying to find the wave functions fundamentally. This is just the tedious details of the normalisation. Although those equations are of bigger use than just with wave functions. Let's find the ground state wave function. What is it? We'll call it u0 of x and it is of course x0. It's the function that's defined by this, the amplitude to be at x, if you were in the ground state. Also called this. These are just two notations for the same thing. This satisfies a natty equation because what we know is that a, operating on n minus one, is equal to nought. The ground state is defined to be, it comes into the world as the state, the one and only state, that the operator, the destruction operator a kills stone dead. So if we multiply this equation on the left by x, we still have a valid equation. And this, and let's write in what a is. So a is m omega x plus ip. In principle, it's over the square root of 2 m h bar omega, but because I'm about to put this stuff equal to 2 m h bar omega, I'm about to put this stuff equal to nought. I can neglect the factor on the bottom. I can multiply through both sides of the equation, the factor on the bottom, and get rid of the garbage, right? Clean things up. Why don't we write this? I mean, we're more or less committed to writing this in the position represent. Well, I mean, we have it there, right? This tells me that m omega x, m omega x, x nought, because this is the position operator. This is an observable. I can imagine that it operates backwards on this. This is an eigenfunction of that operator. So those are both operators. That's not an operator. This operates on this, if I want, backwards. This is its eigenfunction. We get an x, the number, times this. This then meets that and produces this, which is our wave function. And then we also have plus i x p nought, and the momentum operator was defined by saying that this thing is minus i d by dx of x. This was the definition for any wave function of how p operated. With an h bar. Thank you. Thank you, absolutely. So I need to write this stuff down. I need to write some more stuff down. This is equal to zero, provided I put in the m omega x, x. If I rewrite that in wave function notation, it becomes m omega x u nought of x is equal to, and let's clean this stuff up. Sorry, it's not equal to plus h bar d by dx of u nought of x equals nought. So what is this? This is a first order linear differential equation, and it's in fact an ordinary differential equation because there are no other variables than x present. I've written this as a partial derivative. Consistency with what we do in more in three dimensional cases, I suppose. But it is a first order linear differential equation. There is no sort of differential equation that's more user friendly that we encounter. We solve such equations by using an integrating factor. Just to get this into standard form, I would actually write this as d u nought by dx plus m omega over h bar x u nought. This is the standard form for a first order linear equation, which you should remember from Professor Ross's course or whatever in the first year. It has an integrating factor, which is e to the integral of the coefficient of the linear of the constant term, the not derived term, this term, e to the integral m omega over h bar x dx, which is clearly e to the m omega x squared over 2 h bar. That's its integrating factor, and the equation is then, if you multiply by the integrating factor, the equation says that d by dx of the integrating factor times u nought is equal to the right side, which happens to be zero ergo. This quantity here is a constant ergo u nought is equal to e to the minus m omega x squared over 2 h bar, which I want to write as e to the minus x squared over 4 l squared. Why do I want to do that? I want to do that because the probability associated with x, the probability of finding your particle at x, which is equal to u nought mod squared, will then be e to the minus x squared over 2 l squared. This is a normal distribution with dispersion l. The reason I want to get this into this form is in order, because I can identify that as the Gaussian width of the distribution, the width of the Gaussian distribution. Also, once I say that this is the probability distribution from my knowledge of statistics and stuff, I know what the normalising constant has to be. I know that it's 2 pi l squared, root 2 pi l squared, because that's the factor you need to normalise a Gaussian distribution. Oops, I'm missing a minus sign, crucial. So what does l have to be in order that this form is the same as that form? I think a simple algebra to show that l has to be h bar over 2 m omega, but l squared has to be that. Let's check that this has the right dimensions. One should always be checking one's dimensions in physics. This has the dimensions of momentum times distance, so this has dimensions of m, x, p. This is m, x, v. This is what I should have said, momentum times m, v, x. What am I talking about? Yes, m, v, x, right. m, v, x is the dimensional structure here over the dimensions here. Indeed, this thing has dimensions of length squared, so l is indeed length, so it's okay. What have we learnt? We've learnt that the ground state of a harmonic oscillator has a wave function, which is a Gaussian, with a characteristic width given by this. Crucially, we've already studied distributions of particles which are Gaussian, have wave functions which square up to Gaussians, and we now know what the probability distribution is for momentum. From what we did before, we will have that the probability of momentum will also be a Gaussian. It will be e to the minus p squared over 2, should we call it sigma p squared, over some square root 2 pi sigma p squared. Remember, we had the uncertainty principle, which said that the dispersion in x times the dispersion in momentum is equal to h bar over 2 for the Gaussians. This was a result we established when talking about the free particle. That tells me that sigma p is equal to h bar over 2l. There's some characteristic momentum that the particle has. In the ground state, a particle is not stationary. It is moving with a characteristic amount of energy. It's also not at the bottom of the potential well. It has a characteristic amount of average potential energy. We've come to the conclusion. We have an example here of one of the most important, perhaps the most important prediction of quantum mechanics, which is the existence of zero point energy. The basic issue is that if we're trying to minimise the energy of the particle, which clearly the ground state by definition does minimise the energy of the particle, is the state with the lowest energy by definition. If we're trying to minimise this energy, we want to get the potential energy to be as small as possible. That clearly means moving towards the origin and getting as close to the origin as you can. But because of the uncertainty relationship, because the more narrowly confined you are in real space, the more uncertain your momentum has to be. If you restrict yourself too much in position to be too much at the bottom of the potential well, you will have a large uncertainty in your momentum and you'll have kinetic energy. In practice in the ground state, there's a compromise between being reasonably close to the origin and having a reasonably small kinetic energy. Because of the uncertainty relation, quantum mechanical systems in their lowest states have a finite extent spatially, even though really it's a point particle, there's a finite extent in which you'll find the particle, and a finite kinetic energy. I hope you see that this is a specific example of a totally general phenomenon. We have to expect to occur always when we are considering particles trapped in some kind of potential wells. This is enormously important because it's exactly this physics which determines the size of atoms. Atoms here are typically pretty close to there, near enough in their ground states, and the size of these atoms is determined by the electron. If the atom gets any smaller, and indeed you take a piece of steel or something and you stuff it into a press and squeeze the thing down, it will get smaller, but it'll resist violently, you'll have to do work, you'll have to increase its energy to make it smaller, and what happens is that in order to make it smaller in real space you have to give it more kinetic energy and the uncertainty principle, and that's the work that you do squashing it down, and you can see that idea worked out quantitatively in one of the later chapters of the book. The size of atoms is determined by this zero-point energy business and the uncertainty principle, and interestingly the mass of protons and neutrons is not entirely but is overwhelmingly accounted for by the kinetic energy of the quarks and gluons inside there, which are moving around relativistically, they're in a very deep potential well, and because they're very narrowly confined into 10 to the minus 15 metres, in order to be confined, even though they're fairly massive particles into this very small space, they have to have a lot of kinetic energy, and the mass associated with that kinetic energy accounts for most of the mass of us, that's what it mostly is. So this zero-point energy phenomenon is extremely general and enormously important, and here we have the simplest, the classical example. In fact, you see if we say H is equal to 1 over 2M of P squared plus M squared, omega squared, x squared, and we put in the uncertainty relation, we say that x squared, if we say that P squared is equal to H bar squared over 2x, let's get this right, sorry, over 4, because I've squared it, over 4x squared. So for the ground state, x squared is essentially the uncertainty in x, so it's associated with the uncertainty of momentum in the same way. If you stuff that into this, you put this relationship in, you find that H is 1 over 2M is going to be whatever it is, H bar squared over 4x squared plus M squared, omega squared, x squared. This now is a function, maybe I shouldn't call it x squared, maybe we should call it L actually, perhaps that would be more helpful. So I'm saying that x squared is on the order of L squared and P squared is on the order of this. So here we have a function of L and a minimum, if you ask yourself, what's the, what value of L does this function have a minimum? The answer is it's that value that we gave up there from quantum mechanics. So it's really true that L is being chosen to minimise the energy given the constraints imposed by the uncertainty principle. Okay, so that we found the ground state wave function, it would now be useful, I guess, to show how we should calculate the, sorry, my notes are a bit out of order here. Yep. So what we want to do, let's get the first excited state as an example. So first excited state, excited state wave function. So u1 of x, which is by definition x1. What do we know? We know that 1 is equal to 1 over the square root of 1 times a dagger working on the ground state. So if I bra-through by x, that tells me that u1 of x, which is equal to x1, is equal to 1 over the, sorry, the 1 over the square root of 1 doesn't need to be written anymore, but this is 1 over the square root of n plus 1, n here is 0, of a dagger, which is n omega x minus ip, yes, over the square root of 2 m h bar omega. Now what we want to do is, it's helpful actually to find out what is to rewrite this in terms of l, this characteristic length there. So let's just say what a dagger is in terms of l. 2 m h bar omega, well where's, so if you take that equation up there, that defines l, and you multiply both sides by the square root of 2 m omega, and then you multiply through by a square root of h, you find, so I need to write this down, so l is equal to the square root of h bar over 2 m omega, multiplying through by this, I find that the square root of 2 m omega is equal to the square root of h over l, if I multiply through by h bar, that's a bar by the way, sorry, if I multiply through by h bar, I find that the square root of 2 m h bar omega is in fact equal to h over l. So this factor here is equal to h over l, so I can say that a dagger is equal to m omega x on the bottom is h over l, so an l here and an h bar there, minus I need a, I have an h bar on the bottom, and an l on the top times p. Let's keep working, this stuff is looking remarkably like l all over again, if I would multiply this by 2 on the top and 2 on the bottom, then this would become 1 over l squared, so this is equal to, the l squared would cancel this, and I would have that this was x over 2l minus, now let's put this into the position representation, in the position representation this is minus i h bar d by dx, so the, yes, so the i's get together and make a minus sign, and the minuses cancel each other, so we have an overall minus sign, and this becomes l d by dx. A dagger was billed as being dimensionless, is it dimensionless, yes, because they have an x over l and an l over x. So this is a handy formula for future reference. So let's find out what, sorry I shouldn't have written that complicated expression up there, it wasn't helpful, that u1 of x is equal to this baby, this animal working on, what does it work on? It works on u0 of x, but what is u0 of x? It's business end is e to the minus x squared over 4l squared, and under here I have to have a 2pi l squared to the quarters power. This factor comes in, I said what the normalisation constant was, so just to see where that comes from, I needed p of x to have this, so we obtained u as, I should have said a constant k times this, right, there was an arbitrary constant in this, arbitrary constant of integration, which is in fact going to be the normalising constant, so I know now that the wave function is, behaves like this, which gives me a probability that looks like this, the correct normalisation of the probability is this, so what I need to do now is to say that in order to get things to work out well, I should replace that k with a quarterth power of what's inside here, so when you square it up, you find the right normalising constant of the probability. So we've got the ground state now at last properly normalised, it has that quarter power, and this, now this is going to come out because we've paid proper attention to normalisation, this will come out correctly normalised, and it's, what happens is very simple, when we do this differentiation, we're going to bring down a minus x over 2l squared, and this l will cancel that, and we'll be, and cancel this, and we'll have an x over 2l coming from here, we've got an x over 2l coming from there, so the whole thing at the end of the day, is 1 over this 2pi l squared, one quarterth power, x over l e to the minus x squared over 4l squared, that's the first excited state wave function. To find the second excited state wave function, we'd use this self-same operator, I'm not going to do this, but let's just see what it would look like, u2 would be x over 2l minus l d by dx, 1 over 2 factorial, sorry, 1 over the square root of 2, that's the 1 over the square root of n plus 1, operating on u1, which is x over l e to the minus x squared over 4l squared over 2pi l squared to one quarterth power, and you can see that what's going to happen is we're going to get an x squared term times e to the garbage, we're going to get from this differentiation, we're going to get an x term from this, from the diff, we'll get various things, we're basically going to get an x squared term, and when differentiating away this, we'll get a term in x to the nothing, and when we bring this down, we'll get another x squared term as well, and we multiply this in x squared, so we're going to get terms in x squared and x to the nothing times e to the minus thingy. So this is going to be a polynomial of degree 2. It goes by the name of a Hermite polynomial. It doesn't matter. And every time we use this operator, we're going to get a more and more elaborate polynomial. Can you see that that's what's going to be the consequence? Our wave functions are all going to be this Gaussian that came with the ground state, and then they're going to be times polynomials, which are going to be of order n. So the general state, un of x, is going to be a Hermite polynomial, hn of x, e to the minus x squared over 4l squared. I'm not paying proper attention to the normalisation at the moment. And it's straightforward to find out what these are. You only have to differentiate. You don't have to do any clever, any things. Just differentiate, and they will all drop into your lap. Something important to notice is that the ground state wave function is an even function of x. E to the minus x squared. The first excited state wave function is an odd function of x because this operator is odd. It has one power of x in both places. It changes sign if you turn x to minus x. This one is going to be an even function of x because we're going to apply another odd operator to a thing that's odd, and we'll end up with an even result. Ground state is an even function of x. First excited is an odd function, and the second excited. In fact, un of x is even if n is even, an odd otherwise. We'll meet this phenomenon in other cases. It's very often the case that the ground state is even, and the first excited is odd, and the next one is even, and the next one is odd, and so on like that for similar reasons. Quantum mechanics has its own jargon for this. It says that this is an even parity state. Parity just means is it even, or is it odd, the wave function, and this is an even parity state. We'll have more to say about parity in a general context later on when we're covering the material in Chapter 4. Sorry. This one is odd, excuse me. I was not sure which line I was on. This one is even. The ground state is even. The first excited state is odd, and this has a parity. We say this has a parity minus 1 to the n so that, you know, that jargon is used sometimes. I wouldn't worry about it. It turns out to be useful to know. We'll find it's useful to know whether your wave function is even or odd. It enables you to short-circuit various computations. Okay. We can do, yeah, all right. We're out of board there. Let's, yeah, let's have a go at this. Let's work out the expectation value in the nth excited state of x squared. So we want, it will be nice now to build some connection to classical physics. Can we connect these results to classical physics? Classical physics, as I've said several times, is all about expectation values. From quantum mechanics to classical physics occurs through expectation values. So let's work out this expectation value, which is going to enable us, for example, to work out what the mean potential energy is because the potential energy is proportional to x squared in the nth excited state. Now, how do we work this out? What we do is we observe that A, now we had a dagger. We went to some trouble. Here we go, here we go, here we go. Here is, here is A dagger. A is going to be L, sorry, is going to be x over 2L plus i L upon h bar p. And A dagger we've got there, we've got there, is going to be x over 2L minus i Lp over h bar. So if you add these two equations, you discover that A times L, sorry, L A plus A dagger is equal to x. So a handy, this is a very handy relationship, expresses the x operator as a sum of annihilation and creation operators times L. I've just added these two equations, the momentum of cancel on each other. These have added up to give us an x over L. Here we've had A plus A dagger. I've multiplied through by L. So when I want to work out this expectation value, I can replace each of those x's with this thing here. L squared comes out because it's only a number A plus A dagger squared N. Let's multiply this out. It's L squared into N A squared plus A dagger squared plus A A dagger plus A dagger A. Now, what's this? A squared applied to that is proportional to N plus 2, sorry, N minus 2, because each of these A's takes away a unit of excitation. So A squared times that is proportional to N minus 2, but N minus 2 is orthogonal to that, so that makes no contribution to this expectation value. Similarly, a dagger squared on this produces some multiple of N plus 2, which is orthogonal to that, so that doesn't contribute. So these two terms don't contribute to the expectation value. These terms, Jollywell do, contribute to the expectation value. We know that A on this produces root N minus root N times root N times N minus 1, and this produces root N. Let me write this out. So this is going to be L squared of N A. What's A dagger on this is going to produce root N plus 1 of N plus 1. This A dagger working on that will produce root N plus 1 times N plus 1, and then, and now I've left the other A to be done, and here we're going to say that this is equal to N A dagger times the result of A working on that, which is root N times N minus 1. It's the square root of the largest integer occurring in the equation, and then this A is going to produce a root N plus 1 times N, which will couple with this and produce 1, so this is going to be L squared. This is going to lower N plus 1 down to N, which will produce a 1 when it meets this, and we'll have another of these square roots, so we're going to have N plus 1, and when we use this on this, this N minus 1 is going to be raised back to N, which will produce a 1 when it meets this, but we'll get another square root of N in the process, so we'll have plus N. In other words, it's equal to 2N plus 1L squared. My time is up. We've discovered something interesting, which is the expectation value of X squared is 2N plus 1L squared. Did we already know about it? No, we didn't, in a certain sense. Yes, we did. For the ground state, N equals 0, we knew that the probability distribution was a Gaussian, and we knew that the dispersion of that Gaussian, the expectation value of X squared, was L squared. We've now discovered how the dispersion increases when we add excitations and the probability distribution gets broader, but tomorrow I want to connect this to classical physics.