Okay. So these this is we were at work on the harmonic oscillator, which, as I said, is the single most important dynamical system in physics, a model for a huge number of physical applications. We had this Hamiltonian, which is which is basically P squared plus X squared. The crucial thing is that it's quadratic. It's obviously quadratic in P, but especially it's also quadratic in X. That's his peculiar, peculiar characteristic of its of its potential energy.
We went the plan for what we're trying to do is find the stationary status. In other words, we're trying to find the states here which are eigen functions of the Hamiltonian, because they will enable us to to follow the dynamics of the system. Oh, they're 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 the we evaluated what we found to it. This thing had two properties.
First, it a dagger a is very nearly the Hamiltonian a dagger ray plus a half h bar omega is the Hamiltonian first important property. Second important property that it's commutative with its emission joint is. Is minus one when done this way round. So using those properties we said, suppose we have a state, a stationary stage of energy.
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 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 age by 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 a h bar omega e plus two bar omega. And as we can see, e plus any amount, any number of bar omegas, any number of omegas are. 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 we start with now 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, h, a, and then we have to add in the commentator a comma h. All operating on e still, we yesterday this is absolutely a repeat of what we did yesterday. This is H.A. Now we replace that H as advertised up there by a dagger, etc. We observe.
So this should become the this is going to be replaced by a dagger, a plus a half h bar omega. We can forget about the half because we're inside a commentator and a half commits with everything. We can take the bar omega outside the commentator and then what we're looking at is plus a comma, a dagger, a h bar, omega, close brackets, e we use our usual rules for taking the commentator of products, which which is to say it should be the commentator of this with this without that standing idly by.
And then in principle the commentator of this with this with this standing idly by with the second commentator vanishes. And this commentator, by the result up there is equal to plus one. Okay, because the turned around there a dagger is minus one. So this commentator a comma, a dagger must be plus one. So this is saying this becomes H plus a H bar omega. E So we have an a e 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 the E minus. This H bar omega goes to the other side and becomes a minus H bar omega e is equal to h e, which establishes on the face of it what we might have hoped, which is that the stage you get that you have after you use a on this stationary state e is another stationary state with an energy lower than E by each power omega.
So a dagger raises your energy. It increases the excitation of the harmonic oscillator and evidently lowers the energy of the harmonic oscillator. Now, you have to at this point start to worry, because by the previous rhetoric it would follow that if there is E, then there is also E minus, h, bar, omega and so on down through E, minus n, h bar omega. 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, an x squared should be positive. You can't trust such things in quantum mechanics, but what we can do is we can work out and we can we can say, look, e is the expectation value of the Hamiltonian in the state. E Obviously because H on E is equal to E comes outside and and this on this is one.
So that's a self-evident equation. Replace what's inside here. This becomes over to m of e p p e plus m squared omega squared e. X x. E. And because these are observables and therefore emission, I can put a dagger there if I want to. It 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 it's delivered to m, which is boring and this is manifestly greater than or equal to zero. Even in quantum mechanics, there can't be any argy bargy. This has to be greater than equals 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 three, we use a. So there's a problem. And the problem is the assumption. So what let's just look carefully at what we've established here. What we've established is, is this equation here? This equation is copper is absolutely copper bottom.
It's beyond it's beyond criticism. It's true. It was obtained by totally legitimate operations. And. It. It establishes that this thing is an eigen function 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, the lowest energy you will and you apply a simply kills it. It produces a cap of no length at all. Then we haven't got we won't have a state of even lower energy.
And since it's clear that this chain of operations it is not it's it's logically impossible to go on creating states in 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 then. So. 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, so. That's the ground state. We must have. But a on a zero equals nought. Right? So I'm using this symbol now to indicate the lowest. Energy. So let's. So what does this equation mean to give, to give, to give?
More precise meaning to it? What we so what we say we say is that this case, as you get here, has no length squared. So let's just evaluate the model length squared. So so we're saying that A is zero month squared, which is equal to E, a dagger. A E is nothing but this thing. We have it somewhere up there. It's g ust. Yeah, it's just in range. H is equal to a dagger plus a half h bar. So this thing is this is the expectation value of H over H by omega minus a half.
So and this one, these ones have zeros on it. Sorry. These one have zeros 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 this equation is valid for. And so what we're discovering is that E0 over H Bar Omega is equal to a half.
In other words, the ground state energy is 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 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 a way functions of the corresponding Afghan state stationery states.
But let's move over here. So let's look at let's ask about wave functions. Of the stationary states. Oops. Now we've found the allowed images. Actually. But for 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 is that a dagger on and I'm going to use a new notation.
Right. We're going to say what previously I'd call E is going to go to N. So this is the state. With E equal n plus a half. H Bar Omega. Right. We could. We could write an end in here. But everybody writes just end. It's just saves energy. It works well. Right now. We've discovered what the energies are and that they're labelled by an integer n nought. One, two, three, four. So this is and equals nought, one, two, and so on.
It makes sense to have a nice compact notation into and to call our states, the stationary states, the state nought. The ground state, the state one. The first excited state. The state to the second excited state. And what we've established is that a on N is some constant I'll call it K times N plus one. Right, because we've discovered that when we used A on the state e we got the state. E We got, we got we got a state which was an Oregon catch of H with eigenvalue e plus h bar omega,
which would be in the new notation the state n plus one. If if this e was n plus a half h for omega, this will be n plus three halves h bar omega. And the issue arises. So what value does this thing have? So just just to be clear, we so when we have we've had relationships like this, we had a corresponding one up there. Yep. There we go. We arrived yesterday. We derive this equation and that equation establishes that this thing is an organ of age.
It doesn't establish that as a properly normalised dog and kids of age, I want it to be properly normalised. And so I'm asking about what's the normalisation constant I have to use after I have applied a dagger? It's easily found. Found because we just take the mod square of this side of the equation, of the monster of that side of the equation. So taking the mod square. So taking. Mud square on both sides.
We get an a a dagger and is equal to k mod squared times one because that's by definition going to be correctly normalised. We 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 this just becomes k squared. We just trying to get the thing normalised. We don't care about the face. And 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 comma, a dagger. Whoops. Commentator. So that's. That's that rewritten. This is the Hamiltonian minus is the Hamiltonian overage bar omega. Where is it? Where is it? Where is it? Yeah, it's up there. It's the Hamiltonian minus. It's the Hamiltonian overage bar, omega minus a half. So this is H overreach bar omega minus a half.
That's that, that's this thing. And what's this? This is plus one, I think because other the commutation, the other way round was minus one. So this is plus one. So that's that that h on n is equal to n plus a half h bar omega by definition. Right? So this on this produces n plus a half by omega divide by the omega. And we have n plus a half take away a half. We have an add one, we have n plus one. So this is in fact equal to one plus one. So what? Yeah. So. So. So what follows this?
So comparing. 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. Yeah, that's right. Square root of N plus one. Of of a degree. This is a very important equation. If we would repeat this, if we would go through this rigmarole.
The same logic using a on end being some other constant times n minus one we would we would find that n minus one is equal to one over the square root of n of a operating one n. We know that operating on n is going to produce n minus one. It depletes the energy by 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 that, after the lecture.
It's a precise repeat of this logic here, and it's worth remembering these rules because this is such an important relationship and 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, right? So in this equation, the biggest number appearing is N plus one. That's what you use in this equation. The biggest number that appears is N, that's what you use.
So these are these are two very important equations which physicists remember. Okay. So so what we have to do now we're trying to find a way 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? It's we will call it use zero of X. And it is, of course, x nought.
It's the it's the function was defined by this, the amplitude to be it x. If you were in the ground state also called this right, these are just two notations for the same thing. And this satisfies a nasty equation because what we know is that a operating on nought is equal to nought. Right. 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 an omega x. Plus I.P. In principle, it's over the square root of two M by Omega. But because I'm about to put this stuff equal to nought, I can I can 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 in the position represent. Well, no, I mean, we have it there. Right. This tells me that m omega x and omega x x nought because this is the position operator we this is an observable I can imagine that it operates backwards on this. This is the eigen function of that operator. So those are both operators. That's not an operator. This this operates on this. If I want backwards, this is its eigen function.
So I 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 the x of x. All right. This was the this was the definition for any wave function of how of how P operated with an H. Thank you. Thank you. Absolutely. So I need to write this stuff down. I need to write. And most of this is equal to zero. This is equal to zero.
Provided I put in the m omega x x. If I rewrite that in in wave function notation becomes m omega x, you note of x is equal to and it's clean this stuff up I get a sorry it's not equal to plus h bar d by the x 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.
The next present I've written, this is a partial derivative consistency with what we do in more than three dimensional cases, I suppose, but it is a first order linear differential equation. There is no sort of differential equation that's more friendly, user friendly that we encounter. We solve such equations by using an integrating factor just to get this into standard form. I would write this as I would actually write this, as do you note by the x plus and omega over h bar x u nought.
This is the sort of 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 well, the not derived term this term each of the integral am omega over h bar x d x, which is clearly e to the omega x squared over to h bar.
That's its integrating factor. And the equation is then if you multiply it by the integrating factor, the equation says that the x of the integrating factor times you nought is equal to the right side, which happens to be zero. Ergo, this quantity here is a constant. Ergo you note is equal to e to the minus and omega x squared over two, which I want to write is e to the minus x squared over four 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 you nought mod squared will then be e to the minus x squared over to l squared. So this is this is a normal distribution. With dispersion. L So the reason I want to I want to get this into this form is in order because because I can identify that is the Gaussian width of the distribution, the width of the Gaussian distribution.
Also, once I say that this is the probability distribution for my knowledge of statistics and stuff, I know what the normalising constant has to be. I know that it's two pi l square root to pile squared. Right, because that's the factor you need to normalise a Gaussian distribution. Oops, I'm missing a minus sign. Crucial, right. Crucial. So what does l have to be in order that this form is the same as that form?
It. It's, I think, simple algebra to show that l has to be, uh, h bar over to omega. What 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 and momentum times distance. So this has dimensions of. M. X. P. A. So. So this is sorry. This is m x v says what I should have said. Momentum times and the x right. Talking about yes and the x right. And the x is the dimensional structure here of.
Dimensions here. So indeed this thing has dimensions of length squared. So L is indeed the length. So it's it's. It's okay. So what did we learn? We've learned that the ground state of a harmonic oscillator has a wave function, which is a Gaussian with a characteristic width given by this. And crucially, what we've already studied, right? We've already studied distributions of particles which are Gaussian, have way functions which square up to Gaussians.
And we now know what the amplitude distribution would be or the probability distribution is for momentum from what we did before. We will have the probability of the momentum will also be a Gaussian. It'll be each of the minus P squared over two, should we call it sigma p squared over some square root to pi sigma P squared. And remember, we have the uncertainty principle which said that the dispersion in x times, the dispersion in momentum is equal to h for over two for the Gaussians.
This was a result we established when talking about the free particle. So that tells me that sigma P is equal to h bar over two L so, so, so there's some, there's some characteristic width, uh, there's some characteristic momentum that the particle has. Now this. This set in the ground state. A particle is not stationary, it is moving with it with a characteristic amount of energy. It it's also not at the bottom of the potential. Well, it has a characteristic amount of average potential energy.
So 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, 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 there is isn't 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 if you restrict yourself in 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. So. So in practice in the ground state, there's a compromise between being being reasonably close to the origin and having a reasonably small kinetic energy. So because of the uncertainty relation, quantum mechanical systems in their lowest states have a finite extent spatially.
Even though they even though really it's a point particle, but because of the there's a there's a finite extent in which you'll find the particle and a finite kinetic energy. And this is a totally. 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.
And this is this is enormously important because it's exactly this physics we will see that is exactly this physics which determines the size of atoms. Electrons. So atoms here are typically pretty close to their near enough in their ground states. And the size of these atoms is determined by the electron if if the atom gets any smaller and indeed you, you know, you take 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 will resist violently. You'll have to do work. You 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 by 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.
It so the size of atoms is determined by this 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 relativistic li.
They're in a very deep potential. Well and because and very narrowly confined right into ten to the -15 metres in order to be confined, even though the fairly massive particles into this very small space, they have to have a lot of kinetic energy and that and the the mass associated with that kinetic energy accounts for most of the mass, you know, 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 the if we say H is equal to one over two M of P squared plus m m squared, omega squared, x squared. And we put in the uncertainty relation, we say that that x squared if we if we say that x squared is sorry, if we say that p squared is equal to h while squared over two x gets right sorry over four because I've squared it over for x squared.
Right. So this is so for the ground, say x squared is essentially the uncertainty in x. So it's associated with the uncertainty momentum. In the same way if you stuff that into this, you put this relationship in, you find that H is one over two M is going to be whatever it is h by squared of a for x 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 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 the minimum. If you if you ask yourself what, 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 the that L is being chosen to minimise the energy given the constraints imposed by the uncertainty principle.
Okay but we so that we found the ground state way function. It would now be useful I guess to show how we should calculate. The. Sort of a natural order here. Yeah. So what we want to do, let's get the, let's get the first excited state as an example. So first excited state. Excited state wave function. So you won of X, which is by definition x one. What do we know? We know that one is equal to one over the square root of one.
Time's a dagger working on the ground state. So if I bra through by X, that's that. That tells me that you won of x, which is equal to x one is equal to one over the sorry the one of the square. One doesn't need to be written any more, but this is one over the square root of N plus one. And here is note of a dagger. Which is. And Omega X minus II minus IP. Yes. Over the over the square root of two and 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 to m h bar omega. Well was. So if you if you if you take that equation up there that defines l. And you multiply both sides by the square root of two 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 to omega multiplying through. By this I find that the square root of two and omega is equal to the square root of h of l if I multiply through by h bar. That's bar by the way. Sorry I multiply through by h bar. I find the square root of 2 a.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.
This is equal to an omega x on the bottom is h over l so an L here and an h bar there minus i. I need to. 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 El all over again. Right? If I would multiply this by two on the top and two on the bottom, then this would become one of l squared. So this is is is is equal to the L squared would would cancel this and I would have that. This was X over to L minus.
Now let's put this into the position representation. In the position representation. This is minus h bar D by the x. So the. Yes. So the eyes get together and make a minus sign and the minuses cancel each other. So we have an overall minus sign. The H bars cancel and this becomes l d by the x. 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 you one of x is equal to this baby, this animal working on. What does it work on? It works on you. Zero of x. But what is you? Zero of x? Its its business end is easier than minus x squared over four l squared. And under here I have to have a two pie 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 set a constant k times this. Right? There was an arbitrary constant in this 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 the quarter power of what's inside here. So when you square it up. So when you square it up, you find the right normalising constants of the probability. So. 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 we've paid proper attention to normalisation.
This will come out correctly normalise. And it's what happens is very simple. We when we do this differentiation, we're going to bring down a minus X over over to L squared. And this L will cancel that and we'll be in and cancel this and we'll have an X over to L coming from here we've got an X over to L coming from there. So the whole thing at the end of the day is one over this two pi l squared. One quarter of power x of l e to the minus x squared over four l squared.
That's the first excited state wave function. To find the second excited state wave function would use this selfsame operator. Not going to do this, but let's just see what it would look like. You two would be x over two l minus l d by the x one over two factorial. Sorry, one over the square root of two. That's the one over square root of n plus one operating on you one which is x of l e of the minus x squared over four l squared over two pi l squared two one fourth is power.
And you can see that what's going to happen is we're going to get an X squared term times. These are 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 will 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 and it's going to be squared. So we're going to get terms in x squared, an x to the nothing times e to the minus thingy. So this is going to be. A poly polynomial of degree to. It goes by the name of a hermit polynomial. It doesn't matter. And every time we use this operator, we're going to think we're going to get a more and more elaborate polynomial. Can you see that? That's that's what's going to be the consequence?
Well, wave functions are all going to be this Gaussian that came 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 my polynomial h and of x e to the minus x squared over four l 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. Don't 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 right e to the minus x squared. The first excited state wave function is an odd function of x because this operator is odd, right? It has one power of x in both places is numbers. It changes sine.
If you turn x to minus x, this one is going to be an even function of x because you because we're going to apply another odd operator to a thing that's odd and we'll end up with an even result. So. So the ground state. Well, ground state. And isn't even a function of X. First excited. It's an odd function. And the second excited. So in fact you kn of x is even. If N is even. An odd. Otherwise. We'll meet this phenomenon in other in other cases that it's it's very often the case.
The ground state is even in the first exciting resort and the next one is even in the next one is odd and so on like that for similar reasons. And quantum mechanics has its own jargon for this. It says that this is an odd parity state sorry, an even parity state. This isn't even parity state parity. Parity just means is it even or is it all the wave function and this isn't 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 four. So I think both. Sorry. First, though, 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 is is has a parity. We say this has a parity minus one to the end 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 old and then enables you to short circuit various computations. Okay. We can. Yeah. All right. Round a border and. Let's. Yeah. Let's have a go at this. Let's work out the expectation value in the excited state of x squared. So we want and it will be nice now to build some connection to classical physics.
Can we connect these results to two classical physics? Classical physics? As I've said several times, it's all about expectation values. The connection 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 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. Yeah, we got. We got. We got. Here is. Here is a dagger. A is going to be L3 is going to be x over to l plus i l upon h bar p and a dagger. We've got there. We've got there is x over to L minus i, l p over h bar. So if you add these two equations, you discover that a times l sorry, a plus a 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. Right, I've just added these two equations the momentums, the momenta of cancelled 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 Xs with this thing here.
L squared comes out because it's only a number and I have an into a plus eight dagger squared and. Let's multiply this out. It's l squared into an x squared plus a dagger squared plus AA dagger plus dagger a. Now what's this a squared applied to that is proportional to when plus two. Right. Because sorry n minus two because each of these A's takes away a unit of excitation. So a squared times that is proportional to n minus two with n minus two 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 two, which is orthogonal to that. So that doesn't contribute. So these two terms don't contribute to the expectation value. These terms jolly well do contribute to the expectation value. We know that a on this produces root and minus root n times, root n times. What we've at times end minus one and this produces routine.
Let me write that out. So this is going to be l squared of rn. Hey now wants a dagger on this is going to produce root and plus one of end plus one. Right this this a dagger working on that will produce route end plus one times and plus one. And then uh. And now I've left the other way to be done. And here we going to you 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 one.
Remember that's the square root of the largest integer occurring in the equation. And then this a is going to produce a root and plus one times N which will couple with this and produce produce one. So this is going to be L squared, right? This is going to lower N plus one down to n, which will produce a one when it meets this. And we'll have another of these square roots. So we're going to have N plus one.
And when we use this on this, this N minus one is going to be raised back to N, which will produce one when it meets this. But we'll get another square root of N in the process. So we'll have plus n. So in other words, it's equal to two N plus one L squared. So my time is up. What I want to do. So we've discovered we've discovered something interesting, which is the expectation value of x squared is two and plus one l squared.
We already knew. I think about what? Did we already know about it. No, we didn't. In a certain sense, yes, we did. We use for the ground state and equals nought. We knew that it was a we knew the probability distribution was a Gaussian and we knew that the dispersion of that Gaussian, i.e. the expectation value of X squared was L squared. We've now discovered how the how the dispersion increases when we when we add excitations and the probability distribution gets broader.
But tomorrow, I want to connect this to classical physics.