 We have an assignment due. There will be one more assignment on the topic of the simple Hermione oscillator, which we will distribute later on. And then we will have an example. The exam will cover the material through the similar Hermione oscillator, which we might finish today. We might finish tomorrow. I mean, probably we won't finish tomorrow. But we could finish it on Thursday. And then we will have an exam. And, well, I haven't quite decided about the exam. I wanted to get your opinion. So the question is, what would you prefer? Would you prefer an exam of the type of the first exam? Or would you prefer a take on the exam of the first exam? Yeah, absolutely. Well, you're just going to decide to be like horrendous, right? No, horrendous. Yeah, that's kind of a guy. No, the idea would be that there would be a lot less time pressure. The take on the exam, the idea is not to be of the difficulty, say, of the homework assignment, but something that you have more time. But we'll see, you know, a little bit more involved, you know. But if that has positive and minuses, obviously, right? So think about it. And you can send me an email for your feedback. You can just do one of those polls again. I'll just get your feedback. Right. So we're talking about the Symbol for Modern Composite. In our last lecture, we were talking about different representations of the quantum problem with respect to different bases. So we could express our representation in terms of what are called the number states, which is a countably infinite set of vectors that span the Hilbert states. The number states are the eigenstates of a dagger a. They are the eigenstates of the number operator. And of course, they're also the energy eigenstates, right? And they form a resolution of the identity. And alternatively, we could look at representations with respect to the span of the states that are eigenstates of the position momentum operator. OK. Those are, those eigenvalues are continuously infinite. But nonetheless, they form a basis for the space. They form another resolution of the identity. And we, of course, look at position momentum representations. Those are the weight functions, right? So we could talk about the state in terms of its expansion in number basis or its expansion in terms of position and momentum states. OK. So we have different representations of operators. The raising and lowering operators with respect to the number basis are matrices. They're infinite dimensional matrices. But matrices nonetheless, same for position momentum. In the basis and the representation of position momentum, of course, then position and momentum operators have representations as multiplication by constants and derivatives, those operators, then are the wave mechanical operators that one is typically introduced to in introductory quantum mechanics. Yeah. Well, P was T gather by a say. You are correct. No, that's no. It is correct. That's a statement. It's because the i doesn't mean no. That's correct. Yep. That's correct. The reason we know that is that, of course, A is x plus i of T. So you just attract from a dagger and you get the right thing. The energy eigenfunctions, that's to say, the positional momentum representations of the energy eigenstates are easily derivable by just saying, by this defining equation, the ground state is annihilated by the lower operator. So when we look in the position representation, we have a differential equation for the ground state wave function, which we can solve. And that equation tells us that the ground state of the sub-ormonic oscillator is a Gaussian. And at higher energy levels, the excited states are obtained by applying the creation operator to that. And the creation operator applied to that involves taking the value of this, which is x minus, the derivative of respect to x, n times, on this. And what we get are these functions. Those functions are Gaussians that are multiplied by permitt polynomials. The permitt polynomials can be written in this way compactly. In fact, one, there are a number of properties and sort of traditional theories of special function differential equations and so forth. Just study this kind of thing ad nauseam. We don't have so much anymore. But one of the things that's, for example, useful property of permitt polynomials, which is a kind of property you see of all hypergeometric function and so-called generating function. The generating function tells us a way of obtaining the hermit polynomials from a power series. So if I look at, as a function of some parameter t, that Gaussian, one can show fairly easily that this is equal to the following. So the hermit polynomials are the expansion coefficients in the Taylor-Shearer's extent. So it's an intimate relationship, which is basically given this factor between Gaussians and hermit polynomials. The other thing that we mentioned, briefly it's very easy to see, is that the wave functions that are associated with the energy eigenstates in position space are exactly the same as they are in momentum space. And that's not surprising because the Hamiltonian is completely symmetric in position and momentum. And it's just a question of a phase convention that we have. So in momentum space, the energy eigenfunctions are exactly the same with the factor just to keep everything self-consistent of i up front. Of course, we know that because we have a reflection symmetric potential for the simple harmonic oscillator about the origin, the eigenstates of the Hamiltonian are symmetric and anti-symmetric functions. Brown state written which has a zero point energy. And then we have the first excited state, second excited state, et cetera. Let's continue a little bit further along this path. One thing we didn't discuss that I wanted to get to you last time, but we didn't get it far is to ask the question about the uncertainty in position momentum. So let's look at the energy eigenstates. Remember, eigenstates, of course, are not eigenstates in position momentum. They're wave functions that are distributed in space and momentum space. So they have a finite extent, or anything there for a finite uncertainty. Now, just looking at this equation, or could only momentum space, you should be able to read off what the uncertainty is of the ground state. How about h bar over 2? Sorry? h bar over 2? Well, there are some h bars there depending on that. It's not quite the units of x or p. Those are units of the action. I'll enter you out. So what I'm asking is a problem question. We have, in position space, we have this expression. So what that tells me, of course, is that the probability distribution in position is a square wave function, probability density, which is 1 over the square root of s e to the minus x squared. Now, is that a normalized function? Absolutely. It's a Gaussian, and everyone remembers, by heart, the form of a normalized Gaussian. And if you don't, you will memorize it now. And what is the form? If I have a Gaussian with rms with sigma, and what is the form of the Gaussian distributed in x? 2 sigma squared? 2 sigma squared. And this goes like 1 over the square root of 2 plus sigma squared? Yeah. And everyone's going to put that in here. OK. Used to be on the Deutsch Mark when, before the euro, it's like this character, the Gaussian's picture on it. Who is it? It's a very nice one. OK. So with this, what is the mean value of position for the ground state? Zero. And what is the uncertainty? Of course, the uncertainty here is sigma. 1 over 2. Wow. So in the ground state, x squared, we have these relations. OK. Or I should say the rms with the uncertainty itself in dimensionless units is 1 over the square root of 2. So you can read that off. What about a momentum? It's the same, because they're the same weight function. So in dimensionless units, we've got this. What about in dimensionful units with the uncertainty of position? How do I put the units back? x, absolutely, thank you. We have a characteristic, so everything is in units of the characteristic scale. So we multiply that by xc. And then we have to look back and see what the heck xc was. And that was the square root of hr over gamma omega. So that's the uncertainty in position of the ground state and position of x, but that's zero. Just to emphasize that this is for the ground state. And the uncertainty in momentum states is the characteristic momentum, which is equal to square root of h bar times m omega. So we have h bar and omega. So the characteristic scale that we derived just from numerology, basically, dimension analysis, are essentially, up to this factor 2, the widths of the weight function of the ground state. And we know that, generally, that the product of uncertainties must satisfy the Heisenberg uncertainty principle. And what is this, in this case, of this problem? H bar over 2. So the ground state of the harmonic oscillator is a minimum uncertainty weight factor. Does this help you? It is true that the only minimum uncertainty weight function is a Gaussian, but it is not true that every Gaussian is a minimum uncertainty weight factor, as we will see in homework. And this, for the ground state of the harmonic oscillator, is a minimum uncertainty weight factor. Now, what about the higher excited states? What is the uncertainty in position of momentum for them? Well, that's a more tedious thing to do in weight mechanics. I mean, you could do it with the Herbie polynomials and the generating functions and integrals, but it's trivial to do it in the number of representations. So that's how you should do it, OK? So let's do it. So what I seek is the mean position and momentum and the uncertainty for the energy level. That's what I seek, OK? And the way to calculate that more easily is to use the algebra of a Bay-Bagger, OK? So let's do it. Of course, x is a plus a Bay-Bagger over 2, and p is a minus a Bay-Bagger over i. So first of all, just quickly, what is this? Zero. Zero. There is no mean value of a or a dagger for a number state because this raises the numbers. So what that tells me is that for a position eigenstate for a number eigenstate, there is no mean position. Of course, we could have read that off the picture. It's a symmetric potential, and the binding functions squared are all symmetric functions. So therefore, the mean position is always zero in a parity symmetric potential. So that's got to be true. All right, what about the uncertainties? Well, that tells me that the uncertainty in position will to the expected value of the square since the moon itself is zero, right? And that is equal to a plus a Bay-Bagger over root 2 to the power n. Power 2? Sorry, to the power 2, thank you. That was hard. OK, so now here's another trick of the trick. Whenever you're dealing with, it's looking at the algebra of a and a-bagger, and you want to calculate something like an expectation value, the only non-zero terms are the ones that have an equal number of power of a's and a-bagger's, right? Because otherwise, on one side, I'm raising more than I'm lowering. Yeah? But that only applies if n is equal to zero, then you go and add that a-bagger to zero. True, but it doesn't negate what I said. The only terms that are certainly not zero are those. Then in addition, there would be additional ones that are zero. So that means that, generally speaking, the only terms that survive are the ones with equal numbers of a's and a-bagger's. So there's an a, a-bagger, and there's an a-bagger, a. Now, as Steven pointed out, if this were the ground state, then this term would not be there. All the terms that are so-called normally ordered, where the a's are on the left, and the a's are on the right, and the a-bagger on the left vanish. But this is general. So how do you deal with this? Well, the easiest thing to do is to, what we say, is normally order this. Let's put this in the form where a-bagger's are on the left, and the a's are on the right. That's just a jargon, because there is this common normal order, but don't worry about that. Why? Because then, this is the number operator, and we know how it acts on it. So this, if I, this is equal to a-bagger a plus the commutator of a-bagger, which is 1. So this is equal to 1 half of a, this a-bagger a is the number operator plus 1, and then the expected value of x squared is 1 half, which is n plus 1 half. And it's very easy to show, even exactly the same way, that the expected value of p is the same as it should be, because it's got to be symmetric in x and p, for example, eigenstates of the Hamiltonian, because the Hamiltonian is symmetric in x and p. OK? What this says, of course, is that the energy is equally distributed between the kinetic energy and the potential energy, right? The Hamiltonian in dimensionless units is a half x squared. And we know that this is n plus 1 half. So it's not surprising, we even have to go through that. This thing, we could have just read that off by it, as well. All righty. So what this tells me is that the uncertainty in position is square root of n plus 1 half times the square root of h-bagger. So are these minimum certainty-weight packets? Net, right? The product of this uncertainties. This is, I'm sorry, of course, the dimensions back in. So we've given up a little lower case of that. Only for the ground state is this minimum uncertainty state. Very good. So let's talk now about dynamics. So our central player in talking about dynamics is the time evolution operator. Time evolution operator is either minus i over h-r. And the Hamiltonian, I could express in terms of x and p, or in terms of a-dagger a. So if I put in the Hamiltonian, this becomes e to the minus i omega a-dagger a plus 1 half times the number operator. Now the zero point energy is there. Emphasizing the past, absolute value or what the value of the ground state energy is is irrelevant to dynamics. The only thing that affects the dynamical evolution of the system is the relative energies. So typically we ignore this in talking about dynamics. You can put it in there. But it's an overall phase factor that doesn't affect any prediction of any probability because it's never going to affect the relative phase of anything. So you can completely ignore it. It's just where we decide to put the zero energy. All right, so if we wanted to look at time evolution, there's lots of ways we could do it. Let's talk about the Heisenberg equation. And again, there's lots of ways we could figure it out from the Heisenberg equation. Let's look at the Heisenberg equation's motion first of all. So let's look at the, and what is that in this case? Well, I can write it out what that is. I have h's there. I have minus i omega times the commutator of, and what is that commutator? Well, x commutes with x squared and x with p squared. The 2 comes with it out. So this becomes 2i, the commutator of x and p in dimensionless units. So this is equal to omega p. And similarly, the time derivative of momentum is minus omega x. If I wanted to just write this loss again, there's many different ways to do the same thing. The time derivative of a minus i omega, the commutator of a and h, that's equal to minus i omega, the commutator of a with a dagger a with a minus i omega a. So what we see here is that the Heisenberg equations of motion for the Heisenberg oscillator are exactly the equations of motion that we had for the classical canonical chords. Exactly. There's nothing different about them, right? And we can write down the solution. The Heisenberg operator a commutator is a0 e to the minus i omega t and x of t. The solution, we can get that by writing a plus a dagger over 2 is x is 0 plus omega t plus p is 0 sine omega t and p of t is a minus a dagger and 2 is p 0 plus omega t minus x 0 sine omega t. So that's a very important result. The Heisenberg oscillator, in some sense, looks entirely classical. Yeah. This looks like it's relevant in what we were supposed to do with the squeeze operator in the homework. Because the format just seems too similar to be possible. They're similar, but in the squeezing operator isn't related to this kind of operation in something different. But the mathematics is quite similar. In fact, I would say another way of getting this solution instead of solving the differential equation is just to directly time evolve the operator according to the unitary transformation. That's another way we can get exactly this. This is equal to either the plus i n hat omega t, a 0 e to the minus i n hat omega t, and then use multiple commutators. This is equal to the sum n equals 0 to infinity, the multiple commutators. So there's many ways we can get the same thing. So what we see here is that for the simple harmonic oscillator, the expected values follow the classical trajectories. That's an important result. That's unique to the harmonic oscillator and a free particle kind of linear potential. If something is no more than quadratic, then that is the case. So what we have here is that the average x follow the classical trajectory. Brings us to a very interesting topic that we're going to spend the rest of our time on the harmonic oscillator asking about the question of the classical limit. What is it? What do we mean by the classical limit? In what ways can we think about the quantum simple harmonic oscillator like a classical simple harmonic oscillator? You remember last lecture where we were talking about the correspondence mechanism. We talked about the idea that when we go to very high quantum numbers, high energies, where the Roy wavelength was very small compared to the scale of the risk potential change, that there were aspects of the classical trajectory encoded in the energy eigenfunctions according to the WKB approximation. So we said, I'll come back and talk to you over here, get the correspondence. So if I have some harmonic oscillators and I have some highly exciting state, then the wave function is such that it's very deep near the classical turning points, which is kind of what you'd spend for a harmonic oscillator to ask what's the probability to find the position of the harmonic oscillator somewhere. It spends most of its time near the turning points and the middle time where the kinetic energy is large. And so the amplitude of this average looks like the classical wave function associated with this state was proportional to 1 over the square root of a classical kinetic energy at that position, where the classical kinetic energy at that position is, I mean, kato momentum is the square root of the kinetic energy, which is the total energy minus potential to F. So in some sense, there's an aspect of this which contains the classical dynamics in it. But if I prepare the harmonic oscillator, the quantum harmonic oscillator in this wave function, if I somehow were able to create this energy eigenstate, would you say that the predictions of measurements that you would do would be what you would expect, classically? Classically this thing oscillates, right? And classically the position of this guy, or I would just say this classically, what about if I prepared the state of the system in this state? What does this do? Does it oscillate? What is the expected value of position as a function of time? Does it oscillate? It's zero. Why? Well, it's true that the Heisenberg operators follow the classical trajectories, but for this guy, the mean value is zero at time t equals zero for both of these things. So it's zero for all times. This is a stationary state, and for a stationary state nothing, no probability of anything changes as a function of time. So stationary states are not associated with classical trajectories. We need to have a wave packet of different energy eigenvalues in order to have anything change as a function of time. State it another way. So let me first state it this way. So stationary states are stationary. Nothing changes with respect to time. Let's state it another way. If I look in phase space, the states of definite energy lie on the circle. Let's put it in the units. So this is alpha as a function of time. And this thing has a definite energy. The characteristic energy comes out of this state. A stationary state is a state that corresponds to something which is completely uncertain where I am on this circle. But my position momentum is not here or here or here. It's anywhere on this. So if I had a classical probability distribution on phase space, suppose that I prepared my harmonic oscillator, but I didn't tell you what little kick and position I'm going to hit. So you just then would assign a probability distribution. I said it's going to have this energy, but I'm not going to tell you what its initial position and momentum are. Well, then you would say, well, it's on this circle, but I don't know where. That's a classical hidden variable. Missing information, you assign a probability distribution based on your lack of knowledge in which case as a function of time, the position momentum would be the same probability distribution because this whole thing is rotationally symmetric. That probability distribution would not change as a function of time because every point on this curve moves around with angular velocity, omega, and they all stay the same. So that's the classical analog of the quantum stationary state. So if we want to define something that looks more like a classically oscillating particle whose position and momentum oscillate with time, we cannot look at a single stationary state. We need to think about a wave packet which is a superposition of energy by instincts. In fact, we see that already here in this classical phase space. What we have here is that if I have an energy, a state that has a definite energy, it has a completely uncertain phase. Let's say the classical trajectory is one where this phaser, the phase of this oscillator changes as a function of time. So what we see here is that from the quantum mechanical point of view, let me summarize a few things, to see dynamical change in the mean x and p, we need to consider a wave packet of energies. The relationship to this notion of phase is the following. So we see the time evolution operator generates rotations in phase space. So this phaser is rotating at an angular velocity omega. So that means that the number operator is the generator, displacements in the same way that we said momentum was the generator of displacements in position. And position was the generator of displacements in momentum. The number which is related to the magnitude is the generator of displacements in phase. In fact, this comes back to another point that's written here. We had two different ways of expressing this complex amplitude. We can express them in terms of the real and imaginary parts, which were the quadratures, which are the position of momentum. And those we said satisfied when we quantized them, they were the canonical operators and they didn't commute. What about the polar representation? I mean, I can look at this for different energies and there are different coordinates here. I can look at this in polar coordinates. Can I write down a polar representation of the quantum league? The classical amplitude, complex amplitude becomes the annihilation operator. So we had alpha of t became any of t. I'm going to just say alpha at any time. X plus alpha of q. But that representation, you know, alpha is this magnitude type of phase. Can we write the annihilation operator in a polar decomposition? Do you remember the polar decomposition? That horrible problem in the first week of November. What the heck is he making us do this for? Here's the question. So you remember we showed in homework that we should be able to express an operator. It might be natural to think, okay, there is a unitary and then a square root. What is the polar decomposition? This is a to n, this is a to n. This has the form of e to the i phi hat, the phi intermission operator. That's a number operator, which is what we might have guessed. The amplitude is the square root of the number, the magnitude of the amplitude. Then there's a phase. Now, there's a problem with this. What's the problem? Well, in fact, I'm going to put the hat over the whole thing for a reason you'll see why in a moment. Let's look at what this operator is. We know that this operator in the number basis is the square root of n and minus one n. This operator is... What is this operator in the number table? It's n with eigenvalues n increasing square root of n. That's what this is. So it's invertible. One over that is one over the square root. So I can just take the inverse of that. So this tells me that this operator, this thing that was the unitary in the polar decomposition, is equal to n equals 1 to infinity. So right over here you do the i phi hat. They have it on it. It looks kind of unitary because unitary operators take bases to bases. But there's a problem. Let's product of this with its adjoint. If this were unitary, this should be jn of the operator, right? u times u dagger should be 1. What is it equal to? Well, I have the sum n equals 1 to infinity n minus 1 and the adjoint. The problem with that... No, that's okay. This one's fine. This is n equals 1, which is also equivalent to n equals 0 to infinity n to infinity, which is the identity. That's cool. There's no problem with that. However, if I look at this, I get equal to 1 minus the projector ground state, which tells me that these two operators don't commute. Therefore, it's not a normal operator. It's not unitary because of this fact that I can't lower beyond 0. So what that means is that this polar decomposition is kind of bogus. It only holds always for finite dimensions. But for infinite dimensions, I can't exactly do this. I can write it this way. I mean, this is true. This is correct, but it's not unitary. Which means if it's not unitary, there exists no Hermitian operator that I call the phase operator. There is no Hermitian observable corresponding to the phase of the oscillator. There's been umpteen papers written about this and what this means, how to interpret this. But I can say a few things that are important. This doesn't mean that we can't still talk about the measurements of the phase and the fact that in some sense, number and phase of the oscillator are conjugate variables. In the same way that position and momentum are conjugate. So if I come back over here, what we said is that if I had a state that had a definite energy, it has a completely uncertain phase. If I want to make something that has a definite phase, it's going to have to have an uncertain energy. Phase and number are conjugate in the same way that position and momentum are. Let's derive that and explain what we mean by that. So let's, if I look at the commutator of this operator, so that's equal to n minus 1, which is equal to minus, of course we're not holding it. So what that says is that in some sense, the number of operator looks like the generator phase translation, which is equal to IE by E5. For equivalently, I have this kind of life derivative perspective. In the same way that position operator looks like the derivative of momentum in the momentum representation and momentum looks like the derivative of position in the position representation. Now, why isn't this exactly true? Well, it comes back to this picture. You see, I can translate it x and p arbitrarily. But it's the fact that there's this singularity here, this origin. I can't go below n equals 0. It was really just a derivative. I should be able to go through to the other side, but I can't. And that's the mathematical problem. Nonetheless, what this tells me is that in some sense, as long as I'm not near that singular point, that this is approximately, as long as I'm not really close to that singular point. And what that tells me is that there is a kind of number phase uncertainty. If I want to create a quantum state that has a well-defined phase, it must have an uncertain number, an uncertain entity. It must be a wave packet consisting of many different number states. It's the only way I can do it. The state that has a definite number has a completely uncertain phase. It states with definite phase. Well, kind of find the following. Let's define phase states to be an equal superposition of all the number states. This is analogous to the fact that position eigenvector is a superposition of all momentum eigenvectors. And momentum eigenvectors are equal superpositions. Right? Why is it fair to call this thing a phase state? I said there was no phase operator. There is no Hermitian operator that I can consider to be the phase which I could measure. But we do have this operator. Let's see how this operator acts on this state. Now I can rewrite that e to the i phi times that state. So this state, as long as I don't have much of the vacuum and not close to vacuum, this state looks kind of right an eigenstate of that. If we were unitary, then we know the eigenvalues would be phases. But it's not quite unitary. It's got this extra thing that happens near the origin. Nonetheless, this is a set which spans the space and can be considered to be. We can show, nonetheless, if I integrate over all these phases, I get a resolution of 5. Oh, I'm sorry. Yes, thank you. I imagine it's important. So are these, if we have this resolution of the identity, can we make a Hermitian operator? Just out of, I mean why don't I just define for this diagonal in that representation? Well, the problem is that these guys are not orthogonal kets. It's not Hermitian because these guys are orthogonal. This, this, this. Nonetheless, we have a way of thinking about phase because we have this relationship. What this is telling us is that we have a resolution of the identity in terms of positive operators. This is what we discussed in the semester as a POVM, is to say that I can, we have a probability distribution over phase. That's to say the probability distribution of finding a certain phase is equal to 1 over 2 pi of phase states. Even though I cannot obtain this probability distribution by doing a projective measurement, there is no projective measurement, there is no Hermitian operator whose probability distribution is this because these are not orthogonal projectors. So, one of the things that has been a challenge in quantum metrology, the idea of trying to measure, say, the phase of an oscillator as precise as possible within the confines of quantum kets is how do I design an experiment which gives me this probability distribution? It's a non-trivial thing. Nonetheless, it tells us intrinsically about what the phase uncertainty is because that's just related to this probability distribution. This is equal to the mu, that's the average phase, and I can calculate those things with this probability distribution. So, there's one final thing I want to say in respect to this, which I haven't touched on, I haven't said a word about this semester. The time of energy uncertainty principle. I haven't mentioned that. Why not? Well, it's really the same thing as we have over here. I can think about what does it mean to talk about time, energy, uncertainty? Well, what do I mean by time? Well, in the context of an oscillation, a simple harmonic oscillator, I can think about as the time over which particle is oscillating, starting from tomorrow. And that time is just whatever the phase of the oscillation is, by the way, omega. So if I had a phase operator, then I'd have a time operator. In fact, we know that energy is the generator of translations in time. Therefore, we might expect that there's a time operator is the generator of translations in energy. We might expect that. In the same way that we have n is the generator of translations in phase, such that we might expect that phase is the generator of translations in number. But we just showed that mathematically we can't do this because there's a bottom to this. You can't translate a bow, a certain way. And the same thing is true about there's no time operator. There's no permission operator I can call the time operator because I can't translate an energy below the ground state. There's no math and math operator that is a permission operator. Nonetheless, we can still talk about in the same way that we talk about measuring the number of phase, we can have an energy time uncertainty. That's to say if I look at, let me multiply this by hr omega. I'm just going to multiply both sides by hr omega. And then if I say delta n is equal to delta e divided by hr omega, and delta phi is equal to delta t over omega. For the reason we just talked about. Omega delta t, I think. What we get is delta e delta t is greater than or less than or equal to hr omega. It should be squared. No, it's not. You have delta e over hr equal to grade equal to hr omega 4, right? There's an hr over here that cancels this. So time energy uncertainty relation is a slippery business. It's never in the same way as number of phase uncertainty is. But it does have a physical meaning. It just means that it doesn't follow from the same kind of measurements of her mission observables. There is no measurement I can do with projected outcomes that are timed. It just doesn't exist back. But that does not mean that I can't measure time. You do it pretty poorly on this thing. But we measure it in a different kind of way. In the same way that we measure phase, for example, in the interferometer, we can do it with the hr mission operator, but we still can measure it and there are distributions and probabilities and they satisfy certain circumstances. What we're going to talk about on Thursday then is what is the state of the quantum harmonic oscillator that really looks like the classical motion? How do we connect the classical motion of the harmonic oscillator to the quantum energy states? They're clearly not the energy eigenstates. They have an uncertain phase. We want something whose phase oscillates. There's some superposition that we'll talk about on Thursday.