 All right. So there is a first problem of the current assignment is posted on the finished exam for the final problem. That will be posted later today. I don't know. And we are having an exam coming up. And it will cover through the material on the Symbol Hermann Gospel, which we will finish today. And you have the second assignment done as well. And I've gotten some feedback, but I'll wait to hear more as I ask the last lecture what your preference is for a take-home exam for an exam of the sort we had for the first exam. A take-home exam has the advantage of much less time pressure, but it'll be a little bit of a more challenging exam. So there are trade-offs. So get your feedback on that. Just send me some email. Very good. So last time we talked about uncertainty relations related to the Symbol Hermann Gospel. So if we look at the position and momentum of the particle in the Symbol Hermann Gospel, then in the number basis, that is the energy island states, the mean position and the mean momentum is here. And they, of course, have, though, fluctuations in those values that is to say, if we were to measure the position of momentum, we wouldn't necessarily get a definite value. We could get some random value about that mean with fluctuations given by the variances, or we want the RMS as the fluctuation. Given that we calculated just looking at base using the operator algebra and the position momentum variances in the energy level is equal to n plus 1 half, something we could have just seen just looking at what we know, of course, the energy eigenvalue is n plus 1 half h bar omega. And there's equal amounts in this. So we could just read it off just from the energy eigenvalues. And so in the ground state, in these dimensionless units, the uncertainty is in the RMS, it's 1 over root 2. So we have 1 over root 2 in dimensionless units of delta x and delta p in the dimensionless units. If we put the units back in, then as we saw, as we know, that those are, that's a minimum uncertainty state. That's to say delta x times delta p is the minimum value allowed by the uncertainty principle. And we saw that also from the fact that in position of momentum space, the wave function was a Gaussian, a Gaussian with delta x being the minimum of delta x if we have a Fourier transform, a limited Gaussian post, it's a minimum uncertainty. I'm going to come back to this in a moment. I want to, I really meant to say this first, so let's talk about this. There's another kind of uncertainty relationship, one that we've not always talked about, but should be. And that's number of phase uncertainty. So if I just, this is a terrible picture, let me try. You get the picture. These are circles. Sorry. Classically, we can think about different sets of canonical coordinates. We have x and p. That's a set of conjugate variables. We also have what are called shin angle variables. If you've ever taken an advanced course in dynamics, you'll learn about those things. They're really just, in the case of a harmonic oscillator, they are the polar coordinates of phase space. So I have an amplitude and a phase. Those are conjugate variables in classical mechanics, just like x and p. And in the context of the harmonic oscillator, we know that we call these things the quadratures of the oscillator. And this is the amplitude and the phase of the oscillator. Now, of course, when we go to the quantum description, then what we said is that this classical amplitude became the annihilation operator, the lower operator, which has, is a non-hermitian operator. It has a hermitian part and an anti-hermitian part. Now, we could try, and it was a natural thing to do, to think about not a decomposition of this operator in terms of its sort of real and imaginary parts, its hermitian and anti-hermitian parts, but in terms of a polar decomposition, which we studied at the beginning of the semester. It's just an exercise and operator algebra. And so we could think about this. It would be natural to think that this could be written as some kind of amplitude and phase. And an amplitude loosely is the square root A, of course, here is the square root of alpha star alpha. And so you might think, well, this is the square root of A dagger A times some e to the i phi operator, the natural thing to do. Actually, it's not quite. I mean, if you were to do the polar decomposition, of course, there's a right and a left unitary as we studied in that homework assignment. So we could write in the polar decomposition, we would say that this is equal to some unitary times the square root. If this is upright on the left, if we're going to write this A dagger A, then I write it right on the left. And this would be unitary. And a unitary, we know, is of the form e to the i times our hermitian operator, which we would call the phase operator. The problem is that this doesn't work. And in infinite dimensions, things are just different about linear operators. There is no polar decomposition of A in this way. There is no hermitian operator that does this, that allows us to make this operative composition. So this is in truth. So there does not exist a phase operator. We cannot write a canonical set of coordinates quantum mechanically that are the equivalent exactly of amplitude and phase. However, well, that doesn't mean that in some sense, amplitude and phase are not conjugate variables quantum mechanically in the same way that position would have to be. There is a generalization. So what we showed last time was the following that I can decompose A. This is true. That there is an operator. I'll call e to the i with a big hat over the whole thing without putting the hat over the phi, because this is not the exponentiation of a hermitian operator, times the square root of the non-operative. And a dagger, I wrote e to the i phi as an operator that does the lowering. The problem is that this is not unitary. It looks unitary. It looks unitary because a unitary operator takes an orthogonal basis to an orthogonal basis. But the fact that this starts at n equals 1 and doesn't go to n equals minus infinity is why this is not unitary. And you could see that. You can just try, if you look at the commutator of this with this, what it equals is this. And since this doesn't commute with its adjoint, it's not a normal operator. Normal operators commute with their adjoints. This doesn't. So it's not unitary. It has no, it's not a unitary operator. And the reason it's not is because this is, if you look back at this picture, it's this, what goes on here. We can't displace beyond this point. We get to the minimum point, and we're stuck. There's kind of singularity that would be all right. There is an approximate sense in which these guys are conjugated. What we can say is that approximately, just like x. There's a very sophisticated mathematics in which we can kind of derive this based on the theory of POVX, which I'm not going to describe in great detail, but just give you a little flavor for it. And that's the following. We can define phase states, states of definite phase. These states were defined to be states which are the equal superposition of all the number of states. So in some sense, this is conjugated in the same way that when we had a position eigenstate, a position eigenstate is a superposition of all the momenta states with equal weight. An momentum eigenstate is a superposition of all the position eigenstates with equal weight. X and P are conjugated. Number and phase are conjugated. The problem with these states is that they're matter of volume. But just look at that. But look at this for two different guys. So this is equal to the sum of 0 to infinity of minus i and minus n prime. Sorry, part of me. And you can find minus i. Is there a minus or a plus there? Should I get that wrong? This is a problem. So that's not a delta function. It would be a delta function if this was from minus infinity to infinity. But it's not a delta. So these are not orthogonal states. They're not the eigenstates of our permission operator. However, what is true is that these states, you can check this, form a resolution of the identity, which means that they form an over-complete basis in the same way that X and P do. Now, since this is a resolution of the identity, this is a PODM. That is to say, I can look at this as an integral over this guy of positive operators. They're not orthogonal projectors. It's not a projected measurement, because these guys are not orthogonal for different files. But that's OK. We learned that not every measurement is an orthogonal projective measurement. We could do measurements that are associated with non orthogonal positive operators, which would say, now I ask you, what would be the probability of finding, or probability density, a finding 5? Yeah. So this is this. That's what we learned sometime in the beginning of the semester when we had Bayley-Faradu. That's the Born rule with PODM, rather than projectors. And of course, if this was a mixed state, we could say this is this, because that's how you find expectation values in the other mixed states. This is the probability. And so there, in principle, exists a measurement. It's not a measurement on a permission observable. But we learned that not every measurement of quantum mechanics is a measurement on a permission observable. So it's perfectly fine to talk about phase and quantum mechanics in the same way, even though there is no phase operator. There is a measurement we can try to rid up. Now it's not easy, and we don't know how to do it necessarily easily in the lab. But it's a perfectly good thing to think about. And that would be equal to, in this case, if it's a pure state, that's equal to this probability distribution is 1 over 2 pi. Sorry. 1 over 2 pi, the phase states, the sum over n equal 0 to infinity. And it follows from this the fact that this probability distribution is like this, that we get from this we can derive for states. So this is that number and phase are conjugated. That is to say, this is approximately true. I wrote a quarter of your last time, and I meant to have it. So let's go back now to dynamics. So nothing we talked about last time was the dynamical evolution of the system for the simple harmonic oscillator. The fundamental object is the time evolution operator. And that has this simple form. We typically ignore this overall phase. It affects nothing. So we typically just write it like that. And what we showed last time, and it's very easy to show, is that the Heisenberg equations of motion for the quantum operators are exactly the classical differential equations. This is exactly the same thing. And the solution is exactly the same as we have in classical dynamics. So one thing we see here, and again, this goes to this number phase uncertainty relation, is that n, the number operator, is the generator of translations in phase. That's to say, classically, what would happen is this thing would just go around, translate in phase. So that's why n is conjugated to phase. And phase generates translations in n. It's conjugated to n. All right. So this seems to indicate that everything about the simple harmonic oscillator is in some sense classical, because the equations of motion are the same. There's subtleties involved there. For example, if I was in a stationary state with some energy eigenvalue n, then the mean position momentum are for all times 0. And that's not classical. So the stationary states aren't really representing the classical evolution, even though we talked about the WKB approximation and the fact that at high n, we see something about the wave function building up near the turning points in the same way that we would expect for a classical oscillator to be. So there are aspects of the classical dynamics that say something about the stationary states. But stationary states don't capture classical evolution. In fact, we kind of know that from the number phase uncertainty principle. n has a definite, I mean, an eigenstate of the number oscillator as a definite n. So it has a completely uncertain phase. In some sense, a stationary state is a state that's in some sense equally distributed around the whole circle with a completely uncertain phase. It has a definite energy, which means it's on this circle. But we don't know where the phase is. Its phase is completely uncertain. So now we come to really the subject of today's lecture, which is, so, we want to ask ourselves, is there some state of the quantum system that in some way we would call quadrilateral? That's to say, it has features that we expect from the classical dynamics. So what would we expect from a quadrilateral? So we seek a quantum state such that the expectation value, say, of x and p, written either in the Schrodinger picture or the Heidegger picture, I don't care, follow the classical trajectory. We might also want not only that the mean value, but what we would like if this is really as close as possible to classical, we would also kind of like it to be the case that the fluctuations in x and p at all times are minimal. That's to say, we want something that's as close as possible as a point in phase space that goes around like this with the minimal possible uncertainty in it. That we would call a quasi-classical state. A state such as the wave packet in the center of the wave packet moves around in phase space along the classical trajectory and the uncertainty is minimal. That we would call a quasi-classical state. Does such a state exist? How would we find it? We can guess. Let's unquantize. What do I mean by that? Well, we quantized by saying the classical amplitude alpha, which we had mistaken, when to a quantum operator. We seek a state that in some sense, its x value is the classical x and its p value is the classical p. We seek a simultaneous eigenfunction of x and p. Now we can't do that because that violates the uncertainty principle. We can't find a state that has both a definite position and a definite momentum. But we want to find a state that is as close as possible to a state that has a definite x and a definite p. Let's look for a state which is an eigenstate of A. If it was an eigenstate of A, it would have an x and a p. So let's look for a state that is an eigenstate of A with eigenvalue alpha. That, in some sense, would have both an x and a p. This is written, this is the x plus i p hat on this state. Is this? Now, there's no guarantee that such a state exists. Why don't I know? Why can't I just say, of course, every operator has eigenvectors, doesn't it? It's not a revision. In fact, it's not only not a revision. It's not normal. A and A dagger don't communicate. Their commutator is one. And we know that the only operators that are guaranteed to have a complete set of eigenvectors are normal operators. And those are the permission operators and the unitary operators. But this is neither permission nor unitary. So we don't know that such a thing exists. In fact, there are no, I will show that thing to you in a moment. But we might have asked for this. And in fact, there are no eigenvectors of A dagger. They don't exist. There are in mathematics. There are eigenvectors, A, but not of A dagger. So I'm going to look for those states in a moment. Well, if such a state exists, if a set of states exist here, what would their properties be? We'll move in a moment what those states are. But why are these states for quasi-classical states? Let's see if they have the properties that I asked of them. What is the expected value of the position? Well, by definition, I said alpha for quasi-classical states is an eigenvector of the annihilation of the group with eigenvalue alpha. So how am I going to evaluate that? Anybody have a suggestion? Do you form the creation of an ally? Sure we can. But there's no, you know, a larger point, see, but let's do it. So this acts on this and gives me alpha. But this acting on that, I don't know. However, yes, but oh, excellent. I can act it to the left, because this is true. So this is alpha plus alpha star root 2, which is acting what we have classically. This is root 2 times the real part of alpha, which is x. The x, if we wrote alpha as root 2 x plus i b. And similarly, in the case you now use the imaginary part, which is b. That's what we expect. And if I wanted to look at this as a function of time, I could do it in the Heisenberg fiction. That's the expected value of position as a function of time. That's how you do it. And we just solve that. This is equal to the expected value of x 0 cos omega t plus b 0 sine omega t. This is the classical trajectory. The mean value of positional momentum as a function of time follows the classical trajectory. The mean value of this wave packet follows the classical trajectory. In some sense, that's always true, because the Heisenberg equation is a motion. It's just that these guys have in the post of the stationary state where these were 0, the mean value at 0 was 0. And then for the mean value at 0, it was always 0. These guys can have any value you like. What about their uncertainty? So let's look at the uncertainty in x, the variance. Well, once again, to do this, we write this in terms of a and a dagger. But we have to ensure that all the a's are on the right and all the a daggers are on the left. Because we know how a dagger acts on this and we know how a acts on that, but we don't know it the other way around. So let's do it. What have we got here? We've got to normally order it, which means put all the a's to the right and all the a daggers to the left. How am I going to do it? Use the computation. Use the commutator. So this thing is equal to a dagger a plus the commutator, which is more. So what do I got here? This acts on this twice as alpha squared. This acts on this twice as alpha star squared. And then I got twice alpha star alpha plus 1. So this is alpha squared plus alpha star squared plus twice alpha star alpha plus 1, which of course is equal to plus a. And this is just x squared, the mean value of x squared. That's what we just found. So what is this? What is it in here? A half. Exactly. And what is it? What's the minimum value? A half. In fact, the only reason this came about was because of the commutator. If there wasn't a commutator, there would be no quantum uncertainty, right? We would just come back to itself. And similarly, delta p squared, you could do the same thing as a half. It's a minimum certainty state. Delta x, delta p is a half. So indeed, this state has exactly the property that we want. It is a state which has its mean value localized at the classical point in phase space and has a uncertainty, which is the minimum uncertainty product, the equal amounts of uncertainty in x and p. And as a function of time, what happens? This thing moves around. Just like the classical trajectory, it's got a little look. It's technically known as the uncertainty bubble. And it just moves around at angular velocity only. These are the quasi-classical states. By the way, these states were examined and brought to the fore most in detail in quantum optics by Orich Lauer in the 1960s. And in the context of quantum optics, and now they're almost always simply called the coherent states, the term coherent is kind of technical. It has to do with up-co-coherence. It's not important why they're called coherent states, but they are typically called coherent states. So they're quasi-classical states for the harmonic oscillation. Yes, sir? So the only superposition of member states that's unaffected by the Lorentz operator, does that have to be an infinite? Like how important is that to have an infinite number of states? So we're going to look at that in a moment. It has a normalizable state. It is a well-defined state. It has probability amplitude over all number states. But it falls off very rapidly. So it's localized in number. We'll talk about that in a moment. The state that is a complete equal superposition of all number states is the phase state. So the phase state you could think about is this state, a state which is localized at a given in faith that has a completely indefinite number. It goes off to infinity. It's localized and has a perfectly well-defined phase, but a completely uncertain number. Completely uncertain energy. It has a definite time, but a completely uncertain energy. This is the time energy that's certain to you. But that always does this a lot nicer than that. Yeah, OK. Now, do you think states exist? There's no guarantee they do. But thank you, and let me show you how. We already know of one state, a state which is an eigenstate of the annihilation of energy. Can you even tell me a state that's an eigenstate of the annihilation of energy? It's beside that. But there's a general state. There's one you've seen, a ray on the state. The ground state, excellent. What if it goes to zero? It's eigenvalue is zero, huh? Exactly. So a acting on this is zero times that, right? That's zero. So this is an eigenvalue. This state is a coherent state. The ground state is a quasi-classical state. It's a state whose mean value is zero. So in this case, alpha equals zero. That is to say, x equals to e is zero. That's the ground state. That's a state that would be, classically, it would be the origin of space space. That would be the state that would do nothing. Quantumly, it's a state which is a uncertainty bubble around the origin. So this is the quasi-classical version of the state that just has no position, no amendment. It's at the bottom of the world. Classically, it would just be at the bottom of the world. Normally, it has its kind of fluctuating. Now, of course, that's a little bit of a confusing picture because it's really a stationary state. But let me just ask you the following question. Suppose I have x and p. And suppose that I have a probability distribution of initial complex amplitudes, initial x and p, that is either mutually symmetric. What would happen? Well, the whole thing would go around at omega t. And it would be stationary. I couldn't tell the difference. So if I had a classical distribution that's to say, I'm not going to tell you what x and p are. I'm going to prepare with my little ball p-hammer and pull it. But I'm not going to tell you what it is. And I'm going to tell you that I'm going to uniformly distribute it around the origin. Well, then at any time, what you would see would be completely stationary. You wouldn't tell the difference. It would all go around. So this is actually equivalent to a classical distribution of possible x and p's that are as a mutually symmetric around the origin. That's a stationary state, classically. OK, so that's one. We're done. We've got none of that. We've just got one of these five console states. How can I make another? Here's the trick. So we call that d. And with the dimensions in there, that was equal to either the i over h for x of p hat minus p x. No, that's what the squeezing operator is. The squeezing operator had a dagger squares and a net square. This was the displacement operator. Remember, we have that either the minus sign p or p x hat minus p hat. Remember that we had that either the i x p hat is a translation in position. That is to say momentum generates translation in position. And e to the plus i p x hat is a translation in momentum. That is to say the position operator translates in momentum. This operator has both. This operator had the following, well, I could write this also in dimensionless units. Characters, momentum characters, the position x hat now have capital p capital x minus capital x. And what is the product of the characteristic position times the characteristic momentum? h r. x times p is h r. This is why not putting the 2s makes this pretty. Otherwise, we'd have a 2 there. And this had the property we showed in homework that this had the property that if you did the unitary transformation on x, it translated into x. And if you did the unitary transformation on p, it translated into p. Or dimensionlessly, it does the same thing. So now I ask you. Let's write it in terms of dimensionless stuff. What is this? k. This is equal to x plus ip, a plus alpha, right? Because x gets translated by the real number x, and p gets translated by the real number. So you have x plus ip, that's alpha. So this is equal to a plus alpha. It translates in phase space. This is a phase space translation. It takes something and translates it by the complex amplitude alpha. So in fact, we can express the displacement operator in terms of the complex amplitude instead of the real number x rs. d and p, if we write that out again, that's equal to i alpha minus alpha star over root 2 times i, that's p, a plus a dagger over root 2. That's x minus alpha plus alpha star over root 2. That's x, a minus a dagger over i root 2. That's p. That's just rewriting the thing. And a little bit of algebra not much shows that this is a dagger plus alpha star over root minus. That's another way of writing the phase space displacement operator. We often write it as p as a function of alpha. So why is this useful? Well, suppose I take the ground state. Let's come back over here and take this ground state, which is a minimum uncertainty state. It has uncertainty and x of p, which are the minimum. And displace it by some alpha. If I do that, is it an eigenstate of the annihilation operator with eigenvalue alpha? If it was, that would be great, because that would just be my state localized at that point with that minimum uncertainty. So let's do it. Let's consider the phase space displacement operator on the ground state. Consider. What is if I apply, I claim that this is equal to alpha. That's to say, this is an eigenstate of the annihilation operator with eigenvalue alpha. How do we prove that? Any suggestions? Yeah, it's a good one, apply the annihilation operator. Let's just do it. So let's take this and apply it to the displacement ground state. We know about the displacement operator and what do we know about it? We know this. So what are we going to do? We can't just apply the data, because that's not the identity. I mean, if we had an equation on both sides, so we can't just willingly apply the operator to life. But what do we know? It's unitary. Yeah, exactly. We know it's unitary. So I can apply the dagger. That's cool. I can do that. That's the identity. And now what is this? That's that. It displaces it by alpha. This then is equal to d on alpha, a plus alpha, right? This is a constant. It comes out. And what about this? Zero. Voila. The displaced ground states are quasi-classical states. So we all start rotating around the origin? Exactly. So if at time equal to zero, I take my particle on the spring and I pull it, and I give it a push or a pull or whatever, or I take my hammer, then I give it an x and a p and let it go. And then it goes, right? What it means, of course, is that the wave packet goes up. The wave packet, which was the Gaussian at the origin, is now a Gaussian here, and it goes to the turning point, and then it turns around. And it was a classic Gaussian. Now, we can see that kind of wave packet picture. We showed this sort of in the Heisenberg picture. Let's look at it in the Schrodinger picture. Let's say, let's look at the time evolution of the wave function given an initial condition that at time t equals zero, the state is this. That's it going to get around. So let's say at time t equals zero, the state is a covariant state. See, the Hamiltonian is a simple harmonic oscillator Hamiltonian, and I wanted to know what is the state at a later time. How am I going to do it? Any suggestions? You did a problem like this in homework with the squeeze state, apply the time evolution operator, right? So this guy is the time evolution operator on that, right? And for the simple harmonic oscillator, now I guess it's going to raise to just here, wasn't it? That is the minus i omega t n hat. Any suggestions? You did something just like this with the squeeze state. The times the acting on the ground state. That's cool. Well, this is the exponentiation of something, right? Can I just put a db dagger on the left side? Yeah. I couldn't do that. Yeah, I could do that. But I think it will be easier. What you said suggests even is it's a fine solution, but it's easier to put this in here. Let me just call this u, and I'm going to stick in a u dagger. Now, u dagger acting on the vacuum is what? It should be u dagger u. It should be, yep, that would be better. So what's u acting on the vacuum? What do I do? Yeah, it doesn't do anything to it. It's a stationary state. So this is equal to u t d in an alpha u dagger. Now here's the trick. We have d over there. So I'm going to write it out. e to the alpha a dagger minus alpha star a. Within the exponent, right? u e to the a u dagger is equal to e to the u a dagger. So this is equal to e to the alpha u a dagger minus alpha star. It does. So we've got to be a little bit careful here, but you're absolutely right. What do we know? What did we know from time evolution? A dagger to a dagger e to the i omega t. So we know from the time evolution of that trick, right? But we got the u and the u dagger on the wrong spot. What's the sign? What's the sign? Because you know that u dagger of t is equal to u of minus t. So this is e to the plus i omega t a. And this is e to the minus i a dagger. And so what we see here is that this is equal to e to the alpha minus i omega t a dagger minus acting on it. The ground state. And what's that? What is this operator? Well, it's not the squeeze operator. It's this operator. It's the displacement operator. But it's d at a function at alpha t, which is just, in other words, the time evolution just makes, still an eigenstate of the annihilation operator. But with the new eigenvalue corresponding to the time evolved classical phasor, that is to say this thing is just a classical phase. So as I was saying, this coherent state just is a new coherent state, which whose mean value follows the classical trajectory. Beautiful. I love it. So now let's come to the question of number and phase uncertainty that a coherent state cannot be an eigenstate of the number operator because it has a well-defined phase. A state with a well-defined phase cannot have a well-defined number. What about the number? So is this an eigenstate? If I look at the number operator on this, it's not an eigenstate, right? Because that's a dagger a acting on that, which is alpha a dagger on that, which is not an eigenstate. So alpha is not an eigenstate of the number operator. What is the mean value? This acts on this alpha, this acts on this alpha star. It's what you expect. It's the classical amplitude squared. So the mean number is alpha squared. But what is the fluctuation in the number? It's not an eigenstate of the number operator. It has fluctuations around the mean. It has uncertainty in the number. I don't know what is the uncertainty in it. That's the expected value of a squared minus the mean value squared. How am I going to evaluate this? What would be a dagger a times a dagger a? That's what it is. Now what? This is alpha squared. That's fine. That's this times squared. That's a dagger a on the outside. The alpha squared turns on the outside. Well, I can't do that, right? I have to be careful because I can't. No, you can look on the right and then on the left. Sure. Then do one minus a dagger a. Sure. So that's the key. The key here is to always normally order it. Commute these things so that, right? This over here is a dagger a plus one. So this is equal to a dagger squared a squared alpha plus one times that, which is... And this is alpha squared squared. And this is alpha squared. And then just alpha squared. Exactly. Which is equal to the mean value of n squared. And then that's n. Sorry. Thank you. All right. So the point here, what we see is that for a coherence state, we don't have the eigenstate of number. There's number fluctuation. The mean value of n is equal to alpha squared. The uncertainty in n is equal to the square root. Because the square root is equal. So given this fact, do you know of a probability distribution whose fluctuations, this is correct, whose fluctuations are equal to the square root of the mean? Plus one. Plus one. This is corresponding to a plus one distribution. So how do we see that? That this is plus one. We want to find the probability distribution in the number. I want to know the probability distribution of having a certain number of excitations of my oscillator. How do I find that? Well, what is it by definition? This is... I have... I wonder what is the probability to find the particle at the level n given that it's in a coherence state? I just want to get to one more. So I want to know what is the expansion of the coherence state in the number basis? What's the probability of having that? So let's look at those amplitudes. Well, that's going to be complicated. Because the way we define alpha, alpha is the displacement operator acting on the vacuum, right? That's how we... This is either the alpha a dagger minus alpha star a. Now here's a trick. Remember, we have this is equal to... We can separate these guys when any of the commutes make their commutes. Right? And that is the case here. So the displacement operator using this rule is e to the minus 1 half alpha squared e to the alpha a dagger e to the minus alpha star. So with that said, the coherence state is equal to e to the minus alpha squared e to the alpha a dagger e to the minus alpha star a. Now what? Well, a acting on this gives me zero. So every term in this will be e to the zero. So that's one. This I can use in Fridays with Taylor series. You know that when you apply the creation operator or the raising operator n times, you get the nth eigenvalue. I can say this n over the square root of n factorial. And that is that. So this is equal to the sum n equals zero to infinity alpha to the n over the square root of n factorial e to the minus alpha squared. And this is... So what is the probability distribution? It's the square of that. The probability of seeing n and citations is the square of that amplitude, which equals alpha squared to the power n e to the minus alpha squared over n factorial. Alpha squared is the mean n. This is equal to e to the minus the average n average n to the power n over n factorial. Do you recognize that probability distribution? That's a Poisson distribution. So for a coherent state, we don't have a definite number. We have number fluctuations. The number fluctuations go like the square root of the mean. That's Poissonian. Those fluctuations relative to the mean become smaller and smaller as the mean gets bigger and bigger, it becomes more and more classical. But nonetheless, they're always there. There's always... What about the phase? Let's just conclude with the phase. Now that's complicated because we have to calculate this thing. So if I put that in over here, that's equal to the sum and equals zero to infinity. Then I have this distribution over here. I have e to the minus alpha squared over two over two the square root of n factorial alpha e to the minus i5 to the power n. Now, what we can say though is that this sum is going to have these very rapid oscillations except where the phase of this cancels the phase of that. As I say, if I wrote alpha at some phase at zero, then this sum is going to be very big. When the phase is away from this phase, this is going to have lots of oscillations and those terms will tend to cancel. So we kind of expect this probability distribution to be peaked around a phase of alpha. In fact, it is. And if I were to draw this picture here, say here's my coherent state, it has a mean n. It has this uncertainty bubble has a width square root of two. Then I ask you, what is the width in phase? Now, it's going to depend, the width in phase is going to depend on n, right? The bigger it is, it's way out there in the same, it's going to be much smaller. So what is the uncertainty? Just geometrically. Roo two, square root of n. Mean value of n. So that tells me that for a coherent state of fluctuations in phase, my uncertainty phase are equal to a half the mean value of squared. But that's equal to a half times the uncertainty of n. So this is a minimal uncertainty wave packet. That's the same graph story. That's to say delta n, I know those factors are two here. Maybe it's twice, I should call phase twice that. But the point here is that this is, you know, of order. It's a minimal uncertainty packet. So the quasi-classical states are toward states and thinking about the fact that in the harmonic oscillator we can create wave packets that go just like the classical trajectories. And they are way, they are important particularly in thinking about quantum fields. Because if I wanted to think about the state of a field which is an oscillator, the field states which are classical like, classical fields are coherent states. The state of the quantum magnetic field that comes out of a radio antenna or out of a laser beam is described by a coherent state. So that concludes our discussion on the simple harmonic oscillator. We'll have some problems on that.