 The main results of the presentation I gave in the last lecture on the direct-algebraic method for solving the hydrogen atom in minus ulcer this morning. The direct-algebraic method for solving the harmonic oscillator I think you're probably familiar with most of this. The essential ingredients of the creation and annihilation operators, there are matrix elements with respect to the energy in the state. The expression of the Hamiltonian in terms of the creation and annihilation operators also solved with the position of momentum in terms of the A's and A's daggers, just the inversion of the definitions, which is also sometimes useful. All right. Now, I'd like to begin today with the subject of wave functions. In particular, let's talk about the ground state wave function, zero is the ground state. And if I form this camera product with the position I can ask, we definitely cannot consider it to be the ground state wave function called side zero of S. The question is, what is side zero of S? Well, this follows easily from the fact that the annihilation operator interacting with the ground state annihilates in just gives you zero. The annihilation operator, however, is X plus I t over the square root of 2. And if we put that in the wave functional language, X plus I t goes over into the X, of course, as both of them have to emphasize these are operators. X goes into the multiplication by X, T goes into minus I d dx. So excuse me. So as an operator, acting on the wave function, this goes into X plus d dx. And therefore, the ground state wave function satisfies the differential equation that X plus d dx acting on the side zero of X is equal to zero. It's a simple differential equation. And it's easy to solve. And if you solve it, and if you normalize the result, you find that side zero of X is 1 over pi in the one-quarter power times e to the minus X squared over 2, pi stuff is the normalization. And this is the ground state wave function, which is, I guess you know, as a Gaussian function, a function of X, it's a Gaussian function centered before it connects to the ground state of the chromatic oscillator, side zero of X, here like this. Now, one of the things we're going to be interested in, many of our attention to today, is expectation values of position momentum, as well as versions, which are expectation values of first moments, both of the interesting and second moments, which are related to the dispersions. And in particular, the Q piece for the ground state is easy to show that the expectation value of X and the expectation value of momentum here, both equal to zero, which is before the ground state, for the ground state. And one can also show quite easily that the dispersion delta X is equal to dispersion also in momentum of P, which is 1 over the square root of 2. This means, obviously, the dispersions are equal, and it also means that the product of delta X times delta P is equal to 1 half. This is all the units in where h bar is equal to 1. So this would be h bar over 2 at ordinary units. And this is the minimum value of delta X and delta P, which are allowed by the uncertainty principle. So the result of this is the ground state of the harmonic oscillator is actually a minimum uncertainty wave packet. It's a Gaussian wave packet. All minimum uncertainty wave packets in terms of delta X and delta P have to be Gaussian, but this is a particular example of it. And so that's an interesting conclusion also. While I'm talking about momentum, by the way, it's interesting to take the Fourier transform of a size 0 in order to obtain a momentum space wave function, phi 0 of P. That's straightforward. It leads to the same 1 over the pi over the 1 quarter normalization, and then it becomes p to the minus p squared over 2. The result is, as you see, that for the ground state anyway, the momentum space wave function has the same functional form as the configuration space wave function. This is an example of a wave function which goes into itself in a Fourier transform. All right. Now, so those are some basic facts about the ground state. Now, let's take a look at the excited states. Let's see, maybe before I do that, let me just provide some motivation for where we're going here. The harmonic oscillator is a particularly useful system for exploring and understanding the correspondence in classical mechanics, to understand how far it goes and what needs to be changed by going to quantum mechanics. Makes the relationship between the two particularly clear. And that's really going to be the main theme that we'll be following in the rest of the lecture today, is to understand the relation in classical and quantum mechanics. Here's just one thing to say, is that because this ground state of the harmonic oscillator has these dispersions, this is obviously related to the statistical nature of a quantum measurement. We can visualize the ground state of the harmonic oscillator in quasi-classical terms in the following way. If I plot the classical phase space x and p, then they can think of the ground state of the harmonic oscillator as a kind of blob or a distribution, which is centered in the origin in base space. It's centered at the origin because the expectation, the average value is x and p are 0. And, but on the other hand, it has some spread around it because of non-zero dispersions. Spread of course is necessary by quantum mechanics because x and p don't commute. And so they can't be measured simultaneously with infinite precision. This blob that I've drawn here is not intended to be a rigorous representation of ground state. It's really intended just for the purposes of intuition to give you an idea, represents in some way the statistics of measurement that come out of the ground state rate function. One could make this more precise by wanting the Wigner function, which was actually a part of last week's homework problem. But I won't go into that. I'll just use this for the illustrative purposes. All right. Anyway, that's what kind of the ground state looks like. Now, if we're thinking about the classical limit of the harmonic oscillator, or if that matter ain't one of the problem hydrogen atom, one is struck by the fact that in classical mechanics you have particles moving on orbits. And in quantum mechanics you've got these energy eigenfunctions. At least that's what people talk about most of the time. And an energy eigenfunction doesn't look at all like a particle moving in an orbit. And the question is, why? What's the difference in how we understand this? Well, just for the case of the harmonic oscillator, let's shift the tension down to the excited states in, say in is one of the excited states. And let's consider the expectation values of position and momentum for the excited states, just to see, because expectation values of position and momentum are ordinary numbers, and they could sort of look like classical quantities. So let's consider the expectation value of position. Let's say in an energy eigenstate. And I'll put a hat on the x to emphasize that that's an operator. Similarly, we can do this for the momentum. Look at the average value of momentum in an energy eigenstate. By the way, my use of the hats here is that I tend to put hats on something to indicate that it's an operator when there's some danger of confusion between the operator and something else like an eigenvalue or a classical quantity. And I tend not to do it when there's no danger of confusion. That's the rule for the hats for the time being. Well, in any case, these expectation values are easy to work out. Because the easiest way to do it is to write in terms of a's and a's and a's and a's and a's and a's and if you do that, you'll find quite easily that both these expectation values are zero. So as far as expectation values are concerned, an energy eigenstate doesn't look at all like a classical particle in an orbit. If you have a classical harmonic oscillator and you give it some definite energy, it's going to swing back and forth within amplitude. So the question is, where is that classical time dependence in the quantum problem? It's not appearing here in the eigenstates at all. In fact, the energy eigenstates of quantum mechanics are what they call stationary states. That means the probability density in configuration space is independent of time. And so any expectation value you take of any operator is going to be independent of time in an energy eigenstate. Therefore, there is no time evolution at all. In particular, it stays zero for all time. So it's disappointing if you're trying to make a connection with a classical orbit. Now, how can we understand this from an intuitive standpoint? Here's one way to do this. Let's go back to the classical problem. Here's a classical phase space like this. The orbit of the classical harmonic oscillator in phase space is a circle, something like so. And let's say we've got some initial condition called x0p0, which lies on this circle. Then the motion is a clockwise motion around the circle, this classical motion. So a clockwise motion around the circle. Let's suppose in a later time, we've got x of t and p of t. Another point in the circle, when there's an angle in between, and that angle is actually equal to the time. Or if we want to restore ordinary units, then the time that's replaced by omega t is basically the phase angle of the harmonic oscillator. Simply speaking, this looks like the hands of a clock even go through the right direction clockwise. All right. Now, so what this suggests is, OK, so the question is, why do we have expectation values at zero for the energy eigenstate? Here's an intuitive way of understanding it. It's related to the time, energy, and uncertainty relation. It says that the product of delta e is equal to, or greater than 8 to the h bar over 2. This looks just like the delta x, delta p uncertainty relation for position of momentum. We haven't really talked about time energy uncertainty relations yet, but we did talk about delta x and delta p. And in fact, the mathematics is deriving this is quite similar to what she used for the delta x, delta p uncertainty relations. But the physical interpretation is different, because t is not an operator. Whereas x and p are, at least in the non-oldivistic theory, that's how it is. And so the interpretation of this uncertainty relation actually depends on the context, and it's very various from one application to another. But in the present application, it means the following. Is that if we have an energy eigenstate, then delta e is equal to 0, because in an energy eigenstate, the energy is exactly low. And in some sense, that means that delta t must be less than infinity. The sense in which that's true is that the idea is that instead of having something that's localized on a classical orbit, which you could tell the time if you followed that at a point in terms of 2 o'clock, it's like in order to make a measurement of energy, precise measurement of energy so the energy is known exactly, the measurement process must extend over an infinite amount of time. That's another way of saying it. So if we look at the classical orbit and we try to measure over an infinite amount of time and blur our time clock, what happens is this particle gets smeared out over the entire orbit in a uniform manner. And if we do that, then even the classical average value is a maximum theory of 0, because it averages to 0 over this oscillation. And that, in some sense, is the reason why we have these expectation values. What it shows is that if we want to get some quantum object that looks like the classical motion, we can't use energy eigenstates. We've got to use a linear combination of energy eigenstates, then those different terms, and the expansion will have different time evolutions, and we'll get a non-trivial time evolution for the quantum state. So in particular, let's think about putting a minimum uncertainty in the white packet, let's say in some different position in the phase space, some different expectation value besides 0 and 0, different from what we have in the ground state of the harmonic oscillator. All right, that's kind of an introduction to where I want to go for the rest of the lecture. I'm sorry I have to cover this up now. But that's the location. So now this takes me to the next topic that I want to tell you about, which is coherent states. First of all, the definition of a coherent state in the context of mechanical oscillators, like we're talking about now, a coherent state is a wave function, which is, first of all, satisfies the fact that this version in both position and momentum are equal, and they're equal to 1 over the square root of 2. And this means in particular, these are minimum uncertainty, these are minimum uncertainty wave packets. This would be h bar over 2 higher stored ordinary units. But for simplicity, I'm going to get rid of the physical constants by choice of units in this convention earlier, so it's called 1-hash. All right, so coherent states are all minimum uncertainty wave packets. And also, as far as the expectation value position in momentum are concerned, these are allowed to be anything. But in fact, the expectation values of X and P are parameters of coherent state. And you can see in particular that the Brown state and the harmonic oscillators, an example of coherent states, what are these expectation values are 0 and 0. But what we like to do then is to take this, in fact, take this Brown state and the harmonic oscillator, and think of them in a space like this, and we like to move it to other places in the base space to change these expectation values. Now, the nice way of doing that is to use translation operators. The translation operators we introduced before over the top of the board, excuse me for covering things again, let me draw your attention to them. These are the translation operators that, as we defined previously, the first slide here is the definition of the translation operator. This is in one dimension, and the A here is the displacement for the translation goes into the violation operator. I'm sorry, it's the same symbol. The second line gives the effect of the translation operative wave functions, and the third line gives the expression of the translation operator in terms of its generator, which is the momentum. Putting a hat here to indicate that's the momentum operator, the A here is an ordinary number or a C number, as the rat would say, multiplying p as well as displacement. Anyway, these are the things we've covered before. All right. Now, the first question I'd like to address is how does the translation operator affect expectation values? So let's suppose we've got a state psi, and this has an expectation value called Lx0, which is just psi sandwiched from x hat. Let's define a new state called psi prime, where it's just a translation operator, T of A, acting on psi. Well, let's consider the expectation value position operator with respect to the prime state. How does that change compared to the expectation value of the O prime state? Well, you can easily see by just substituting in here the definition of psi prime, this is the same thing as psi times T of A dagger times x hat times T of A times psi like this. And so working on the answer, it's going to involve working out this conjugation relation of the position operator by means of the translation operator. So let's make a place over here where we're going to write the answer, T of A dagger x hat T of A is equal to, and we'll fill this in in just a moment. The easiest way to work out this conjugation relation is just to use this second line here for what translation operator suits the wave functions. We'll sketch it for you. The wave function is psi of x, so we want to apply T, translation operator, first in multiplication by x and then the initial translation. So first of all, if I apply T of A, to this, this comes up with psi of x minus A. And if I then apply x hat, that's multiplication by x, so this goes there when the x times psi of x minus A. If I then apply T of A, well, it's really T of A dagger. But these are unitary operators. So T of A dagger is T of A inverse. And that's the same with T of minus A. So I'm going to write it this way. T of minus A, which is T of A dagger. So here I use this middle rule here except I change the psi of A, so it's x psi of x plus A. So x here has to be replaced by x plus A. It goes in the x plus A, the multiplying of psi of x. And this is the same thing as an operator x hat plus A acting the way function psi, which is then evaluated at x. And so this is a sequence here that shows you the effect of these three operators. And it shows you that the product of these three operators is the same as x hat plus A right here. So there you are. There's the conjugation relation. And now if I take expectation values of both sides, which is speculated by the initial state psi here, you can see this turns into the original expectation value of x hat 0 plus A. So expectation value is shifted by forward by A under a translation. Now this is pretty obvious because the translation I've already done is that the way function is just pushes it down to the x-axis. And so expectation value is having shifted in the same way. Anyway, while we're at it, let's also look at momentum. What does T of A do to the momentum operator when we conjugate? And the answer is it doesn't do anything because T of A is a function of momentum. That's so it can use with it. So I can bring the T of A dagger past the p or it cancels itself out and just take p. This is a position space translation operator and it doesn't do anything to the momentum or only to position variables. What does it do to dispersions? Delta x and delta p, what happens to them? It's delta x going to under translation. Well, the dispersion, we just learned the statistics. It's a measure of the dispersion about the mean. It's really a difference between the mean and the actual value of x as an average of the square root of the difference square root is the RMS difference. And the translation affects the x, but it also affects expectation values in the same way. And so the difference between the two cancels is out when you do a translation. Translations don't affect dispersions. So delta x goes in the delta x and delta p goes in the delta p. Now, this is important because it means that if you have a minimum uncertainty wave pack, that the translation operator maintains the minimum uncertainty condition that doesn't change at all. The only thing that changes is the expectation values. All right. And so, in particular, we took the ground state of the harmonic oscillator in this space-based picture or image. I'm just going to blog here at the board and here's the ground state zero right there. And we apply a translation operator to it to get the state of t of a acting on the ground state zero. We can think of this as a blog here that's centered in a position that positions x equals a on the x-axis. In fact, it will just be that Gaussian wave bunch which is just that Gaussian root value of x-axis. That's all it is. But let's look at it in a face-face and think of it this way. So by doing this and using t of a, you can change the expectation value of x. That's not quite what we wanted, though. We wanted to change the expectation value of both. See, do I have it somewhere here? Both position x and y now because we're interested in starting some kind of wave packet out of the orbit here in an arbitrary initial. I sort of have a p zero. So to take care of the momentum translation, allow me now to introduce momentum translation operators. Think I can erase this part here to make this u table right next to the first one. Allow me to define. This is just a definition. I'll call it s of v. I'll use s just to make it a different letter from t. But this is a translation operator momentum. And this is defined as a definition of it. It's e to the plus i v times x hat. v here is just a number and it's going to be the momentum displacement and the x hat is the operator. It sort of looks like this except x and v have been swapped and the minus sign has been answered by the number used. So let's see what s of v does to wave functions. So first of all, if I take s of v and let it act on a wave function psi and evaluate it at x. Well, s of v is a function of the operator x hat which is just a multiplication by x on an x space wave function. So this is just the same thing as e to the i v x times psi of x. Both by psi of x by a position dependent phase factor. What does this do to momentum space wave functions? Let's call this psi prime of x. What does it do to momentum space wave functions? Let's write phi prime of p to be the momentum space wave function corresponding to psi prime of x. Well, that's the Fourier transform and it will be x over square root of 2 pi h bar and then it's e to the minus i p x h bar is 1 versus just 2 pi square root of 2 pi h bar. e to the minus i p x and then we need to multiply by psi prime of x which is in the line above here and I'll write e to the i v x times psi of x when we place these psi prime of x like this. Now you can see that this is the same as the Fourier transform of psi of x except the parameter p has been replaced by q minus b. So this is the same thing as phi of q minus b where phi is the Fourier transform of the original psi. So the net effectiveness is that I'll write this this way. So s and b hacking on a momentum space wave function as a function of p is equal to phi of q minus b. Notice how it's analogous to the position space translation operator. Notice that this is an effect that moves the momentum space wave function down the momentum axis by displacement of p. Likewise, one can easily show that s and b can act a lot in a momentum of the state p, and that's the key plus b. This way we have a parallel now between parallel set of coordinates in position and momentum displacements. So if we took our Brown state wave function and we're displacement in x, we can then follow that by displacement in p that's called as s of b times t of a acting on zero. We can filter it as a blob in face space which is centered on position coordinates a, b like this. Before I say more about this, let me go back to the question of expectation values which I've covered for the translation operators t of a over this panel. What does this new momentum space displacement operator, what does that do to expectation values? Well it's an obvious analog of what t of a does. It works like this. I kind of like put it over near this so I'm not going to erase this. It works like this. Because that s of b, dagger s of b, s of b dagger x times s of b so we conjugate the position operator by the momentum displacement. That just does nothing to it. The reason is that s of b is a function of x hat, so it commutes from there. That's just like this formula over here with t of a of p. If we take s of b dagger using the particle momentum operator and it goes into itself plus the displacement which just indicates how momentum space expectation values are shifted forward. And then finally we have the effect on dispersions, deltas and deltaki, both of them going into the cell just to say they did not change for the same reason that they did in the case of the X-paces displacement. That's a yes question. No, the momentum position translation operators do not commute and that's the effect going to be by next point. The just one remark before I enter that is that by using these products, these operators, we can in effect take this ground state with a harmonic oscillator which is an example in the current state and move its expectation values ground without changing and thus maintaining the minimum uncertainty wave packet property of these sort of equations in these states. So by doing this we can create what we call coherent states with any expectation values of the position momentum in which there's a two parameter camera that can effect cover all the base space. However, because of this point that was raised here a minute ago there is an issue we need to deal with which is that getting with this point a d in base space direction and then p direction what if we went the other direction applied momentum translation operators first and then position operators second would we get the same answer? The answer is no because these operators t of a and s and p don't commute with each other. That's no surprise because they're exponentials of operators namely p hat and x hat that don't commute. So there's a difference between these two paths and getting there. Turns out the difference between the two paths is only a phase factor and some people would say that's not very important but nevertheless I'm going to devote some attention to giving a more symmetrical way of getting the final location that hits around the dual to the space factor and the manner which is symmetrical under exchanges of x and p. This is the idea of what we're doing now about creating coherent states and the basic idea of coherent states is that there are minimum and certain wave packets based based parameterized by the expectation values of position of momentum. By the way before I go further with coherent states let me say that in current physics I think there's two main uses for coherent states one of them is really conceptual or theoretical if they're used for exploring the classical limit. The reason is that a coherent state has a minimum dispersion to an x and p we know because the uncertainty principle is that you can't make measurements of position of momentum with infinite precision simultaneously as you can in classical mechanics. However a coherent state is as close as you can come to that and so in some sense it's the quantum object and some people think of it this way it's the quantum object that this is as close as possible to a single point of a definite classical state. And so the use of all kinds of studies, very common in literature nowadays to use them for this purpose is used for studies trying to understand the classical limit of quantum theories. There's a second purpose of coherent states which is in the case of quantum objects they are they are quantum states that can actually be produced with lasers and they're different than these energy eigenstates which in the case of lasers are states of a definite number of photons coherent states have a variable number of photons that's not an eigenstate and a number operator but in the case they're interesting and experimentally for things that one can do with them I'm mainly here going to concentrate on the use of them for understanding classical correspondence alright in any case the main problem I want to address now is the non-computativity of these operators S of A and S of B excuse me, S of B and T of A in order to work around this I'm going to find a new operator which is called a Heisenberg operator and it's just kind of a combination of translations and positions of momentum taken at the same time we'll call it W it's parameterized by both A and B and this is a defined and behavior with I times B Bx hat minus AP hat so it looks like the T of A and S of B are the exponents of mind that's all it is however this is not equal to E to the I Bx hat times E to the minus I AP which is the same thing as S of B times T of A it's not equal to it because X and P don't commute if X and P were ordinary numbers which do commute then the usual rule of explanations would give you an equality but because it ordering matters this is actually not true there's some other stuff that has to be taken into account however the Heisenberg operators can be expressed in terms of these S and T operators it's not too complicated and the way to get from one to the other is to use what is what's called Glauber's theorem which I'll now tell you about Glauber's theorem has rather restricted conditions on it so it's not applicable in many circumstances but this happens to be one of them so it's useful for this longer by the way I see physicists in Harvard won a Nobel Prize for this and worked on quantum optics which they did in the 1960s alright it is all related to this kind of stuff alright Glauber's theorem says this is that if you have two operators A and B which commute with their commutator then there's a nice identity that says that e to the a times e to the b is equal to e to the a plus b that's all you'd have if they commuted but then there's an extra term which is one half the commutator of A and B so in particular if they do commute the commutator is zero you're just going to be usual rules of exponentials alright let me go over here and prove this give me the proof let's define, this proof will follow what we did in the first Homer problem one of the first Homer problems first Homer set let's define an operator which depends on a parameter lambda which is equal to e to the lambda a times e to the lambda b and we'll get a differential equation for it I think for edge a df to the lambda is equal to a df to the lambda b plus e to the lambda a times b times b times e to the lambda b like this by the way let's note that f of zero is equal to the identity operator it's kind of an additional condition for this differential equation we're going to get now in this line here for df to the lambda I brought these operators a and b down to the left when I did the differentiation in this expression for this derivative allow me to insert here between the b and e to the lambda b we put in e to the minus lambda a times e to the plus lambda a which of course cancels out and the reason for doing that is that we get on the right hand side here in this term e to the lambda a e to the lambda b and likewise in this term e to the lambda a e to the lambda b in the same term and that's just the same thing as f itself so effectively we've factored out after the right of this and so this becomes this expression of a plus e to the lambda a times b times e to the minus lambda a the whole thing multiplied on to f so this is our differential equation now at this point we haven't made any assumptions about a and b all a and b but again going back to that first homework exercise this product here which is exponential is conjugated around b this can be developed into a power series of iterated commutators in particular it's equal to b plus the commutator of here's a multiply by lambda so it's lambda the commutator of a with b plus lambda squared between factorial times the iterated commutator of a is the commutator of a with b plus dot dot dot infinite series like this and that's also almost true however while our theorem says we've been assumed that a and b can use with their commutator in particular that means that this lambda squared term goes to zero in fact all the higher order terms are all the iterated commutators for zero and so we're just left with two terms of this series and so this gives us then to rewrite the differential equation as the f of lambda is equal to a plus b plus the lambda times the commutator of a with b times f of lambda now if this differential equation were numbers instead of operators it would be easy to solve it would be that f of lambda would be equal to lambda times a plus b plus lambda squared number two times the commutator of a with b this is assuming f of zero is equal to one but we have to put a question mark on this until we can check to see that this is true even for the case of non-commuting operators well in fact it will be true for non-commuting operators if the derivative of the exponent commutes the exponent itself the derivative of the exponent is a plus b plus the lambda times the commutator of a with b and you can see that you commute a plus b in operator commutes with itself so a plus b commutes with a plus b a plus b commutes with a commutator because that's what we're assuming commutator commutes with a plus b and the commutator commutes with itself so the derivative of the exponent commutes with the exponent and so in fact the solution is correct this works and this is the solution for this function that the lambda now finally all we need to do is just set lambda equals to one and it turns into the derivative which is stated over here that's the proof of the derivative and it appears so much better proof anymore let's erase that let's apply the derivative so let's do this let's let a be equal to i v x hat and let's let b be equal to minus i a key hat using these products of these displacements in the momentum and position then the commutator of a with b is equal to i times minus i is one more case a times lower case p all right down what's left is the commutator of x hat with p hat but that's equal to i as far as one here so this just becomes i a b commutator at least to operators is constant it's a c number c numbers can mean with everything so in particular they can mean with a and b and so the conditions of lovers pair are matched and the result of this is that this operator e to the i v x hat times e to the minus i a key hat is equal to e to the i v x hat minus a p hat combined together in the x component plus i a b over 2 that's the result of the product today momentum times a position displacement you see it's a Heisenberg operator in the first term we find up here and then a correction term which is a face factor in fact since this is a face factor it commutes with everything and I can just put it as e to the i as a separate face factor here and bring it back over to the other side and if we do this then we get this result is that the Heisenberg operator w a b is equal to s of b times t of a or even the face factor first e to the minus i a b over 2 times s of b times t of a and so this then it allows us to express the Heisenberg operator in terms of position momentum splices alright now having done that let's now define a general coherent state which is going to be parameterized by its expectation values a and b these are expectation values in position momentum this will be defined as the Heisenberg operator w of a b acting on the ground state and if we work out the wave function of this let's call it psi a b of x which is equal to the scalar product of x with the state a b like this it's straightforward to work this out by just applying the position momentum of translations we'll find that this is the same normalization whatever pi is in one quarter e to the minus x minus a squared over 2 that's the x shift in that when you get plus i b x that's the momentum shift and then minus i a b over 2 which is this extra face factor here and so this then is the wave function they compare and state minimum uncertainty wave packet with given expectation values of x and p think of it as a blob in base space and by using the Heisenberg operators it's symmetrically treated in position momentum that's where this extra face goes from now go back to our original problem then which is we're back to this board what we'd like to do and understand the classical rhythm is to take the initial condition for the quantum harmonic oscillator which is the coherent state centered in some position x zero and p zero I've been calling this a and b but now I'll call it x zero and p zero because we'll think of it as an initial point on a classical orbit like this in the classical base space and we'd like to do the time evolution of classical mechanics maybe before I do that let me say something about the time evolution of classical mechanics while I already showed you what it is just a rotation by an angle but there's a convenient way of expressing this that I'd like to tell you about let us to introduce complex coordinates in the classical base space let's introduce a coordinate I'll call z which is defined to be x plus i p with a square root of two and except for the square root of two this is really identifying the base plane x p plane with a complex plane so it's the real and imaginary parts of z square root of two is just convenient to see here you might notice however that this is a class of definition now so the x is a piece of the square root of zero which you may notice that this is a classical formula which is exactly the same as the definition of the unilation operator in quantum mechanics so the z is the classical analog of the unilation operator likewise let's introduce the complex conjugate of z which is equal to x minus i p over the square root of two I'll note the complex conjugate with a bar here so that's like the a dagger alright now if we do this then this x zero p zero can be translated into a complex initial condition called z zero and this x of t and p of t can be translated into a complex final position called z of t and it's easy to see what the solution of the classical equations is it's in z of t it's equal to it's just the phase factor which is e to the minus i t multiplying on the z to the zero in complex variables the time of evolution can become really simple by the way I might mention that this is a classical analog the solution of the Heisenberg equation is the motion of quantum mechanics for the unilation operator if you solve those which you find this is an a of t is equal to e to the minus i t times a is zero this is in the Heisenberg picture exactly the same formulas no wonder the equations are exactly the same too except they're reinterpretation as soon as alright anyway this is a useful way of talking about the classical solution in terms of these complex in terms of these complex coordinates now if I may go back to the where do I have it these Heisenberg operators are they here allow me to make some further slight changes in notation this is really pretty simple let's first of all let's take a and replace it by x zero and b and replace it by p zero because we're going to be thinking about these displacements as moving us to some additional condition and then we've got a Heisenberg operator which is parameterized by x zero and p zero and just as by the way in notation we can also write this as w of z zero so z zero is the complex coordinate x zero plus i of b zero over the square root of two just a different way of parameterizing the same operator but similarly as far as the coherent state itself is concerned which I call a b over here let's call x zero p zero is w of x zero p zero applied to the ground state zero and let's also just define this to be z zero for short again parameterizing the coherent state also by the same complex number complex coordinate in the phase space so let's do that now then here then is the problem that we'd like to solve as we like to say suppose the psi initial time zero is equal to this initial coherent state the problem is what is the psi a later time t equal to this is an initial value problem for the Schrodinger equation by the way according to the error test relations which we went through before the expectation values follow the classical motion they do because the Hamiltonian is a quadratic function the potential is a quadratic function and so in particular we know that whatever this solution is the expectation values of x and p will follow this is the same classical orbit here however, that still leads up in the question about whether the initial coherent state remains a coherent state does it remain a minimum and certainly weight packet actually the dispersion only involves the second moment what we'd really like to do is to solve Schrodinger equation and get the whole solution and the quantum problem so it's actually more than just looking at expectation values at least we know something already about the solution excuse me, from the error test relations alright, so anyway to go back and this is going to be the problem we'd like to address so the answer of course is psi of t is the unitary time evolution of u of t acting on this initial coherent state z zero and this of course is the Hamiltonian given the minus i t h what if h had an operator acting on z zero the strategy I'm going to use for solving this problem is to decompose the initial coherent state that location z zero is the linear combination of harmonic oscillator eigenstates and then we'll evolve that forward in time by the usual method of basically normal load at the time so the first task is to take the initial coherent state and to decompose it into an energy eigenstates so here's what we've got we've got z zero then is equal to the same thing as x zero p zero which is the same thing as w of x zero p zero acting on the ground state which is the same thing as e of i p zero x hat minus x zero p hat applied to the ground state that's the Heisenberg operator now allow me to do this let's go from the x zero p zero or z zero as ordinary numbers and the x hat and p hat are operators let's express these in terms of their corresponding complexified versions so the x hat is equal to a plus a dagger over the square root of two p hat is equal to a minus a dagger over i the square root of two x zero is equal to z plus z bar over the square root of two this is the classical correspondence and then we've got p zero is equal to c minus z bar over i the square root of two this is just using defining relations to go over from x's and p's to a's and a's daggers and one of the operators or z's and z bars in the classical picture I should put knots on the z's here because it referred to the initial conditions at zero and p zero and if you do this and you transform this exponent in terms of this in terms of e to the power of z zero z zero times a dagger minus z zero bar times a applied to the ground state it's just the algebra to transform that exponent for these equations if you do it in terms of this so this is what we need to evaluate now for this purpose I'm going to use Flouder's theorem again let's consider the product z zero a dagger times even the minus z zero bar times a and let's let our first operator a be z zero a dagger capital A z zero times annihilation operator a dagger capital D is equal to minus z zero bar times annihilation operator a