13. Quantum Dynamics (continued)
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Zwiebach, Barton
In this lecture, the professor continued to talk about the Schrodinger equation and discussed Heisenberg dynamics.
 The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. OK, so let me go back to what we're doing. What the plan for today is as follows. We're going to look at this unitary time evolution and calculate this operator, U, given the Hamiltonian. That will be the first order of business today. Then we will look at the Heisenberg picture of quantum mechanics. And the Heisenberg picture of quantum mechanics is one where the operators, the Schrodinger operators, acquire time dependence. And it's a pretty useful way of seeing things, a pretty useful way of calculating things as well, and makes the relation between classical mechanics and quantum mechanics more obvious. So it's a very important tool, so we'll discuss that. We will find the Heisenberg equations of motion and solve them for a particular case today. All this material is not to be covered in the test. The only part, of course, the first few things I will say today about solving for the unitary operator, you've done in other ways, and I will do it again this time. So what we have is, going back to what we were saying last time, we postulated unitary time evolution. We said that psi at t was given by some operator U of t, t0, psi at time t0. And then we found that this equation implied a Schrodinger equation with a Hamiltonian given by the following expression, i h d u d t of t, t0, u dagger of t, t0. So that was our derivation of the Schrodinger equation. We started with the time evolution. We found that whenever we declare that states evolve in time in that way, they satisfy a first-order time differential equation of the Schrodinger form in which the Hamiltonian is related, is given in terms of U by this equation. And we talked about this operator. First, we showed that it doesn't depend really on t0. Then we showed that it's Hermitian that has units of energy. And as you may have seen already in the notes, there's a very clear correspondence between this operator and the way the dynamics follows with the ideas of Poisson brackets that are the precursors of commutators from classical mechanics. So that's in the notes. I will not go in detail in this. Many of you may have not heard of Poisson brackets. It's an interesting thing. And if you read it, that will be good enough. So our goal today is to find U given h. Because as we mentioned last time, for physicists, it's typically more easy to invent a quantum system by postulating a Hamiltonian and then solving it than postulating a time evolution operator. So our goal in general is to find U of tt0 given h of t. That's what we're supposed to do. So the first thing I'm going to do is multiply this equation by U. So I will have, by multiplying this equation by a U from the right, I will write first this term iH dU dt of tt0 is equal to h of t U of tt0. So I multiply this equation by a U from the right. This operator is unitary, so U dagger U is 1. That's why this equation cleaned up to this. Now, there's no confusion really here with derivatives. So I might as well write them with a normal derivative. So I'll write this equation as d dt of U tt0 is equal to h of t U of tt0. This equation, you should be able to look at that equation and say, I see the Schrodinger equation there. How? Imagine that you have a psi of t0 here and you put it in. Then the right-hand side becomes h of t at t1 psi of t. And on the left-hand side, this psi of t0 can be put inside the derivative because it doesn't depend on t. Therefore, this becomes iH by d dt of psi of t. So the Schrodinger equation is there. OK, so now let's solve this. We will consider three cases. Case one, h is time independent. So doing this sort of quickly, so h of t is really h. Like that, no explicit time dependence there. So what do we have? iH bar, let's write dU dt is equal h times U. And we try to write the solution of the form U is equal to e to the minus iHt over h bar times U0. Does that work? Well, we can take dU dt and iH. So we get iH. When I take dU dt, I have to differentiate this exponential. And now in this exponential, this full operator h is there. But we are differentiating with respect to time. And h doesn't depend on time. So this is not a very difficult situation. You could imagine the power series expansion. And h, as far as this derivative goes, is like if it would be even a number. Wouldn't make any difference that it's an operator. So the derivative with respect to time of this thing is minus iH over h times the same exponential. Moreover, the position of this h could be here or could be to the right. It cannot be to the right of U0, though, because this is a matrix, a constant matrix that we've put in here as a possible thing for boundary conditions. So so far, we've taken this derivative. And then the i's cancel, the h bar cancels. And you get h. But this whole thing is, again, U. So the equation has been solved. So try this, and it works. So having this solution, we can write, for example, that U of t, t0 is going to be e to the minus iHt over h bar, some constant matrix. When t is equal to t0, this matrix becomes the unit matrix. So this is e to the minus iHt0 over h bar times U0. And therefore, from here, U0 is the inverse of this matrix, which is nothing else but e to the iHt0 over h bar. So I can substitute back here what U0 is, and finally obtain U of t, t0, is e to the minus iH over h bar, t minus t0. And this is for h time independent. And that's our solution. There's very little to add to this. We discussed that in recitation on Thursday. This unitary operator, you've been seeing that from the beginning of the course in some sense, that you evolve energy eigenstate. If this acts on any energy eigenstate, h is an energy. If you act here on an energy eigenstate, since the energy eigenstate is an eigenstate precisely for h, you can put just the number here. That is, e to the, say, alpha h on a state psi n is equal to e to the alpha en psi n if h on psi n is equal to en on psi n. So the function of an operator acting on an eigenstate is just a function evaluated at the eigenvalue. So this is a rule that you've been using for a really long time. OK, so when h is time independent, that's what it is. How about when h has a little time dependence? How do I call a little time dependence? A little time dependence is an idea designed to make it possible for you to solve the equation, even though it has some time dependence. So you could have Hamiltonians that are time dependent, but still have a simplifying virtue. So h of t is time dependent. But assume that h at t1 and h at t2 commute for all t1 and t2. So what could that be? For example, you know that the particle in a magnetic field, a spin in a magnetic field, is minus gamma B dot the spin. And you could have a time dependent magnetic field, B of t times the spin. I'm not sure this is the constant gamma that I usually call gamma. But it may be. Now then, if the magnetic field is time dependent, but imagine its direction is not time dependent. So if its direction is not time dependent, then for example, you would have here minus gamma Bz of t times Sz. And the Hamiltonian at different times commute, because Sz commutes with itself. And the fact that it's time dependent doesn't make it fail to commute. So if you have a magnetic field that is fixed in one direction, but changes in time, you can have a situation where your Hamiltonian is time dependent, but still at different times it commutes. And you will discuss such case, because it's interesting. But later on, as we do nuclear magnetic resonance, we will have the more interesting case in which a magnetic field rotates. And therefore, it's not that simple. So what happens if you have a time dependent Hamiltonian that actually commutes? Well, the claim is that u of tt0 is given by a natural extension of what we had before. You would want to put exponential of minus iht. But the reason this worked was because the derivative with respect to time brought down an ih over h bar. So one way to fix this is to put t, t0, h of t prime dt prime. So this is an answer to try this. Look at this. If the Hamiltonian were to be time independent, you could take it out, and then you would get t minus t0 that brings you back to this case. So this looks reasonable. So let me call this quantity r of t. And then you notice that r dot of t, the derivative of this quantity with respect to time, well, when you differentiate an integral, the upper argument, you get just the integrand evaluated at the time represented by the upper limit of integration. So this is h of t. And now here comes a crucial point. You're trying to differentiate this u is really e to the r. And you're trying to differentiate to see if the equation holds du dt. So what is du dt? It would be d dt of 1 plus r plus r r plus 1 3 factorial r r r. And now what happens? You differentiate here, and the first term is r dot. Here you would have 1 1 1 1 1 half r dot r plus r r dot. And then 1 over 3 factorial, but 3 factors r dot r r plus r r dot r plus r r r dot. But here is the claim r dot commutes with r. Claim r dot and r commute. Why is that? Well, r dot depends on h. And r is an integral of h as well. But the h is a different times commute anyway. So this must be true. There's no place where you can get a contribution, because r dot is like an h. And here's an integral of h. So since the Hamiltonians are assumed to commute, r dot commutes with r. And this becomes like a normal derivative of an exponential in which you can move the r dot to the left everywhere. And you're differentiating the usual thing. So this is r dot and times the exponential of r. So actually, that means that we've got pretty much our answer, because r dot is minus i over h bar h of t. And e to the r is u. So we've got du dt equal this, which is the same as this equation. So the only reason a derivative with respect to time would not give the usual thing is if r and r dot fail to commute. And they don't. So you can put the r dot here. You can put the r dot on the other side, because it commutes with r. But it's better here. And therefore, you've got this very nice solution. So the solution is not that bad. Now finally, I want to discuss for a second the general case. General case. So that's case. Oh, there was a 1, a 2, a 3. h of t general. What can you do? Well, if h of t is general, there's not too much you can do. You can write something that will get you started doing things, but it's not obviously terribly useful. But it's interesting anyway that there's a way to write something that makes sense. So here it is, u of t and t naught. I'll write the answer and explain how it looks. And then you will see that it's OK. It's interesting. But it probably is not the most practical way you can solve this problem. So here it is. There's an acronym for this thing. t is called the time-ordered exponential. This operator does something to the exponential function. So it's a definition. So I have to say what this time-ordered exponential is. And it's the following. You take the exponential and just begin to expand. So 1 minus i over h bar, or I'll put like this, plus minus i over h bar integral from t naught to t of dt1 h of t1. So far, so good. I've just expanded this. Now, if I would continue expanding, I would get something that doesn't provide the solution. You see, this thing is the solution when the Hamiltonian at different times commute. So it's unlikely to be the solution when they don't commute. In fact, it's not the solution. So what is the next term here? The next term is you think of the exponential as you would expand as usual. So you would have here plus 1 half of this thing squared. So I will put something and then erase it. So maybe don't copy. 1 half minus i over h bar squared. And you would say, well, t naught to t dt prime h of t prime, t naught to t dt double prime h of double prime. Well, that would be just an exponential. So what is a time-ordered exponential? You erase the 1 half. And then for notation, call this t1 and t1. And then the next integral, do it only up to time t1. And call this t2. So t1 will always be greater than t2 because t2 is integrated from t0 to t1. And as you integrate here over the various t1s, you just integrate up to that value. So you're doing less of the full integral than you should be doing. And that's why the factor of 1 half has disappeared. This can be continued. I could write the next one would be minus i over h bar cubed i over h bar cubed integral t naught to t dt1 h of t1, integral t naught to t1 dt2 h of t2. And then the next integral goes up to t2. So t naught to t2 dt3 h of t3. Anyway, that's the time-ordered exponential. And I leave it to you to take the time derivative of that and at least to see that the first few terms are working exactly the way they should. That is, if you take a time derivative of this, you will get h times that thing. So since it's a power series, you will differentiate the first term, and you will get the right thing, then the second term, and you will start getting everything that you need. So it's a funny object. It's reassuring that something like this exists. But in general, you would want to be able to do all this integrals and to sum them up. And in general, it's not that easy. So it's of limited usefulness. It's a nice thing that you can write it, and you can prove things about it, and manipulate it. But when you have a practical problem, generally that's not the way you solve it. In fact, when we will discuss the rotating magnetic fields for magnetic resonance, we will not solve it in this way. We will try to figure out the solution some other way. But in terms of completeness, it's kind of pretty in that you go from the exponential to the time-ordered exponential, and I think you'll see more of this in 806. So that's basically our solution for h. And for the unitary operator, u in terms of h, and what we're going to do now is turn to the Heisenberg picture of quantum mechanics. Yes, questions? Why does our loss of alpha h have to be h? Why? Yeah, because that's really a property of integrals, d dx integral up to x from x0, g of x prime, dx prime, is just equal to g at x. This is a constant here, so you're not varying the integral over in this limit. So if this limit would also be x dependent, you would get another contribution. But we only get a contribution from here. What's really happening is you're integrating up to x, then up to x plus epsilon subtracting, so you pick up the value of the function at the upper limit. Yes? So what happens to the t? Is it like the vector slash? What happens to this t? That's just a symbol. It says time-ordered the following exponential. So at this stage, this is a definition of what t on an exponential means. It's not, let me say, t is not an operator in the usual sense of quantum mechanics or anything like that. It's an instruction. Whenever you have an exponential of this form, the time-ordered exponential is this series that we've written down. It's just a definition. Yes? So when we have operators in differential equations, do we still get unique solutions? If we have what? So when we have operators in differential equations, do we still get uniqueness substitutions? Yes, pretty much. Because at the end of the day, this is a first-order matrix differential equation. So it's a collection of first-order differential equations for every element of a matrix. So everything, it's pretty much the same as you have before. If you know the operator at any time, initial time, with the differential equation, you know the operator at a little bit time later. So the operator is completely determined, if you know it initially, and the differential equation. So I think it's completely analogous. It's just that it's harder to solve, nothing else. One last question. So let's say that we can somehow find this unitary operator, and then we have a differential equation, and then we somehow, let's say, get a wave function out of it. Is there, what is really the interpretation of that wave function? Because of the problem? Well, it's not that we get the wave function out of this. What really is happening is that you have learned how to calculate this operator given h. And therefore, now you're able to evolve any wave function. So you have solved the dynamical system. If somebody tells you at time equals 0, your system is here, you can now calculate where it's going to be at a later time. So that's really all you have achieved. You now know the solution. When you're doing mechanics and they ask you for an orbit problem, they say, at this time the planet is here. What are you supposed to find? x as a function of time. You now know how it's going to develop. You've solved equations of motion. Here it's the same. You know the wave function at time equals 0. If you know it at any time, you've solved the problem. OK. So Heisenberg picture of quantum mechanics. Heisenberg picture. So basically, the Heisenberg picture exists thanks to the existence of the Schrodinger picture. Heisenberg picture of quantum mechanics is not something that you necessarily invent from the beginning. The way we think of it is we assume there is a Schrodinger picture that we've developed, in which we have operators like x, p, spin, Hamiltonians, and wave functions. And then we are going to define a new way of thinking about this, which is called the Heisenberg picture of the quantum mechanics. So it all begins by considering a Schrodinger operator, as hat, which is s is for Schrodinger. And the motivation comes from expectation values. Suppose you have time dependent states. In fact, matrix elements, one time dependent state alpha of t, one time dependent state beta t. Two independent time dependent states. So you could ask, what is this matrix element of a between these two time dependent states? The matrix element. But then armed with our unitary operator, we know that as is here. And this state beta at time t is equal to u of t, comma, 0, beta at time 0. And alpha at t is equal to alpha at 0, u dagger of t 0. So the states have time dependence. But the time dependence has already been found, say, in principle, if you know u dagger. And then you can speak about the time dependent matrix elements of the operator as, or the matrix element of this time dependent operator between the time equals 0 states. And this operator is sufficiently important that this operator is called the Heisenberg version of the operator s. It has time dependence, and it's defined by this equation. So whenever you have Schrodinger operator, whether it be time dependent or time independent, whatever the Schrodinger operator is, I have now a definition of what I will call the Heisenberg operator. And it is obtained by acting with the unitary operator u. And operators always act on operators from the left and from the right. That's something that operators act on states from the left, they act on the state. But operators act on operators from the left and from the right, as you see them here. Is the natural, ideal thing to happen. If you have an operator that's on another from the right only or from the left only, I think you have grounds to be suspicious that maybe you're not doing things right. So this is the Heisenberg operator. And as you can imagine, there's a lot of things to be said about this operator. So let's begin with a remark. Are there questions about this Heisenberg operator? Yes? Sorry, can you tell me about the Schrodinger operator? You have to speak louder. Is this Schrodinger operator related to the Hamiltonian? No, no, it's any Schrodinger operator. This could be the Hamiltonian. This could be x hat. It could be sz. Could be any of the operators you know. All the operators you know are Schrodinger operators. So remarks, comments. OK, comments. 1, at t equals 0, A Heisenberg becomes identical to a Schrodinger. Schrodinger at t equals 0. So look why, because when t is equal to 0, u of t of 0, 0, is the operator that propagates no state. So it's equal to the identity. So this is a wonderful relation that tells you that at time equals 0, the two operators are really the same. Another simple remark, if you have the unit operator in the Schrodinger picture, what is the unit operator in the Heisenberg picture? Well, it would be u t0 dagger 1 u t0. But 1 doesn't matter. u dagger with u is 1. This is a one Schrodinger. And therefore, it's the same operator. So the unit operator is the same. It just doesn't change whatsoever. OK, so that's good. But now something interesting also happens. Suppose you have Schrodinger operator c that is equal to the product of A with B, two Schrodingers. If I try to figure out what is ch, I would put u dagger, avoid all the letters t0. It's supposed to be t0, cs, u. But that's equal u dagger, as, bs, u. But now, in between the two operators, you can put a u dagger, a u u dagger, which is equal to 1. So As u u dagger bs u. And then you see why this is really nice. Because what you get is that c Heisenberg is just a Heisenberg times b Heisenberg. So if you have c Schrodinger equal a Schrodinger b Schrodinger, c Heisenberg is a Heisenberg, b Heisenberg. So there's a nice correspondence between those operators. Also, you can do this for commutators so that you don't have to worry about this thing. So for example, if a Heisenberg, a Schrodinger with b Schrodinger is equal to c Schrodinger, then by doing exactly the same things, you see that a Heisenberg with b Heisenberg would be the commutator equal to c Heisenberg. Yes? That argument for the identity operators being the same in both pictures. If the Hamiltonian is time independent, does that work for any operator in the communities with the Hamiltonian? Hamilton, this time, for the time independent. More than any. Because then you can push the operator just through the exponential of the Hamiltonian. Yeah, we'll see things like that. I'm not sure if I would say we could discuss that maybe a little later. But there are some cases, as we will see immediately, in which some operators are the same in the two pictures. So basically, operators that commute with the Hamiltonian, as you say, since h, since u involves the Hamiltonian, and this is the Hamiltonian, if the operator commutes with the Hamiltonian, you can move them across and they are the same. So I think it's definitely true. So we will have an interesting question, in fact, whether the Heisenberg Hamiltonian is equal to the Schrodinger Hamiltonian. We'll answer that very soon. So the one example that here I think you should keep in mind is this one. You know this is true. So what do you know in the Heisenberg picture that x Heisenberg of t times p Heisenberg of t, commutator, is equal to the Heisenberg version of this. But here was the unit operator. And therefore, this is just i h bar times the unit operator again, because the unit operator is the same in all pictures. So this commutation relation is true for any Heisenberg operator. Whatever commutation relation you have of Schrodinger, it's true for Heisenberg as well. OK, so then let's talk about Hamiltonians. Three Hamiltonians. So Heisenberg Hamiltonian, by definition, would be equal to U dagger t0 Schrodinger Hamiltonian times U of t0. So if the Schrodinger Hamiltonian, actually, if hs at t1 commutes with hs at t2, if the Schrodinger Hamiltonian is such that for all t1 and t2, they commute with each other, remember, if that is the case, the unitary operator is anyway built by an exponential, it's this one. And the Schrodinger Hamiltonian's commute, so as was asked in the question before, this thing commutes with that, and you get that they are the same. So if this is happening, you get that they are the same. If this is happening, the two Hamiltonians are identical. And we'll have the chance to check this today in a nice example. So I will write this as saying that the Heisenberg Hamiltonian as a function of time then is equal to the Schrodinger Hamiltonian as a function of time. And this goes if hs of t1 and hs of t2 commute. OK. Now, I want you to notice this thing. Whenever I suppose the hs of t is some hs of xp and t, for example. Now you come and turn it into Heisenberg by putting a u dagger from the left and a u from the right. What will that do? It will put the u dagger from the left, u dagger on the right, and then it will start working its way inside, and any x that it will find will turn into a Heisenberg x. Any p will turn into Heisenberg p. Imagine, for example, any Hamiltonian, there's some function of x. It has an x squared. Well, the u dagger and u come and turn this into x Heisenberg squared. So what I claim here happens is that h Heisenberg of t is equal to u dagger h Schrodinger of xp t u. And therefore, this becomes h Schrodinger of x Heisenberg of t, p Heisenberg of t, and t. So here is what the Heisenberg Hamiltonian is. It's the Schrodinger Hamiltonian where x's and p's or spins and everything has become Heisenberg. So the equality of the two Hamiltonians is a very funny condition on the Schrodinger Hamiltonian, because this is supposed to be equal to the Schrodinger Hamiltonian, which is of xp and t. So you have a function of xp and t, and you put x Heisenberg, p Heisenberg, and somehow the whole thing is the same. So this is something very useful, and we'll need it. One more comment, expectation values. So this is three, I'll comment number four, on expectation values, which is something you've already, it's sort of the way we began the discussion, and I want to make sure it's clear. So four expectation values. So we started with this with alpha and beta, two arbitrary states, matrix elements. Take them equal and to be equal to psi of t. So you would have psi t, As psi t is in fact equal to psi 0, A Heisenberg psi 0. Now that is a key equation. You're doing expectation value at any given time of a Schrodinger operator, turn it into Heisenberg and work at time equals 0. It simplifies life tremendously. Now this is the key identity, it's the way we motivated everything in a way, and it's written in a way that maybe it's a little too schematic, but we write it this way, we just say the expectation value of A S is equal to the expectation value of A H. And this, well, we save time like that, but you have to know what you mean. When you're computing the expectation value of a Schrodinger operator, you're using time dependent states. When you're computing the expectation value of the Heisenberg operator, you're using the time equals 0 version of the states, but they are the same. So we say that the Schrodinger expectation value is equal to the Heisenberg expectation value. We write it like in the bottom, but we mean the top equation, and we use it that way. So the Heisenberg operators at this moment are a little mysterious. They are supposed to be given by this formula, but we've seen that calculating U can be difficult, so calculating the Heisenberg operator can be difficult sometimes. So what we try to do in order to simplify that is find an equation that is satisfied by the Heisenberg operator, a time derivative equation. So let's try to find an equation that is satisfied by the Heisenberg operator rather than a formula. You say, well, this is better, but the fact is that seldom you know U, and even if you know U, you have to do this simplification, which is hard. So finding a differential equation for the operator is useful. So differential equation for Heisenberg operators. So what do we want to do? We want to calculate I h bar d dt of the Heisenberg operator. And what do we get? Well, we have several things. Remember, the Schrodinger operator can have a bit of time dependence. The time dependence would be an explicit time dependent. So let's take the time derivative of all this. So you would have three terms. I h bar d U dagger dt A s U plus U dagger A s d U dt plus with an I h bar plus U dagger I h bar d A s d t. I'm sorry. Yeah, d A s d t and U. So well, you have these equations. Those were the Schrodinger equations we started with today. The derivatives of U are the derivatives of U dagger. So what did we have? Well, we had that I h bar d U dt was h U h Schrodinger times U. And therefore, I h bar d U dagger dt. I take the dagger of this. I would get a minus sign. I'll put it on the other side. It's equal to U dagger h s with a minus here. And all the U's are U's of t and t naught. I run out of this thick chalk. So we'll continue with thin chalk. All right. So we are here. We wrote the time derivative, and we have three terms to work out. So what are they? Well, we have this thing. I h bar this. So let's write it. I h bar d d t of A s of A Heisenberg. I'm sorry. I'm sorry. It's equal to that term is minus U dagger h s A Schrodinger U. The next term, plus I h bar d U dt on the right. So we have plus U dagger A s h s d U dt. So U, well, that's not bad. It's actually quite nice. And then the last term, which I have very little to say, because in general, this is a derivative of a time dependent operator, partial with respect to time. It would be 0 if A s depends just say on x or p or s x or any of those things has to have a particular t. So just leave this as plus I h bar d A s d t Heisenberg. The Heisenberg version of this operator, using the definition that anything, any operator that we have a U dagger in front, a U to the right, is the Heisenberg version of the operator. So I think I'm doing all right with this equation. So what did we have? Here it is. I h bar d dt of A Heisenberg of t. And now comes the nice thing. Of course, this thing, look at it. U dagger U. This turns everything here into Heisenberg. H Heisenberg, A Heisenberg. Here you have A Heisenberg, H Heisenberg. And what you got is the commutator between them. So this thing is A Heisenberg commutator with H Heisenberg. That whole thing. And then you have plus I h bar d A s d t Heisenberg. So that is the Heisenberg equation of motion. That is how you can calculate a Heisenberg operator, if you want. You try to solve this differential equation. And many times, that's the simplest way to calculate the Heisenberg operator. So there you go. It's a pretty important equation. So let's consider particular cases immediately to just get some intuition. So remarks. Suppose A s has no explicit time dependence. So basically, there's no explicit t, and therefore this derivative goes away. So the equation becomes I h bar d A s d A h, of course. d t is equal to A h Heisenberg sub h of t. And you know the Heisenberg operator is supposed to be simpler, simple if the Schrodinger operator is time independent, is identical to the Schrodinger Hamiltonian. Even if the Schrodinger operator has time dependence, but they commute, this will become the Schrodinger Hamiltonian. But we can leave it like that. It's a nice thing anyway. Time dependence of expectation value. So let me do a little remark on time dependence of expectation value. So suppose you have the usual thing that you want to compute. How does the expectation value of a Schrodinger operator depend on time? You're faced with that expectation value of A s, and it changes in time, and you want to know how you can compute that. Well, you first say, OK, the I h bar d d t, but this thing is nothing but psi 0 A Heisenberg of t psi 0. Now I can let the derivative go in. So this becomes psi 0 I h bar d A h d t psi 0. And using this, assuming that A is still no time dependence, A has no explicit time dependence, then you can use just this equation, which gives you psi 0 A h H H psi 0. So all in all, what have you gotten? You've gotten a rather simple thing, the time derivative of the expectation value. So I h bar d d t. And now I write the left-hand side as just the expectation value of H Heisenberg of t. Now the left-hand side has the A Schrodinger expectation value, but we call those expectation values the same thing as a Heisenberg expectation value. So this thing becomes the right-hand side, and the right-hand side is the expectation value of A Heisenberg, H Heisenberg, like that. And just the way we say that Heisenberg expectation values are the same as Schrodinger expectation values, because you could as well write it if you prefer as d d t of H Schrodinger is equal to expectation value of H Schrodinger with H Schrodinger. It's really the same equation. This equation we derived a couple of lectures ago, and now we know that the expectation values of Schrodinger operators are the same as the expectation values of the Heisenberg counterparts, except that the states are taking a time equals 0. So you can use either form of this equation. The bottom one is one that you've already seen. The top one now looks almost obvious from the bottom one, but it really took quite a bit to get it. One last comment on these operators. How about conserved operators? What are those things? A time-independent, A s, is said to be conserved if it commutes with the Schrodinger Hamiltonian. If A s commutes with H s equals 0. Now you know that if A s with H s is 0, A h with H h is 0, because the map between Heisenberg and Schrodinger pictures has a commutator that is valid in the Schrodinger picture, is valid in the Heisenberg picture by putting h's. So what you realize from this is that this thing, this implies A h commutes with H h, and therefore by 0.1, by 1, you have dA h dt is equal to 0. And this is nice, actually. The Heisenberg operator is actually time-independent. It just doesn't depend on time. A Schrodinger operator, it's a funny operator. It doesn't have time in there. It has x's, p's, spins, and you don't know in general if it's time-independent in the sense of expectation values. But whenever A s commutes with H s, well, the expectation values don't change in time. But as you know, this d dt can be brought in because the states are not time-dependent. So the fact that this is 0 means the operator, Heisenberg operator, is really time-independent. Whenever you have a Schrodinger operator, has no t, the Heisenberg one can have a lot of t. But if the operator is conserved, then the Heisenberg operator will have no t's after all. It will really be conserved. So let's use our last ten minutes to do an example and illustrate much of this. In the notes, there will be three examples. I will do just one in lecture. You can do the other ones in recitation next week. There's no need really that you study these things at this moment. Just try to get whatever you can now from the lecture. And next week, you'll go back to this. So the example is the harmonic oscillator. And it will illustrate the ideas very nicely. The Schrodinger Hamiltonian is p squared over 2m plus 1 half m omega squared x hat squared. OK. I could put x Schrodinger and p Schrodinger, but that would be just far too much. x and p are the operators you've always known. They are Schrodinger operators. So we leave them like that. Now I have to write the Heisenberg Hamiltonian. Well, what is the Heisenberg Hamiltonian? Yes? Yes, I did. Sorry? Yes, I did. Identical? Yes. But I will leave that for a little later. I will just assume, well, I'm supposed to do u dagger u. And as you said, this is a time independent Hamiltonian. It better be the same. But it will be clear if we now write what it should be in general. Have a u dagger and a u from the right. They come here and they turn this into p Heisenberg over 2m plus 1 half m omega squared x Heisenberg. OK. That's your Heisenberg Hamiltonian. And we will check, in fact, that this time independent. So how about the operators x Heisenberg and p Heisenberg? What are they? Well, I don't know how to get them unless I do this sort of u thing that doesn't look too bad, but certainly would be messy. You would have to do an exponential of e to the minus i ht over t with the x operator and another exponential. Sounds a little complicated. So let's do it the way the equations of the Heisenberg operators tell you. Well, x and p are time independent Schrodinger operators. So that equation that I box holds. So i h v x Heisenberg dt is nothing else than x Heisenberg commuted with h Heisenberg. OK. Can we do that commutator? Yes. Because x Heisenberg, as you remember, just commutes with p Heisenberg. So instead of the Hamiltonian, you can put this. This is x Heisenberg p Heisenberg squared over 2m. OK. Well, x Heisenberg and p Heisenberg is like you had x and p. So what is this commutator? You probably know it by now. You act with this on these two p. So x on one, x on the other gives the same on each. So you get a p Heisenberg times the commutator of x and p, which is i h bar, times a factor of 2. So we could put hats. Better, maybe. And then what do we get? We get the h there and the i h cancels. And we get some nice equation that says dx Heisenberg dt is 1 over m p Heisenberg. Well, it actually looks like an equation in classical mechanics. dx dt is p over m. So that's the good thing about the Heisenberg equations of motion. It looks like ordinary equations for dynamical variables. Well, we've got this one. Let's get p. Well, we didn't get the operator still, but we got an equation. So how about p dp dt? So i h dp Heisenberg dt would be p Heisenberg with h Heisenberg. And this time, only the potential term in here matters. So it's p Heisenberg with 1 half m omega squared x Heisenberg squared. So what do we get? We get 1 half m omega squared. Then we get, again, a factor of 2. Then we get 1 left over x h. And then a p with x h, which is a minus i h bar. So the i h bars cancel, and we get dp h dt is equal to the h bar cancel m omega squared x h minus m x. All right. So these are our Heisenberg equations of motion. So how do we solve for them now? Well, you sort of have to try the kind of things that you would do classically. Take a second derivative of this equation. The second x h dt squared would be equal 1 over m dp h dt. And dp h dt would be minus 1 over m times minus m omega squared x h. So d second x h dt squared is equal to minus omega squared x h. Exactly the equation of motion of a harmonic oscillator. It's really absolutely nice that you recover those equations that you had before, except that now you're talking operators. And it's going to simplify your life quite dramatically when you try to use these operators. Because in a sense solving for the time dependent Heisenberg operators is the same as finding the time evolution of all states. This time the operators change, and you will know what they change like. So you have this, and then you write x h is equal to a cosine omega t plus b sine omega t, where a and b are some time independent operators. So you get x h of t. Well, that's the solution. How about what is p, p h of t would be m dx m dx dt. Minus m omega sine omega t a plus m omega cosine omega t b. OK, so that's it. That is the most general solution. But it still doesn't look like what you would want. That's it. No, because you haven't used the time equals zero conditions. At time equals zero, the Heisenberg operators are identical to the Schrodinger operators. So at t equals zero, x h of t becomes a, but that must be x hat, the Schrodinger operator. And at t equals zero, p h of t becomes equal to, this is, zero m omega b. And that must be equal to the p hat operator. Actually, we have already now a and b, so b from here is p hat over m omega. And therefore, x h of t is equal to a, which is x hat cosine omega t, plus b, which is p hat over m omega sine omega t. And p h of t is here. A is, p h of t is m omega b, which is m over s p. So it's p hat cosine omega t minus m omega x hat sine omega t. So let's see. I hope I didn't make mistakes. p hat minus m omega x hat sine omega t. Yep. This is correct. This is your whole solution for the Heisenberg operator. So any expectation value of any power of x and p that you will want to find its time dependence, just put those Heisenberg operators and you will calculate things with the states at time equals zero. It will become very easy. The first thing I want to do is complete the promise that we had about the, what is the Heisenberg Hamiltonian? Well, we had the Heisenberg Hamiltonian there. And now we know the Heisenberg operators in terms of the Schrodinger one. So h h of t is equal to p h 1 over 2 m p h squared. So I have p hat cosine omega t minus m omega x hat sine omega t squared plus one half m omega squared. So x h squared, so x hat cosine omega t plus p hat over m omega sine omega t squared. So that's what the Heisenberg Hamiltonian is. So let's simplify this. Well, let's square these things. So I have 1 over 2 m cosine squared omega t p hat squared. Let's do the square of this one. We would have plus 1 over 2 m m squared omega squared sine squared omega t x squared. And then we have the cross product, which would be plus or actually minus 1 over 2 m. The product of these two things, m omega sine omega t cosine omega t, and you have p x plus x p. OK, I squared the first terms. Now the second one. Well, let's square the p squared here. What do we have? One half m omega squared over m squared omega squared sine squared of omega t p squared. The x plus one half m omega squared cosine squared omega t x squared. And the cross term plus one half m omega squared over m omega times cosine omega t sine omega t x p plus p x. A little bit of work. But what do we get? 1 over 2 m, and here we must have 1 over 2 m, correct, 1 over 2 m sine squared omega t p squared. So this is equal 1 over 2 m p squared. And here you have 1 half m omega squared, so it's 1 half m omega squared. Cosine and sine squared is x hat squared. And then here we have omega over 2, and here omega over 2. Same factors, same factors, opposite sines. Very good. Schrodinger Hamiltonian. You confirm that this theoretical expectation is absolutely correct, and what's the meaning? You have the Heisenberg Hamiltonian written in terms of the Heisenberg variables, but by the time you substitute these Heisenberg variables in, it just becomes identical to the Schrodinger Hamiltonian. All right, so that's all for today. I hope to see you in office hours in the coming days. Be here Wednesday, 12.30. Maybe 12.25 would be better, and we'll see you then. All right. Thank you.
 The following content is provided under a Creative  Commons license.  Your support will help MIT OpenCourseWare  continue to offer high quality educational resources for free.  To make a donation or to view additional materials  from hundreds of MIT courses, visit MIT OpenCourseWare  at ocw.mit.edu.  OK, so let me go back to what we're doing.  What the plan for today is as follows.  We're going to look at this unitary time evolution  and calculate this operator, U, given the Hamiltonian.  That will be the first order of business today.  Then we will look at the Heisenberg picture  of quantum mechanics.  And the Heisenberg picture of quantum mechanics  is one where the operators, the Schrodinger operators,  acquire time dependence.  And it's a pretty useful way of seeing things,  a pretty useful way of calculating things as well,  and makes the relation between classical mechanics  and quantum mechanics more obvious.  So it's a very important tool, so we'll discuss that.  We will find the Heisenberg equations of motion  and solve them for a particular case today.  All this material is not to be covered in the test.  The only part, of course, the first few things  I will say today about solving for the unitary operator,  you've done in other ways, and I will do it again this time.  So what we have is, going back to what we were saying last time,  we postulated unitary time evolution.  We said that psi at t was given by some operator U of t, t0,  psi at time t0.  And then we found that this equation implied a Schrodinger  equation with a Hamiltonian given by the following expression,  i h d u d t of t, t0, u dagger of t, t0.  So that was our derivation of the Schrodinger equation.  We started with the time evolution.  We found that whenever we declare  that states evolve in time in that way,  they satisfy a first-order time differential equation  of the Schrodinger form in which the Hamiltonian is related,  is given in terms of U by this equation.  And we talked about this operator.  First, we showed that it doesn't depend really on t0.  Then we showed that it's Hermitian  that has units of energy.  And as you may have seen already in the notes,  there's a very clear correspondence  between this operator and the way the dynamics follows  with the ideas of Poisson brackets  that are the precursors of commutators  from classical mechanics.  So that's in the notes.  I will not go in detail in this.  Many of you may have not heard of Poisson brackets.  It's an interesting thing.  And if you read it, that will be good enough.  So our goal today is to find U given h.  Because as we mentioned last time, for physicists,  it's typically more easy to invent a quantum system  by postulating a Hamiltonian and then solving it  than postulating a time evolution operator.  So our goal in general is to find U of tt0 given h of t.  That's what we're supposed to do.  So the first thing I'm going to do  is multiply this equation by U.  So I will have, by multiplying this equation by a U  from the right, I will write first this term iH dU dt  of tt0 is equal to h of t U of tt0.  So I multiply this equation by a U from the right.  This operator is unitary, so U dagger U is 1.  That's why this equation cleaned up to this.  Now, there's no confusion really here with derivatives.  So I might as well write them with a normal derivative.  So I'll write this equation as d dt of U tt0  is equal to h of t U of tt0.  This equation, you should be able to look at that equation  and say, I see the Schrodinger equation there.  How?  Imagine that you have a psi of t0 here  and you put it in.  Then the right-hand side becomes h of t at t1 psi of t.  And on the left-hand side, this psi of t0  can be put inside the derivative  because it doesn't depend on t.  Therefore, this becomes iH by d dt of psi of t.  So the Schrodinger equation is there.  OK, so now let's solve this.  We will consider three cases.  Case one, h is time independent.  So doing this sort of quickly, so h of t is really h.  Like that, no explicit time dependence there.  So what do we have?  iH bar, let's write dU dt is equal h times U.  And we try to write the solution of the form U  is equal to e to the minus iHt over h bar times U0.  Does that work?  Well, we can take dU dt and iH.  So we get iH.  When I take dU dt, I have to differentiate this exponential.  And now in this exponential, this full operator h is there.  But we are differentiating with respect to time.  And h doesn't depend on time.  So this is not a very difficult situation.  You could imagine the power series expansion.  And h, as far as this derivative goes,  is like if it would be even a number.  Wouldn't make any difference that it's an operator.  So the derivative with respect to time of this thing  is minus iH over h times the same exponential.  Moreover, the position of this h could be here  or could be to the right.  It cannot be to the right of U0, though,  because this is a matrix, a constant matrix  that we've put in here as a possible thing  for boundary conditions.  So so far, we've taken this derivative.  And then the i's cancel, the h bar cancels.  And you get h.  But this whole thing is, again, U.  So the equation has been solved.  So try this, and it works.  So having this solution, we can write, for example,  that U of t, t0 is going to be e to the minus iHt over h bar,  some constant matrix.  When t is equal to t0, this matrix  becomes the unit matrix.  So this is e to the minus iHt0 over h bar times U0.  And therefore, from here, U0 is the inverse of this matrix,  which is nothing else but e to the iHt0 over h bar.  So I can substitute back here what U0 is,  and finally obtain U of t, t0, is e to the minus iH  over h bar, t minus t0.  And this is for h time independent.  And that's our solution.  There's very little to add to this.  We discussed that in recitation on Thursday.  This unitary operator, you've been  seeing that from the beginning of the course in some sense,  that you evolve energy eigenstate.  If this acts on any energy eigenstate, h is an energy.  If you act here on an energy eigenstate,  since the energy eigenstate is an eigenstate precisely for h,  you can put just the number here.  That is, e to the, say, alpha h on a state psi n  is equal to e to the alpha en psi n if h on psi n  is equal to en on psi n.  So the function of an operator acting on an eigenstate  is just a function evaluated at the eigenvalue.  So this is a rule that you've been using for a really long time.  OK, so when h is time independent, that's what it is.  How about when h has a little time dependence?  How do I call a little time dependence?  A little time dependence is an idea designed  to make it possible for you to solve the equation,  even though it has some time dependence.  So you could have Hamiltonians that are time dependent,  but still have a simplifying virtue.  So h of t is time dependent.  But assume that h at t1 and h at t2 commute for all t1 and t2.  So what could that be?  For example, you know that the particle in a magnetic field,  a spin in a magnetic field, is minus gamma B dot the spin.  And you could have a time dependent magnetic field,  B of t times the spin.  I'm not sure this is the constant gamma  that I usually call gamma.  But it may be.  Now then, if the magnetic field is time dependent,  but imagine its direction is not time dependent.  So if its direction is not time dependent,  then for example, you would have here minus gamma Bz  of t times Sz.  And the Hamiltonian at different times commute,  because Sz commutes with itself.  And the fact that it's time dependent  doesn't make it fail to commute.  So if you have a magnetic field that  is fixed in one direction, but changes in time,  you can have a situation where your Hamiltonian is time  dependent, but still at different times it commutes.  And you will discuss such case, because it's interesting.  But later on, as we do nuclear magnetic resonance,  we will have the more interesting case in which  a magnetic field rotates.  And therefore, it's not that simple.  So what happens if you have a time dependent Hamiltonian  that actually commutes?  Well, the claim is that u of tt0 is given  by a natural extension of what we had before.  You would want to put exponential of minus iht.  But the reason this worked was because the derivative  with respect to time brought down an ih over h bar.  So one way to fix this is to put t, t0, h of t prime dt prime.  So this is an answer to try this.  Look at this.  If the Hamiltonian were to be time independent,  you could take it out, and then you  would get t minus t0 that brings you back to this case.  So this looks reasonable.  So let me call this quantity r of t.  And then you notice that r dot of t,  the derivative of this quantity with respect to time,  well, when you differentiate an integral, the upper argument,  you get just the integrand evaluated at the time represented  by the upper limit of integration.  So this is h of t.  And now here comes a crucial point.  You're trying to differentiate this u is really e to the r.  And you're trying to differentiate  to see if the equation holds du dt.  So what is du dt?  It would be d dt of 1 plus r plus r r plus 1 3 factorial r r r.  And now what happens?  You differentiate here, and the first term is r dot.  Here you would have 1 1 1 1 1 half r dot r plus r r dot.  And then 1 over 3 factorial, but 3 factors r dot r r plus r r  dot r plus r r r dot.  But here is the claim r dot commutes with r.  Claim r dot and r commute.  Why is that?  Well, r dot depends on h.  And r is an integral of h as well.  But the h is a different times commute anyway.  So this must be true.  There's no place where you can get a contribution,  because r dot is like an h.  And here's an integral of h.  So since the Hamiltonians are assumed to commute, r dot  commutes with r.  And this becomes like a normal derivative  of an exponential in which you can move the r dot to the left  everywhere.  And you're differentiating the usual thing.  So this is r dot and times the exponential of r.  So actually, that means that we've got pretty much our answer,  because r dot is minus i over h bar h of t.  And e to the r is u.  So we've got du dt equal this, which  is the same as this equation.  So the only reason a derivative with respect to time  would not give the usual thing is if r and r dot fail  to commute.  And they don't.  So you can put the r dot here.  You can put the r dot on the other side,  because it commutes with r.  But it's better here.  And therefore, you've got this very nice solution.  So the solution is not that bad.  Now finally, I want to discuss for a second  the general case.  General case.  So that's case.  Oh, there was a 1, a 2, a 3.  h of t general.  What can you do?  Well, if h of t is general, there's not too much you can do.  You can write something that will get you started doing things,  but it's not obviously terribly useful.  But it's interesting anyway that there's  a way to write something that makes sense.  So here it is, u of t and t naught.  I'll write the answer and explain how it looks.  And then you will see that it's OK.  It's interesting.  But it probably is not the most practical way  you can solve this problem.  So here it is.  There's an acronym for this thing.  t is called the time-ordered exponential.  This operator does something to the exponential function.  So it's a definition.  So I have to say what this time-ordered exponential is.  And it's the following.  You take the exponential and just begin to expand.  So 1 minus i over h bar, or I'll put like this,  plus minus i over h bar integral from t naught  to t of dt1 h of t1.  So far, so good.  I've just expanded this.  Now, if I would continue expanding,  I would get something that doesn't provide the solution.  You see, this thing is the solution  when the Hamiltonian at different times commute.  So it's unlikely to be the solution when they don't commute.  In fact, it's not the solution.  So what is the next term here?  The next term is you think of the exponential  as you would expand as usual.  So you would have here plus 1 half of this thing squared.  So I will put something and then erase it.  So maybe don't copy.  1 half minus i over h bar squared.  And you would say, well, t naught to t dt prime h of t prime,  t naught to t dt double prime h of double prime.  Well, that would be just an exponential.  So what is a time-ordered exponential?  You erase the 1 half.  And then for notation, call this t1 and t1.  And then the next integral, do it only up to time t1.  And call this t2.  So t1 will always be greater than t2 because t2 is integrated  from t0 to t1.  And as you integrate here over the various t1s,  you just integrate up to that value.  So you're doing less of the full integral  than you should be doing.  And that's why the factor of 1 half has disappeared.  This can be continued.  I could write the next one would be minus i over h bar cubed  i over h bar cubed integral t naught to t dt1 h of t1,  integral t naught to t1 dt2 h of t2.  And then the next integral goes up to t2.  So t naught to t2 dt3 h of t3.  Anyway, that's the time-ordered exponential.  And I leave it to you to take the time derivative of that  and at least to see that the first few terms are working  exactly the way they should.  That is, if you take a time derivative of this,  you will get h times that thing.  So since it's a power series, you will differentiate the first  term, and you will get the right thing, then the second term,  and you will start getting everything that you need.  So it's a funny object.  It's reassuring that something like this exists.  But in general, you would want to be  able to do all this integrals and to sum them up.  And in general, it's not that easy.  So it's of limited usefulness.  It's a nice thing that you can write it,  and you can prove things about it, and manipulate it.  But when you have a practical problem,  generally that's not the way you solve it.  In fact, when we will discuss the rotating magnetic fields  for magnetic resonance, we will not solve it in this way.  We will try to figure out the solution some other way.  But in terms of completeness, it's  kind of pretty in that you go from the exponential  to the time-ordered exponential, and I think  you'll see more of this in 806.  So that's basically our solution for h.  And for the unitary operator, u in terms of h,  and what we're going to do now is turn to the Heisenberg  picture of quantum mechanics.  Yes, questions?  Why does our loss of alpha h have to be h?  Why?  Yeah, because that's really a property of integrals, d dx  integral up to x from x0, g of x prime, dx prime,  is just equal to g at x.  This is a constant here, so you're not  varying the integral over in this limit.  So if this limit would also be x dependent,  you would get another contribution.  But we only get a contribution from here.  What's really happening is you're integrating up to x,  then up to x plus epsilon subtracting,  so you pick up the value of the function at the upper limit.  Yes?  So what happens to the t?  Is it like the vector slash?  What happens to this t?  That's just a symbol.  It says time-ordered the following exponential.  So at this stage, this is a definition  of what t on an exponential means.  It's not, let me say, t is not an operator  in the usual sense of quantum mechanics or anything like that.  It's an instruction.  Whenever you have an exponential of this form,  the time-ordered exponential is this series  that we've written down.  It's just a definition.  Yes?  So when we have operators in differential equations,  do we still get unique solutions?  If we have what?  So when we have operators in differential equations,  do we still get uniqueness substitutions?  Yes, pretty much.  Because at the end of the day, this  is a first-order matrix differential equation.  So it's a collection of first-order differential  equations for every element of a matrix.  So everything, it's pretty much the same as you have before.  If you know the operator at any time, initial time,  with the differential equation, you  know the operator at a little bit time later.  So the operator is completely determined,  if you know it initially, and the differential equation.  So I think it's completely analogous.  It's just that it's harder to solve, nothing else.  One last question.  So let's say that we can somehow find this unitary operator,  and then we have a differential equation,  and then we somehow, let's say, get a wave function out of it.  Is there, what is really the interpretation of that wave  function?  Because of the problem?  Well, it's not that we get the wave function out of this.  What really is happening is that you have learned  how to calculate this operator given h.  And therefore, now you're able to evolve any wave function.  So you have solved the dynamical system.  If somebody tells you at time equals 0, your system is here,  you can now calculate where it's going to be at a later time.  So that's really all you have achieved.  You now know the solution.  When you're doing mechanics and they ask you for an orbit  problem, they say, at this time the planet is here.  What are you supposed to find?  x as a function of time.  You now know how it's going to develop.  You've solved equations of motion.  Here it's the same.  You know the wave function at time equals 0.  If you know it at any time, you've solved the problem.  OK.  So Heisenberg picture of quantum mechanics.  Heisenberg picture.  So basically, the Heisenberg picture  exists thanks to the existence of the Schrodinger picture.  Heisenberg picture of quantum mechanics is not something  that you necessarily invent from the beginning.  The way we think of it is we assume  there is a Schrodinger picture that we've developed,  in which we have operators like x, p, spin, Hamiltonians,  and wave functions.  And then we are going to define a new way  of thinking about this, which is called the Heisenberg  picture of the quantum mechanics.  So it all begins by considering a Schrodinger operator,  as hat, which is s is for Schrodinger.  And the motivation comes from expectation values.  Suppose you have time dependent states.  In fact, matrix elements, one time dependent state alpha  of t, one time dependent state beta t.  Two independent time dependent states.  So you could ask, what is this matrix element of a between  these two time dependent states?  The matrix element.  But then armed with our unitary operator,  we know that as is here.  And this state beta at time t is equal to u of t, comma, 0,  beta at time 0.  And alpha at t is equal to alpha at 0, u dagger of t 0.  So the states have time dependence.  But the time dependence has already  been found, say, in principle, if you know u dagger.  And then you can speak about the time dependent matrix  elements of the operator as, or the matrix element  of this time dependent operator between the time equals 0  states.  And this operator is sufficiently important  that this operator is called the Heisenberg version  of the operator s.  It has time dependence, and it's defined by this equation.  So whenever you have Schrodinger operator,  whether it be time dependent or time independent,  whatever the Schrodinger operator is,  I have now a definition of what I will call the Heisenberg  operator.  And it is obtained by acting with the unitary operator u.  And operators always act on operators from the left  and from the right.  That's something that operators act on states  from the left, they act on the state.  But operators act on operators from the left and from the right,  as you see them here.  Is the natural, ideal thing to happen.  If you have an operator that's on another from the right only  or from the left only, I think you  have grounds to be suspicious that maybe you're not  doing things right.  So this is the Heisenberg operator.  And as you can imagine, there's a lot of things  to be said about this operator.  So let's begin with a remark.  Are there questions about this Heisenberg operator?  Yes?  Sorry, can you tell me about the Schrodinger operator?  You have to speak louder.  Is this Schrodinger operator related to the Hamiltonian?  No, no, it's any Schrodinger operator.  This could be the Hamiltonian.  This could be x hat.  It could be sz.  Could be any of the operators you know.  All the operators you know are Schrodinger operators.  So remarks, comments.  OK, comments.  1, at t equals 0, A Heisenberg becomes  identical to a Schrodinger.  Schrodinger at t equals 0.  So look why, because when t is equal to 0, u of t of 0, 0,  is the operator that propagates no state.  So it's equal to the identity.  So this is a wonderful relation that tells you  that at time equals 0, the two operators are really the same.  Another simple remark, if you have the unit operator  in the Schrodinger picture, what is the unit operator  in the Heisenberg picture?  Well, it would be u t0 dagger 1 u t0.  But 1 doesn't matter.  u dagger with u is 1.  This is a one Schrodinger.  And therefore, it's the same operator.  So the unit operator is the same.  It just doesn't change whatsoever.  OK, so that's good.  But now something interesting also happens.  Suppose you have Schrodinger operator  c that is equal to the product of A with B, two Schrodingers.  If I try to figure out what is ch, I would put u dagger,  avoid all the letters t0.  It's supposed to be t0, cs, u.  But that's equal u dagger, as, bs, u.  But now, in between the two operators,  you can put a u dagger, a u u dagger, which is equal to 1.  So As u u dagger bs u.  And then you see why this is really nice.  Because what you get is that c Heisenberg is just  a Heisenberg times b Heisenberg.  So if you have c Schrodinger equal a Schrodinger b Schrodinger,  c Heisenberg is a Heisenberg, b Heisenberg.  So there's a nice correspondence  between those operators.  Also, you can do this for commutators  so that you don't have to worry about this thing.  So for example, if a Heisenberg, a Schrodinger with b  Schrodinger is equal to c Schrodinger,  then by doing exactly the same things,  you see that a Heisenberg with b Heisenberg  would be the commutator equal to c Heisenberg.  Yes?  That argument for the identity operators  being the same in both pictures.  If the Hamiltonian is time independent,  does that work for any operator in the communities  with the Hamiltonian?  Hamilton, this time, for the time independent.  More than any.  Because then you can push the operator just  through the exponential of the Hamiltonian.  Yeah, we'll see things like that.  I'm not sure if I would say we could discuss that maybe  a little later.  But there are some cases, as we will see immediately,  in which some operators are the same in the two pictures.  So basically, operators that commute with the Hamiltonian,  as you say, since h, since u involves the Hamiltonian,  and this is the Hamiltonian, if the operator commutes  with the Hamiltonian, you can move them across  and they are the same.  So I think it's definitely true.  So we will have an interesting question, in fact,  whether the Heisenberg Hamiltonian  is equal to the Schrodinger Hamiltonian.  We'll answer that very soon.  So the one example that here I think you should keep in mind  is this one.  You know this is true.  So what do you know in the Heisenberg picture  that x Heisenberg of t times p Heisenberg of t, commutator,  is equal to the Heisenberg version of this.  But here was the unit operator.  And therefore, this is just i h bar times the unit operator  again, because the unit operator is the same in all pictures.  So this commutation relation is true for any Heisenberg operator.  Whatever commutation relation you have of Schrodinger,  it's true for Heisenberg as well.  OK, so then let's talk about Hamiltonians.  Three Hamiltonians.  So Heisenberg Hamiltonian, by definition,  would be equal to U dagger t0 Schrodinger Hamiltonian times U  of t0.  So if the Schrodinger Hamiltonian, actually, if hs at t1 commutes  with hs at t2, if the Schrodinger Hamiltonian is such that for all  t1 and t2, they commute with each other, remember,  if that is the case, the unitary operator is anyway built by an  exponential, it's this one.  And the Schrodinger Hamiltonian's commute, so as was asked in the  question before, this thing commutes with that, and you get  that they are the same.  So if this is happening, you get that they are the same.  If this is happening, the two Hamiltonians are identical.  And we'll have the chance to check this today in a nice example.  So I will write this as saying that the Heisenberg Hamiltonian as a  function of time then is equal to the Schrodinger Hamiltonian as a  function of time.  And this goes if hs of t1 and hs of t2 commute.  OK.  Now, I want you to notice this thing.  Whenever I suppose the hs of t is some hs of xp and t, for  example.  Now you come and turn it into Heisenberg by putting a u dagger  from the left and a u from the right.  What will that do?  It will put the u dagger from the left, u dagger on the right,  and then it will start working its way inside, and any x that it  will find will turn into a Heisenberg x.  Any p will turn into Heisenberg p.  Imagine, for example, any Hamiltonian, there's some function of x.  It has an x squared.  Well, the u dagger and u come and turn this into x Heisenberg  squared.  So what I claim here happens is that h Heisenberg of t is equal  to u dagger h Schrodinger of xp t u.  And therefore, this becomes h Schrodinger of x Heisenberg of t,  p Heisenberg of t, and t.  So here is what the Heisenberg Hamiltonian is.  It's the Schrodinger Hamiltonian where x's and p's or spins  and everything has become Heisenberg.  So the equality of the two Hamiltonians  is a very funny condition on the Schrodinger Hamiltonian,  because this is supposed to be equal to the Schrodinger Hamiltonian,  which is of xp and t.  So you have a function of xp and t, and you put x Heisenberg,  p Heisenberg, and somehow the whole thing is the same.  So this is something very useful, and we'll need it.  One more comment, expectation values.  So this is three, I'll comment number four, on expectation  values, which is something you've already, it's sort of the way  we began the discussion, and I want to make sure it's clear.  So four expectation values.  So we started with this with alpha and beta, two arbitrary  states, matrix elements.  Take them equal and to be equal to psi of t.  So you would have psi t, As psi t is in fact  equal to psi 0, A Heisenberg psi 0.  Now that is a key equation.  You're doing expectation value at any given time  of a Schrodinger operator, turn it into Heisenberg  and work at time equals 0.  It simplifies life tremendously.  Now this is the key identity, it's the way we motivated  everything in a way, and it's written in a way  that maybe it's a little too schematic, but we write it  this way, we just say the expectation value of A S is  equal to the expectation value of A H. And this, well,  we save time like that, but you have to know what you mean.  When you're computing the expectation value of a  Schrodinger operator, you're using time dependent states.  When you're computing the expectation value of the  Heisenberg operator, you're using the time equals 0  version of the states, but they are the same.  So we say that the Schrodinger expectation value is  equal to the Heisenberg expectation value.  We write it like in the bottom, but we mean the top  equation, and we use it that way.  So the Heisenberg operators at this moment are a little  mysterious.  They are supposed to be given by this formula, but we've  seen that calculating U can be difficult, so calculating  the Heisenberg operator can be difficult sometimes.  So what we try to do in order to simplify that is find an  equation that is satisfied by the Heisenberg operator, a  time derivative equation.  So let's try to find an equation that is satisfied by  the Heisenberg operator rather than a formula.  You say, well, this is better, but the fact is that  seldom you know U, and even if you know U, you have to do  this simplification, which is hard.  So finding a differential equation for the operator is  useful.  So differential equation for Heisenberg operators.  So what do we want to do?  We want to calculate I h bar d dt of the Heisenberg  operator.  And what do we get?  Well, we have several things.  Remember, the Schrodinger operator can have a bit of  time dependence.  The time dependence would be an explicit time dependent.  So let's take the time derivative of all this.  So you would have three terms.  I h bar d U dagger dt A s U plus U dagger A s d U dt plus  with an I h bar plus U dagger I h bar d A s d t.  I'm sorry.  Yeah, d A s d t and U.  So well, you have these equations.  Those were the Schrodinger equations we started with  today.  The derivatives of U are the derivatives of U dagger.  So what did we have?  Well, we had that I h bar d U dt was h U h Schrodinger times  U.  And therefore, I h bar d U dagger dt.  I take the dagger of this.  I would get a minus sign.  I'll put it on the other side.  It's equal to U dagger h s with a minus here.  And all the U's are U's of t and t naught.  I run out of this thick chalk.  So we'll continue with thin chalk.  All right.  So we are here.  We wrote the time derivative, and we have three terms to work  out.  So what are they?  Well, we have this thing.  I h bar this.  So let's write it.  I h bar d d t of A s of A Heisenberg.  I'm sorry.  I'm sorry.  It's equal to that term is minus U dagger h s A Schrodinger U.  The next term, plus I h bar d U dt on the right.  So we have plus U dagger A s h s d U dt.  So U, well, that's not bad.  It's actually quite nice.  And then the last term, which I have very little to say,  because in general, this is a derivative of a time dependent  operator, partial with respect to time.  It would be 0 if A s depends just say on x or p or s x  or any of those things has to have a particular t.  So just leave this as plus I h bar d A s d t Heisenberg.  The Heisenberg version of this operator,  using the definition that anything, any operator that we  have a U dagger in front, a U to the right,  is the Heisenberg version of the operator.  So I think I'm doing all right with this equation.  So what did we have?  Here it is.  I h bar d dt of A Heisenberg of t.  And now comes the nice thing.  Of course, this thing, look at it.  U dagger U. This turns everything here into Heisenberg.  H Heisenberg, A Heisenberg.  Here you have A Heisenberg, H Heisenberg.  And what you got is the commutator between them.  So this thing is A Heisenberg commutator with H Heisenberg.  That whole thing.  And then you have plus I h bar d A s d t Heisenberg.  So that is the Heisenberg equation of motion.  That is how you can calculate a Heisenberg operator,  if you want. You try to solve this differential equation.  And many times, that's the simplest way to calculate  the Heisenberg operator.  So there you go.  It's a pretty important equation.  So let's consider particular cases immediately  to just get some intuition.  So remarks.  Suppose A s has no explicit time dependence.  So basically, there's no explicit t,  and therefore this derivative goes away.  So the equation becomes I h bar d A s d A h, of course.  d t is equal to A h Heisenberg sub h of t.  And you know the Heisenberg operator is supposed to be  simpler, simple if the Schrodinger operator is time  independent, is identical to the Schrodinger Hamiltonian.  Even if the Schrodinger operator has time dependence,  but they commute, this will become the Schrodinger Hamiltonian.  But we can leave it like that.  It's a nice thing anyway.  Time dependence of expectation value.  So let me do a little remark on time dependence  of expectation value.  So suppose you have the usual thing that you want to compute.  How does the expectation value of a Schrodinger operator  depend on time?  You're faced with that expectation value of A s,  and it changes in time, and you want to know how you can  compute that.  Well, you first say, OK, the I h bar d d t,  but this thing is nothing but psi 0 A Heisenberg of t psi 0.  Now I can let the derivative go in.  So this becomes psi 0 I h bar d A h d t psi 0.  And using this, assuming that A is still no time dependence,  A has no explicit time dependence,  then you can use just this equation, which gives you psi 0 A h H H  psi 0.  So all in all, what have you gotten?  You've gotten a rather simple thing, the time derivative  of the expectation value.  So I h bar d d t.  And now I write the left-hand side as just the expectation  value of H Heisenberg of t.  Now the left-hand side has the A Schrodinger expectation value,  but we call those expectation values the same thing  as a Heisenberg expectation value.  So this thing becomes the right-hand side,  and the right-hand side is the expectation value of A Heisenberg,  H Heisenberg, like that.  And just the way we say that Heisenberg expectation values  are the same as Schrodinger expectation values,  because you could as well write it if you prefer as d d t of H  Schrodinger is equal to expectation value of H Schrodinger  with H Schrodinger.  It's really the same equation.  This equation we derived a couple of lectures ago,  and now we know that the expectation values of Schrodinger operators  are the same as the expectation values of the Heisenberg counterparts,  except that the states are taking a time equals 0.  So you can use either form of this equation.  The bottom one is one that you've already seen.  The top one now looks almost obvious from the bottom one,  but it really took quite a bit to get it.  One last comment on these operators.  How about conserved operators?  What are those things?  A time-independent, A s, is said to be conserved  if it commutes with the Schrodinger Hamiltonian.  If A s commutes with H s equals 0.  Now you know that if A s with H s is 0,  A h with H h is 0,  because the map between Heisenberg and Schrodinger pictures  has a commutator that is valid in the Schrodinger picture,  is valid in the Heisenberg picture by putting h's.  So what you realize from this is that this thing,  this implies A h commutes with H h,  and therefore by 0.1, by 1, you have dA h dt is equal to 0.  And this is nice, actually.  The Heisenberg operator is actually time-independent.  It just doesn't depend on time.  A Schrodinger operator, it's a funny operator.  It doesn't have time in there.  It has x's, p's, spins, and you don't know in general  if it's time-independent in the sense of expectation values.  But whenever A s commutes with H s,  well, the expectation values don't change in time.  But as you know, this d dt can be brought in  because the states are not time-dependent.  So the fact that this is 0 means the operator,  Heisenberg operator, is really time-independent.  Whenever you have a Schrodinger operator, has no t,  the Heisenberg one can have a lot of t.  But if the operator is conserved,  then the Heisenberg operator will have no t's after all.  It will really be conserved.  So let's use our last ten minutes to do an example  and illustrate much of this.  In the notes, there will be three examples.  I will do just one in lecture.  You can do the other ones in recitation next week.  There's no need really that you study these things at this moment.  Just try to get whatever you can now from the lecture.  And next week, you'll go back to this.  So the example is the harmonic oscillator.  And it will illustrate the ideas very nicely.  The Schrodinger Hamiltonian is p squared over 2m  plus 1 half m omega squared x hat squared.  OK.  I could put x Schrodinger and p Schrodinger,  but that would be just far too much.  x and p are the operators you've always known.  They are Schrodinger operators.  So we leave them like that.  Now I have to write the Heisenberg Hamiltonian.  Well, what is the Heisenberg Hamiltonian?  Yes?  Yes, I did.  Sorry?  Yes, I did.  Identical?  Yes.  But I will leave that for a little later.  I will just assume, well, I'm supposed to do u dagger u.  And as you said, this is a time independent Hamiltonian.  It better be the same.  But it will be clear if we now write what it should be in general.  Have a u dagger and a u from the right.  They come here and they turn this into p Heisenberg over 2m  plus 1 half m omega squared x Heisenberg.  OK.  That's your Heisenberg Hamiltonian.  And we will check, in fact, that this time independent.  So how about the operators x Heisenberg and p Heisenberg?  What are they?  Well, I don't know how to get them unless I do this sort of u thing  that doesn't look too bad, but certainly would be messy.  You would have to do an exponential of e to the minus i ht  over t with the x operator and another exponential.  Sounds a little complicated.  So let's do it the way the equations of the Heisenberg operators tell you.  Well, x and p are time independent Schrodinger operators.  So that equation that I box holds.  So i h v x Heisenberg dt is nothing else than x Heisenberg  commuted with h Heisenberg.  OK.  Can we do that commutator?  Yes.  Because x Heisenberg, as you remember, just commutes with p Heisenberg.  So instead of the Hamiltonian, you can put this.  This is x Heisenberg p Heisenberg squared over 2m.  OK.  Well, x Heisenberg and p Heisenberg is like you had x and p.  So what is this commutator?  You probably know it by now.  You act with this on these two p.  So x on one, x on the other gives the same on each.  So you get a p Heisenberg times the commutator of x and p,  which is i h bar, times a factor of 2.  So we could put hats.  Better, maybe.  And then what do we get?  We get the h there and the i h cancels.  And we get some nice equation that says dx Heisenberg dt  is 1 over m p Heisenberg.  Well, it actually looks like an equation in classical mechanics.  dx dt is p over m.  So that's the good thing about the Heisenberg equations of motion.  It looks like ordinary equations for dynamical variables.  Well, we've got this one.  Let's get p.  Well, we didn't get the operator still, but we got an equation.  So how about p dp dt?  So i h dp Heisenberg dt would be p Heisenberg with h Heisenberg.  And this time, only the potential term in here matters.  So it's p Heisenberg with 1 half m omega squared x Heisenberg squared.  So what do we get?  We get 1 half m omega squared.  Then we get, again, a factor of 2.  Then we get 1 left over x h.  And then a p with x h, which is a minus i h bar.  So the i h bars cancel, and we get dp h dt is equal to the h bar cancel  m omega squared x h minus m x.  All right.  So these are our Heisenberg equations of motion.  So how do we solve for them now?  Well, you sort of have to try the kind of things that you would do classically.  Take a second derivative of this equation.  The second x h dt squared would be equal 1 over m dp h dt.  And dp h dt would be minus 1 over m times minus m omega squared x h.  So d second x h dt squared is equal to minus omega squared x h.  Exactly the equation of motion of a harmonic oscillator.  It's really absolutely nice that you recover those equations  that you had before, except that now you're talking operators.  And it's going to simplify your life quite dramatically when you try to use these operators.  Because in a sense solving for the time dependent Heisenberg operators is the same as finding the time evolution of all states.  This time the operators change, and you will know what they change like.  So you have this, and then you write x h is equal to a cosine omega t plus b sine omega t,  where a and b are some time independent operators.  So you get x h of t.  Well, that's the solution.  How about what is p, p h of t would be m dx m dx dt.  Minus m omega sine omega t a plus m omega cosine omega t b.  OK, so that's it.  That is the most general solution.  But it still doesn't look like what you would want.  That's it.  No, because you haven't used the time equals zero conditions.  At time equals zero, the Heisenberg operators are identical to the Schrodinger operators.  So at t equals zero, x h of t becomes a, but that must be x hat, the Schrodinger operator.  And at t equals zero, p h of t becomes equal to, this is, zero m omega b.  And that must be equal to the p hat operator.  Actually, we have already now a and b, so b from here is p hat over m omega.  And therefore, x h of t is equal to a, which is x hat cosine omega t,  plus b, which is p hat over m omega sine omega t.  And p h of t is here.  A is, p h of t is m omega b, which is m over s p.  So it's p hat cosine omega t minus m omega x hat sine omega t.  So let's see.  I hope I didn't make mistakes.  p hat minus m omega x hat sine omega t.  Yep.  This is correct.  This is your whole solution for the Heisenberg operator.  So any expectation value of any power of x and p that you will want to find its time dependence,  just put those Heisenberg operators and you will calculate things with the states at time equals zero.  It will become very easy.  The first thing I want to do is complete the promise that we had about the, what is the Heisenberg Hamiltonian?  Well, we had the Heisenberg Hamiltonian there.  And now we know the Heisenberg operators in terms of the Schrodinger one.  So h h of t is equal to p h 1 over 2 m p h squared.  So I have p hat cosine omega t minus m omega x hat sine omega t squared plus one half m omega squared.  So x h squared, so x hat cosine omega t plus p hat over m omega sine omega t squared.  So that's what the Heisenberg Hamiltonian is.  So let's simplify this.  Well, let's square these things.  So I have 1 over 2 m cosine squared omega t p hat squared.  Let's do the square of this one.  We would have plus 1 over 2 m m squared omega squared sine squared omega t x squared.  And then we have the cross product, which would be plus or actually minus 1 over 2 m.  The product of these two things, m omega sine omega t cosine omega t, and you have p x plus x p.  OK, I squared the first terms.  Now the second one.  Well, let's square the p squared here.  What do we have?  One half m omega squared over m squared omega squared sine squared of omega t p squared.  The x plus one half m omega squared cosine squared omega t x squared.  And the cross term plus one half m omega squared over m omega times cosine omega t sine omega  t x p plus p x.  A little bit of work.  But what do we get?  1 over 2 m, and here we must have 1 over 2 m, correct, 1 over 2 m sine squared omega  t p squared.  So this is equal 1 over 2 m p squared.  And here you have 1 half m omega squared, so it's 1 half m omega squared.  Cosine and sine squared is x hat squared.  And then here we have omega over 2, and here omega over 2.  Same factors, same factors, opposite sines.  Very good.  Schrodinger Hamiltonian.  You confirm that this theoretical expectation is absolutely correct, and what's the meaning?  You have the Heisenberg Hamiltonian written in terms of the Heisenberg variables, but by the time you substitute these Heisenberg variables in,  it just becomes identical to the Schrodinger Hamiltonian.  All right, so that's all for today.  I hope to see you in office hours in the coming days.  Be here Wednesday, 12.30.  Maybe 12.25 would be better, and we'll see you then.  All right.  Thank you.
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