 So we arrived yesterday at a statement of the fundamental dynamical principle of quantum mechanics, the time-dependent Schrodinger equation here. I didn't yet justify, I didn't offer any justification for this at this time, the justification or arguments which will make this seem plausible are about to come, but they have not yet come. However, we already, at the end of yesterday's lecture, realized that if a psi, if the state of our system is such that the energy is well determined, the result of measuring energy is certain, then the time evolution of that state is absolutely trivial. The state evolves only in that its phase increments at a rate with a frequency, an angular frequency, e upon h bar, which is typically very large, because h bar is very small. And we went on from that result there to show that the solution to this equation for any state of psi takes this form, that the state of the system at an arbitrary time is this sum over the states of well-determined energy, where an, as ever, is en evaluated. So if you evaluate these coefficients, that time t equals 0, or indeed at any time, at your convenience, then just by inserting into this sum, this ordinary sum which we've seen few times of psi expanded in terms of a complete set of states, if we just insert these exponential factors, bingo, we evolve the state according to that equation. So that makes these states of well-defined energy of crucial operational significance. And the result of that is that we spend a great deal of time solving the defining equation of these states, which is that h en is equal to en en. These states of well-defined energy are by construction eigenstates of the Hamiltonian operator. And this is the time-independent Schrodinger equation to distinguish it from the time-dependent up there. So only states of well-definite energy solve the time-independent Schrodinger equation. Any state, the state of our system, always has to solve the time-dependent Schrodinger equation. So what we want to do now, the next item on the agenda fundamentally is to provide some justification, make it seem plausible that that is the correct equation of motion. And a good way to do that is to link back to classical mechanics. So recall, classical mechanics, classical physics is that limit where the uncertainty in the values of dynamical variables is sufficiently small that it isn't necessary to calculate the whole probability distribution. It's enough to know what the expectation value of the probability distribution is because the true value will be very close to the expectation value, and we all really don't distinguish between the expectation value and the true value when doing classical physics. So what we want to do is calculate d by dt of the expectation value of some observable, I think we're calling observables q. And let's put an ih bar in front of here just in order to simplify the algebra. So we're trying to calculate this. The rate of change of the expectation value of something, this something might be position. So this might be x, this might be momentum, it might be energy, it might be whatever you want to know, might be angular momentum, whatever you want to know about the system. This would tell you the rate of change of the classical value of that variable because the classical value is the expectation value. So what is that? Well, this is just a product. So obviously, it comes into three parts, ih bar d psi by dt times q times psi plus, let's take the last bit now, plus psi dq by dt, sorry, these things probably should be whether we make these things partial derivatives or total derivatives is of no significance. This is clearly a total derivative because this thing doesn't depend on anything except time, whether we make these partial or total is very unimportant. So here we have an ih bar rate of change of psi. So this can immediately be replaced by h of psi. Here we have the bra of psi and the original equation, we can take the Hermitian adjoint of the original equation when it becomes minus ih bar d by dt of the bra of psi, because it's bra of psi equals of psi h. Remember the rule when we take Hermitian adjoints is we reverse the order of the symbols and we take the adjoints of the individual bits. h is an observable, so it's a Hermitian operator. So h equals h dagger. So I can pop those into here and here. So the first two terms give me, including the ih bar. This one is going to be minus of psi h. So that is ih bar d by dt of the bra of psi times q times of psi. This one, including the ih bar, this is going to be plus of psi qh. And then we have a trailing bit here, which will be plus ih bar q, sorry, psi dq by dt of psi. And these two can be handily combined together into a commutator, because there's a minus sign here. Otherwise the order of the q and the h are swapped around. So this can be written as psi commutator q comma h of psi. And then this is plus ih bar. So this result goes by the name of Ehrenfest theorem. And it's one of the more important results of quantum mechanics. In most of our applications, we can forget about this, because what is this last term here? It's the expectation value of the rate of change of your observable. So if the observable was, for example, x, it would have no rate of change. x is x is x. It's always the same operator. If your observable were p, it would have no rate of change, because the momentum is momentum is momentum. It doesn't change our understanding of what momentum is and doesn't change from moment to moment. If it were angular momentum, it doesn't change from moment to moment. So this term here usually falls away. And so if dq by dt equals 0, which is the normal state of affairs, then we have the ih bar d by dt of the expectation value of q, which is often written like this in shorthand notation. We've left out either psi. Just leave the angle brackets, which means the expectation value. In any state whatsoever, is equal to the expectation value of the commutator q, comma, h. OK. What does that tell us? It tells us immediately that if an observable commutes with the Hamiltonian, if q commutes with h, that being the same thing as q, comma, h equals 0, then clearly the expectation value of q is always constant. And physicists are always very excited by quantities which are constant. So what do we call it in classical physics? We'd call it a constant of motion. Famously, Newton said that if you didn't go for messing with particles, the momentum was constant. Or maybe he said the velocity was constant, right? So velocity is a constant of motion. We now would reinterpret that as momentum is constant. We often know that the angular momentum of the Earth is almost constant, insofar as it's not acted on too much by the moon, and so on and so forth. So physics is full of constants of motion. They're very important. In quantum mechanics, there's a special notation around here, which is to say that we say that q, the eigenvalue of this is a good quantum number. So if somebody says that angular momentum is a good quantum number, they're simply saying angular momentum is a constant of the motion. We can go further than that, though, because so we've shown that the expectation value of q is constant. We could show easily that the expectation value of q squared is constant, too. It's very straightforward, which is obviously the case, because q squared is an observable. If q commutes with the Hamiltonian, then q squared has to commute with the Hamiltonian. So q squared is also a constant of the motion. That means that if you start, so this implies that if initially, if at t equals 0, psi is one of the states of well-defined q. So qr, one of the eigenstates of the operator q, so we know for certain that the result of a measurement will be qr, then what does that mean? That means that at t equals 0, the expectation value of q is obviously equal to qr. And the expectation value of q squared is equal to qr squared, which implies that the variance of q, that implies that the variance of q, which is as ever defined to be the expectation value of q squared minus the expectation of q itself squared, vanishes. So this variance expresses, of course, the fact that there's no uncertainty in the value that we get. But since this is a constant of the motion and this is the constant of the motion, if this variance vanishes at t equals 0, it vanishes at all times. And the only way that that can happen is if your system stays in the state that it was in originally. So if it was in a state of well-defined q at the beginning of t equals 0, it'll be in the state of well-defined q at all subsequent times. So that's the meaning of this good, that's why people talk about good quantum number. If I know at some particular time that the angular momentum of this isolated body is h bar or whatever, then I know that at all later times, it's also h bar. It's a quantum number worth knowing because it's always valid information. So we're very interested in quantities in observables that commute with the Hamiltonian. Let's move over here for the next point. Obviously, OK, so we're interested in things that commute with the Hamiltonian. And a trivial observation is that h comma h equals 0. This is a bad place, isn't it? I'm not sure there's anything we can do about it. So the Hamiltonian commutes with itself, which means that the expectation value of h is a constant if dh by dt equals 0. So now we need to come back to this point that I said usually it's going to be the case that the partial derivative of q with respect to time, I've lost it, it's up there somewhere, the partial derivative of q with respect to time vanishes. Now the Hamiltonian is an interesting case where there isn't necessarily the case. There are very important circumstances in which the Hamiltonian does explicitly depend on time. The expression for the energy depends on time. For example, if you put a particle in a time-varying magnetic field, the expression for the Hamiltonian, which the Hamiltonian is going to depend on the magnetic field, depends on time. And in those circumstances, the energy of the particle is not going to be constant. And the reason is you're working on it. That time depends on the Hamiltonian reflects the work that you're doing on the particle. But if the Hamiltonian is independent of time, that will reflect you're not doing any work. And then the expectation value of h equals constant is conservation of energy. So this condition, it will turn out, is intimately connected to whether or not you're working on the particle. Now let's have a look at the rate of change of the expectation value of any observable when we are in a state, when psi happens to be a state of well-defined energy. So these states of well-defined energy, we've explained that they're the key to solving the central equation of the theory. So let's ask ourselves a little bit about those states. So the amplitude, let's have a look at q, e. This is the amplitude to determine q, given that we're in the state e. Well, this is an eigen, this is an eigen. So let's give it, let's give it its n, qn. So this is the amplitude to find the value qn if you would measure with the observable q, given that you're in a state of well-defined energy e. Let's work out the time derivative of this. i h bar d by dt of this quantity is equal to, this is so, it's very specific, d qn by dt. I wonder if we should turn this off. Is there any way we can turn it down? We do the same thing. This has to be i h bar thing. We get from the Hermitian adjoints of the equation. And we get the time-dependent Schrodinger equation. So we get a minus qn h. So that is that. And then we put e onto it. And this is going to be qn stands by, while that, we put it in the i h bar, produces an h, e. Nothing very much, right? Because h works on e to produce e, because that's an eigenstate h. e times the ket e, compared to this, reduces minus e times qn e. And this h produces an e. And so we get a plus e times qn e. So these two terms cancel. So we've discovered the rate of change of the amplitude to have the value qn is constant. The rate of change vanishes. This amplitude, qn e, any q. In order to obtain the result, we didn't make any restriction. Any restriction whatsoever of what the observable q was. So the remarkable fact is that in these states of well-defined energy, if your system has well-defined energy, all its properties, the expectation value will ever be observable. And by thinking about, in fact, then follows with a couple of extra steps, the probability distribution of measuring any observable whatsoever is completely constant, nothing of change. So that leads to e being called stationary state. These states really are forever. They are completely eternal. They're completely unchanged. They are not of this world. And in particular, you can never get a system into a state of well-defined energy, because if it's going into there, it implies that something's changing. But they never change. Kind of a remarkable one. So now we have a new topic, representation. We'll bring us much closer to the real visible world, as far as we're reaching it. So far, we've talked about we use abstract representations a little bit like the energy representation. Let's say we've assumed that our observable has discrete spectrum. So the spectrum is made up of discrete numbers. The thing about the position operator, this is the thing made up of the operation, which encodes the states of well-defined position down the x-axis. It has a spectrum, which is usually continuous. It runs from minus infinity to infinity. It's not discrete, it's continuous. And this requires some adjustment. So we have been writing, let's divide the board, we have been writing that psi is equal to sum a n, shall we say, e n. We use the energy representation. Now we're going to write that psi is equal to an integral. That sum over the discrete set of possible, the discrete values numbers in the spectrum becomes an integral over the possible values in the spectrum. So that reminds me of infinity to infinity. Of some amplitude, the amplitude to be at x times the state of well-defined definitely being at x. So this is the state that the system is in and where it is at x, when I particle whatever is at x. And this is the amplitude to be at x, right? Amplitude to be, and this is the state of being at x. We used to have that e m, e n equals delta m. What are we going to have now? Let's draw this thing through my x prime. And we're going to have the integral d x prime, sorry, d x. x prime x of psi of x. This side, oops. This side is clearly equal to psi, sorry, sorry, sorry. What do we want? What we want is that this thing vanishes, except when x prime is equal to x, because if it's definitely at x, sorry, if it's definitely at x, then it certainly isn't at x prime if x prime is different from x. So this thing here is nothing except when x equals x prime. And it must be non-zero, and presumably rather large. I think we can fill this in. This is, what is this? By definition, this is the amplitude to be x prime. And I've already said that that is the amplitude to be at x prime. Sorry, x. So it's clear that this thing here is psi of x prime, the amplitude to be at x prime. So we have the amplitude of psi of x prime, that function is equal to dx of x prime x of psi of x. And I hope that you, I'm sure you've already met this relationship. This can also be written as delta of x prime minus x psi of x. So we have, so that result up there has been generalized, or is morphed into the state that the x prime to be x is equal to a drag function, rather than running the delta, delta of x minus x prime. We used to have that psi was the sum of the a n squares, was 1, by conservation probability, that this was the sum of the probabilities to get the value en, since you had to get some value, that sum of probabilities had to be 1. Now what do we have? What does this turn into? Um, this, this turns, this relation here turns into psi of psi should be 1. And how do we, how do we express it? Like this, we say that this is the integral of dx of psi x x prime. So where we're, here we're using the idea, sorry, that we used to have that the sum en, en is the identity operator. Now we have that the integral dx is the identity operator. So every, all of these sums are turning into integrals. So this relationship becomes this because this is the identity operator, slugging to there. And this is what is the amplitude to be at x. So we already call that the wave function of psi of x. This is the complex conjugate of that. Therefore the complex conjugate of this. So this becomes the integral dx of mod of psi squared. So the integral of mod of psi squared should be 1 in this continuous range. So these are just natural transformations of what we've already done in the discrete case to the continuous case. The one more thing that we need to write down, we used to have, we used to have that if psi was the sum en en and phi was psi en en, then the complex number of phi psi was the sum en star en sum of n. So that's a result we have had. What's the analogous thing here? The analogous thing here is going to be the phi of psi is going to be the integral. Into here we stick an identity operator, the integral dx of ket x, yx. So this becomes the integral dx. So we stick an identity operator into that cap of phi x or x of psi. This is what we have been calling the wave function of psi of x. It is the amplitude to be at x, which appears as a function of x. It is a function of x. It's a complex number, depending on psi of x. And by analogy, we should call x psi, x phi, sorry, we should call the wave function phi x. That being so, this becomes the complex conjugate of it. It becomes an integral of x. So both of these things of course are functions of x. So this is precisely a transformation of that with the sum where n has been replaced with x. And the sum of n becomes an integral of x. This is the stuff we have to do with the spectrum of x is continuous and not discrete. Let's just do a little practice with this by asking ourselves, how does the operator x work on an arbitrary state of psi in this representation? So what we want to know is, the thing to do is to ask ourselves, what wave function represents that? That is to say, what is this complex number as a function of x? So here's the operator x. Here is an arbitrary value of x. I would like to know what the amplitude v at x is for this state that you get when the operator x works on the integral of the wave function x. When you see an operator x, the obvious thing to do is stick into here an identity operator made up of the eigenfunctions of x. So in order to understand what this is, what we do is we slide into here one of these identity operators. x is busy, so my identity operator is going to have to be a sum of x prime, some new value, some independent value of x. x, x, x prime, x prime psi. So here's the identity operator along with that. Now, like this relatively straightforward, because this is an eigenfunction of that operator with this eigenvalue. That's the definition of this answer here. So when x meets this, it produces simply x times x prime to the number times the ket x prime. So this becomes the integral, the x prime, or x prime. There's the eigenvalue popped out. Then we have x, x prime, and this we recognize to be the wave function of psi evaluated on x prime. But this we recognize now, we've seen that this is the derived delta function of x minus x prime. So when we do the integration over x prime, this ensures that we get no contribution except for that little second when x prime is equal to x, well, and we get the value of the integrand evaluated with these eigenlines turned into x. So this is equal to x of psi of x. So at the end of a long story, what have we discovered? We've discovered that the wave function associated with the result of using the operator x on some state is simply x times the wave function of the original state. We can express that if the way to remember that is to say that the operator acts of wave functions by order of multiplication. So you don't usually go through this kind of performance. You know what's going to happen when you do it. But that's the logical basis for this statement. Let's introduce another very important operator, the momentum operator. Now I'm going to make an unsubstantiated claim about what this operator, how this operator looks in the position representation. I don't expect you to think, aha, that makes sense. It doesn't make sense. It's a complete leap in the dark. We will understand later, considerably later, why these operators take the form that they do. But I hope soon to build some confidence in some kind of sense that what we're going to do just right now is completely in the dark. I'm going to say, well, let's investigate the operator. I have to know from a mental point of view, but let's investigate it, which is defined thus. Now let's just make sure we understand fortunately what's happening here. An operator fundamentally is something which turns a state into another state. When we're in the position representation, we are, when we're in the position representation, we are, when we're working with functions, our states are represented by their wave functions, which is the amplitude v at x. So the x operator, the x operator has to turn a wave function of psi into some other function and look, x of psi is another function. It depends on x in a different way than the way psi does. So similarly, this operator, I claim is a momentum operator without any basis, I'm going to try again. This momentum operator is turning the wave function of psi into its derivative, and derivative is a function different from the function we first thought of. So indeed, it's turning a function into a function. So that's kind of, it means it is at least a valid operator. Is it a mission? Is it a mission? And if it is a mission, it certainly can't be the momentum operator. So let's check that out. So let's write down the complex number phi p hat of psi. Let's evaluate this using this Hocus Pocus here, right? So this is the integral dx. What am I going to do? I'm going to put an identity operator just in here made up of x's, right? Why? Because I know I've defined p in terms of what happens when it has an x on the left of it. So this is going to be phi x x p hat psi. And now we can turn this into wave function language. This is the complex conjugate of the wave function phi. So this is the integral dx. So we can now integrate by parts. We're integrating momentum infinity to infinity. So we can integrate by parts this partial derivative to get this partial derivative of psi and onto the phi. So what does that give me? We get a square bracket term. We have a phi star, let's put the minus i h bar outside some vast bracket. So we're going to have a square bracket term now. We're going to have a phi star of psi minus infinity. And then we're going to have minus the integral dx d psi, sorry, psi d phi star, close the big bracket. So we can operate under the assumption that this thing vanishes. Now this is a rather hairy, don't press too hard as to whether this really does vanish. But the general idea is that the amplitude to find your particle at the edge of the universe is 0. So we dispose of this on the grounds that it is the amplitude to find the particle infinitely far away. We're going to say that that's 0. We will actually be working, you'll see, quite soon with some wave functions where that doesn't vanish. And this is an example where physicists are rather fast and loose. And but fortunately, this doesn't lead to any bad effects. So this we put in the pin. And this we can see is more or less what we want. Let's just tidy up a bit. So what is this that survives, including this minus i h bar? So we're going to have a minus. Let's leave the minuses out. dx of psi. And then we're going to have an i h bar d phi by dx. So I think I take the star of all that. Then I think I need a minus sign. Because this minus will cancel on this minus. So we'll have an i h bar times this stuff with psi star. If I take that psi, that star, and put it around the whole caboodle, including the i, I'll need an additional minus sign to cancel the minus sign that will arise when that star is evaluated on here. And I can say now that that is the integral dx of psi star minus i h bar d phi by dx star. So if I take the star, this star, completely outside the whole thing, then psi will need a star, which will be canceled by the global star. And otherwise everything will be OK. And what is this that we have in here in direct notation is psi p hat phi. And that star sits outside. What's inside the star is, by definition, this. So there we are. The answer is, it is Hermitian. Provided we get rid of that surface term, that minus infinite, the square bracket. One more. OK, let's calculate the commutator x hat comma p hat. We're going to calculate it like this. So what we're going to do is calculate the action. So all we know at the moment is the action of p hat on any wave function. And we know the action of x on any wave function. So I want to work with wave functions, which means I put a bra at this end here, a bra x at this end here. And what does this give me? So this is going to be, obviously, x x hat p hat psi minus x p hat x hat psi. No prizes for that. Now, this we've discovered x on this. We've discovered that x on any wave function, on any object, gives you x times the wave function you first thought of, the wave function you were operating on. What is the wave function that this produces? The wave function that this produces is minus i h bar d psi by dx. So we really have to take that and multiply it by x. And then that's what you get there. This is a complex number depending on x. This is a complex number depending on x. So p on this is a certain wave function. It's this. And then x hat on that produces x times that wave function. So that's it. So here, same stuff. x on this is going to produce x of psi, x of psi. And then p on that is going to produce minus i h bar d by dx of that stuff. And there's a minus sign floating here. I mean, this minus sign is this. That minus belongs to the p operator. So I think you can see that the x, the psi by dx terms. When you differentiate out this product, you'll get two terms. You'll get x d psi by dx, which will cancel on this because of the two minus signs. And you will also get an upside times the derivative of x with respect to x, which is obviously 1. So we're going to get i h bar of psi of x, which can also be written as i h bar x of psi. So what have we learned? What we've learned is that for any state of psi whatsoever, we never said what it was, the wave function associated with x p of psi is simply i h bar times the wave function of psi. So that means that we can now write down an operator statement that x hat comma p hat is equal to i h bar. So the commutator of these two operators is a constant, small constant, but constant, and a commutation relation of this sort is called a canonical commutation relation. We will meet other relations of this type where the commutator of two operators is equal to i h bar, and they will be declared canonical as well. The word, this canonical, of course, comes from classical mechanics, Hamiltonian mechanics. And this arises because in classical mechanics, momentum is canonically conjugate, quote, unquote, to x. Right. So now that we've done that, let's, you know, just got time, I think, to do this, let's apply Ehrenfest's nice theorem to, so I'll begin, let's work out this. I h bar d by dt of the expectation value of x. What do you think this should be? The rate of change of the expectation value of x should be the speed, right? Don't I say that right? The rate of change of the expectation value of x should be the speed, velocity, whatever. So we're hoping that this turns out to be i b, which should be i p upon m, if we're doing this right. According to Ehrenfest, what's this equal to? It's equal to psi x comma Hamiltonian, psi, right? That's Ehrenfest's theorem. Concrete example of, application of. So in order to go further, we need to say, so what's h? H is the energy operator. What do we know about the energy of a particle that's moving in, possibly, with some potential present, right? So the energy should be a half, classically, see Callie, if we're doing this classically, I should replace this with an energy, should be a half m v squared plus the potential energy depending on x, which could also be written as the momentum squared over 2m plus the potential energy, right? Because p, classically, is m v. So let's suppose that we can carry this forward into the quantum domain and say that the Hamiltonian operator is the momentum operator over 2m plus v, the function v evaluated on the position operator. Then we're going to have that ih bar db dt of the expectation value of x hat is going to be the expectation value of x commuted with p squared over 2m plus v. But we know that, so this, but this commentator can be broken down into a sum of two commutations. The commutator of x with p and the commutator of x with v. But v is a function of x, and therefore, x, the position operator, is going to commute with this. So we're going to have that x comma v equals naught because v is a function of x. So what we're left with is the expectation value of p, p, sorry, expectation value of the commutator of x with p, p, psi over 2m. I can take the 2m out of the commutator because it's just a number. And I can express p squared as pp. But we discussed probably yesterday how we took the commutator of a product. We used a rule in alligators to differentiation of a product. So this is equal to psi onto x comma p commutator p standing idly by plus p standing, the first p standing idly by while x commutes with the second p. But we've discovered that this animal is ih bar. This is ih bar, and this is ih bar. ih bar is just a boring number, so it can come out front. So that becomes ih bar over 2m. And then we have p plus p, which is 2p. So I can rub out that 2 times. So what have we discovered? We've discovered that we can cancel this ih bar on the right side with what we had on the left side and say that d by dt of the expectation value of the position is, in fact, equals the expectation value of the momentum, what I claims the momentum anyway, over m, which is exactly what we were hoping for. So we've recovered the relationship between velocity and momentum, which in classical in Hamiltonian mechanics is a rather, those of you who've done S7 will realize that the connection between momentum and velocity is not as simple as elementary Newtonian mechanics would lead you to believe. It can be quite subtle, and it's determined by this, which is one of Hamilton's equations. What we've done is derived one of Hamilton's equations, which supersede Newton's laws of motion in classical physics. So we derived from quantum mechanics a classical result, which was already known. But this is the justification for Hamilton's equations. Because this is true, that Hamilton's equation is true. And we'll leave it on that. And tomorrow morning, I'll start by deriving the other of Hamilton's equation, which is analogous to f equals ma.