 All right, last time I was explaining time reversal and classical mechanics. We take a trajectory of a particle in three-dimensional space and reverse time are, if that's the considered the trajectory you get in the random movie backwards. We find that if the question we ask is whether the time-reversed trajectory also satisfies Newton's laws, the answer is it depends on the nature of the forces. The force, for example, is an electric force of a charged particle, then the time-reversed motion does satisfy Newton's laws, and instead physically allow motion. That's because the acceleration on the left-hand side involves only a second event, so the two minus signs are changing the sign of time cancel out. Whereas if the forces are due to magnetic fields, because there's a d cross d force, the force is a velocity-dependent force, when they change the sign on time, it changes the velocity. And so the right-hand side changes side of the left-hand side where the acceleration doesn't. So it's a basic rule then that motion in an electric field is time to personal invariant, whereas motion in a magnetic field is not. Now, by making these statements, it depends on what you call how you define the system. Here, we're thinking of the electric magnetic field as just being given external fields in which the particle is moving. There was a similar issue you would recall in regard to carrier. However, if we include the charges in the currents which produce the electric magnetic fields in the definition of the system, then there's a different situation. As you know, charged densities basically give rise to electric fields, whereas charged currents give rise to magnetic fields. Charged currents involve the velocities of the particles. So in the Trimber-Buschel, it's fairly clear that the charged density goes into itself in a change in the motion of the particles whereas the charged current goes into sign into the properties of densities and currents under Trimber-Buschel. And the result of this is that electric fields go into themselves under Trimber-Buschel, whereas magnetic fields change sign. And these are part of the transformation laws which are involved in understanding at a deeper level than I've given you here, understanding the transformations of the magnetic field under Trimber-Buschel. In any case, if we apply these transformation laws to the fields in the right-hand side, in other words, if we include the charged currents as a definition of my system, then we see the equations are mapped into themselves under Trimber-Buschel. The Trimber-Buschel invariance is restored. So there's two lessons here. This is that Trimber-Buschel invariance depends on the definition of the system. But another one is that the electromagnetic forces are actually varying under Trimber-Buschel, including electric electromagnetic fields themselves. All right. So this is classical mechanics. Now, what about quantum mechanics? Let's suppose we've got a, I'll do the same kind of game. Let's suppose we have a solution to the equations of motion. And let's say it can be specific that it's still less particle moving in three dimensions. So we've got some time-dependent wave function like this. Well, we'll suppose if the solution of the Schrodinger equation, time-dependent Schrodinger equation, h bar d dt psi, is equal to a Hamiltonian times psi. Well, then we'll totally go with chews. Let's repeat kind of what we did here but at a quantum level. First, let's take an electric field. If we did, then the Schrodinger equation, of course, is minus 6 bar squared over 2 and plus 2 squared. And then plus q times the scalar potential pi of r multiply it on the psi like this. So this is the Schrodinger equation for a particle or a different electrostatic field. Now, again, repeat, let's suppose psi of r dt is a solution of this equation. The question that we first want to ask is, what if we just change the direction of time? Is psi of r common minus d? Is that also a solution? I'm just repeating the same type of question that we asked earlier in the class of the level. And the answer in this case is no, unlike the classical case, because Schrodinger equation is first order in time, not second order. And if we just change the sign of time, then the left hand side changes sign, but the right hand side does not. And so if this is our definition of the time of our state in one mechanics, then it is not a solution of the original Schrodinger equation. However, we can fix this up and obtain a genuine solution if we replace psi of r common minus t by its complex conjugate. This is a new way function obtained from the old one. And the new way function does satisfy the original Schrodinger equation, because in changing the sign of time, there is one minus sign introduced from the left hand side, but I'm thinking the complex conjugate of the Schrodinger equation is a second logic. This is the i here. And so time reversal invariance is restored, or as we say, time reversal invariance holds for the emotion of a charged particle in the lecustatic field, just like in the classical case. However, it's necessary to make this the definition of the time reversal solution. Similarly, if we have a magnetic field, let's take this as the definition of the time reversal solution. But let's suppose there's a magnetic field now, then the Schrodinger equation involves the usual thing. It's the momentum minus psi h by 3 minus q over c times the vector potential squared of y psi, like this. And you can now see that even with this definition of the time reversal state, it does not satisfy the original Schrodinger equation, assuming the first way function does satisfy it. The reason this is not taking a complex conjugation, the minus i gradient will turn into a plus in itself, but the q over c a doesn't. So these two terms suffer an opposite sign for the change of the sign. And so the Hamiltonian operator is not going to over into itself. And so it's just like in the classical case. We have time reversal invariance for lecustatic and emotional lecustatic fields with not in the case of magnetic fields. We're going to be talking about a time reversal operator, sorry, time reversal operator in a bar mechanical system, which we call theta. This operator does not involve time itself. It's rather it's an operator that maps cats into cats with state factors into the cell. So I write it like this, that's what it means. It's a mapping. And what we will do is interpret the following. If we take the state as a function of time, let's say this is a solution of Schrodinger equation, we replace t by minus t. And then we apply this time reversal operator. We will regard this as a time reversal solution, which I'll call psi r of t. In wave function language, psi r of t is just the complex conjugate of the original wave function t replaced by minus t. But in general language, our time reversal operator will satisfy this equation. In fact, in a special case of a spinless particle in 3D, if I write this out of the wave function language, it's just what I had in the previous form, is that psi of r of t is going to go over into a psi of r minus t complex conjugate, which people just defined to be the time reverse state of psi of r and t. So this is what theta does in that particular example. And actually, from this particular example, spinless particle in 3D, you can see already that the time reversal operator cannot be a linear operator. Because a linear operator can never take a wave function into its complex conjugate. We'll always take wave functions into themselves or into new wave functions, but never with a complex conjugate. So in fact, it's an anti-linear operator. I'll say more about that in a minute. Now, for general systems, we're going to derive the form of the time reversal operator very much like what we did with parity by writing out a list of possible demands that would satisfy in general terms. So the postulates are this. Is that first of all, the theta dagger times theta should be equal to 1. This is just like we did in parity in the logic because we want the symmetry operation to preserve probabilities. The second requirement is, well, the second and third requirements actually come from what we do in classical mechanics. But I'm going to go back a moment to this definition of the time-reversed state right here. If we evaluate this equation, let's take this one here, a ket language. If we take this ket equation and evaluate it at t equals 0, it becomes very simple, just this, is that theta acting on the original state of time as 0 is equal to the time inversed state of time 0. In other words, the theta operator, the time-reversal operator, maps the initial conditions of the original state into the initial conditions of the time-reversed state. Now in classical mechanics, if you're doing initial conditions, let's say it's consistent in the initial position of momentum. So let's talk classical mechanics here. If you time-reverse the initial position at the initial conditions, it doesn't do anything to the position. But in a sense of change of the velocity, it will change the momentum also. And so this should go over into r0 and minus d0. That's the effect of time-reversal under time-reversal, this is classical time-reversal, on the initial conditions. But given that our theta in quantum mechanics is supposed to map the initial conditions of the original state into the time-reversed state, what we expect this is that we expect certain conjugation elections on position of momentum operates following the classical pattern. We usually expect that if you conjugate the position vector in time-reversal, it ought to just go into itself. But if you conjugate the momentum operator in time-reversal, it should go to the minus itself, corresponding to the change in the velocity. Now, so these are our basic postulates that our time-reversal operator should satisfy. It follows from number three and four that if you take r cross d, which is orbital angle momentum, and you conjugate that by time-reversal, that it should also go over to minus itself, because r doesn't change sign, but p does, so there's an overall minus sign. I'll remind you that angle momentum is even of a parity and it's out of the time-reversal. And this statement, which holds here for orbital angle momentum, we will generalize and assume that it actually holds for all types of angle momentum, so that theta, j, theta, and dagger, where j is an angle momentum of any system under consideration, should go over to minus itself. It's a logical extension of what holds for orbital angle momentum. So these are requirements of the time-reversal operator. Now, given these requirements, it follows one can show rather easily that the time-reversal operator cannot be a linear operator. We could just work in one dimension. Let's take x and p and take the commutation relations, which is xp minus px, and this is, of course, equal to ih1. And now, allow me to take, I mean, allow me to take theta and multiply it on this side, and you take the dagger and multiply it on the other side. If equations one were true, they would be an ordinary unitary operator. So for the sake of argument, let's suppose theta is just an ordinary unitary operator with these conjugation relations of numbers two and three. Well, then it's gonna leave the x's, x variables alone, but it'll change the sign of p. So the left-hand side of this equation, we get minus the commutator, minus xp, minus px changes the sign of that. But the right-hand side is just a c number, and so theta's unitary, it just goes right through the c number, and you get the ih bar again on the other side. Obviously, these two statements are in conflict. What about the left-hand side changes the sign and the right-hand side changes the sign? Oh, this should be a theta dagger, by the way, on this side. However, if the time reversal operator's anti-linear, then it will change the sign of the ih here, and we pull it through it. I'll say something more about the properties of anti-linear operators in a moment, but it induces a complex conjugation here. So for an anti-linear operator, we get a minus sign there, and the result of this is that the canonical commutational relations are preserved. And so we conclude that from these requirements on this operator, that it must be an anti-linear operator, in fact, an anti-unitary operator. All right. Now, so far, talking about symmetries, I've always said that we require the symmetry operators to be unitary in order to preserve probabilities, but now we're talking suddenly about anti-unitary operators, anti-linear operators. And so I'm gonna tell you something about a theorem which is proved by Wigner in fact, in the old days, the 1930s or so. You can start with the general question about symmetries and quantum mechanics. Wigner posed the question in the following way. He said, suppose we have an operator, and suppose we have a mapping that takes the cut space into itself, and we think of it as a mass of state factors and the state factors, and this is gonna be some kind of a timer whistle off, excuse me, this is gonna be some kind of a symmetry operator. I don't quite look at it in the name, but it could be any symmetry operator. And we will impose the requirement that the measured probabilities are the same for all possible measurements that can be made, both before and after the symmetry is applied. So if we have, let's say, two initial states, psi and phi, and we apply the symmetry, the symmetry here, and let's call the new state, psi prime and phi prime, then what Wigner requires is that the scalar product, psi prime and phi prime, absolute value squared, or even absolute value is dot, should be equal to the absolute value of the original scalar product. And this has to be true for all psi and phi. Now the reason that this is a big measure of requirement is that what you actually measure experimentally are not amplitudes, they're not scalar products like this. Instead, one measures probabilities. And it's necessary to carry out further analysis to extract what the amplitudes, not underlying amplitudes are. You wanna call what we did in the case of that, starting Yarlak apparatus with all the amplitudes that we got from the probabilities. And so what Wigner requires is the probabilities be concerned along the square of this to become probabilities under the symmetry operation. And then by more careful analysis, certainly we've got a carry out in these lectures, Wigner is able to show that such a symmetry operation must either be a linear operator, excuse me, in fact, not just linear, should be a unitary operator, or else an anti-unitary operator. Those are the requirements. Those are the only types of operations that can satisfy this. There's actually phase factors involved in the argument which I'll walk over, but this is the basic conclusion on it. If you read about this theorem in a Messiah's book, and Messi, as they say, book on quantum mechanics is a reference for it, or else in Stephen Weinberg's book about volume water field theory, that's also as a discussion of this theorem. In any case, for our purposes, what it means is this, some symmetries are implemented by unitary operators and some by anti-unitary operators. Now, in fact, the only, wait, yeah, just a second, in fact, the only anti-unitary symmetry that we will encounter in this course is time reversal, and that's what we're talking about now. All the others, rotations, parity, and the others that we'll see too, are actually implemented by means of unitary operators. Yes, what was the question again? So, so could you tell me what exactly is the definition of anti-unitary? Yeah, I'll get to that in just a minute, because we haven't really talked about them much yet in the course. All right, in fact, that's my next, that's my next, that's my next topic, which is to talk about what anti-linear operators are. So, before we talk about anti-unitary operators, let me address the question of anti-linear operators in general. If I do know a linear operator by L, this is a mapping of the Hilbert space one to itself, and an anti-linear map, in other words, it maps kets into kets, and an anti-linear operator is also a mapping from kets into kets. The difference between them, let me, let me denote linear operators by L and anti-linear operators by A for the purpose of this lecture. The difference between them is, is how they behave on linear combinations of vectors. If I take a linear combination of C1 and C1, let's say C2 side two, where C1 and C2 are complex numbers, and I apply a linear operator to it, then if you get the usual rule, it's that it distributes this way. It's C1 times L-actimates sine one, plus C2 times L-actimates sine two, so when the linear combination goes over like that. If we do this for an anti-linear operator after the same linear combination, what it turns into is C1 complex conjugated times A-actimates sine one, plus C2 complex conjugated times A-actimates sine two. In other words, the anti-linear operator causes a complex conjugation of the expansion coefficients when you have a linear combination of vectors. Now, one of the ways of summarizing this is to say that an anti-linear operator does not commute with a ordinary number, or an AC number, which when it is being viewed as a multiplicative operator that's on the right, if you take the category multiplied by a complex number, that's a simple type of linear operator. But an anti-linear operator does not commute with that. In fact, the C is a complex number. What we have is an A times C, particularly with C-complex conjugated times A. And I'll put a box around this because this is one, that is the first one of the rules which apply to anti-linear operators, which are not the same as the case of linear operators. There's actually three rules, and I'll now move on to the second one. Here's the second rule that has to do with the action of operators on tests. Let's review the situation for linear operators first. Supposedly, linear operator is given, so we know what it does to tests. Let's just say a size is state vector, then L-actimates on size, no, we know what that does. Now, when you're back in the first week, I get a lecture on this, the question was, given that we know what L does to the ket vector, it's a question of what does it do to bra vectors? Let's suppose that phi is a bra, and we want L to act on it. The notation is we put the L to the right of the bra and we can get this action on it to the left. Now, the idea, so we want a definition of this thing. Now, the idea is that, the idea is that since phi is a bra, L acts on it, and it's going to produce another bra. The whole thing that I've written here is actually a bra. Well, what is a bra? A bra is a mapping from linear mapping from the ket space on the complex numbers. This has to be linear and this is the definition of where X bra. So the phi L here is supposed to be a bra, and therefore it's a linear map on kets, and therefore I can take it and allow it to act on some ket side, and the question is, I'm trying to define what this thing is in parentheses, but I'm going to define it by saying what it does to kets. Well, here's what it does to kets, because this is the bra phi times L acting on psi where I put parentheses like this to show that on the right-hand side, the L acts on the right to the side. It's given that the action of L on kets, so this part is known, and therefore this makes the definitions of the action of that bra on an arbitrary ket side. Okay, that's the logic of it. However, so it's like the L's acting on the left here and the right there. However, once this definition has been written down, you see that the direction to which L acts doesn't make any difference. The answer is the same, independent of it, and so it's customary to drop the parentheses and just write it in this way, and we can say that in the matrix element that L can act either to the right or to the left and it won't make any difference in the answer. Now, let's try the same thing for bras, excuse me, for any linear operators and see if it works. Let's suppose that the action of an N linear operator on kets is known, and let's suppose we wish to define the action of the N linear operator on the bra. Let's try the same thing we did with linear operators. Let's try this and say that since the, since A acting on the bra psi is supposed to be another bra, let's specify what it does to a arbitrary ket, and let's just say that this is psi with A acting on phi like this going in the other direction. And the question is, does this give us the definition of A acting on psi? We do know what A does to phi, and that was understood that, well, A acting on phi, and this is known for all states, so the action of A and the kets is known. So the right-hand side becomes known, but it certainly gives us a number. Thus, this bra, which we're trying to define, certainly is a mapping from kets on the numbers, just like I said here, that's what a bra's supposed to be. However, the question is, is it a linear map? The answer is no, it's not, unfortunately, because A acting on psi, let's see, that's the other way to say this, is that, yes, so what we're trying to define is A acting on psi. And the question is, is whether this thing, which we want to be a bra, or whether that's the linear map acting on phi? Well, if I replace phi by linear combinations, you see, when I let A act on it to the right, I get complex conjugates of those coefficients, and the result is, is that this is actually, this is a, this is a, indeed, an axon on kets that reduces complex numbers, but it's an anti-linear operator, not a linear one. And so this doesn't work as a definition of A acting on bra's. However, we can fix it up if we complex conjugate the right-hand side, because then the map does become linear, and then this thing really comes, it becomes a, becomes a genuine bra. And so this is the second property which the anti-linear operator is different from linear was. It looks like this, I'll write this on, is that A acting to the left on psi with parentheses indicated with direction of action times phi is equal to psi with A acting to the right on phi, and then brackets around this, parentheses indicated with direction of action, parentheses on this, and complex conjugates. This is the second rule which differs from that of ordinary linear operators. I don't really need the arrows here, because the parentheses tell me which way it's going. I'm just trying to emphasize the arrows telling me that this has to act on that. Parentheses indicate that in some of these. I can take the arrows out, let go, and you can clean back in if you wanna remember which way the A is acting. So this is a complicated-looking formula and it's hard to remember because of all the parentheses and brackets and stuff, but it's much easier to remember if you put it in words. Here's what you say. As you say, in the case of a linear operator, it doesn't matter if it's in a matrix element, whether the linear operator acts to the right or to the left, and so we don't need to specify which direction it goes. In the case of any linear operator, however, it does matter if the A acts to the left or the A acts to the right. What happens is that if you change the direction which A acts, you're gonna have to complex conjugate the matrix element. That's the easiest way of remembering this. Change the direction in which the A acts in the complex conjugate the matrix element. Now there's a third property that we need to discuss and which is, can be regarded as one of the parentheses between linear operators and any linear operators. This has to do with the definition of the permission to conjugate. Again, let's review the case of linear operators. So the linear operators have a linear map of the ket space element itself. So the L-acumen state psi is known for all states psi. The question is, how do we define the permission conjugate, L-dagger? L-dagger is also a mapping, supposed to be a linear mapping of the ket space element itself, and so therefore we need to define the L-dagger size equation. Given that we know what L is, L-acumen size, what we know is L-dagger size. Well, the answer to this is that L-dagger acting on psi is equal to, what we do is we take raw psi, let L act on that from to the right, which we know how to do, because that's what we just discussed over here, and then take the dagger at that. And by counting things, you can see psi, psi mapping it into its bras and ending linear operation, but then taking the dagger again makes it overall linear. It's the L-dagger as a linear operation. It's actually the state for the original one. Now, if you put this in a more familiar form, if I take both sides of this equation, and I multiply by some arbitrary raw, like this, to take the scalar product. So the right-hand side, so let's multiply on the left, like this, let's write it out. On the left-hand side, we find a scalar product with the L-dagger side, including a parenthesis parenthesis. And this is equal to the phi scalar product a state I'll call being a specimen I call this alpha, like this, where alpha is equal to psi L quantity dagger. Now, phi alpha is the same thing as alpha phi complex conjugate. This becomes the same thing since alpha is equal to this, it's equal to this dagger, that under the dagger, and I get this psi L, in this case, after this, right, multiply by 1 to the phi, with an overall complex conjugate. Well, for linear operators, it doesn't matter which way to ask, so I can drop the parenthesis, so the n rule of this is going to be phi, all the way to the psi, is equal to the psi, all the phi complex conjugate. It becomes this result, this is again, just to review a linear operate, this is oftentimes taken actually as a definition to call a dagger, this has to hold for all states by its side. It's also a special case of the rule that if you take a mission conjugate in this expression, you just write it backwards, baggering a complex conjugate, everything you get the right answer. All right, now, let's ask how this same argument works in the case of a linear operator. The basic result is that it does work, it works just the same, but because any linear operators, it makes a different switch way to act, you have to be careful about that. So here's the situation again, just run through this same argument, the challenge, except for any linear operators. Again, we have an ending linear operator, which is a map that takes our cat space into itself, and we wish to define a dagger, what city it comes from. We do it like this, just as we did before, a act on the side is known. And so we define a dagger acting on the side, this is known, and we wish to define a dagger acting on the side, and we write this as a side graph, with a acting from the right, and this whole thing is dead. Previously in item number two, we defined what happens when a acts on the graph, so the right hand side is defined. Now again, if I multiply both sides of this with a graph phi like this, and let's let alpha, let's let alpha be, let's see how do I do this. Yeah, let's let alpha be equal to, let alpha be equal to a dagger, a dagger psi. And let it fly after the left hand side, this becomes phi together product of alpha. And that's the same thing as alpha standard product of phi complex conjugate. But alpha converted into a graph is the same thing as psi A acting like this, multiplying onto phi, and the whole thing is complex conjugate. Okay, and the other hand, the left hand side can be written this way, is phi with a dagger acting on the side like this, which is to show you that it's going to act this way. All right, so let me go back to this. So that's the main result. Let me go back to the graph board, put it below the other two items, because this is the third rule where we have to be careful with any linear operators. It works like this, it says let's see how do I do this. The left hand side is this way, because that is phi, the standard product of A dagger acting on the side with parentheses to show you the A-axis of the side, is equal to psi A parentheses showing the A-axis of the left times phi, and then the alpha is complex conjugate. All right, this third rule is also messy right now, and it's also hard to remember for the same reason. The fact is that the max bracket notation is not so good for any linear operators, and in fact, I would be tempted to use a better notation, except I don't want to add any more notation for you. But if you say this in words, it's not too hard to remember. Recall that in taking the dagger of an expression of quantum mechanics, you reverse the order of everything, and you dagger everything in sight, as you write them down, the dagger of complex numbers understood to be the same as the ordinary complex conjugate. Well, the rule number three is basically that the same rule still applies, even in the case of any linear operators, to accept you also have to reverse the parentheses. So here, let's take the right hand side here. This star here is the same name as dagger in this expression, because it's just a number. But in order to find out, in order to apply this rule, we write it backwards, first, phi ketchup comes to phi bra, which you see here, then there's a parentheses which is reversed, and the A-gis gets reversed and daggered, and the broad turns into the sci-cap, and then there's the final parentheses. But you see, we just apply the usual rules except you're also including the parentheses in the process of doing it. So those are the three rules by which any linear operators differ from linear operators. Time reversal is the only any linear operator that will ever use in this course. It's the main one that people use. So you don't have to use these rules all that much, but you have to use them for time reversal. All right, now, so this has been as a story of any linear operators. Now I want to say something about anti-unitary operators. An anti-unitary operator is an obvious definition to say A is anti-unitary and A dagger times A is equal to one. Let's notice a simple rule here, which is that if you take the product of two anti-linear operators, you get a linear one. So if A is anti-linear, so is A dagger, the product of them is linear. So it makes sense to equate this to one, which is a linear operator. I can show you that, that's the definition, that's all there is to it, really. But let me, in reference to Bigner's theorem, let me show you that anti-linear, or excuse me, anti-unitary operators preserve probabilities in the sense of a Bigner's statement. So let's say we have two states psi, we're gonna end up this way. Two states psi and phi. And let's define prime states, which are obtained by applying an anti-unitary operator to them like this. So these are the new states under the operation. Now let's consider the scalar product, psi prime and phi prime. And let's try expressing in terms of the original scalar product of psi and phi. Using the definition of psi prime and phi prime, we take the broad, sorry, controversial conjugate of the first equation and it becomes bra psi times a dagger. And I put parentheses to show now that it acts to the left. And then for the phi prime, I just copy it and it's a times five. And I'll put the parentheses in the show that acts to the right. It's a scalar product of those two states. Here the a dagger acts to the left and the a acts to the right. Now what I'm gonna do is reverse the direction in which a dagger acts and use rule number two, which says I didn't have to conjugate, I had to complex conjugate to make it solid. So this becomes psi and then a dagger a, I have to run five, where now both a and a dagger act to the right. So it looks like this, except I need to put a star in the whole thing, like my rule number two. On the other hand, if a is anti-unitary, then the a dagger a product becomes just one. And so this whole thing turns into psi by complex conjugate. And what you see then is that under an anti-unitary operator, the scalar product score is in a complex context. They're not invariant, they're square invariant, and those are the measurable probabilities. So this is the criterion which is used to figure this theory, all right. There's a little more about anti-linear operators. If I have a particular anti-linear operator, it's oftentimes convenient to write it in what I'll call the LK decomposition, in which K is an anti-linear operator. And of course, a is also, we're assuming a is a given anti-linear operator. And L is a linear operator. And the idea is that you choose K to be a particularly simple anti-linear operator, one that's easy to work with. The idea is that K takes care of the anti-linearity and then the linear operator L takes care of everything else. Now, the type of, so we want to let this question of what are some simple anti-linear operators? Here's the usual type that is considered. Let's say that decapital Q stands as just in shorthand to some complete set of commuting observables. Now, the practices could be a whole list of observables, and some of them could have a continuous vector and some could have a discrete. That would be the case in position of S sub Z, that we've been talking about, for example. However, for simplicity, let me just imagine it's just one variable and it's got a discrete index. So in the eigenstates of this complete set of commuting observables, again, we've got the idea of just a simple example. So these are the eigenstates of the operators in this complete set. Now, I'd like to now define an operator which I'll call K, K sub Q, it's one of these K type operators that's associated with this complete set. And it's defined in this way, is that it acts on one of these basis states and just maps it into itself. That's why it's a simple operator. Now, this equation, if you don't think about it too hard, looks like it implies that KQ is equal to one, because you're actually, at least in our, these are all written in the ends of a form of basis, so you map in all the basis vectors into these cells. However, KQ equals one is not only not true, in fact, it's not even meaningful. The reason is that K is an anti-linear operator and one is a linear one, and it doesn't make any sense to say one is equal to the other. So in fact, that statement is not true. On the other hand, if I square K, K squared Q, KQ squared, why does KQ square, it's the square of an anti-linear operator which is linear, and you can easily see that that is equal to one. So KQ here is kind of the square root of the identity operator, but it happens to be an anti-unitrack operator. Now, let's suppose we have an arbitrary state and we expanded this linear combination with coefficient Cn, these basis states, and like this. Then if we apply our KQ to the state psi, it's equal to the sum on the end, and because of property number one up there, remember when you apply it anti-linear operator, linear combinations, you have complex conjugated coefficients. So this turns into Cn star times n. What we see is this KQ operator, which I'm advertising as being a simple n-linear operator, has the effect of complex conjugating the expansion coefficients in the corresponding basis that's used for this complete set. By the way, this depends also not on just the, just the observables of the complete set, but also the phase conventions for the eigenstates. Because if you change them, you can change the phases of all of these expansion coefficients. Well, in any case, this is a simple rule. This operator, just a complex conjugates the expansion coefficients. The expansion coefficients I'll remind you of what we call the wave function. So in other words, what KQ does is that it complex conjugates the wave function and the Q representation. And in particular, if you're dealing with a spinless particle in a three-dimensional space for the wave function psi of r, and it completes that in the community observables of the three components of the position operator, then the operator will talk about all case of r. What does it do? Well, it acts on the position eigenstates and just maps them in themselves. What does it do to a wave function psi of r? When you're KQ or KR, in this case, I'm replacing capital P by r there. What does it do? It maps it into its complex conjugate. We saw earlier that the complex conjugate was needed in doing time reversal on a spinless particle moving in a molecular static field. So we're getting close to time reversal here. So this K operator in the position representation just means complex conjugating position space wave function. It's very simple to have a rule. However, if I've chosen a different identity that completes that, such as momentum, talking about the operator K sub p, you would find that it's a different operator. It has a different effect on state vectors than this KR. So these K operators are dependent on the representation each being used. So it's fairly easy to show, if I may go back to the general notation, it's fairly easy to show that this KQ operator is actually anti-linear, it's obviously anti-linear, but we're assuming it's anti-linear. It's a property that I'm missing here. But it's actually easy to show that it's anti-unitary. Let me do that here. This is now because we're continuing now with this general KQ notation. It works like this. Suppose we take KQ dagger and apply it to KR and we wish to work out what this is equal to. To show that it's anti-unitary, we need to show that KQ times KQ daggers is one or the KQ daggers inverse of KQ. That's gonna be the goal. So let's work this out. I'm gonna show you how you can use these three rules up here to work this out. The first thing I'll do is insert a resolution of the identity of the left. So we sum on M and we've got M and an honored product like this. And then we've got KQ dagger applied again. But since KQ is anti-unitary, or it was anti-linear, we need to be careful which direction it acts. So let me put parentheses here to show that it acts to the right and the end. Now then, we can write out first the outer product again like this, or sum on M, maybe the first part of the outer product. The rest of this is now a matrix element. And I will now use rule number three, which says that I can reverse the whole thing if I reverse the parentheses and put a dagger or everything and put a star on it. So this becomes, that brought M on the left, KQ now without an adder, parentheses around all of that, and then kept M on the right, okay? Now, yeah, so sum on M, we're gonna put this into A. Okay, that's a good idea. Oh yes, okay, so yes, now here's what, yes, okay, M versus L. So here's what we do next is this KQ is now acting to the left you see. Next we use rule number two, which says that if you change the direction which KQ acts, you'd have to complex conjugate the matrix element. So this becomes a sum on N, just copying the first part of this, and then we have N on the left, KQ in the middle, and M on the right, except now KQ's been acting to the right as indicated by the parentheses, and I better put a big star around the whole thing. Now KQ on M acting on the right, and the definition of KQ is is it just makes the basis vectors alone. So this means I just dropped the KQ. Then I have the matrix element N and M which is a product for delta. That's real, so the star doesn't do anything to it. The sum just turns into the single sum which is in itself. And the result is, is that KQ that acts on the end of the accident, and it does the same thing as KQ itself did. You must do the same operators. So the conclusion is, is that the KQ, KQ divided by this, this equation satisfies the property of KQ dagger is equal to KQ. It's also clear that if I square it, I get one, KQ dagger squared is equal to one, that's a linear operator of course, because it's a pi KQ that's a second time. And so this means the KQ dagger is also equal to KQ inverse, and the result is KQ is not only anti-linear, it's actually anti-unitary. So these K operators are always anti-unitary operators. No. This step here? Yeah, I'm going to get it left and see if you can get it right. Yes, you might, you should. I reversed everything and I'm using rule number three, so you're right, this should have been a start. And what that means is this is non-start. Yes, thank you, so that was a mistake, but the answer comes out all right. All right. Now, we haven't yet defined with the anti, what the time reversal operator is, but we did say something about what the time reversal evolution's supposed to be. Let me go back to that and let's talk about the general Hamiltonians and the time evolution. So if we have the time evolution, which I'll write in that language like this, then we'll define the time reversal motion, psi r of t, as given by the time reversal operator acting on psi and minus, I'll remind you that psi itself is not involved in time, it just amounts to kets and the kets. So this is an equation which involves time dependence in both sides and it's supposed to hold for all time. It really just defines the time reverse, the time reverse state as a function of time. Now, let's suppose the initial state satisfies the Schrodinger equation for some Hamiltonian h. Let's ask, does the time reverse state also satisfy the Schrodinger equation? So to do that, we'll compute i h bar d dt applied in this time reverse state psi r of t. I can take the psi r of t with theta psi of minus t and plug it in here. Before I do that, however, let's make a substitution. Let's take tau equal to minus t, which is the meaning. And so this becomes i h bar times d dt applied to capital theta applied to psi of tau. Now, however, d dt is the same thing since t is equal to minus tau. d dt is saying it's minus d dt tau. So allow me to do this, let me give this a minus d dt tau like this. Now, next, I want to move this time reversal operators to the left of everything. I can pull the time reversal operator through the d dt tau because the tau or the time t are just parameters in the state vector and they're not operators on the moment space. So the theta just goes through that part. Also the time tau is real. However, the minus i is at this complex and so pulling the theta through it causes the complex conjugate. And the result is that I get theta multiplied times i h bar partial respect of tau now acting on psi of tau. However, tau here is just a done variable so what comes after the theta here is actually the left-hand side of the Schrodinger equation. And so this is the same thing as theta times h acting on psi of tau. Psi of tau, however, if I take theta dagger, I'll move that through by theta dagger, psi of tau, which is the same as psi of minus t, is then theta dagger acting on psi reverse of t. And so, let's say psi of tau. So this becomes equal to theta, yes, becomes theta h theta dagger times psi of t, psi r of t, the time reverse state psi r of t like this. And so going back to the beginning and looking at the end, what you see is the time reverse state satisfies the Schrodinger equation, but with a different Hamiltonian, which is the original Hamiltonian conjugated by the time reversal operation. Now, if it should happen that the Hamiltonian can use the time reversal so that theta times h times theta dagger is equal to h, then it follows that the time reversal is also a solution to the original Schrodinger equation, no change in the Hamiltonian. If Hamiltonian satisfies this tradition, then they say it can use the time reversal or that it obeys or respects time reversal. Time reversal would probably not be interesting except that there are a lot of Hamiltonians who can be of a time reversal, and in fact, for a long time, it was believed that all realistic Hamiltonians could be of a time reversal. This, it turns out, is not exactly true because the similar situation that Perry will recall. And I'll go ahead to that next time and tell you more about specific Hamiltonians when they do and don't be of a time reversal with the physical implications of that r and various aspects of that. So that's all. So before you go, let me remind you, I'll send out an email as soon as I get back to my office, but next Tuesday, next week is Thanksgiving week, right? So we're gonna have three lectures next Tuesday, regular one on Monday and Wednesday, and then an extra one on Tuesday night. And nobody's complaining about 6 p.m., so I'm gonna, where I've had 6 to 7 p.m., is that all I have for the money here on the next Tuesday night? All right, so then I'll go ahead and schedule it at that time and send out an email this morning in front of me.