 So we'll discuss that more as it approaches. Let me review this little bit and then we'll take a question that you might have. So let me review a little bit of what we were talking about. So we discussed a very important idea, the distinction between coherent superpositions or quantum superpositions of different alternatives and statistical mixtures of this mode. That's the language we use and it has an important operational meaning. So if we think about, for example, a spin-1-2 particle, as well we've really discussed it, to date. So if we have a pure state, we've said that could be described by this ket, which generally could be in a coherent superposition of the two basis states. And we know from the Born rule how to calculate the probability of finding one of the outcomes in a projective measurement. And that's given then by the squares of these amplitudes. And if we look at, for example, the probability of finding a spin-up along x, as we've said from the very first lecture, there's an important contribution to this probability which doesn't follow from classical logic. So this probability has pieces in it that we would argue from the point of view of logic. There's some probability to find spin-up along x given that the spin was spin-up along z. Times the probability it really was spin-up along z. Plus the probability that we would find spin-up along x given that the spin was really spin-down along z times the probability that it is spin-down along z. And then there are these terms. These terms which have no classical, logical analog. They are the quantum interference terms between those two alternatives. And that's the hallmark of the fact. The fact that these interference terms are there is the hallmark of the quantum superposition. The fact that these two alternatives were not locally real. We just didn't know which one they were. They exist in some quantum limbo state that we don't really understand and that has impact on this probability. For example, if those two probability amplitudes were, say, equal to one over root two, then the probability of finding spin-up along x would be one. We would definitely get that outcome. And we can think about that as there is constructive interference between those two alternatives. In contrast, if the amplitudes were equal but there was a minus sign there, then the probability to find spin-up along z would be zero. There would be complete destruct interference between those two alternatives. So that's the superposition, the coherence of the superposition, and that's being contrasted with what we call the statistical mixture. What we say is that the preparer, we imagine this scenario, which is a possible thing that one can do and one does do in some, say, quantum communication protocols where the preparer has the ability to send you, the receiver, to be called Bob, not to be called Alice, because they're A and B, and they're the names you'll receive. So Alice prepares either this or that, and she doesn't tell you what she's doing, but she does so with a coin flip, that coin that may or may not be a fair coin, it has some probabilities. What we showed last time is that we could describe the state of the system in that case, not as a quantity reposition of these two alternatives, but as a statistical mixture, which is represented mathematically by this thing we call the density operator or we can call it the state operator. It is our representation of the state of the system. It's the thing that we use in order to calculate probabilities of measurement outcomes. And the way we do that, the generalization of the Born rule to the case of statistical mixtures is that the probability is given by the diagonal matrix element or the expectation value for light of the density operator in the state of consideration. So if we wanted to find the probability of finding spin up along x, this is the rule, okay? And doing that, we find this, and that's just the classical logic piece. It doesn't have the interference. So what distinguishes the statistical mixture from the quantity reposition is the lack of interference between those two alternatives, okay? That's the key point. And you can, any state of the system, whether it's a coherence to reposition or a statistical mixture, is a density operator. So the density operator is the most general way of writing a quantum state. So you can have all possibilities, quantum or classical kind of statistics, all right? And so, for example, a pure state is a state which is just given by this outer product between a ken and a bra for some particular state in Hilbert space. And I just wrote this out as an example in the context of writing it in this basis, so I expand it side in this basis, which I can always do. And when I do that, I'm going to write out all the terms, you see explicitly that this density operator has diagonal elements which look like the statistical mixture of spin up and spin down, which is just like this. But then it has octagonal pieces and those octagonal pieces are the things that we call the coherences. They express the potential for observing interference between those alternatives. Does co-operation be negative? Sure. I mean, in this particular case, they're generally complex numbers. They generally have real unimaginary parts. And they generally, they're positive or negative. I mean, for example, what we showed, excuse me, pardon me, this boo-boo, how we mistake. That should have been that. In this case, these coherences would be negative. Right? This would be minus a half and minus a half. And that was the case where we had destructive interference. It could be pure imaginary. It could be anything, any complex numbers. But the trace of the density operator for a pure state must equal one. Right. We'll get to that in a moment. So these are real numbers. Because they are the absolute squares of complex numbers. How long does it have to equal one to the state's volume? It's normalized, indeed. We'll get to that in a moment. So what we have here is that we said that we have just so statistical mixtures, we could have whatever we like. We could say we can have an ensemble of possible pure states, of pure states side by side. They don't think we could have as many as we like to prepare with probability. So we have this picture of the preparer sending us a pure state but not telling us what she's sending us. She just tells us the probability she's going to do it. And the state we would assign if that were the case, would be this. If she sends, if she tells us she's definitely going to send one state with probability one and all the others with probability zero, it's a pure state. Otherwise it's a mixed state. So this is generally what we call a mixed state. Not all the probabilities are there only. If any, the probabilities aren't one and zero for all the rest of it, it's mixed. Otherwise, it's called a pure. And moreover, we have the, so in a basis, let's just talk, we're talking generally now, not just about three and a half, let's just say we have a heat dimensional elements, then, and let's just say we have a basis, the alpha, to be. Then we can look at the matrix elements of this operator. So just to emphasize the facts, the diagonal elements, the probability, thine, the alpha, you know, projected measurement, that's the more rule. And the off diagonal elements are what we call the coherences. They are the, they in some sense represent the potential to see interference between alternatives, the alpha and the beta. That's how much interference you might see in some, and again, if I look at this from the point of view of the statistical mixture over here, then this is equal to sum over all the members of the ensemble in this guy. That's that, right? So I'm just waiting, this would be the more rule for a particular pure state, but I don't know which pure state she sent me. So I have to average over that. Clear? And that I could write if you like, as this side, or this particular I form I is equal to the sum over alpha c alpha to the right guy. The off diagonal element here, data, that is equal to the sum over, so let me just write this as the ensemble average of that for all those guys. That means average over the ensemble. Now I have this c alpha on here, and this is c data stock, which is, if I write this, each of these complex numbers as having a magnitude at a phase is, I just wrote each of those complex numbers as having a magnitude at a phase. And so, again, to emphasize the key point of the coherences they have something to do with the phase relation. Coherence and phase are intimately related. I mean, you know that from, maybe you know that from the notion from wave interference and classical wave theory, say in optics. And we talk about a coherent coherent field. That has a well-defined phase. And the fact that the phase is well stabilized means that you can see interference between different wave paths. But if the phase is jiggling around in some way and not well-defined, then you don't see interference between the waves, right? And that's what this is saying over here. This coherence is the ensemble average of this old guy over all the members of the ensemble. And if the phase relationship is not well-defined between all members of the ensemble, then it will tend to average out to zero. So there's this intimate relationship between the well-defined phase between the probability amplitudes and whether or not you can see interference between those alternatives. So let's say a few more things now about the density operator. What are its properties from a mathematical perspective? What can we say? Well, if we look at this operator over here, the density operator written this way, this writing again, firstly it's clear that this is her mission. Why is that clear? A mission operator is one which is selfish. Why is that obvious? Because P is real. These are real numbers. We have an ensemble of possible states with sums of one. And when you take the editor to this, this goes back to itself. So it's her mission operator. The other thing I can say is that it's a positive operator. It's one where the expected value of rho is not negative for all possible factors. Is this the case? Sure, because this is equal to the sum over i, P i, and this is a positive number, or it's something that's not negative, it could be zero. So, it's a weird thing. We call them positive operators where we should realize you call them not negative operators because it could be zero. That's just the jargon that we use. All right. And as that means, of course, that means that two things. If it's a her mission operator, then we need a normal operator. And so that means that rho can be dangle. Its eigenvalues are non-negative. Those are two things that have to be hitting follow. Right? So that what that means is that in dimension D d eigenvalues and its eigenvectors can always be chosen. Right? Discuss in the first week or so class that her mission operators have can be diagonalized. It has D, in this case real non-negative and that we can always choose that to be orthonormal. If they're for different eigenvalues we can always, if they're not, we can always find their orthogonal basis within the degenerate space. So let's call the eigenvalues of rho to the eigenvectors R and we call them u lambda and the eigenvalues. And we can always express rho in the basis of its eigenvalues. In the basis of the eigenvectors I should say. So what can you tell me about this representation? How can I interpret this? Well, it looks just like that. Okay. So rho is always can be thought of as a statistical mixture of its eigenvectors with mixing probability given by its eigenvalues. Now here's an important point. These guys need not be the same as these guys. Okay. Now let me explain why. I can tell you what I'm mixing. Here I am, I'm Alice. I'm going to send you a little spin up along Y and I'm going to send you a little spin down along 4, 5, 3, I'm going to send you a little bit of spin up along this axis. I can send you whatever I like. And the vectors I send you don't have to be orthogonal vectors. And I can send you a million of them. Okay. I can send you a million different possible spins. And I can tell you beforehand so this ensemble of decomposition doesn't have to be an eigen decomposition. It could be whatever it likes. One thing you learn from this is that there is not one unique ensemble for a given state. In other words, there are many, many, many different ensembles. In fact, an uncountably infinite number of different possible ensembles which all lead to the same density operator. You could now tell the difference. That's one of the things you're going to see in your homework. And that's because the classical probability and the quantum probability are kind of getting mixed up. So the ensemble decomposition is not unique. Now we can ask what are the conditions on the ensemble such that the same density matrix. And we can quantify that. We can say these two ensembles would give the same state. It is a very positive thing. Now, of course, the eigen decomposition is unique up to degeneracies. So if all the eigenvalues are non-degenerate there is a unique way of thinking about this as a statistical mixture of the eigenmetrics. Yeah? Is this related to a previous question where you can keep back lines that you continuously rotated into that or you could change any into a diagonal matrix. That's not necessarily that matrix. And the reason is if they're not orthogonal. Okay? So the example I gave here goes back over here. This density matrix is in its diagonal representation. Okay? But let me come back over here and say I suppose this is a probability here that you're homework. Let's say I have rho is a half spin-up along z and a half spin-up along y. Okay? That's a possible perfectly good density matrix. A perfectly good density operator. That is not these are not the eigenvectors of this operator. If I look for example what happens if I act this on this? Well I get in this case a half spin-up along z plus a half spin-up along y times that, which is 1 over 2. Right? And so I don't get that the same vector. That's not an eigenvector. In fact you can see it also in a matrix representation. Suppose I wanted to express this operator as a matrix. As a 2 by 2 matrix in the basis that of choice. Here's the basis. Let's say spin-up has been down along z. So what does this operator look like in that basis? A matrix. 0, 0, 0, 0. Everyone see that? This is in the basis so in the basis up and down along z that's the projector. What does this projector look like in that basis? Well that's a little trick here. So the way to do it is there's lots of ways to do it. One way to do it is to write this as rows and column vectors. So this vector if you remember is 1 over root 2 that's the column that oh actually sorry this is that's the wrong list again. So it's a superposition of spin-up along let me do this more slowly. Because what the heck was we doing this? Let's sign it. This let m be this I want to write a matrix representation of this in the basis. There's lots of ways to do this. How do we do it? Well we can insert it in the set right on the side or we can just write out the matrix elements we can say this is this column, column do it in a somewhat systematic way that we can remember anything else. Why in terms of z basis? That's what we're doing. There's lots of ways to do it. Just write it out systematically if you remember how to write an operator in basis. There's lots of ways and I'll show you the other way just so because we have different approaches and I don't think we can change that depending on which is the most convenient thing. So what we remember is that we have to look this up or remember it which I do remember because there's just a few of them that you can remember so with that said this is equal to this operator over here this is equal to this is equal to this and this is equal to that right? So this is equal to what? This is equal to a half on the diagonal ok? This is the square root of those opportunities and then I have this times the column here and this ok? So and as I was saying just a moment ago I could write this and this is what was being suggested and this is how I originally didn't do it because if it's just an outer product of two vectors I can I don't have to go through all those tedious steps I could just say this guy is equal to this is equal to 1 over root 2 I over root 2 and then have the bra and then I take the outer product of doing it you have to learn how to use these different techniques and which one is the most rapid convenient ok? So what was the point of this exercise? The point of this exercise is this is not an diagonal matrix so you know we write this out this is this plus this it's not an diagonal matrix even though it's it's a statistical matrix of these two things it's not the diagonal what does that mean? but coherences between what? they're coherences between spin up and spin down in the z basis and that's because this state although it doesn't show interference between these alternatives it shows interference between these alternatives and that's what's being captured in the diagonal elements see if you have a question so I mean when I first saw it the state the odd angle would be zero but that's not true so we're going to there's a degree of mixing this we're going to talk about that for a completely mixed state so there is a continuum of possibilities perfectly pure they have odd angle element and then in some basis and then we have maximally mixed and those states have no of that in any basis and then you have partially mixed partially pure as we said you can always think about it as a statistical mixture of its eigenvectors and in some ways you can up to degeneracies now let's see what I want to say let's see sort of bringing out a pure state of partial mixed states matrix I mean how can you tell what's rich by just looking at the matrix good question we'll get there very soon it's an excellent question it's an excellent question I would say one thing you can't necessarily talk about looking at the matrix because that's a basis dependent property and we want something that's independent of basis it's a property operator not the representation in a basis it's important to understand that distinction between basis dependent properties and basis independent properties we'll get there in a moment alright do do do do do do okay yeah oh yeah that was a point I forgot to mention that was the point you raised earlier normalization so density operators are permission operators they're positive permission operators we can also normalize them and the same way that we said that when we talked about cats we said we typically normalize them so that we can easily calculate the probabilities okay so we're normalization chosen so that this equals the probability of one trip okay and so that what we mean then is that if we have a basis I guess I was calling it alpha so what we want if we have a basis the alpha then it should be the sum of all the leads choose the probability to find one of the alternatives should be one right but that's how it states normalize and as we saw this is just the sum of the bag which has a name it's called the trace so this is the trace over so we choose our density operators to be normalize by setting their trace to be one now the appointment of course is that we can always normalize afterwards okay so if some reason we had some operation where we got a density matrix that wasn't normalized we could always re-normalize in the same way that when we did projected measurements or measurements in general on cats we had to re-normalize right in our bays and stuff like that so we just find out what the trace is and divide by it so that the trace is one after alright so now I want to say something about pure versus mixed states so for pure state we said there is some cat such that row is then just the projector on that cat right what can you tell me about the eigenvalues and eigenvectors of this state so remember it's a cat and it's a normalized cat if this is a normalized right so the normal of this that matters much but I just wanted to remind you that is this a normalized that's the trace of this equal how do you tell well remember we proved in problem sets that the trace of an outer product is equal to the inner product that's a very very useful tool and I always think about it as outer products with inner products and you always put a little to remind myself that I'm clicking them together when I do the trace I make the same so what is the trace of this it's just that no it's equal to 1 as well because the trace of a so let's say we have a mixed state and let's say I wrote it as I can write it in any ensemble decomposition I like I'll write it in there's some ensemble decomposition I don't care whether it's the eigen decomposition or any other decomposition I want what is the trace of this mixed state that's the trace of that this is some of the traces traces of linear so I can bring it inside this sum that's one of the things we proved and the trace of this is 1 and so this is the sum of all the probabilities in that with 1 so for a mixed state the trace of 1 is as well a state the state is normalized mixed for pure now the question I really for a question I wanted to get to is what are the eigenvalues and eigenvectors of pure states so here's my state it's pure what is can you see one eigenvector eigenvalue in one what about the eigenvector to psi psi so psi is an eigenvector that's obvious because if I operate this on this I get it back right and the eigenvalue is 1 so that's one of the eigenvectors and that's one of the eigenvalues what are the other eigenvectors if it's a d-dimensional space what are the other eigenvectors and eigenvalues well let's just say we had a basis and this was one element of the basis and then we have all the other d-1 basis elements all of which are orthogonal to this what happens when I hit row pure on those you get 0 so all the other eigenvalues are 0 so for a pure state and we can see that over here this is the general form of the density matrix or the density operator written in its decomposition in terms of its eigenvectors and eigenvalues and if it only had one term if it's a pure state then one of these eigenvalues is 1 and the rest are 0 then it just has that one term so for a pure state and that's important there's only one eigenvalue that's 1 and the rest are 0 so for a pure state let's put that on the front board it's important not for the size but somehow we have this eigenvalue is 1 eigenvalues or eigenvector other eigenvalues are 0 and it doesn't matter but choose any eigenvectors we like that are orthogonal complement to this anything that's orthogonal to that is an eigenvector it doesn't matter it gets weighted by 0 so question what is the matrix representation of this pure state in the basis of its eigenvectors what does it look like one of the diagonal places wherever psi is in the basis we just write that one as the first basis element and then all the others some sign doesn't matter and it's back and it's 0 everywhere else now this comes to a point z that you were raising this is a pure state it's at the most coherence that you can have but it's the off diagonal matrix elements are 0 this is a point I'd like to re-emphasize the it's not right to talk about the off diagonal matrix elements it's off diagonal matrix elements with respect to a basis if it's a pure state that means in some basis it will have very big off diagonal matrix elements it doesn't mean that in every basis it will have big off diagonal matrix elements and in fact in the basis of its eigenvectors there's 0 because there's no interference between those different alternatives because there's only this alternative good so now I want to come to the question that Steven raised, he's not here at the moment but maybe he'll come back let's see and that is how do we distinguish because we can't just look at the matrix and look at it unless we're somehow and he says about matrices generally and know whether this was a pure state or not because this one I can see it's a pure state because it's got one on one diagonal and zero that's pure but generally if I looked at that matrix I wouldn't know offhand how pure it makes that work if I add those things together so what is it a defining feature of the density operator that is a function of the state itself and not the basis with which we use to represent the state so let's consider the square of the density let's just consider that well I can always do a function of an operator by expressing it in terms of its bag of representation if it's a permission operator and then just do the square on the eigenvalues that's what we learned so this is the sum over all the eigenvalues of the density matrix so what is the trace of rho squared well that's one so this is equal to the sum of the values of the square so what can you tell me about that if it's a pure state what is the trace of rho squared one so if it's a pure state the trace of rho squared well what if it's mixed well if it's a mixed state what can you tell me about the eigenvalues I guess I should have said that over here what can you tell me about that well there are positive numbers because we said it's a positive operator and it's normalized and it's normalized the trace means it's the sum of the eigenvalues the trace of rho is equal to the sum of the eigenvalues and that sum has to equal one so what can you tell me about each of the eigenvalues they have to be less than one right they can be zero for a mixed state all of these are less than one but they're not necessarily the probabilities they're not necessarily they are the probabilities of a mixing in the eigenvalues but they're not necessarily the probabilities in the original mixing okay so the eigenvalues which are properties of the operator are all less than one if it's a mixed state so what does that mean about the trace of rho squared he's talking about this sum it's got to be less than one so the tell-tale sign of whether or not the density operator corresponds to a pure state or a mixed state is by looking at the trace of rho squared I mean you can look at the eigenvalues but the quick way to do it is to look at the trace of rho squared and this has a name this is called the purity of the density operator you can call also the mix in this but it's typically called the purity and so if it's a pure state then the trace of rho to the purity is one and if it's a mixed state the purity is less than one if it's a statistical mixture of some kind so what if it was maximally mixed go to question what does it mean to be maximally mixed excellent question, I said mixed let's do it let's find that out that's exactly where I was at so in max what do we mean by a maximally mixed state if it's maximally mixed that is what we mean by that is an equal statistical mixture of the orthogonal states that says mixed as you can be for example 50% spin up along z 50% spin down along z that's completely mixed so all the lambdas would be equal? indeed so what are all the lambdas in this case the quick excellent point so what are the lambdas in this case they're all equal and what they have to be they have to sum to 1 1 over d excellent so the lambdas are 1 over d for all lambdas so how could I write then a representation of the density matrix for maximally mixed well to sum over all the eigenvalues each eigenvalue times the eigenvector and the eigenvectors could be anything I like that are form of basis now since it's independent of lambdas bring this outside the sum and so that's 1 over d the sum over lambdas but this is familiar object what is that that's the identity so a maximally mixed state is just proportional to the density I mean to the identity it is 1 over d times and moreover this state it looks the same in all bases this is always what the matrix looks like which means this off-diagonal matrix is r0 absolutely 0 in all bases so the matrix is in 0 but it's off-diagonal matrix 7 it's r0 in all bases if it's maximally mixed so it doesn't matter whether I thought about it as so here is the point for a point I'm Alice I'm going to send you either spin up along x or spin down along x I'm going to do that with a fair point I'm Alice number 2 I'm going to send you spin up along z with probability or spin down with z with probability you couldn't tell the difference you would get random results no matter what you did your result would appear random to you it would be completely a random 50-50 no matter what measurement you did whereas from Alice's point of view she kind of knows if you do a z-measurement but you don't so those are exams of 2 ensembles with your statistics of exactly the same so now what is the trace of rho squared so for a pure state this is 1 for a mixed state it's less than 1 but what's the smallest possible value of purity you can have 0 no it's a good guess 1 over d so let's let's just look at this we have to do this sum over all the eigenvalues I'm saying for dimension d it's not, it's dimension d well it would be the sum of 1 over d squared and then you have them d times so they'll just be d over d squared again which is d 1 over d excellent so the minimum value of this the purity at smallest is 1 over d rho is max so you can't get 0 the purity is always something but that's just the number we get but the off diagonal elements are absolutely 0 for a maximum mixed state so the purity is a measure of how mixed the state is now I should just before there's something I meant to say earlier but I hadn't but let me just say it now some of these concepts are not this is not purely a quantum mechanical idea this is something that you might be familiar with if you study optics and that's the notion of partial polarization and polarization suppose I have the line that's coming out of slide and I looked at it as polarizing and analyzing if I looked at at say vertical half the intensity would come through if I looked at it horizontal no true no we come through but it's not a coherent superposition of vertical and horizontal it's a statistical mixture of those it's not polarization at 45 degrees because if I put it at 45 degrees still half would come through so it's completely mixed polarization now partially polarizing the light that reflects if you studied optics you know that when you reflect some part of it reflects differently than the other and it would be partially polarized so these concepts are classical wave concepts as well where they become quantum is when we think about them as events and probabilities rather than fractions of these okay next so another thing we want to talk about here is so we have we can calculate probabilities if we have a density operator but what about expectation values so what we said here was that given or so this is the probability to find eigenvalue a associated with eigenvector a of some term so this is the problem I could rewrite this in a way that is useful this is equal to the trace of rho with that projector because so another way of writing this probability is the trace of rho with the projector on that this is the projection operator now if I have a remission observable decompose it in terms of its spectral decomposition in terms of its eigenvalues now I can even allow those things to be degenerate that projector could be onto a degenerate subspace associated with that eigenvalue it could be one deep projector like this or it could be that way so the expected value of a eigenvalue well that's equal to the sum over all the eigenvalues times the average value by definition that's what we mean by the expectation value it's the average value and this is equal to this that's what we just said and the trace is linear so I can bring it outside the sum rho doesn't depend on A so I can be and this is equal to this I brought rho trace outside the sum and then I could rho outside the sum and this is the observable A so the general way of expressing the expectation value when I don't have a pure state but I have a general state which might be mixed is to look at the trace of the observable with the state okay and the boardable can be expressed as by a projective measurement the probability I'm going to see that outcome is the trace of rho with the projector can anyone guess if I have a general POVM with POVM elements what is the probability of seeing an outcome what do you think my trace of rho exactly terrific I mean when we had it as a side we wrote it like this now this is the generalization so now you know the most general form of Born's rule that this is the board rule this is really really it it takes into account the fact that your state might not be pure and it takes into account that your measurement might be preprojected that's the most general form and that's the most modern way of saying side star side in the last two plus whatever minutes I wanted to say a couple of one thing about now time evolution so what we said was when we had a if we have a closed quantum system pure states are mapped a unitary operator I'm going to change my notation a little bit I'll fix it in the notes because that's the notation I prefer to use let's say we have the state at time t zero I want to know what is the state at some later time t they're related by a unitary operator which I'll just put these two okay t greater than t zero so this is a unitary operator how does this look just from the point of view of the density operator so the state at time t zero was this the state at time t is a pure state and that state is given by u in depth and this in between was is this so we have for this evolution on a pure state that the state at a later time is this unitary map on this okay, the unitary transformation for a general mixed state which could be in any in some of the confidence in my life then what is the state at a later time if it's a closed quantum system well each one of these guys evolves like this and then that could be brought outside the sum so it's the same thing so if I had a closed quantum system then this is the evolution of the state no matter what it started at okay, this is what we call unitary evolution now what about if I have a closed quantum system what happens to the purity how does the purity change as a function of time it doesn't and how do we do that the trace okay, we want to find what is the trace of the state at a later time that's what we'd like to know so let's plug it in that's equal to the trace of u rho at t0 u dagger times the same thing again and now u dagger u is 1 right? so this trace of rho at time t rho squared at time t is the trace of u rho u dagger we know because we need the homework that you can do the cyclic permutation right? so that's the trace of u dagger u rho squared and that's the identity yay, yay when you have unitary evolution the purity is unchanged and that's what we'd expect that's sort of how we define the fact that the evolution of the close quantum system was unitary because we wanted to say pure states are pure states but moreover whatever mix-it-ness you have that mix-it-ness is unchanged and that's because we're neither losing information nor gaining information okay, what we'll talk about briefly to conclude this discussion next time is that of course not all the evolutions present are purity because for non-states we have to get a thermal state when we take the silk ground that's coming out of the other so we want to talk briefly about that then we have to deal with these