 OK, we were mainly, if not that everything was focusing on bipartite entanglement, OK? So today, we will continue with the bipartite entanglement, and we will see that many of the things there can be generalized to multi-partite systems. Can you hear me? Better now? OK. OK, so what we were discussing yesterday was, first of all, a separability problem. So we were looking at mixed states, and then we moved on to pure states. Sorry. And we were convincing ourselves that entanglement theory is a resource theory where the free operations are LOCC. So everything that we can do locally and the free states are the separable states. So just to remind you, LOCC means that we can apply local operations on LA systems. You can apply some operations. You can communicate some result to Bob. He can then, depending on this outcome, make some measurement, communicate the result back, and so on. OK, so this is what we call LOCC. And entanglement theory considers this for free. So this doesn't cost anything. If you want to do something more fancy than that, then you need entanglement. We were saying yesterday also that you can measure entanglement. And if I'm not wrong, this was point B. And we saw, for instance, that one measure of entanglement for pure state, for pure bipartite state, would be the entropy of the reduced density matrix. So now, the next question that we should address is, how can we sort states? So this was actually C. So now we are at D. How can we sort states or states according to their entanglement that they contain? Well, we know already this answer because we have discussed yesterday that if I can go from some state via LOCC to some other state, then the entanglement in that state has to be larger equal to the entanglement of phi for any entanglement measure E. So this means that LOCC induces some kind of order in the set of states. So if I can go from phi to phi via LOCC, it means that psi is a better resource. So I get some kind of ordering. The ordering, however, is only partial. And this is important. It's only partial because there are states where, so there are pairs of states, psi and phi, such that you can neither go from psi to phi via LOCC, nor can you go the other way around. So nor can you transform phi to psi via LOCC. So this means that those two states here are not comparable. You cannot go from one to the other. So this means that you cannot say that this is a better resource than that or the other way around. So you cannot order them. And this is why we have a partial order in contrast to a total order where all two elements are related. Partial order means that it's not for the whole set of states. But for some of them, you have that. And so what we have learned yesterday is that LOCC induces such a partial order. And regarding entanglement theory as a resource theory, this is the most natural choice. It's the only thing that we can say, because LOCC is for free. And so this is why this state is for sure more a better resource than this state here. Now, what we have also seen yesterday is that this implies also that if two states are LU equivalent, which I write like this. And this means that, as we said also yesterday, that I can write, can find in a bipartite case. So I have two local unitaries. The U1 and U2 are unitaries. And they act on the first and the second system, respectively. And so I can transform, say, psi into phi with local unitaries. If this is the case, then we have seen that e psi has to be equal to e phi for all of them. Because simply because I can go from psi to phi with local operations. And I can go from phi to psi with local operations. I can undo the local unitary, because it's unitary. And so this means that we have this inequality in both directions, so we have an equality for whatever entanglement measure we have. OK, now you might wonder whether the opposite is also true. If this is true for all e, is it true that then the two states must be a LU equivalent? That's a very interesting question, not so easy to answer. Especially, we need some more information. We need some more definitions and knowledge about entanglement, monotones, and so on to be able to answer that. And if we have time, then we will come back to this question. So is the opposite also true? OK, so far we know. And this is important that LOCC induces this partial order. And this is also the reason why one studies LOCC, because this is a way of telling whether a state is a better resource than another state. So we get an order in the whole Hilbert space, which is, in general, very large, as you know. So this also means that, of course, we have to study this LOCC. So let's discuss what LOCC is. OK, so very weak. And roughly, we were talking about this already yesterday. So let's do it more precise. Because I have a system. I have a part A that holds some system. Part B that holds some other system. Now, what is the most general transformation that they can do if they are only allowed to act? So if they're not allowed to use entanglement? Well, Alice, without loss of generality, would start. And she applies some P of M. So she applies some generalized measurement with some number of outcomes. She obtains some outcome. Let's say that she obtains outcome I. So she has implemented AI. And she transmits this information to Bob. Now Bob uses this information. And depending on I, he applies some measurement, which I denote now by Bi, and obtains some outcome J. So he applies this generalized measurement, and he obtains some outcome J. Now he transmits this information again back to Alice. Now Alice, depending on I and J, chooses a new P of M and gets some outcome K, and so on, and so on. So this continues. Now, the problem is that these have to be, first of all, local measurements. So we have to make sure that they obey the completeness relation. And second of all, one can show that this doesn't end. So there are protocols. There are problems where you actually need infinitely many rounds of classical communication. So this makes this very difficult. This is why starting LOCC is so difficult to do. Let's write down what I mean by why is it difficult. Let's write down what is the whole operation. So what is lambda? What would be the corresponding CP map? So this lambda LOCC corresponding to our protocol would be what? If I act on some state row, I would have what? Well, I start with row. I first apply AI. Then Bob applies B, J, I, depending on the outcome of Alice. Then Alice applies A, I, J, K. Bob applies B, I, J, K, L, and so on, and so on. There's a tensor product. We go on, and we have here a sum over all I, J, and so on. And here's our state row. And the whole thing, dagger. So this is our transformation. This would be our map. Now, there are two things that we see. One is that given such a map to show that this is LOCC, it's very difficult. Because you have to show that each of these operators here factorizes in operators where each of them is a complete measurement. So AI has to fulfill the condition that AI, dagger, AI is equal to the identity if I sum over I. The same holds true for AI, J, K. If I sum over K, this has to be the identity, and so on. So I have all these conditions because these must be valid measurements. This is what they do here. So even though you see that these operators here factorize, so I can write this as a sum of some operators AI tilde, tensor, BI tilde. Because I can call this whole operator here AI tilde, and this operator BI tilde. So it will for sure be of this form, which looks pretty innocent. The problem is that for each of these operators, I have to make sure that it factorizes in those. So this looks simple. And these maps is what is called separable operations. It's called separable because you write a map as a tensor product of local operators. And recall the definition of separability. So this is the same thing, but for operators, for maps. So separable operations are those which can be written in the form that we have up there. So it's just a sum of AI, tensor, BI, so some local operators, row AI dagger, tensor, BI dagger, where this has to be a trace preserving map. So this means that the whole operators have to obey this condition of completeness. So this has to be true. So separable maps are defined like that. Any separable map can be written like that and the other way around. All our LOCC operations, as we see, are contained in a set of separable operations. This is an important message. So LOCC is contained at the moment we have it like this. Now, if everything that I can write like that can be implemented locally, so if every separable operation would be LOCC, then our life would be much easier. Because then we just need to deal with those maps and not with these nasty conditions here. These nasty conditions come from the LOCC condition. But the problem is that this is not true. So LOCC is strictly contained in set. It has been proven that there are separable operations which cannot be realized with LOCC. So this means that if we want to study LOCC, then there is no way that we just study, or it's not directly, that we just study the much easier set of operations, the separable operations, but we really have to deal with this complicated structure here. And again, this is not only complicated because I have here some products. This is complicated because this doesn't end. So this goes up, goes on until infinite. So there are protocols where you need infinitely many rounds. And so now I have a product of infinitely many operators, and I need to check that I can write it as a product of operators that obey all these conditions. That's, of course, a mess. It's very difficult. Yes, it has to be realizable with LOCC. So Alice does something here, so she gets all the possible outcomes. So because here I'm talking about really deterministic transformations, I will come to the stochastic transformations later on. So this is really the set of LOCC, is this set of maps with the additional conditions that this is really realizable, like I explained before. Yes, please? That's a very good question. But the answer is because separable operations cannot be So there are separable operations which cannot be realized with LOCC. So if entanglement theory from its very origin is a resource theory. So we want to call a state entangled if it cannot be prepared locally. But this implies that the free operations are LOCC and not set. But good question. Yes, so if this was the case, then things would be much easier. But it's not the case. So now the question is, of course, well, if we want to order our states, so we want to get some partial order, we want to say what is a better resource, then what should we do? We have to study LOCC, but LOCC is very difficult to do. Fortunately, for the bipartite case, one can show that LOCC is very simple. So in the bipartite case, if I study LOCC transformations among pure states, among pure bipartite states, which is what we consider here. We consider transformation from side to side because we want to sort our states in the Hilbert space, our pure states. Because if I consider these, then one can show a very nice theorem, which tells us that all this mess here can be replaced in that case by something very simple, which is just Alice does some operation, as we had it before. She sends the outcome to Bob. And Bob, depending on the outcome, applies just a unitary, and that's it. So we don't have to consider classical communication back to Alice. We don't have to consider that Bob also does some measurement. Just needs to apply a unitary that depends on the outcome of Alice. So this is a theorem by Juan Conglo and Sandoval Bescu. I don't remember now which year, but I don't remember. And this means that everything, or roughly speaking, it means that everything that Bob can do can be done up to local unitaries. OK, so this means that this nasty kind of structure mathematically boils down to something that is very simple, because here our maps simply look like this. So this is just some of AI. This is the measurement that Alice does, and the unitary by Bob. That's it. So I only have to check, then, that these AIs, this set, for this completeness relation. So this is very easy to do. So this means that in the bipartite case, we are very lucky, because these kind of transformations, the ones that we have to study in order to study entanglement become very simple. So this means now what? Well, I want to use that now, because I want to go back to the original problem. I want to study which state is more entangled than the other. So how can we use that? Let's talk about the state transformations. So does anyone know what the outcome will be? So when can I go from a state psi to a state phi? Can you speak up? I cannot. Can you speak up? I cannot hear you. Maturization? Yes, exactly. So this is Nielsen's criterion, which tells us when it is possible. So this problem has been solved. So we know in the bipartite case, when is it possible to go from psi to phi? Now, what I would like to do is to tell you how you can derive that using this result here, because this is something that I will also use later on for the bipartite case. And it's going to be very simple to derive it. So let's see when I can go from psi to phi. So I have a state psi. Let me just draw this picture also. Now, what do I do? Well, L is applied some measurement, right? So I get here some a1 tensor identity applied to psi. This would be the first outcome. The second outcome would be a2 tensor identity applied to psi, and so on, and so on. Now, if I want to reach the state phi deterministically, it means that at some point of this protocol, I have to end up in the state phi for all the outcomes, for all the branches of this protocol. They must be all transformed to psi. But now, I know that the LOCC protocol in the bipartite case looks that simple, and I'm using that now. So this means that the states that I have here must already be a u equivalent to phi, because there's still a local unitary that Bob has to apply in order to get the state phi. OK, so let's write this condition. So I have that ai tensor identity applied to my initial state psi has to be up to some proportionality factor and a unitary that is applied by b to my final state phi. For each of the outcomes, I have to obtain the final state, because I'm considering here deterministic transformations, not with a certain probability, but all of them have to end up in the same state phi. OK, let's write this condition. So the first thing, we have seen yesterday that a bipartite state has the Schmidt, so there exists the Schmidt decomposition. And this means that I can always write a bipartite state up to some local unitaries as some diagonal matrix applied to the phi plus state. So the phi plus state is, as we had yesterday, just a sum over ii. This diagonal one is simply the one that contains the square root of the Schmidt coefficients. If you apply that to this state, you see that you end up with any state that you want. OK, so I write my psi like this. And now I use here instead of a d, because I also want to include the local unitaries, I use here some operator g. Now, of course, the same is true for my final state phi. And I write it as an h tensor identity applied to phi plus. OK, now I look at this condition here. So let's write it down. We have ai applied to g tensor identity, psi has to be square root of pi h tensor the unitary phi plus. So I have now on both sides, I have the same state phi plus, OK, which I was just using the Schmidt decomposition. OK, so now let me consider the case where h is invertible, which just means that I have here, so the Schmidt coefficients are not 0. If you want to consider also the other case, you have to consider something like a pseudo inverse. Let me skip that for a moment. So h is invertible. This means that I can multiply with h minus 1. So I simply do that. OK, it's just rewriting the equation. And now I'm using that the phi plus state has a symmetry, because you might know that this is true. So if I apply an a tensor identity to the phi plus state, then this is the same as identity tensor a transpose applied to the phi plus. OK, so you're familiar with that? If not, then just write it down. It's a simple exercise to prove that this is true, and it's something very important. In fact, this simple relation, which can be proven very easily, implies this very interesting result. OK, so using this relation, you can easily prove that this is true. And we are going to use that now in order to show which transformations we can do to obtain deterministically in other states. So what I do now is, from here, I get that if I consider any unitary, then u tensor, u conjugate applied to phi plus, will be phi plus. But for any unitary, this is true. This simply follows from here. Just multiply with another u or a, and you get this relation. And so this means now that if you look at this expression, this means just multiply with u i dagger here. I'll get u i dagger, and get identity. This means that this operator here has to be just a conjugate of this operator here. So I have that my, and I'm not sure you don't see this. So I have that h minus 1 ai g has to be equal to the square root of pi times u i dagger conjugate, so u i transpose. And this is actually an if and only if, because you can prove that this is an if and only if. This is the only symmetry that you have from a state phi plus. And this is why you need this to be true. OK, now let me recall what these operators are. So this is the one that comes from psi. This is the one that comes from phi. These are the Schmitt coefficients of psi and the phi. And this is our measurement that we do. So for the measurement, we still have to impose the completeness relation. So I have to write now ai as a function of the other operators, which is easy, because this is just square root of pi h u i transpose g minus 1. And I have to impose that the sum of ai dagger ai is that entity. But now what is that? If I insert this expression, then I get the sum of pi. Now ai dagger, so I have here g minus 1 dagger u i conjugate, h dagger h u i transpose g minus 1. And this has to be equal to that entity. Now you see that I'm summing here over i, and this g does not depend on i. So I can multiply with g dagger. So I get g dagger g has to be equal to the sum of pi u i conjugate h dagger h u i transpose. This is a necessary and sufficient condition now. So I can go from psi to phi, if and only if, there exists a probability distribution, pi, and unit there is u i, such that this relation holds. But now what is this relation? Does this sound familiar to any one of you? We have heard the answer already before, but this is exactly the condition that the maturation is fulfilled. So this is exactly equivalent to the lambda, so the vector that contains the eigenvalues of g dagger g is maturized by the vector that contains the eigenvalues of h dagger h. So let me summarize this. What we have proven here is that I can go from psi, which I write as g applied to the phi plus to a state h applied to the phi plus, if and only if these operators g and h fulfill these conditions. So if and only if there exists a pi probability distribution and u i unitary, such that this is true. Now this relation is something very well known in matrix analysis, but also in probability theory. And it's equivalent to this maturization condition on the eigenvalues of our Hermitian operators, the ones that we have here. So this problem is solved. So we know that there exists a probability distribution and unitary is u i, such that this is true if and only if the eigenvector containing so the sorted, but this is implicit. So the sorted eigenvalues of g dagger g maturized by the sorted eigenvalues of h dagger h. So this tells us, just a second, this tells us now that we have solved the question, when can I go from psi to phi? Because now I just have to translate what is this g as a function of psi, right? But this g is just the square root of the reduced density matrix. So this thing here actually is just the reduced density matrix of our state psi. And because of that, we have that this is true if and only if lambda psi, so the Schmidt vector of psi is maturized by lambda phi. Yes, please. Where do you have unitary maps? No, no, no, no. Yes, yes, yes. OK, so we have obtained now Nielsen's criterion, which says that I can go from psi to phi if and only if the Schmidt vector of psi is maturized by the Schmidt vector of phi. Let me write this as a conclusion from what we have seen. This is Nielsen's criterion. And this is a necessary and sufficient condition, so it's indeed a criterion. So psi can be transformed to phi if and only if lambda psi is maturized by lambda phi. Yes, yes, yes. So I'm always talking here about deterministic. So this means deterministic transformation. Of course, in a deterministic transformation, I also have branches. But deterministic means that at the end, all these branches end up in the same state. So I don't have here a state phi prime with a certain probability. I get this probability 1 this state at the end. But of course, I have branches. I might have branches. OK, now what is this relation actually? So this means that the following. This means that if I take the sorted eigenvalues, the sort of Schmidt coefficients of psi, and I sum up the smallest ones. So I take L going from some k to the dimension of the Hilbert space. And this has to be larger or equal to the same expression for phi. So you might also recognize these expressions here, because these are some very well-known entanglement monotones that we will discuss a little bit later again. OK, so but this is a relation that is very easy. It's very easy to check also. So you just take the Schmidt coefficients of your state, psi and phi, and you compare the relation. This is a set of inequalities. You have d minus 1 inequalities. You compare if this is true. If it is true, then you know that the state psi can be transformed to phi. And if not, you know that it's not possible. No, this depends on how you, so which ones you sum. So if I start here now from 1 to k, then I would need to write it the other way around. But I start from k to d. So I'm summing the smallest ones. OK. So what would be an example? So for instance, let's consider the state phi plus. And let's consider two qubits. OK, so what is then the Schmidt vector of our phi plus? Everybody knows it, so I'm considering two qubits. What are the Schmidt coefficients of the phi plus state? Up and down? Yeah, Schmidt vector of phi plus is? For qubits, yes, for qubits. Half, half, exactly. OK, so I'm wondering now whether I can go from phi plus to some state a 0, 0 plus b 1, 1. OK, so this is some other state in the Schmidt decomposition. And I assume now that the a, with others of generality, I assume that a is larger than b. So the state, the Schmidt vector of this state would be a b. So now, when can I go from this state to that state? Well, if and only if this is true. So what is this condition now? So I start with k equal to 1. If I start from 1 and I sum up all of them, then I get identity for both of them. I get 1 for both of them, because it's a normalized state. So I start from 2. So I take the smallest Schmidt coefficient. So I take the second 1. And I have to check whether 1 half is larger or equal to b. OK, so comparing the two smallest Schmidt coefficients. But of course, I sorted it. This is sorted. So a is larger than b. And they have to sum up to 1. So clearly, b is smaller than 1 half. So this is obviously always fulfilled. So this means, as always, and therefore, we have that the phi plus state can be transformed to any other state of that form via ILOCC, because it's true for all of them. No, it's a square of them. So the Schmidt coefficients are the square of the square roots. Why? So I have the smallest Schmidt coefficient of my phi plus is 1 half. And the smaller Schmidt coefficient of my phi state is b. And b has to be smaller than 1 half, because this has to be sorted. And they have to sum up to 1. So it is like that. OK, and now, so what we see is that it doesn't matter, right? Because it doesn't matter which state this is, actually, because this is the Schmidt decomposition of an arbitrary state, right? So from the phi plus state, I can go to any other state. So what we see from here is that some very important message, namely that from studying ILOCC transformations among pure states, we learned that the phi plus state can be transformed to any phi. But this means that the phi plus state is the optimal resource. Because given the phi plus state, I can go to any other state. So the entanglement of the phi plus state is larger equal to the entanglement of any other state. So if I want to do an experiment, then clearly I just prepare the state phi plus, because if I want to use another state, I just do this LOCC, which is for free. So from here, we see that the phi plus state is maximally entangled. It's the optimal resource. So it's nothing better than that. Now you might wonder, why do I say the phi plus state and not the psi minus state? Why do I say the phi plus state? Is the psi minus state also an optimal resource? Because they are all a U equivalent, exactly. So of course, as I said yesterday, we care about a U equivalent classes. Because of course, I could write here any state that is a U equivalent to phi plus. But I don't need to, because I know that all these states have the same entanglement. So I just pick one representative. I could have picked any other, but I pick one representative and say that this is the optimal resource. Of course, all the U equivalent ones are also an optimal resource. But I don't need to consider them, because this is the same thing, where this phi plus or psi minus has met, that this is locally unitary equivalent. But this state allows me to go to any other with LOCC. And this is why it is the optimal resource. And the optimal resource for entanglement theory means maximal entanglement. But now we have proven that phi plus is not only maximally entangled, because there is some measure which is optimized by phi plus. But we have proven that for this state, it holds the entanglement of whatever measure you choose. This is larger equal to e of phi for all phi and for all entanglement measures. This is why I can say this is the optimal resource. Oh yes, yes, you're right. Yeah, you're right. OK, so this is an important message. And we will come to this when we talk about multipartite systems, because there we will see that it's not going to be that easy. Right, OK, so now from the theorem that I had here, which was telling us that these complicated LOCC transformations boils down to something very simple in the bipartite case. We have some more consequences. And let me stress here that this is only true if I talk about LOCC transformations among pure bipartite states. Whenever you relax some of these conditions, either not pure or not bipartite, it doesn't hold. It's just for pure bipartite states. You have this simple characterization of LOCC transformations. And from there, it follows. And write it now for short just that this LOCC is equivalent to Alice doing some operation. And Bob doing some unitary, depending on the measurement outcome. From here, we get the following. We get that for bipartite pure states, we have that LOCC is equal to LOCC finite. So I use only finitely many steps. That's obvious. It's obviously a consequence of this thing here, because this is finite. There are no more rounds. There's one round. Alice does something. Sends the classical communication to Bob. Bob applies the unitary. That's it. One round. So clearly, everything that I can do, even if I consider infinitely many rounds, I can do with finitely many rounds. I can actually do it with one round. So this is clear that this is true. And now I wrote it on the wrong side. Sorry. Let me put it here. And what has been proven for this case for the bipartite transformation is that this is also equal to the separable operations. Remember that I was telling you that the separable operations are mathematically much more tractable than the LOCC. In this case, it turns out to be the same. Can be proven to be the same. And on this side here, we have something else, which we called all deterministic protocol. For short, all that. And this is the following. This is a protocol where I start with my state psi. I do some transformation like we did there. So I get some a1 psi. Here, I get some a2 psi, and so on. And that means that in each step, I do a deterministic step. So this means that all these outcomes are LU equivalent. I might then go on, do something further. But in each step, they are LU equivalent for all the possible outcomes that I have. Now, we have seen in the bipartite case, this is always like that, because we end here. So Alice does some measurement. And then Bob only has to apply a local unitary to obtain the same state phi. So these states are all LU equivalent. So different local unitaries, but all local unitaries. So in the bipartite case, we have this. So we see that in the bipartite case, this all deterministic protocol is equal to LOCC finite, finitely many rounds, which is equal to the general LOCC, also allowing for infinitely many rounds. And this is equal to the separable. In the multipartite case, what happens is, I mean, the worst that can happen. So whatever can go wrong here, because this makes it very simple. So this is the reason why everything is really very simple in the bipartite case. In the multipartite case, everything that can go wrong goes wrong. So first of all, in the multipartite case, this will not be true. So you have protocols where you cannot do it with all that. You need a probabilistic step. This we don't know, and this is also not true. So this is what makes it much, much harder in the multipartite case. And this also shows very clearly the big difference between bipartite and multipartite. We will come to this later on. I just wanted to tell you here that the reason for why starting the bipartite case at the end is really very simple, because you have here everywhere an equivalence. So you can consider the most simplest protocol which there is in order to deal with the most difficult ones which there is. Now the reason why I also write this equation is because, of course, when you want to study this now in the multipartite setting or for mixed states, if you want to study LOCC, then because of this complicated mathematical structure that we saw, there is no way that you study it directly. So I don't know how you could possibly do that, because you have to take infinitely many rounds into account. So what you want to do is to consider a set that contains LOCC, which is the separable set, and something that is contained in LOCC. So then you have a bound on both sides, and you might be able to prove that the bounds are the same, which means that you also know it for LOCC. So this is the general idea. OK, we'll come back to this when we talk about multipartite states. Are there questions so far? OK, so then let's go on to the next chapter, which is entanglement measures and entanglement monotones. So let's talk how we can actually measure entanglement. So we were now using it always, just as a word, basically, and having in mind some examples. But let's look at the definitions. So entanglement measures, which I will for short call enmesse and entanglement monotones. OK, so first, what is the definition also in the case of mixed states? Enmesse and monotones for mixed states. So the definition, what is the definition of an entanglement measure? We have seen that already. You recall? When I can call a functional entanglement measure, it matches a resource, no? Exactly. So the condition is that E is an entanglement measure if E of rho is larger or equal to E of lambda rho for all. Let me write lambda LOCC. Lambda LOCC, this is a map that I can realize with LOCC for all states rho. And all lambda LOCC being contained in LOCC. As you were saying correctly, this is the free operation. So this I can do. This doesn't cost me anything. So this can only decrease entanglement. And now if I have an entanglement measure, then it has to obey this condition, nothing else, just this condition. OK, this makes perfect sense. It's very physical. This is exactly what we want to have. Now there is another very useful concept, which is entanglement monotones. Does any one of you know the definition of entanglement monotones? Somebody was mentioning it yesterday already. So entanglement monotone is something that does not increase on average. So this means that E of rho is larger equal to the sum of Pi E of sigma i for all rho and ensemble Pi sigma i, such that rho can be transformed via LOCC to this ensemble. OK, so here indeed I consider different outcomes. I get this probability Pi to state sigma i. And what I'm asking for now is not only that E, I should not say not only, but it's not that E of rho is larger than E of the sum, which would be the lambda of light to rho. But I'm asking that it's not increasing on average. OK, so it's something different. This is non-increasing on average. In the literature, this condition here is called monotonicity condition. So it's monotonic under LOCC. This condition here is called strong monotonicity condition, which might be a bit confusing because this is not necessarily stronger. We will see that this is stronger in case we consider pure states, or in case we consider convex functions. But otherwise it's not necessarily stronger. Nevertheless, this is called the strong monotonicity and the monotonicity condition. This has historical reasons for why it's like that. OK, so now we have these two definitions. We will see that both of them are very useful. This is really the one that comes from the physical picture. So this is what we have. This is our resource theory, and we cannot increase with LOCC. This is exactly what that means. This definition is slightly different because this means that we cannot increase on average. But we will see that these entanglement monotons are very useful. They can lead to entanglement measures, and they can be used in many different contexts. OK, so unfortunately my time is already over. Are there some questions? Yes, please. Yes, so LOCC is different than ZEP. So there are examples, even for bipartite mixed states, going from a mixed to a pure state. This is a result by Eric Chittambar and I think Juan Conlo, that you can achieve. So that proves that you can do it with ZEP, but you cannot do it with LOCC, even bipartite. This is a state transformation going from a mixed state to a pure state. Well, this is contained in that. So you're asking for an explicit example for mixed to mixed in the bipartite. Sorry? OK, so it's unanswered. So for mixed to mixed, it's unanswered. It's very difficult. What is answered? And we will, I might talk about this briefly, is from pure to mixed in the bipartite case. Are there more questions? Do I still have some time more? Do I still have some time? No.