We are back to this analysis scenario. The goal of any sample is to share these five process states. To share this state, we assume that two parties are assembled. On any side at least generate entire state this guy and keep the first one and send the second one to both sides. Because of the interactions with the environment or some error, the state shot between our sample is not this guy but some non-maximum Penn State. So far we have assumed that pure states, but still not maximum interest between our example and here to make it more general. Also sometimes we do not know fully the errors or noise appearing in quantum channel. Therefore, maybe assume that state shared between any sample is in general as a mixed state. Now the question is, given this guy and we want this tier or transformed this one to five plus by LLCs. Maybe is less than m. I mean, scenario wise, this is exactly the same to the previous one when any sample share non-maximum pure state. This problem contain bit more complicated problem, which is transformation of mixed state into pure state. Mixed state means some Haskell noise appeared in given quantum system. Then pure state does not contain at all Cascade noise. Then a transformation from this guy into here. This transformation means removal of Cascade noise contained in this air, misdeeds and is it possible? Because it is very natural or physical process that system prepared in pure state is going to transform to a mixed state. This is very natural, but the other way round is not natural. Is it possible to achieve this transformation? In fact, this is possible. Then I'm going to show you misstating timeout manipulation. For QB States, assume that state between any sample. They can perform the local operation which is given rho and tensor and trigger. By doing that, the state is a mixture of maximum occurs day and the rest. Which means taking the integration or this can be written in terms of discrete transformations.. This can be written as a integration of the states of unit tabs satisfying the major condition. Then over there we obtain this state shared between any sample, and then we have a single parameter F. This one, we write down load. The properties that this one is close to maximally stay in the values given by F and F is called a singlet fidelity. Then we have a parameter F and F and F. For any quantum state, we perform this operation. This operation is called a twirling. Then we come to this guy, a single parameter, rise to state rho F. Then all parties between any sample, pairwise they share rho F. Then what any sample, do, they take just two copies. Then perform the NOT gates like this. The pair in the first resistor, rho F in the pair in the first resistor is identical rho F as well. We can assume the same copies because they took to keep it generated from the same device and the state the cubiture disputed over the same environment, and therefore two parties share the same state here and then perform measurement and the second register. Then Alice, Bob they make an outcome i or j. Then what any sample do is to keep the first pair whenever i equals j. If i equals j, then we have the rejecting state rho F. But then because of the actions here, we would have different rho F and if i is not equal to j, if any sample have different measurement outcome in the second pair, then they discard. This is probabilistic. With some probability, Alice and Bob succeed to keep the pair in the first resistor, and this is this guy otherwise they discard. Then what is the relations between F and F prime? This is interesting, F prime, I suggest you can follow the lines and you can compute the F prime and [inaudible] is the F plus. That's the relation between F and F prime. Let's see what it means. We have here F and F prime. Then here one and then. This line means F prime is F and this one is something like this. This point is one over four, and we have this line. Suppose we start from two copies with the F at some point here. We start from F here, and then the same F is. But then taking two copies we will have flow F and run the circuit sacrificing this copies and then suppose we make success. The measurement outcome in the second register are the same so we have this guy. Then F prime must be here according to this graph. This is F prime. Then take two copies which after a successful rounds, Rho F prime, and Rho F prime, and then performing the same procedure, and there we perform measurement and suppose we get measurement, outcome the same. Then which means start from F prime here and then run the protocol, then if the outcomes are the same as well, then they are successful, then this is the value [inaudible] assemble and get F two prime, and they keep doing etc. Then what happened is that this guy converges to this point where F prime is one. This point corresponds to maximum [inaudible]. Taking two copies, then a coherent operation between copies and etc, then sacrificing some number of copies, and they can keep the entangle state, which is very close to a five plus state. In this way, it is possible to distill or extract this state from missed entangled state. If we start from this point, then what happen instead, is that it comes to here and then back and that's it. Say that this point is invariant from this one, if we start from entangled state having fidelity F larger than 1.5, it converted to a five plus state, otherwise, it is not possible. Any state Rho F with F larger than 1.5, it is possible that this guy can be transformed by LOCC into maximum cluster. Then we call this state is distiller. Then what does that mean? What does the condition F equal to 1/2 means in this point? In fact, this Rho F is entangled if F is larger than 1/2. This means that Rho F cannot be prepared by LOCC if F is larger than 1/2, if F is smaller or equal to 1/2, then Rho F can be prepared by LOCC, so is separable. Rho is entangled for this one, and then we have this condition. Therefore Rho F entangled, then it is possible to distill to maximal impulse day. If this is possible, then the state must be entangled. That's the conclusion we could have from this node here. There are two lessons, and the one. It is in fact possible to transform mixed state to pure state, and state to process by LOCC is highly non-trivial, but it happened that this is possible. Then for two qubit state, as long as the state is entangled, it is possible to distill entanglement from a given state. In this sense, all entangled two cubic state or all two cubic state distillable.