Let me get back to this scenario again. This arrow is the scenario. Two part is Alice, and Bob. They initially share many copies of Psi. Psi is given here. We also put this angle between 0, and Pi over 4, and from the asymmetric competition, and etc. We do not lose any generality by considering this state. Then we consider the single-copy manipulation. A here, just manipulates the quantum set on a single-copy level. General quantum dynamics in terms of cross operators. That we have constructed a quantum operation to transform state Psi into five plus in a single-copy level. The optimal conversion rates we found is this number, 2 sine square Theta. What is the maximum internal state and the entanglement of this guy? Maybe quantified by sine Theta. The next question is, is it really, say, the highest rate we can imagine for converting state Psi into five plus? The reason that I'm asking this question is that if we think of these single-copy manipulation, there could be some more general manipulations, there could be some more general local operation. This local operation means just operation on A side only. Maybe the most general operation on A side is not just single-copy level, but n copies coherently. Then we can consider say coherent operation on n qubit here together. Then this may allow to crop more number of five plus state here. Then this is the, say, general entanglement manipulation. The answer I'm going to show you is, in fact, n to m is about the entropy, which is [inaudible] of entropy of cosine square, and sine square. That is the case of single-copy level. This is the case of coherent ones n copy level together. I will show you, how we can all achieve this number. Then consider n qubits. Say this guy is given n, so n pairs. Alice is with n qubit, and Bob is with n qubit. How does the state looks like? Well, it's clear that this tensor over n. But that's just two copies, how it looks like, which means cosine square Theta. Two copies. The cosine square Theta, and 00, and 00, and 00. This guy is from the first pair, so A_1, and B_1, and this guy is A_2, and B_2. The first pair, and second pair. Cosine Theta, and sine Theta. Then here we have a 00, 11, and 11, and 00. This guy is for the first pair, and second pair; first pair, and second pair. Then sine square, and 11, and 11. Now, I mean, we can just rewrite this part as A part, and B part because it is easy to show the manipulation of the qubit on A side. Then applying this one, exactly the same states, and 00 now, but this is A_1A_2, and 00, and B_1B_2. Then cosine Theta, and sine Theta. I am interested in this first A side, and then B side here. Then 01, and 01, which is A_1A_2, and here B_1, and B_2. In the same A, we have 10, and 10, so 10, and 10. Then sine square Theta and 1 1, which is A part and B part. We can keep writing second copy, a third copy etc, and overall we can write down this n cubed or 2n cubed state. But n pairs, it can be written as cosine n power n. N here is 000 and all a sign, this is n cubed and this b cubed, n cubed as well. Then cosine n minus 1 sine Theta. But here we have a collection of these guys, have a single one, so 0 0 1 0 0 1 and etc, then we have 1 0 0 1 0 0. We have basically n terms, n vectors here. We can keep doing, and after all, we can write down n sine power n here an. This is a state on state of early sample, so n pairs. Now, if you see here on A side, what happened is that, actually 0 appears with probability cosine squared Theta and one appears sine square Theta. Then this system is repeated and there are lots of collections. We have learned the law of large numbers or information theory or compression and the motivation is the following. I have a coin, it give zero with a probability cosine square and give one with a probability sine square and I just threw once. What is the value zero or one? I have no idea. I just know the prior probability from the prior probability I can make my guess. But then I repeat throwing the coin many times. Then suppose I repeated n times, so I have n sequences. Then I do know the most probably sequences are n times cosine scale zeros and n times sines square ones because this is just from the probability. For instance, the probability that my sequence is all zeros is very low. For the same reason, the probability that we have all ones is also very low. I can make a guess. The most probably sequences, and this is in fact true. This is called the law of large numbers. I cannot make a right guess about my single coin, but repeating many times, I can guess about the sequence. The most probably sequences are called typical sequences. That sequences will survive with a high probability, which is very close to one. They call typical sequences. We can write down here. This is define the typical sequence, meaning that the nb sequence, I write here n and the probability they survive is quite high. They are bounded in fact, say p(x^n). It is bounded lower minus n h of x plus Epsilon. This upper bounded by 2 minus n h x. If we take n very large number then Epsilon converge to zero. In the asymptotic limit we will have the probability of the probable sequence is about 2 to the minus nhx. We can drive the fact that for any sequence when n becomes very large and the probability that this sequence is in the typical sequence is a sufficiently high. We can write down this way. This is a lesson from, we can say, law of large numbers or a lesson from information theory. I want emphasize that the surviving sequences have equally probable, and the probabilities about this one, and this is called asymptotic equipartition property. The surviving sequences, they are same, equally probable. Then how big is this error? Then the number of elements, each element has this probability key, and therefore each element, if we count the, the size is about two to the H_x, and H_x is the binary entropy with this probability. Then we can define some operations on a site. According to this information theory, we can define P_n and epsilon. This is projection into the space spanned by a typical sequences. I will write here X_n. According to this property, for any quantum states any copies, then the probability that the state leave in this subspace is very high. Almost one. Therefore, what we do is to apply this transformation to this guy, 2 to the n, this guy. Then we have lots of sequences here. This all ASI and ASI, and ASI and ASI. Lots of sequences here, all of the possibilities. But then according to this probability, is not that all of them will survive. Some of them will survive with high probability, and they are in fact this typical sequences. Applying this one to here, this will happen, some probability. But according to this one, is fairly high, so, I can say one minus theta. This is the probability that this project can succeed. It's almost one, and we will have some state here. What is this states applying this one here, and then all of them are equally probable. Therefore, this guy, right you're here, is sum of all typical sequences, and there it is the same. A_x, A sine X and P sine. How many are there? These numbers, and all of them are equally probable with this probability. Therefore we have the constant here to the n and Hx. That's the states that Alice and Bob share. In fact, this one, we wrote A Part D Part, A Part D part, and now we recombine them to write the state pairwise, so, pair to pair. Then we have each element here, then this is in fact five times m. The relation between M and this guy can be found. This is precise is the relation between this number and here, and m is in fact row two, and this number. Two to the n and Hx and Hx is this number is cosine scatheta and sine scatheta. Then log and now we have here at this point. Rate between n given a number of this non-maximum intensity and the number of maximal intensity we will have at the, at the end of the protocol. The ratio is given by this entropy M here. We can summarize. Alice and Bob can do more general operations than single copy of manipulation. Alice can project this many qubit into typical sequences or the subspace spanned by a typical sequences. This will happen probably, very high probability. Just I'm performing there. Then I'm in density is shared between our example is this guy. This is in fact the asymptotic limit that asymptotic limit of the conversion rate. Convergent that this state can be transformed to 5 plus pi at OCC.
