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Let a three-qubit state shared between Alice, Bob and Charlie stationed at distant laboratories be $$\psi_{ABC}=\frac{\sqrt{2}}{\sqrt{3}}|000\rangle+\frac{1}{\sqrt{3}}|111\rangle.$$

How to evaluate the maximum probability of transforming the state to a three qubit maximally entangled state by local operations and classical communication only?

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    $\begingroup$ You can erase charlie out of the problem by measuring his qubit in the X basis and doing a fixup Z operation dependent on the outcome, which would reduce this to the two-party problem. $\endgroup$ – Craig Gidney Nov 15 '18 at 21:08
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    $\begingroup$ @CraigGidney How would the answer be related to a 3-qubit maximally entangled state? $\endgroup$ – Norbert Schuch Nov 15 '18 at 23:41
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    $\begingroup$ I'm not sure the general solution to the 3-qubit problem is known. For this case a protocol with success probability 2/3 is easy to find, and it is likely optimal. --- EDIT: Probably optimality can be proven by merging two parties - for the bipartite case, the optimal protocols are known. $\endgroup$ – Norbert Schuch Nov 15 '18 at 23:42
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    $\begingroup$ @NorbertSchuch Ah, good point. I just assumed "maximally entangled" meant EPR pair and overlooked the "three". Is there actually a unique maximally entangled three qubit state according to all measures? $\endgroup$ – Craig Gidney Nov 16 '18 at 0:45
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    $\begingroup$ @CraigGidney no, but for each class of entangled states, there is a maximally entangled representative. So in this case, you have to be talking about GHZ because it’s impossible to convert to a W state under LOCC. $\endgroup$ – DaftWullie Nov 16 '18 at 6:34
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The first question that we have to deal with is what is meant by "maximally entangled" in this context. There's no single straightforward notion. In particular, for three qubits, there are two inequivalent classes of entangled state that cannot be interconverted by SLOCC (stochastic local operations and classical communication). Each has a maximally entangled representative: $$ |W\rangle=\frac{1}{\sqrt{3}}(|001\rangle+|010\rangle+|100\rangle)\qquad |GHZ\rangle=\frac{1}{\sqrt{2}}(|000\rangle+|111\rangle). $$ In this case, we're obviously talking about conversion to GHZ; conversion to W is impossible.

A straightforward protocol to achieve the conversion is to introduce the POVM $$ M_1=\frac{1}{\sqrt{2}}|0\rangle\langle 0|+|1\rangle\langle 1|\qquad M_2=\frac{1}{\sqrt{2}}|0\rangle\langle 0| $$ such that $M_1^\dagger M_1+M_2^\dagger M_2=\mathbb{I}$. If Alice performs this measurement and gets answer 1, then she has created the desired state, $$ (M_1\otimes\mathbb{I}\otimes\mathbb{I})|\psi\rangle=\sqrt{\frac{2}{3}}|GHZ\rangle, $$ which succeeds with probability 2/3. The other outcome gives a separable state, so there's no hope of getting anything useful out of it.

How do we know that this is the best we can do? Imagine a second protocol of two players, Alice and Bob. They initially share $(\sqrt{\frac23}|00\rangle+\frac{1}{\sqrt{3}}|11\rangle)$ and wish to create $(|00\rangle+|11\rangle)/\sqrt{2}$. We know that their greatest probability of success is 2/3 thanks to work by Vidal. Now, one specific strategy that Alice and Bob could follow is for Bob to introduce an extra qubit in the $|0\rangle$ state, apply a control not targeting this new qubit, controlled from his qubit in the entangled state. This leaves them with $|\psi\rangle$. Then they apply the optimal tripartite protocol for converting to $|GHZ\rangle$ before Bob repeats the same controlled-not again. The overall success probability cannot be higher than 2/3, and the only probabilistic step in there is the protocol we're interested in, so that protocol cannot succeed with probability greater than 2/3. Hence our original solution must be optimal.

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  • $\begingroup$ we are interested in conversion to GHZ state. You are right. $\endgroup$ – Adex Nov 16 '18 at 7:45

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