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Apparently $\vert W \rangle$ was first reported (and the naming convention first adopted) by Dür, Vidal and Cirac in this preprint on May 26, 2000 (version 1 of 2). This is supported by the footnote on page 4 of this preprint on June 25, 2000 (version 3 of 3, this footnote did not appear in the earlier versions), which states (in part) Very recently Dürr ...

4

The metric you're looking for is known as entangling power. Here are some references: Entangling power of two-qubit gates on mixed states (Guan et al., 2014) Entangling power of two-qubit unitary operations (Yi Shen & Lin Chen, 2018) I will expand on this later.

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You can calculate the probability of a given answer $\pm 1$ to each measurement by evaluating $$\langle B|\frac{I+(-1)^{\eta_1}\vec{n}_1\cdot\vec{\sigma}}{2}\otimes\frac{I+(-1)^{\eta_2}\vec{n}_2\cdot\vec{\sigma}}{2}|B\rangle$$ Thus, the probability of equal measurement outcomes is $$\langle B|\left(\frac{I+\vec{n}_1\cdot\vec{\sigma}}{2}\otimes\frac{I+\vec{... 3 Simply start by writing out everything$$ |B_{00}\rangle_{13}|B_{00}\rangle_{24}=\frac12\left(|00\rangle_{13}|00\rangle_{24}+|00\rangle|11\rangle+|11\rangle|00\rangle+|11\rangle|11\rangle\right) $$Let me rearrange each of these terms$$ \frac12\left(|00\rangle_{12}|00\rangle_{34}+|01\rangle|01\rangle+|10\rangle|10\rangle+|11\rangle|11\rangle\right).  Now ...

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This is not much of an answer, but is probably too long for a comment... I don't believe that there's a canonical way of doing this. You'd be best off understanding why you're asking the question, and what you want to get out of it. From there, you tailor how you're going to measure it. But multipartite entanglement is a really messy problem, even just for ...

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Intuitively, you can rotate $\vec{n}_1$ to $Z$. As $Z$ axis has two antipodal points $|0\rangle$ and $|1\rangle$, let $\vec{n}_1$ have two antipodal points $|b_0\rangle$ and $|b_1\rangle$. Now the Bell state can be rewritten as $\frac{1}{\sqrt{2}}(|b_0b_0\rangle+|b_1b_1\rangle)$. Now in this new basis, the calculation shall be much easier. To be precise, ...

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