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4

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{...


1

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, ...


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 ...


2

There's a mistake. It's incorrect even for $a=0,b=0,U=I$. In this case the correct formula is $$ |\phi\rangle \otimes |B_{00}\rangle = \frac{1}{2} \bigg(|B_{00}\rangle \otimes |\phi\rangle + |B_{01}\rangle \otimes Z|\phi\rangle + |B_{10}\rangle \otimes X|\phi\rangle + |B_{11}\rangle \otimes XZ|\phi\rangle\bigg) $$ but their formula swaps $X$ and $Z$ in the ...


2

In my understanding, the key part for entanglement to increase capacity is to have a suboptimal channel. Suppose the input of you channel can take value in the set $X$, and note $G(X)$ the graph where nodes are possible inputs and there is an edge between to inputs if the range of their corresponding output is overlapping, i.e. When going through the channel ...


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