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Consider the four Bell states $$ |\psi^{00}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle), \hspace{2mm} |\psi^{01}\rangle = \frac{1}{\sqrt{2}}(|00\rangle - |11\rangle),\hspace{2mm} |\psi^{10}\rangle = \frac{1}{\sqrt{2}}(|01\rangle + |10\rangle),\hspace{2mm} |\psi^{11}\rangle = \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle), $$ and denote by $\psi^{ij} = |\psi^{ij}\rangle\langle\psi^{ij}|$ the corresponding density matrix. If we group the Bell states into two mixed states, it is easy to see that $$ \rho_0 = \frac{1}{2}\psi^{00} + \frac{1}{2}\psi^{01}, \quad \rho_1 = \frac{1}{2}\psi^{10} + \frac{1}{2}\psi^{11} $$ are perfectly distinguishable by local operations and classical communication (measure in $Z$-basis locally and compare results), and similarly it is easy to see $$ \sigma_0 = \frac{1}{2}\psi^{00} + \frac{1}{2}\psi^{10}, \quad \sigma_1 = \frac{1}{2}\psi^{01} + \frac{1}{2}\psi^{11} $$ are perfectly distinguishable by local operations and classical communication (measure in $X$-basis locally and compare results). Is it similarly easy to see $$ \xi_0 = \frac{1}{2}\psi^{00} + \frac{1}{2}\psi^{11}, \quad \xi_1 = \frac{1}{2}\psi^{01} + \frac{1}{2}\psi^{10} $$ are perfectly distinguishable by local operations and classical communication? I have tried to come up with a measurement, but without success. I would appreciate any help!

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The states $\sigma_0$ and $\sigma_1$ can be seen to have the form $$\sigma_0 = N[\mathbb P(e_1+e_4) + \mathbb P(e_2+e_3)], \\ \sigma_1 = N[\mathbb P(e_1-e_4) + \mathbb P(e_2-e_3)],$$ where $N$ are normalisation constants, $\mathbb P(\psi)\equiv |\psi\rangle\!\langle\psi|$ is shorthand notation to refer to projections, and $$ e_1\equiv |00\rangle, \quad e_2\equiv|01\rangle, \quad e_3\equiv|10\rangle, \quad e_4\equiv|11\rangle.$$ If an operator is a sum (with equal weights) of two (or more) orthoprojections, it can also be written as a linear combination of said projections upon an arbitrary unitary operation. Better said, if $A=\sum_i \mathbb P(u_i)$ with $\langle u_i,u_j\rangle=\delta_{ij}$ , then $A=\sum_i \mathbb P(Uu_i)$ for any operator $U$ that is unitary when restricted to the span of $\{u_i\}$. In particular, this means in our case that we can write our states (ignoring normalisation constants) as $$\sigma_0 = \mathbb P(\underbrace{(e_1+e_4)+(e_2+e_3)}_{=(|0\rangle+|1\rangle)\otimes(|0\rangle+|1\rangle)}) + \mathbb P(\underbrace{(e_1+e_4)-(e_2+e_3)}_{=(|0\rangle-|1\rangle)\otimes(|0\rangle-|1\rangle)}) \\ %= \mathbb P((|0\rangle+|1\rangle)\otimes(|0\rangle+|1\rangle))\\ \sigma_1 = \mathbb P(\underbrace{(e_1-e_4)+(e_2-e_3)}_{=(|0\rangle-|1\rangle)\otimes(|0\rangle+|1\rangle)}) + \mathbb P(\underbrace{(e_1-e_4)-(e_2-e_3)}_{=(|0\rangle+|1\rangle)\otimes(|0\rangle-|1\rangle)}).$$ So the gist is that you can find a way to write $\sigma_i$ as sum of projections over orthogonal product states.

Why is this possible at all? To better understand it, focus on $\sigma_0$. The considerations above tell us that, for any pair of complex numbers $\alpha,\beta$ with $|\alpha|^2+|\beta|^2=1$, we can write $$\sigma_0 = \mathbb P(\alpha(e_1+e_4) + \beta(e_2 + e_3)) + \mathbb P(\bar\beta(e_1+e_4) - \bar\alpha(e_2 + e_3)).$$ The question is whether there are coefficients $\alpha,\beta$ such that the vectors in this projections are product states. Focusing on the first one, we are asking whethere $$(\alpha,\beta,\beta,\alpha)^T = (a,b)^T\otimes (c,d)^T$$ for some $a,b,c,d$. While the solution is easy (and already known) in this case, let's keep things general for the sake of argument. We can figure out when $(\alpha,\beta,\beta,\alpha)$ is a product state by computing the entropy of the corresponding reduced density matrix. Equivalently, by computing its eigenvalues. These eigenvalues are easily seen to be (again, up to multiplicative constants) $|\alpha\pm\beta|^2$, which immediately tells us that we get product states whenever $\alpha=\pm\beta$, compatibly with the solution we already knew.

So far, so good. Now with all of this in mind, let's get back to the question at hand. We can write $\xi_i$ as $$4\xi_0 = \mathbb P(e_1+e_4) + \mathbb P(e_2-e_3) \\ = \mathbb P(\alpha (e_1+e_4) + \beta(e_2-e_3)) + \mathbb P(\bar\beta (e_1+e_4) - \bar\alpha(e_2-e_3)), \\ 4\xi_1 = \mathbb P(e_1-e_4) + \mathbb P(e_2+e_3) \\ = \mathbb P(\alpha(e_1-e_4) + \beta(e_2+e_3)) + \mathbb P(\bar\beta(e_1-e_4) - \bar\alpha(e_2+e_3)).$$ Focusing on the first projection of $\xi_0$, we thus need to find out whether there are $\alpha,\beta$ such that $(\alpha,\beta,-\beta,\alpha)$ is a product state. The eigenvalues of its reduced density matrix can be seen to be $1\pm 2\operatorname{Im}(\alpha\bar\beta)$. We need one of these two eigenvalues to vanish, which is only possible if $\operatorname{Im}(\alpha\bar\beta)=\pm1/2$, which is the case with $\alpha=1/\sqrt2$ and $\beta=\pm i/\sqrt2$.

All this to say that we can write $$\xi_0 = \mathbb P((|0\rangle-i|1\rangle)\otimes(|0\rangle+i|1\rangle)) + \mathbb P((|0\rangle+i|1\rangle)\otimes(|0\rangle-i|1\rangle)) \\ \equiv |+i\rangle\otimes|-i\rangle + |-i\rangle\otimes|+i\rangle,$$ and similarly $$\xi_1 = |+i\rangle\otimes|+i\rangle + |-i\rangle\otimes|-i\rangle.$$ Thus the states are perfectly distinguishable measuring in the $Y$ eigenbasis and comparing results, exactly as it was the case for $\rho_i$ and $\sigma_i$ in $Z$ and $X$ bases.

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