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Suppose Alice and Bob share the quantum state $\frac{1}{\sqrt 2}(|x\rangle + (-1)^b |y\rangle)$ for some $x\neq y \in \{0,1\}^2$ and $b \in \{0,1\}$. They both do not know $x,y$, and use some middlemen who wishes to learn $b$. They are allowed to send only classical messages to him (i.e. not quantum states). The marginal density matrix of Alice, for example, is $$ \rho_A = tr_B(\rho_{AB}) = \frac{1}{2}( |x_1\rangle \langle x_1| +|y_1\rangle \langle y_1| + (-1)^b |x_1 \rangle \langle y_1| \cdot \langle x_2 | y_2 \rangle + (-1)^b |y_1 \rangle \langle x_1| \cdot \langle y_2 | x_2 \rangle ) $$ Note that when $x_2 \neq y_2$, then $\rho_A$ does not depends on $b$. Moreover, if also $x_1 \neq y_1$, from symmetry arguments, $\rho_B$ does not depends on $b$. In such case, it seems to me that they can not send any data to the middleman (such as measurements), to help him deduce $b$, as the marginal density matrix, i.e. their personal view of the system, is oblivous of $b$. Is it correct to say so?

Of course when $x_2 = y_2$ for example, then Alice marginal density matrix does depends on $b$, and she can apply Hadamard gate + measure, to obtain exactly $b$, and send it to the middleman.

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Yes, they can do it. The reduced density matrix is not the appropriate tool to use here: it gives you the statistics Alice can obtain while having no information at all about what Bob is doing. But in this case she (or the middleman, in any case) does have information about what Bob is doing, she can know the results of his measurements.

For example, if they both measure in the $\{|+\rangle,|-\rangle\}$ basis, the probability they get both result $+$ is $(1+(-1)^b)^2/8$, which does depend on $b$.

More generally, it is always possible to learn all the amplitudes of the state in this way. What you are asking is if local tomography is possible, and it is, because you can always find a tensor product basis for the shared Hilbert space.

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If Alice and Bob knew the values of $x$ and $y$, there would always be measurements they could perform which would allow the middleman to distinguish the value of $b$. An argument using the reduced density matrix of a single qubit is irrelevant because that neglects the correlations that can arise between measurement results of the two parties, and it is precisely those correlations that you're relying on in some cases.

However, the question states that $x$ and $y$ are unknown. This is far more of a problem. Let's assume we have no prior knowledge of those values, so all possibilities are equally likely. Then, Alice and Bob are trying to distinguish between two possible ensembles, whose best descriptions are joint density matrices $$ \rho_0=\frac{1}{12}\sum_{x\neq y\in\{0,1\}^2}(|x\rangle+|y\rangle)(\langle x|+\langle y|) $$ and $$ \rho_1=\frac{1}{12}\sum_{x\neq y\in\{0,1\}^2}(|x\rangle-|y\rangle)(\langle x|-\langle y|). $$

I believe (although you'll want to check, because I did this quickly, not carefully) that these look like $$ \rho_0=\frac{1}{12}\left(\begin{array}{cccc} 3 & 1 & 1 & 1 \\ 1 & 3 & 1 & 1 \\ 1 & 1 & 3 & 1 \\ 1 & 1 & 1 & 3 \end{array}\right)=\frac{1}{6}I+\frac{1}{3}|++\rangle\langle ++| $$ and $$ \rho_1=\frac{1}{12}\left(\begin{array}{cccc} 3 & -1 & -1 & -1 \\ -1 & 3 & -1 & -1 \\ -1 & -1 & 3 & -1 \\ -1 & -1 & -1 & 3 \end{array}\right)=\frac{1}{3}I-\frac{1}{3}|++\rangle\langle ++|. $$ These clearly do not have support on orthogonal subspaces, so there is no measurement that can guarantee to distinguish between $\rho_0$ and $\rho_1$ even if Alice and Bob could perform two-qubit measurements jointly.

However, you could ask what the best measurement they could do is. This is called the Helstrom measurement. The probability of success is $$ \frac12+\frac14\text{Tr}|\rho_0-\rho_1|=\frac34 $$ Moreover, that differentiation is achieved by measuring both qubits in the $|\pm\rangle$ basis: if you get the $|++\rangle$ answer, you assume you had $\rho_0$ ($b=0$), otherwise you assume you had $\rho_1$ ($b=1$). Since this optimal basis happens to be a tensor product, this is measured by Alice and Bob independently and resolved by the referee.

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