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Suppose I share three Bell states among two participants Alice and Bob and Charlie in the following manner: $$ |\psi\rangle=\left(\dfrac{|0\rangle_1|0\rangle_2+ |1\rangle_1|1\rangle_2}{\sqrt{2}}\right)\left(\dfrac{|0\rangle_3|0\rangle_4+ |1\rangle_3|1\rangle_4}{\sqrt{2}}\right)\left(\dfrac{|0\rangle_5|0\rangle_6+ |1\rangle_5|1\rangle_6}{\sqrt{2}}\right) $$ I want to find out the density matrix for $A$ whos has qubits $(1,4)$ the others have say $(2,3)$ and $(5,6)$. So I first calculate $|\psi\rangle\langle\psi|$ and then take the trace over the particles $(2356)$, Since this is a long computation, we can just notice that the terms whose inner product will be involved are $$\langle0000|\rho|0000\rangle,\langle0100|\rho|0100\rangle, \langle1000|\rho|1000\rangle,\langle1100|\rho|1100\rangle,\langle0011|\rho|0011\rangle,\langle0111|\rho|0111\rangle,\langle1011|\rho|1011\rangle,\langle1111|\rho|1111\rangle,$$ the result I get on calculating this is $$\dfrac{2\left(|0\rangle_1|0\rangle_4\langle0|_1\langle0|_4+|0\rangle_1|1\rangle_4\langle0|_1\langle1|_4+|1\rangle_1|0\rangle_4\langle1|_1\langle0|_4+|1\rangle_1|1\rangle_4\langle1|_1\langle1|_4\right)}{8}=\dfrac{\mathbb{I}\otimes \mathbb{I}}{4} $$ Is this correct? What does this density matrix say about the information that the players have about each others particles?

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The result is correct. You can see it in the other way. You have three two-qubit subsystems $A = \{1,2\}$, $B = \{3,4\}$ and $C = \{5,6\}$. The whole state is the product state on those three subsystems, i.e. the whole density matrix is $\rho_A \otimes \rho_B \otimes \rho_C$. Product state means there are absolutely no correlations between the states on subsystems. Hence you can fully ignore the $C$ subsystem if you are interested only in the state on $D = \{1,4\}$. Also since $\rho_{AB} = \rho_{A} \otimes \rho_{B}$ then it must be $\rho_D = \rho_1 \otimes \rho_4$.

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  • $\begingroup$ What does this density matrix tell us about the measurement outcomes of Alice, say if he measures $|\phi^+\rangle$? What does the first row second column element $a_{12}$ tell us? $\endgroup$ – Upstart Nov 26 '19 at 16:16
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What does this density matrix say about the information that the players have about each others particles?

Nothing, if no further assumptions are made on the initial state. If $A$ and $B$ share a state $\rho$, a reduced state $\rho^A$ doesn't say anything about the information that $A$ has about $B$'s state. Indeed, $B$ can have any state as far as $A$ knows: any shared state of the form $\rho^A\otimes \sigma$ is compatible with $A$'s knowledge, for any state $\sigma$.

If on the other hand Alice knows that the initial state is pure, then knowing that their share of the state is $\rho^A$, $A$ can infer the entropy of $B$'s state, because $S(\rho^A)=S(\rho^B)$. Nothing else can be said though, as any local operation applied to $B$ will not change $A$'s outcomes.

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  • $\begingroup$ Okay but if I already tell that the state shared is say$ |\beta_{00}\rangle_{12}|\beta_{00}\rangle_{34}$ and give $A$ (1,4) partcilces and $B$ (2,3) particles. Then without B's knowledge he can construct the whole state $ |\beta_{00}\rangle_{12}|\beta_{00}\rangle_{34}$ $\endgroup$ – Upstart Nov 27 '19 at 11:22
  • $\begingroup$ @Upstart if you already tell them what the overall state is, they know what the overall state is even without doing any measurement.. what do you mean by "construct" here? $\endgroup$ – glS Nov 27 '19 at 11:25
  • $\begingroup$ do you mean to say that I should keep it to myself what I shared, say $ |\beta_{00}\rangle_{12}|\beta_{00}\rangle_{34}$ , but don't tell them and just give them the particles. $\endgroup$ – Upstart Nov 27 '19 at 11:28
  • $\begingroup$ @Upstart well I don't know, in order to do what? Saying the parties what the overall state is doesn't change anything in the measurement outcomes. $\endgroup$ – glS Nov 27 '19 at 11:33

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