# Does partial tracing a system with three shared Bell states give the identity?

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?

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 $$\rho_D = \rho_1 \otimes \rho_4$$.

• 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? Nov 26 '19 at 16:16

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.

• 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 $|\beta_{00}\rangle_{12}|\beta_{00}\rangle_{34}$ Nov 27 '19 at 11:22
• @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?
– glS
Nov 27 '19 at 11:25
• 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. Nov 27 '19 at 11:28
• @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.
– glS
Nov 27 '19 at 11:33