# Discrimination of Bell states on unkown qubits

Consider the scenario where Alice is given three qubits, and promised two of them are maximally entangled Bell state $\in_R \{\frac{|00+11>}{\sqrt{2}}, \frac{|00-11>}{\sqrt{2}}\}$ and one of them is BB84 state $\in_R \{ |0\rangle, |1\rangle, \frac{|0+1\rangle}{\sqrt{2}}, \frac{|0-1\rangle}{\sqrt{2}} \}$, but she does not which qubit is the BB84 state.

My questions are the follwing

1) Is there a (probabilistic) strategy for Alice to determine what bell state was given to her?

2) Suppose the set of Bell states to choose from was $\{\frac{|00+11>}{\sqrt{2}}, \frac{|00-11>}{\sqrt{2}}, \frac{|01+10>}{\sqrt{2}}, \frac{|01-10>}{\sqrt{2}}\}$. Would that make things different?

Alice only has one copy and has access to all the qubits (I am not talking about LOCC). We allow von Neumann measurements and don't care if the state discrimination is (non)destructive.

• I suppose someone who downvotes is intelligent enough (at-least than the asker) to also give an answer. Any help appreciated. Sep 15, 2015 at 6:10
• People are giving you a hard time (and down votes) because you could find a huge amount of literature about this. That said, I think this is still a reasonable question. Can you make it more clear by explaining what you mean about not destroying the qubits? Sep 15, 2015 at 7:18
• Thanks @DanielSank and Norbert Schuch. I have revised the question, perhaps this makes more sense now? Sep 15, 2015 at 7:45
• Definitely makes more sense now, but what is the BB84 state? Also, why are you first talking about 3 qubits and then of two -- could you clarify this a bit more? Are you given 3 qubits and you want to know which two are in a Bell state? Sep 15, 2015 at 7:48
• @NorbertSchuch well, it could be any state, i just restricted to BB84 for convenience, let's keep it BB84 state for now. The problem I described now is the full blown problem. Earlier, I had restricted myself. And Yes, Exactly, in a nutshell, given 3 qubits, I want to know which two are the Bell state and also what Bell state it is. Sep 15, 2015 at 8:21

Let $\vert\psi_i\rangle$ be the 4 Bell states and $\vert\phi_k\rangle$ the 4 "BB84" states. Then, define a POVM measurement with elements $$P_{ik\pi} = \tfrac{1}{12}\vert\psi_i\rangle\langle\psi_i\vert\otimes \vert\phi_k\rangle\otimes \langle\phi_k\vert\ ,$$ where $\pi$ is a permutation of the three qubits (i.e., the three possible choices for the unentangled qubit, since the Bell part is symmetric under permutation).
Then, it is straightforward to check that $\{P_{ik\pi}\}$ forms a complete POVM, and whenever you obtain outcome $i,k,\pi$, you can be sure to have the corresponding state (while any other state will not deterministically yield this outcome).