# Are Bell states distinguishable through LOCC?

Define $$|\psi^{00}\rangle = \frac{1}{\sqrt2}(|00\rangle + |11\rangle)$$ and $$|\psi^{01}\rangle = \frac{1}{\sqrt2}(|00\rangle - |11\rangle)$$, and consider the state $$|0\rangle\langle 0|^C\otimes |\psi^{00}\rangle\langle\psi^{00}|^{AB} + |1\rangle\langle 1|^C\otimes |\psi^{01}\rangle\langle\psi^{01}|^{AB},$$ where the subsystems are distributed among three parties Alice, Bob and Charlie. Is it possible for Alice and Bob to extract Charlies bit through LOCC?

Essentially, I am asking if Bell states are locally distinguishable. If we consider $$|\psi^{00}\rangle$$ and $$|\psi^{10}\rangle = \frac{1}{\sqrt2}(|01\rangle + |10\rangle)$$, it is easy to see they are indeed locally distinguishable, but I have not been able to come up with a measurement that reveals the phase.

Any help is appreciated.

• Can you clarify the operations that we can perform to determine the phase. What kind of communication is allowed between Alice and Bob? Can each of them apply only single qubit Unitaries or can we apply MultiQubit Unitaries together. – vasjain Jun 19 '20 at 14:24

If you look at these states in the $$X$$ basis, they are $$|++\rangle+|--\rangle,\qquad |+-\rangle+|-+\rangle.$$ Thus, by both measuring in the $$X$$ basis and computing the parity of the answers, you can tell $$C$$'s bit value.

We can distinguish these states using SingleQubit Unitaries and Measurements.

Lets the 2 states be $$|\psi^{00}\rangle = \frac{1}{\sqrt2}(|00\rangle + |11\rangle)$$ and $$|\psi^{01}\rangle = \frac{1}{\sqrt2}(|00\rangle - |11\rangle)$$. Let Alice be in possession of the 1st qubit and Bob be in possession of the 2nd qubit.

Bob applies a $$H$$ gate on his qubit. This would result in the following states $$I \otimes H|\psi^{00}\rangle = \frac{1}{\sqrt2}(|0\rangle \otimes |+\rangle + |1\rangle \otimes |-\rangle) = \frac{1}{\sqrt2}(|0\rangle|+\rangle + |1\rangle|-\rangle) = \frac{1}{\sqrt2}(|+\rangle|0\rangle + |-\rangle|1\rangle)$$

$$I \otimes H|\psi^{01}\rangle = \frac{1}{\sqrt2}(|0\rangle \otimes |+\rangle - |1\rangle \otimes |-\rangle) = \frac{1}{\sqrt2}(|0\rangle|+\rangle - |1\rangle|-\rangle) = \frac{1}{\sqrt2}(|-\rangle|0\rangle + |+\rangle|1\rangle)$$

Now Bob performs a measurement on his qubit. If the initial state was $$|\psi^{00}\rangle$$ then the result of the measurement could be $$|0\rangle$$ or $$|1\rangle$$ and Alice's qubit would be $$|+\rangle$$ or $$|-\rangle$$ respectively.

Similarly, if the initial state was $$|\psi^{01}\rangle$$ then the result of the measurement could be $$|0\rangle$$ or $$|1\rangle$$ and Alice's qubit would be $$|-\rangle$$ or $$|+\rangle$$ state respectively.

Now Alice can apply the $$H$$ gate on her qubit and then perform a Measurement on it. If Bob's and Alice's results were $$(0,0)$$ or $$(1,1)$$ then the initial state was $$|\psi^{00}\rangle$$. On the other hand if Bob's and Alice's results were $$(0,1)$$ or $$(1,0)$$ then the initial state was $$|\psi^{01}\rangle$$.

Hence we can now formulate an Algorithm for distinguishing the phase.

1. Bob applies $$H$$ gate on his qubit.
2. Bob performs a measurement and sends the result of the measurement to Alice.
3. Alice applies $$H$$ gate on her qubit.
4. Alice performs a measurement and sends the result of the measurement to Bob.
5. Now if their results were the same then the state was $$|\psi^{00}\rangle$$ else it was $$|\psi^{01}\rangle$$