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We have 2-qubit state which we know is 1 of 4 Bell states. Can we determine, using unitary transformations and single-qubit measurements, which Bell state do we have, and if we can, how?

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  • $\begingroup$ I wonder who decided that the question is an exercise from a textbook and added corresponding tag. I am not against, but it is not; the question come to my mind than I tried to understand more clearly basic quantum protocols such as superdense coding and teleportation. $\endgroup$
    – kludg
    Commented Jul 7, 2022 at 11:54
  • $\begingroup$ Is this a single shot scenario? Otherwise just do tomography. And, are the unitary transformations single qubit also? $\endgroup$
    – Rammus
    Commented Jul 7, 2022 at 14:37
  • $\begingroup$ @Rammus: I don't really care single- or two-qubit unitary transformations are used. Consider superdense coding as a practical example. Bob encodes two classical bits into a Bell state and Alice decodes them by analyzing the Bell state. We don't care here that Alice and Bob are in different places, each has his/her own qubit and Bob sends his qubit to Alice. Alice recovers 2 bits of information by analyzing whole Bell state, and she knows that her 2-qubit state is one of four possible Bell states. I am surprised that the question appeared to be so hard. $\endgroup$
    – kludg
    Commented Jul 7, 2022 at 15:14

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If two-qubit unitary transformations are allowed then the problem is straightforward. Consider the unitary $$ U = (|0\rangle \langle 0 | \otimes I + |1\rangle \langle 1|\otimes I) (H \otimes I), $$ i.e., a Hadamard applied to qubit 1 and then a CNOT controlled on qubit 1. This unitary maps the two-qubit computational basis to the Bell basis, i.e.,
$$ U |xy\rangle = |\Phi_{xy}\rangle = (I \otimes Z^x X^y) \frac{|00\rangle + |11\rangle}{\sqrt{2}}. $$

Well then, if we're given a Bell-state $|\Phi_{xy}\rangle$ then we just apply the unitary $U^\dagger$ to our state to map each Bell-state to a unique state in the computational basis $|xy\rangle$. After which local measurements in the computational basis can clearly distinguish the 4 possible states.

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The four Bell states for a 2 qubit system are:

$$ B_1 = \frac{1}{\sqrt2} \left( |00\rangle + |11\rangle\right)$$ $$ B_2 = \frac{1}{\sqrt2} \left( |01\rangle + |10\rangle\right)$$ $$ B_3 = \frac{1}{\sqrt2} \left( |00\rangle - |11\rangle\right)$$ $$ B_4 = \frac{1}{\sqrt2} \left( |01\rangle - |10\rangle\right)$$

Applying CNOT to each of the four states:

$$ B_1 \rightarrow \frac{1}{\sqrt2} \left( |00\rangle + |10\rangle\right)$$ $$ B_2 \rightarrow \frac{1}{\sqrt2} \left( |01\rangle + |11\rangle\right)$$ $$ B_3 \rightarrow \frac{1}{\sqrt2} \left( |00\rangle - |10\rangle\right)$$ $$ B_4 \rightarrow \frac{1}{\sqrt2} \left( |01\rangle - |11\rangle\right)$$

Applying H onto the first qubit:

$$ B_1 \rightarrow \frac{1}{\sqrt2} \left( \frac{1}{\sqrt2}(|0\rangle + |1\rangle)|0\rangle + \frac{1}{\sqrt2}(|0\rangle - |1\rangle)|0\rangle \right)=|00\rangle$$

$$ B_2 \rightarrow \frac{1}{\sqrt2} \left( \frac{1}{\sqrt2}(|0\rangle + |1\rangle)|1\rangle + \frac{1}{\sqrt2}(|0\rangle - |1\rangle)|1\rangle\right) = |01\rangle$$

$$ B_3 \rightarrow \frac{1}{\sqrt2} \left( \frac{1}{\sqrt2}(|0\rangle + |1\rangle)|0\rangle - \frac{1}{\sqrt2}(|0\rangle - |1\rangle)|0\rangle\right)= |10\rangle$$

$$ B_3 \rightarrow \frac{1}{\sqrt2} \left( \frac{1}{\sqrt2}(|0\rangle + |1\rangle)|1\rangle - \frac{1}{\sqrt2}(|0\rangle - |1\rangle)|1\rangle\right) = |11\rangle$$

Thus, a final measurement of $|00\rangle$ implies for certain that the entanglement status of the 2 qubits was $\frac{1}{\sqrt2} \left( |00\rangle + |11\rangle\right)$.

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