I am trying to build a circuit that can detect which of the 4 entangled states i am in. The states are \begin{align} |0+\rangle &+ |1-\rangle\\ |0+\rangle &- |1-\rangle\\ |1+\rangle &+ |0-\rangle\\ |1+\rangle &- |0-\rangle \end{align}

What i am confused is both qubits are in different basis and how can i get information about the other qubit which is in a different basis? Won't there be loss of information if we tried using different basis? And also, is such a circuit possible to implement? How do i implement it?


1 Answer 1


First we apply a Hadamard on the second Qubit which takes $|+ \rangle \rightarrow |0 \rangle$ and $|- \rangle \rightarrow |1 \rangle$. Now the states have been transformed to:

$$\frac{1}{\sqrt2}(| 00 \rangle + |11 \rangle)$$ $$\frac{1}{\sqrt2}(| 00 \rangle - |11 \rangle)$$ $$\frac{1}{\sqrt2}(| 10 \rangle + |01 \rangle)$$ $$\frac{1}{\sqrt2}(| 10 \rangle - |01 \rangle)$$

Which are the Bell Basis Vectors. The Bell Basis vectors can each be mapped to a single eigenstate of the 2 Qubit System in the computational basis using an Inverse Bell Transform as follows:

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And so now we are done. Given that our input the the circuit above are Bell States(which we just ensured is the case) we will measure the proper and $x$ and $y$ every time. It should be noted that the order in which the states were provided in the question are not the conventional ordering for the Bell States and so the measurements of $x$ and $y$ would have to be mapped to whatever ordering we are interested in.

  • $\begingroup$ How would you differentiate the different phases when you perform the measurement, since the probabilities are the same? (for the case of 0+ plus 1- and 0+ minus 1- for example) $\endgroup$
    – Iberico
    Commented Sep 27, 2021 at 16:39
  • 1
    $\begingroup$ The probabilities are not the same. The phases are not global and are distinguished by the Inverse Bell Transform. Derive the matrix for the CNOT + Hadamard in the answer, and see their effect on the States after applying a Hadamard to the second bit (i.e. the states I listed in the question). You will very clearly see that the outputs are all different. $\endgroup$ Commented Sep 27, 2021 at 16:45

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