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I wonder how to initialise a Bell-like state, in the circuit model, where instead of standard $|\Phi^{\texttt{+}}\rangle$, the entanglement is in the x-basis. Hence a state $\frac{1}{\sqrt{2}}(|\texttt{++}\rangle + |\texttt{--}\rangle$.

I thought I could just apply an $H^{\otimes 2}$ after the standard protocol: $CX(H\otimes I)$. But by applying a simple circuit equivalence, this turns out to create the standard $|\Phi^{\texttt{+}}\rangle$.

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Your state is $|\Phi^+\rangle$. One way to see this is by expanding each qubit into the computational basis first, then doing the FOIL, and grouping the bases after.

Indeed, to answer the specific question about preparing your state, there’s no need to perform the two Hadamard gates after the CNOT gate.

Remember, entanglement is basis-independent.

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