Consider an unrealistic 2-qubit plus 1 ancilla bit-flip error correction code (images generated by quirk), where I know by some means or other that an error may have happen on qubit0 (represented by the octagon)
First, qubit1 is entangled with q0, then after the potential error, calculate the parity on q2 (the ancilla), and then apply a CNOT to q0.
Now I know that my ancilla MUST be either |0> or |1> because of how it has been prepared, independent of the initial state of q0. Does that mean I can perform the parity measurement classically, and get the same result? i.e.:
Are these two circuits completely equivalent? Or is there some subtlety of collapsing the entangled state that I am missing? Does the initial state of q0 matter? What if q0 is already entangled with some other qubit (not shown)?