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I'm not too great at dealing with superpositions and applying the CNOT gate when superpositions are involved. Can you go through it in detail each gate using math/matrices etc. It's based on the quantum error correction (using repetition codes) part of the Qiskit textbook.

Link to the specific page in the textbook: https://qiskit.org/textbook/ch-quantum-hardware/error-correction-repetition-code.html

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  • $\begingroup$ If you are talking about the 0 to the right of the downwards error of the measurement operator, that’s the index of the classical bit in which the measurement is stored, not the measurement result $\endgroup$
    – epelaaez
    Oct 1 at 5:32
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    $\begingroup$ @epelaaez, you are correct about that. However, it is also true that the measurement result for this circuit will indeed always be 0. $\endgroup$ Oct 1 at 5:39
  • $\begingroup$ Yes, it will be 0, can you explain how though using each gate step by step? $\endgroup$ Oct 1 at 7:29
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You have the initial state $|000\rangle$. The first Hadamard gate on qubit zero sends the register to $\frac{1}{\sqrt{2}}\left(|000\rangle + |100\rangle\right)$. Then, the $\text{CX}$ controlled on qubit zero and target on qubit one takes the state to $\frac{1}{\sqrt{2}}\left(|000\rangle + |110\rangle\right)$. And the next $\text{CX}$ between qubit zero and qubit two takes it to $\frac{1}{\sqrt{2}}\left(|000\rangle + |111\rangle\right)$. At this point we have a maximally entangled $\text{GHZ}$ state, but we are still missing a final $\text{CX}$ gate. This is between qubit one and two, with qubit one being the control. Thus, the final state is $\frac{1}{\sqrt{2}}\left(|000\rangle + |110\rangle\right)$. Therefore, as you can see, qubit three will always be projected into $|0\rangle$ when measured in the $Z$ basis.

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  • $\begingroup$ Thank you so much!!! $\endgroup$ Oct 3 at 1:26

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