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In this question, I assume we use matrix representation to calculate what each output could be? For example, for the middle qubit in state: $$|00\rangle$$ After the first Hadamard and CNOT gate (so the dashed red line), would this be carried out by: $$ |00\rangle((\mathbb{I} \otimes \hat H)(|0\rangle\langle0|\otimes \mathbb{I} + |1\rangle\langle1|\otimes \hat X))$$ of which there is a solution given at this link: https://www.quora.com/What-is-the-matrix-of-a-Hadamard-with-a-CNOT-gate Which shows that we are given a superposition after this operation. My question is what happens after the further CNOT12 and CNOT23 are applied to the state. I assume it is just a further application of the appropriate matrices? And I suppose with this we will be left with a superposition in the output? Is this correct? I am quite new to qubit circuits and I am actually unsure if it is correct that you may have a superposition at the end of a circuit. I thought that CNOT gates were essentially a measurement, and should collapse the superposition into a single state $$|0\rangle \space or \space |1\rangle$$

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    $\begingroup$ CNOTs are not a measurement and can very easily create superposition in the output (specifically: entanglement). I suggest, for now, keep going with trying the calculation yourself. Just remember that each time you move right along a quantum circuit, the gate operation that acts on the state acts from the left. Maybe then you can report how far you get, and we can give you some more directed pointers towards possible improvements (if any). $\endgroup$
    – DaftWullie
    Commented Dec 10, 2021 at 16:36
  • $\begingroup$ Still finding it quite difficult! $\endgroup$
    – Dwye
    Commented Dec 10, 2021 at 20:57

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