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The repetition code encodes $\vert \psi \rangle = \alpha \vert 0 \rangle + \beta \vert 1 \rangle \rightarrow \vert \psi \rangle = \alpha \vert 000 \rangle + \beta \vert 111 \rangle$ using the following circuit:

Circuit

where $$t_0: \vert \psi \rangle = (\alpha \vert 0 \rangle + \beta \vert 1 \rangle) \otimes \vert 0 \rangle \otimes \vert 0 \rangle = \alpha \vert 000 \rangle + \beta \vert 100 \rangle$$ $$t_1: \vert \psi \rangle = \alpha \vert 000 \rangle + \beta \vert 110 \rangle$$ $$t_2: \vert \psi \rangle = \alpha \vert 000 \rangle + \beta \vert 111 \rangle$$.

I am wondering what is the operator used from $t_0$ to $t_1$? I understand that an XOR was applied there to get the desired output but what is the explicit operator (gate-wise or matrix-wise) used?

Furthermore, I tried $U = \mathbb{I} \otimes \mathbb{X} \otimes \mathbb{I}$ and applied it to $\vert \psi \rangle$ which resulted in $$\vert \psi \rangle = \alpha \vert 010 \rangle + \beta \vert 110 \rangle$$

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Operator applied from $t_{0}$ to $t_{1}$ is $CNOT \otimes I$, i.e. \begin{equation} U_1= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ \end{pmatrix} \end{equation} As a CNOT gate produces a entangled quantum state, it is not possible to express it as Kronecker product $I \otimes X$.

Your operator $I \otimes X \otimes I$ means that negation $X$ is applied on $|q_1\rangle$ and identical operator $I$ is applied on $|q_0\rangle$ and $|q_2\rangle$ always. There is no connection established between $|q_0\rangle$ and $|q_1\rangle$.

This operation changes your input qubits to $\alpha|000\rangle + \beta|110\rangle$.

Operation from $t_1$ to $t_2$ is decribed by following matrix: \begin{equation} U_2= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ \end{pmatrix} \end{equation}

Application of $U_2U_{1}$ leads to state $\alpha|000\rangle + \beta|111\rangle$.

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    $\begingroup$ Thank you for the comment. Unfortunately, $(CNOT \otimes I) \vert \psi \rangle = \vert \psi \rangle$, it doesn't change the state to $\vert \psi \rangle = \alpha \vert 000 \rangle + \beta \vert 110 \rangle$ $\endgroup$ – M. Al Jumaily Dec 21 '19 at 1:54
  • $\begingroup$ symbolab.com/solver/step-by-step/… $\endgroup$ – M. Al Jumaily Dec 21 '19 at 1:56
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    $\begingroup$ The gate you’ve written out is $I\otimes$cnot not cnot$\otimes I$ $\endgroup$ – DaftWullie Dec 21 '19 at 6:13
  • $\begingroup$ @M.AlJumaily Al: Apologize for my mistake. Now, it should be all right, I checked my calculation in Octave. $\endgroup$ – Martin Vesely Dec 21 '19 at 8:03
  • $\begingroup$ @DaftWullie: Thanks for checking. I recalculated Kronecker product and expanded answer. Now, it should be all right. $\endgroup$ – Martin Vesely Dec 21 '19 at 8:03

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