Q:

Repetition code encoder circuit

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:

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 $$\vert \psi \rangle = \alpha \vert 010 \rangle + \beta \vert 110 \rangle$$

A:

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 $\alpha|000\rangle + \beta|111\rangle$.

