# Kraus Operator for two-qubit basis

Let A and B each be a single qubit so that $$\mathbb{H_{AB}}$$ is a two-qubit system. In the basis {$$|\uparrow\uparrow>,|\uparrow\downarrow>,|\downarrow\uparrow>,|\downarrow\downarrow>$$, the Kraus operators are defined as $$K_\uparrow=\mathbb{I_A}\otimes<\uparrow|_B$$ and $$K_\downarrow=\mathbb{I_A}\otimes<\downarrow|_B$$. What is $$\mathbb{I_A}$$ here in the form of a matrix. I am confused because when I use $$A=\begin{pmatrix}{ 1 \\ 0}\end{pmatrix}$$, $$B=\begin{pmatrix}{ 0 \\ 1}\end{pmatrix}$$ and $$\mathbb{I_A} = \begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix}$$. I don't get the right answers.

$$K_\uparrow=\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\end{pmatrix}$$

and

$$K_\downarrow=\begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}$$

This calculation that I am trying to do is from the page 14 of the paper: 'Entanglement entropy and non-local duality: Quantum channels and quantum algebras ', arXiv:2207.12436.

Note that $$\langle\uparrow|_B = \begin{pmatrix} 1 & 0 \end{pmatrix}_B$$ is a row vector, so that the tensorial product inside $$K_\uparrow$$ is calculated as a Kronecker product, i.e. $$K_\uparrow = 1_A \otimes \langle\uparrow|_B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 \cdot \begin{pmatrix}1&0\\0&1\end{pmatrix} & 0 \cdot \begin{pmatrix}1&0\\0&1\end{pmatrix} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}$$ Of course, the same procedure applies to $$K_\downarrow$$.
$$K_\uparrow=\mathbb{I_A}\otimes<\uparrow|_B$$
$$K_\uparrow=<\uparrow|_B\otimes\mathbb{I_A}$$