0
$\begingroup$

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.

The answers are given as:

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

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.

$\endgroup$

2 Answers 2

0
$\begingroup$

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 $K_\downarrow$.

$\endgroup$
1
  • $\begingroup$ In the second row if you multiply 0.(1 0) then you get (0 0), not (0 1) right $\endgroup$
    – Lelouch
    Aug 31, 2023 at 18:11
0
$\begingroup$

It's just a notation problem.They are writing

$K_\uparrow=\mathbb{I_A}\otimes<\uparrow|_B $

but doing:

$K_\uparrow=<\uparrow|_B\otimes\mathbb{I_A} $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.