It is known that qubit channels can be written in the form: $$ \begin{align} \Phi(\rho) = \frac{1}{2}\left(I+(T\vec{r}+\vec{t})\cdot\sigma\right)\ \end{align} $$ where $\vec{r}$ is the Bloch vector of the state, $\rho=\frac{1}{2}\left(I+\vec{r}\cdot\sigma\right)$. The map is completely positive and trace preserving (CPTP) if certain conditions are obeyed by the matrix $T$ and real vector $\vec{t}$.

Assume we are given such a CPTP map in the above representation, how can we express it in the Krauss representation, that is, in the form $$ \begin{align} \Phi(\rho)=\sum_k A_k^\dagger \rho A_k \end{align} $$ where the $A_k$ satisfy the condition $\sum_k A_k A_k^\dagger =I$? Obviously there might not be a unique way of doing that but I am looking for one such way.

  • 1
    $\begingroup$ There must be some more cleaver and efficient way to do this, but one immediate way from the top of my head is: since you have access to $T$ and $\vec{t}$, you can just calculate the process matrix $\chi$ (preferably in Pauli basis)and then from that, construct your Kraus operators as linear combinations of Pauli matrices depending on the weights. $\endgroup$
    – FDGod
    Commented Feb 11 at 7:59
  • 1
    $\begingroup$ the general way is to compute the Choi matrix, which directly gives you Kraus operators via its eigendecomposition. See eg quantumcomputing.stackexchange.com/q/21155/55 and links therein. Is this what you're asking? $\endgroup$
    – glS
    Commented Feb 11 at 11:40

1 Answer 1


TL;DR: The given form of the channel is essentially an obscured way of writing down the channel's Pauli transfer matrix. One set of Kraus operators coincides (up to vectorization) with the eigenvectors of the Choi matrix. Therefore, modulo rearrangements of matrix elements, the task amounts to the conversion from the Pauli transfer matrix to the Choi matrix followed by eigendecomposition.

Affine transformation of the Bloch sphere to Pauli transfer matrix

By considering the action of $\Phi(\rho) = \frac{1}{2}\left(I+(T\vec{r}+\vec{t})\cdot\vec{\sigma}\right)$ on each Pauli operator, we see that the channel's Pauli transfer matrix is $$ R(\Phi)=\begin{bmatrix}1 & 0\\ \vec{t}&T \end{bmatrix}.\tag1 $$

Pauli transfer matrix to Choi matrix

Let's denote the three columns of $T$ with $\vec{t}_X$, $\vec{t}_Y$, and $\vec{t}_Z$ and $\vec{t}$ with $\vec{t}_I$. We compute the channel's Choi matrix as $$ \begin{align} J(\Phi)&:=(\Phi\otimes\mathcal{I})(|\psi\rangle\langle\psi|)\tag2\\ &=(\Phi\otimes\mathcal{I})\left(\frac12\sum_{P\in\{I,X,Y,Z\}}P\otimes P^T\right)\tag3\\ &=\frac12\sum_{P\in\{I,X,Y,Z\}}\Phi(P)\otimes P^T\tag4\\ &=\frac{I\otimes I}{2}+\sum_{P\in\{I,X,Y,Z\}}\frac{\vec{t}_P\cdot\vec{\sigma}}{2}\otimes P^T\tag5 \end{align} $$ where $|\psi\rangle=|00\rangle+|11\rangle$ is an unnormalized maximally entangled state of two qubits.

Choi matrix to a Kraus representation

Next, we rewrite the Choi matrix as a sum of rank-one matrices $$ J(\Phi)=\sum_k|\psi_k\rangle\langle\psi_k|.\tag6 $$ This can be accomplished for example by computing the eigendecomposition of $J(\Phi)$. Finally, by rearranging the four components of $|\psi_k\rangle$ we construct the matrix $A_k$ such that $|\psi_k\rangle=(A_k\otimes I)|\psi\rangle$. But then $$ J(\Phi)=\sum_k(A_k\otimes I)|\psi\rangle\langle\psi|(A_k^\dagger\otimes I)\tag7. $$ Comparing to $(2)$ we conclude that $\Phi(\rho)=\sum_kA_k\rho A_k^\dagger$ is a Kraus representation of $\Phi$.


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.