Finding the Kraus Operators of a Quantum Channel from its Choi Matrix

Question

TLDR : What is the form of the projector I need to use to attain a 2X1 vector from $P_i v_k$ with which I can build my Kraus Operator?

I am calculating the Kraus Operators for a Quantum Channel associated to a protocol where the unitary carried out can change in a way which is not parameterised but where I know the final state at the end of this process. As such, I am using the Choi Matrix associated to this channel to find the Kraus operators related to it.

To be more precise, I have an initial state $$ \rho = \rho_S \otimes \rho_B $$ and I know that at the end $$\rho'_S = \mathcal{E}(\rho) = \text{tr}_B\{U\rho U^\dagger\} = \sigma$$ where $\sigma$ is the state I know the form of at the end of the process and $\mathcal{E}$ is the associated map. Since this is a CPTP map it may be expressed as some Kraus representation $$\mathcal{E}(\rho) = \sum_i \lambda_i \Lambda_i \rho_S \Lambda^\dagger_i$$ where $\lambda_i$ are the eigenvalues of the channel and $\Lambda_i$ are the corresponding Kraus operators. By the Choi-Jamiolkowski isomorphism this map may be related to a Choi Matrix defined as The Choi matrix is defined \begin{gather} \Upsilon_\mathcal{E} = \left(\mathcal{E}(\rho) \otimes \mathcal{I}_n\right)|\varphi\rangle\langle\varphi| \end{gather} such that \begin{gather} \mathcal{E}(\rho) = \text{tr}_n\left\{\left(\mathcal{I}_n\otimes \rho_S^T\right)\Upsilon_\mathcal{E}\right\} \end{gather} where $|\varphi\rangle = \sum^{n+1}_{i = 1} |ii\rangle$ i.e. the unnormalised bell state on $n+1$ dimensions.

The product of the eigenvalues $c_k$ and the eigenvectors $|c_k\rangle$ of the Choi matrix $v_k = c_k |c_k\rangle$ relate to the Kraus operators in the following way by Choi's Theorem $$\Lambda_k e_i = P_i v_k$$ where $e_i$ is a basis eigenket corresponding to $\rho_S$ and $P_i$ is a projection onto the subspace of the system.

My issue is here. Making this concrete, let our system be a single qubit and the environment we are tracing over be $n$ qubits. This would mean that our Kraus operators are 2$\times$2 matrices, our basis eigenkets are $|0\rangle$ and $|1\rangle$ and $v_k$ is an $n+1$ row eigenvector.

What is the form of the projector I need to use to attain a $2\times1$ vector from $P_i v_k$ with which I can build my Kraus Operator?

I imagine I will have a situation $$\Lambda_k |0\rangle = \left(|0\rangle\langle0|\mathcal{R}\right)v_k$$ where $\mathcal{R}$ is some $2\times n+1$ matrix which can reshape my projector as desired. But is this correct?

Answer 1

This answer explains how to extract Kraus operators from the Choi state.

Given a CP map $\mathcal E$, the Choi state is $$ \sigma = (\mathcal E\otimes I)(\lvert\Omega\rangle\langle\Omega\rvert)\ , $$ where $\Omega = \sum \lvert i,i\rangle$.

Now consider any ensemble decomposition $\sigma = \sum \lvert\psi_i\rangle\langle \psi_i\rvert$ (e.g. the eigenvalue decomposition), and express $$ \lvert\psi_i\rangle = (K_i\otimes I)\lvert\Omega\rangle\ . $$ (Such a $K_i$ always exist, it is unique, and its entries are basically the expansion coefficients of $\lvert\psi_i\rangle$ in the computational basis.)

Then, $$\mathcal E(\rho) = \sum K_i\rho K_i^\dagger$$ -- this can be seen e.g. by using that the map from $\mathcal E$ to the Choi state $\sigma$ is injective.

(Note that the ambiguity of the Kraus representation is precisely the same as the ambiguity of ensemble decompositions, just as it must be.)

Answer 2

This is just a slightly different wording of what was already said in this other answer.

Let $\Phi:\mathrm{Lin}(\mathcal{H}_A)\to\mathrm{Lin}(\mathcal{H}_B)$ be a completely positive map, and suppose its Kraus decomposition reads $\Phi(X)=\sum_a A_a X A_a^\dagger$ for some collection of linear operators $A_a:\mathcal H_A\to\mathcal H_B$.

Write the Choi representation of $\Phi$ as the positive-semidefinite operator $J(\Phi)\equiv (\Phi\otimes I)(|\varphi\rangle\!\langle\varphi|)$, where $|\varphi\rangle\equiv\sum_i |ii\rangle$. There is a linear isomorphism between $\Phi$ and $J(\Phi)$. You can then verify that $J(\Phi)$ decomposes as $$J(\Phi) = \sum_a \mathrm{vec}(A_a)\mathrm{vec}(A_a)^\dagger,$$ where $\operatorname{vec}(A)$ is the "vectorisation" of the operator $A$, or explicitly, $$A=\sum_{ij} A_{ij} |i\rangle\!\langle j|\iff \mathrm{vec}(A)= \sum_{ij} A_{ij} |i,j\rangle.$$ Vice versa, if a Choi $J(\Phi)$ can be written as sum of unit-rank positive semidefinite operators, $J(\Phi)=\sum_a v_a v_a^\dagger$, then "unvectorising" the vectors $v_a$ you get Kraus operators for the associated map. See Watrous' book for a more thorough discussion on this topic.