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

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?

• You will probably be able to answer this in under a minute @norbert-schuch Feb 16 at 16:48
• I can answer the question "How can I extract the Kraus representation from the Choi state" quickly. On the other hand, I have difficulties with the notation/terminology in your post, so I don't think I would supply in answer in that language (which projector?). Let me know if you are interested in the type of answer I mention. (If you favor I can make a separate post for that.) -- P.S.: Tagging doesn't work like that (maybe it wasn't meant as a tag either.) Feb 17 at 10:06
• @NorbertSchuch I would be very grateful for such an answer. If I'm still uncertain I can always add a comment :) Indeed perhaps my understanding of what this P_i object is - is just unclear and your answer could help with that Feb 17 at 10:08
• Feb 18 at 9:31

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.)

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.