How does the spectral decomposition of the Choi operator relate to Kraus operators?

Question

In Nielsen and Chuang's QCQI, there is a proof states that

Theorem 8.1: The map $\mathcal{E}$ satisfies axioms A1, A2 and A3 if and only if $$ \mathcal{E}(\rho)=\sum_{i} E_{i} \rho E_{i}^{\dagger} $$ for some set of operators $\left\{E_{i}\right\}$ which map the input Hilbert space to the output Hilbert space, and $\sum_{i} E_{i}^{\dagger} E_{i} \leq I$

To figure out the proof of it, I find in this paper that what Nielsen wants to do is the so called Choi-Jamiołkowski isomorphism. But I just don't understand why Choi operator in the paper I mentioned (eq.8) can be written as $M=\sum_{j}|K_{j}\rangle\rangle\langle\langle K_{j}|.$, where $|K_j\rangle\rangle\equiv I\otimes K_j^T|\Omega\rangle$, and $|\Omega\rangle$ is the unnormalized maximally entangled states, i.e., $\sum_n|n\rangle|n\rangle$.

The aim of eq.(8) is that we don't know if $\varepsilon$ is CP, i.e., we only know the correspoinding choi operator is positive semidefinite, and we want to prove the channel $\varepsilon$ is CP. And there comes eq.(8) which states that any positive semidefinite choi operator can be written as the form $\sum_j|K_j\rangle\rangle\langle\langle K_j|$.

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Edit: Three axioms are(I originally thought they are not very related to the main question, so I didn't show it in the main post.):

A1: First, $\operatorname{tr}[\mathcal{E}(\rho)]$ is the probability that the process represented by $\mathcal{E}$ occurs, when $\rho$ is the initial state. Thus, $0 \leq \operatorname{tr}[\mathcal{E}(\rho)\rfloor \leq 1$ for any state $\rho$

A2: Second, $\mathcal{E}$ is a convex-linear $m a p$ on the set of density matrices, that is, for probabilities $\left\{p_{i}\right\}$, $$ \mathcal{E}\left(\sum_{i} p_{i} \rho_{i}\right)=\sum_{i} p_{i} \mathcal{E}\left(\rho_{i}\right) $$ A3: Third, $\mathcal{E}$ is a completely positive map. That is, if $\mathcal{E}$ maps density operators of system $Q_{1}$ to density operators of system $Q_{2}$, then $\mathcal{E}(A)$ must be positive for any positive operator $A$. Furthermore, if we introduce an extra system $R$ of arbitrary dimensionality, it must be true that $(\mathcal{I} \otimes \mathcal{E})(A)$ is positive for any positive operator $A$ on the combined system $R Q_{1}$, where $\mathcal{I}$ denotes the identity map on system $R$.

And the $M$ in the main post is the positive semidefinite choi operator.

Answer 1

Choi operator of a linear map $\mathcal{E}$ is defined as

$$ J(\mathcal{E}) = \sum_{ij} \mathcal{E}(|i\rangle\langle j|)\otimes |i\rangle\langle j|.\tag1 $$

Substituting $\mathcal{E}(\rho)=\sum_k E_k\rho E_k^\dagger$ into $(1)$, we have

$$ \begin{align} J(\mathcal{E}) &= \sum_{ijk} \left(E_k|i\rangle\langle j| E_k^\dagger\right)\otimes |i\rangle\langle j| \\ &= \sum_k (E_k\otimes I)\left(\sum_{ij}|i\rangle\langle j|\otimes|i\rangle\langle j|\right) (E_k^\dagger\otimes I) \\ &= \sum_k (E_k\otimes I)|\Omega\rangle\langle\Omega| (E_k^\dagger\otimes I) \\ &= \sum_k |E_k\rangle\rangle\langle\langle E_k| \end{align} $$

where for an operator $A$ we define $|A\rangle\rangle=(A\otimes I)|\Omega\rangle$ which is easily checked to be the same state as $(I\otimes A^T)|\Omega\rangle$.

Answer 2

1. Let $M\in\mathrm{Lin}(\mathcal Y\otimes\mathcal X)$ be some linear operator whose input and output spaces are both $\mathcal Y\otimes\mathcal X$, for some pair of finite-dimensional Hilbert spaces $\mathcal X,\mathcal Y$. Moreover, suppose $M$ is positive semidefinite: $M\ge0$.

   It being positive semidefinite implies it admits a decomposition of the form $M=\sum_j v_j v_j^\dagger$, with $\{v_j\}_j\subset\mathcal Y\otimes\mathcal X$ a set of orthogonal vectors. In bra-ket notation, this is what you wrote as $M=\sum_i |K_j\rangle\!\rangle\!\langle\!\langle K_j|$. I'll stick with the standard linear algebraic notation here though, because I think it makes the underlying math more transparent.

2. There are several ways to switch between Choi and Kraus representations. One I personally find particularly straightforward is looking at how the components of these objects are related (with respect to some fixed choices of bases for the underlying spaces). If $J(\Phi)\in\mathrm{Lin}(\mathcal Y\otimes\mathcal X)$ is the Choi representation of a map $\Phi:\mathrm{Lin}(\mathcal X)\to \mathrm{Lin}(\mathcal Y)$, we have $$\langle i|\Phi(E_{jk})|\ell\rangle = J(\Phi)_{ij,\ell k} \equiv \langle i,j|J(\Phi)|\ell,k\rangle, \qquad E_{ij}\equiv |i\rangle\!\langle j|.$$ This is equivalent to the relations $$J(\Phi)=(\Phi\otimes I_{\mathrm{Lin}(\mathcal X)})\mathbb P_\Omega, \qquad \mathbb P_\Omega\equiv|\Omega\rangle\!\langle\Omega|, \quad |\Omega\rangle\equiv \sum_i |ii\rangle\in\mathcal X\otimes\mathcal X,$$ and its inverse $$\Phi(X) = \operatorname{Tr}_{\cal X}[J(\Phi)(I\otimes X^T)].$$

3. Now, if $J(\Phi)$ is positive semidefinite, and thus admits a decomposition $J(\Phi)=\sum_a v_a v_a^\dagger$, the above relations immediately imply that $\Phi$ itself must have a specific structure: $$\Phi(X) = \operatorname{Tr}_{\cal X}[J(\Phi)(I\otimes X^T)] = \sum_{a\alpha\beta\gamma\delta\eta} (v_a)_{\alpha\beta} (\bar v_a)_{\gamma\delta} (X^T)_{\delta\eta}\delta_{\eta\beta} E_{\alpha\gamma} \\ = \sum_{a,\alpha\beta\gamma\delta} (v_a)_{\alpha\beta} (\bar v_a)_{\gamma\delta} X_{\beta\delta} E_{\alpha\gamma} \equiv \sum_a A_a X A_a^\dagger,$$ where in the last step we defined the operators $A_a\in\mathrm{Lin}(\mathcal X,\mathcal Y)$ as $A_a\equiv \sum_{\alpha\beta} (v_a)_{\alpha\beta}E_{\alpha\beta}$. This is sometimes stated saying that $v_a$ is the vectorisation of $A_a$: $v_a=\operatorname{vec}(A_a)$. For more details on how vectorisation works, and how it relates to Choi and Kraus, see How does the vectorization map relate to the Choi and Kraus representations of a channel?.

   To connect with the relation you mentioned between eigenvectors of Choi and Kraus operators, you can observe that writing $v=\operatorname{vec}(V)$, for some $v\in\mathcal Y\otimes\mathcal X$ and $V\in\mathrm{Lin}(\mathcal X,\mathcal Y)$, is equivalent to $$v = (I \otimes V^T) |\Omega\rangle,$$ as you can verify pretty straightforwardly by expanding the expressions.

4. Another way to understand the relations between Choi and Kraus representations, without getting bogged down with the indices and seemingly arbitrary relations, is using diagrammatic notation. You can simply observe the following: where the leftmost image is a schematic representation of the channel, which is an object taking two inputs, one living in $\mathcal X$ and the other living in its dual $\mathcal X^*$, and corresponding outputs. Note that this notation is exploiting the isomorphism $\mathrm{Lin}(\mathcal X)\simeq\mathcal X\otimes\mathcal X^*$.

   The first identity is the Kraus decomposition: the blue line connecting the $A$ tensors corresponds to the $a$ index in the equations above. The rightmost picture corresponds to the Choi representation of the channel. Notice how the only thing we are doing is changing which wires we interpret as "inputs" and which ones we interpret as "outputs". The overall object never changes, we are just looking at it from different perspectives.

Another related post that might be of interest is Deduce the Kraus operators of the dephasing channel using the Choi.

Answer 3

This paper is a nice reference on this topic, and presents the results algebraically and graphically