What is the relation between the Choi matrix and the Liouville space (superoperator) representations of a channel?

Question

A.S. Fletcher, P. W. Shor, and M. Z. Win Phys. Rev. A 75, 012338 (2007) says

the Choi matrix for the operation $\mathcal{A}$ is given by $X_A \equiv \sum_k |A_k\rangle\!\rangle\langle\!\langle A_k|$, and the channel mapping $\mathcal{A}:\mathcal{L}(\mathcal{H})\mapsto \mathcal{L}(\mathcal{K})$ is defined by \begin{equation} \mathcal{A}(\rho) = {\rm{tr}}_{\mathcal{H}}[(\rho^{\rm{T}}\otimes I)X_A]. \tag{11} \end{equation}

Here they used the Liouville space representation with $|\rangle\!\rangle \langle\!\langle|$. How do we get to this representation starting from the usual definition of Choi matrix representation of a channel we know from Preskill's notes Eq.(3.71)

$(I\otimes \mathcal{E})\left((|\tilde\Phi\rangle\langle|\tilde\Phi|)_{RA}\right)$ ?

Answer 1

I think there's some confusion here, so let me try to clarify some basic things:

1. Given any quantum channel $\Phi$, you define its Choi representation as the operator $J(\Phi)=(\Phi\otimes \operatorname{Id})\mathbb{P}_m$, where $\mathbb{P}_m\equiv |m\rangle\!\langle m|$ and $|m\rangle\equiv\sum_i |i,i\rangle$ is the (unnormalised) maximally entangled state. It is also common to instead talk about the Choi state, which is the same thing, except you define $|m\rangle$ as the actual (normalised) maximally entangled state. The two definitions only differ by a multiplicative factor, so it doesn't matter which one you use (as long as you're consistent with your notation of course).

2. Given a quantum channel with Kraus representation $\Phi(\rho)=\sum_k A_k \rho A_k^\dagger$, where $A_k$ are the Kraus operators, its Choi (following the definition above) can be written as $$J(\Phi)=\sum_k \operatorname{vec}(A_k)\operatorname{vec}(A_k)^\dagger = \sum_k |A_k\rangle\!\rangle\langle\!\langle A_k|.$$ Note that $|A_k\rangle\!\rangle$ refers to the vector obtained vectorising the operator $A_k$, and $\operatorname{vec}(A_k)$ is an equivalent notation for the same thing. You can directly verify these formulas by applying the general definition of Choi representation to a channel having $A_k$ as Kraus operators.

   It goes without saying, but there is a bijective relation between channels and their Chois, which also means that a channel $\Phi$ has Kraus operators $\{A_k\}$ if and only if its Choi can be decomposed as above in terms of the vectors $\{\operatorname{vec}(A_k)\}$.

3. The channel $\Phi$ corresponding to the Choi $J(\Phi)$ is $$\Phi(\rho) = \operatorname{tr}_2[(I\otimes\rho^T)J(\Phi)]. $$ To see this explicitly, consider the following: $$\operatorname{tr}_2[(I\otimes \rho^T)J(\Phi)] = \sum_{ij} \operatorname{tr}_2[(I\otimes \rho^T)(\Phi(E_{ij})\otimes E_{ij})] = \sum_{ij} \Phi(E_{ij}) \underbrace{\operatorname{tr}[\rho^T E_{ij}]}_{=\rho_{ij}} = \Phi(\rho).$$

You can find some related discussions in How does the spectral decomposition of the Choi operator relate to Kraus operators?.