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)

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

  • $\begingroup$ what are $A_k$ here? I don't fully understand the question: you want to go from the Choi representation, to which representation? The formula you write seems to be the way you express the channel itself in terms of its Choi representation. Are you asking how to derive this expression? $\endgroup$
    – glS
    Commented Jun 27, 2023 at 12:29
  • $\begingroup$ Sorry, I'm quite new to it so I might be mixing up stuff. The second formula is for 'Choi-Jamiolkowski isomorphism' where $\mathcal{E}$ is the channel, $\Phi$ is the maximally entangled state between systems $R$ and $A$. The first formula $X_A = ...$ is the Choi matrix but in terms of sums of superoperators $|A_k\rangle\rangle\langle\langle A_k|$ making up the channel, I think. Is Choi matrix in the first formula, same as 'Choi-Jamiolkowski isomorphism' ? If so how to derive the first one from the second one? $\endgroup$ Commented Jun 27, 2023 at 14:53

1 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?.


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