Let $\Phi\in\mathrm T(\mathcal X,\mathcal Y)$ be a quantum channel, $\Phi:\mathrm{Lin}(\mathcal X)\to\operatorname{Lin}(\mathcal Y)$. We define its Choi representation as the operator $J(\Phi)\in \mathrm{Lin}(\mathcal{Y})\otimes \mathrm{Lin}(\mathcal{X})$ defined by $$J(\Phi) = (\Phi\otimes I) \,d\,\mathbb P_+= \sum_{a,b}\Phi(E_{a,b})\otimes E_{a,b},\tag1$$ where $\mathbb P_+\equiv \lvert +\rangle\!\langle +|$ with $\sqrt d|+\rangle=\sum_i |i,i\rangle$, and $E_{a,b}\equiv |a\rangle\!\langle b|$.

One way to retrieve the map from the Choi, used for example in Watrous, Eq. (2.66), is $$\Phi(X) = \operatorname{Tr}_{\mathcal X}[J(\Phi)(I_{\mathcal Y}\otimes X^T)].\tag2$$ Verifying the equivalence between these two isn't too hard: $$ \operatorname{Tr}_{\mathcal X}[J(\Phi)(I_{\mathcal Y}\otimes X^T)] = \sum_{a,b} \operatorname{Tr}_{\mathcal X} [ \Phi(E_{a,b})\otimes E_{a,b} X^T ] = \Phi(X). $$

More generally, this gives us a way to associate to each bipartite state $\rho$ a map $\Phi_\rho$ such that $J(\Phi_\rho)=d\,\rho$, and if $\operatorname{Tr}_{\mathcal Y}\rho=I/d$, then $\Phi_\rho$ is trace-preserving (and thus CPTP).

In (Horodecki, Horodecki, Horodecki 1998) the authors mention another way to associate a map to a state $\rho$. Writing its eigendecomposition as $\rho=\sum_i p_k \mathbb P_{\psi_k}$, and writing with $\psi$ the operator whose vectorisation is $|\psi\rangle$, i.e. $\operatorname{vec}(\psi)\equiv |\psi\rangle$, we have $|\psi_k\rangle=(\psi_k\otimes I)\,\sqrt d |+\rangle$, and thus $$\rho = \sum_k p_k (\psi_k \otimes I) \,d\,\mathbb P_+(\psi_k^\dagger\otimes I) = (\Phi_\rho\otimes I) \mathbb P_+,\tag3$$ where $\Phi_\rho(X) = d\sum_k p_k \psi_k X \psi_k^\dagger.$

I presume (2) and (3) should be equivalent, provided $d \,\rho=J(\Phi)$. What's a good way to show this equivalence?

  • $\begingroup$ This follows from linearity of the isomorphism. $\endgroup$ Sep 7, 2020 at 8:06
  • $\begingroup$ @MarkusHeinrich can you expand as to which part exactly follows from linearity? $\endgroup$
    – glS
    Sep 7, 2020 at 8:42
  • 1
    $\begingroup$ It is clear how the inverse should look like on rank-one matrices $|u\rangle\langle u|$, or even more generally $|u\rangle\langle v|$. Thus, one only needs to check that Eq. (3) gives the right result for those which is straightforward. The general result follows via a decomposition into those (e.g. SVD for general matrices). But that's basically what you did. $\endgroup$ Sep 7, 2020 at 9:46
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    $\begingroup$ What I actually meant was that inverting is trivial for $|u\rangle\langle v|$, thus for any matrix $M=\sum_k x_k |u_k\rangle\langle v_k|$ the associated superoperator has to be $\Phi(X) = \sum_k x_k U_k X V_k^\dagger$ (operators not necessarily unitary). The only thing to justify is the inversion formula Eq. (2), which you already did in you first post. $\endgroup$ Sep 7, 2020 at 10:12

1 Answer 1


The goal is to start from a bipartite state $\rho$, and find the channel $\Phi_\rho$ such that $J(\Phi_\rho)=d\, \rho$.

The goal is, given $J(\Phi)$, to show that we can retrieve $\Phi$ via (3) rather than (2). In other words, that (2) is equivalent to $$\Phi(X) = d\sum_k p_k \psi_k X \psi_k^\dagger,$$ where $p_k$ are the eigenvalues of $J(\Phi)/d$ and $\psi_k$ the (unvectorised operators corresponding to the) eigenvectors of $J(\Phi)/d$. We, therefore, wish to show that $$d\sum_k p_k \psi_k X \psi_k^\dagger = \operatorname{Tr}_{\mathcal X}[J(\Phi)(I_{\mathcal Y}\otimes X^T)].$$ Writing $J(\Phi)=d \sum_k p_k \mathbb P_{\psi_k}$, we get from the RHS: $d\sum_k p_k \operatorname{Tr}_{\mathcal X}[\mathbb P_{\psi_k}(I_{\mathcal Y}\otimes X^T)]$. We therefore only need to show that, for all $k$, $$\psi_k X \psi_k^\dagger = \operatorname{Tr}_{\mathcal X}[\mathbb P_{\psi_k}(I_{\mathcal Y}\otimes X^T)].$$ This can be proven directly analysing the matrix components, or again applying the previous trick to write $\mathbb P_{\psi_k}=(\psi_k\otimes I)\mathbb P_+(\psi_k^\dagger\otimes I)$.


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