# Equivalence of two ways to recover a map from its Choi state

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?

• This follows from linearity of the isomorphism. Sep 7 '20 at 8:06
• @MarkusHeinrich can you expand as to which part exactly follows from linearity?
– glS
Sep 7 '20 at 8:42
• 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. Sep 7 '20 at 9:46
• 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. Sep 7 '20 at 10:12

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 $$\mathbb P_{\psi_k}=(\psi_k\otimes I)\mathbb P_+(\psi_k^\dagger\otimes I)$$.