How to switch between the definitions of Choi as $(\mathcal{N}_A \otimes id_B)(\phi^+_{AB})$ and $(id_A \otimes \mathcal{N}_B)(\phi^+_{AB})$?

Question

For the channel $\mathcal{N}$ the Choi state can be either defined as $(\mathcal{N}_A \otimes id_B)(\phi^+_{AB})$ or $(id_A \otimes \mathcal{N}_B)(\phi^+_{AB}) $, where $\phi^+_{AB}$ is the maximally entangled state. Given the Choi state with the first convention, how do we get the Choi state with the other convention?

Answer 1

Yes, you get different Choi states this way, but the important thing is that they both encode complete positivity, i.e., $(\mathcal{N} \otimes {\rm id})(|\phi^+\rangle\langle\phi^+|) \geq 0$ if and only if $({\rm id} \otimes \mathcal{N})(|\phi^+\rangle\langle\phi^+|) \geq 0$. The reason for this is that these operators are unitarily equivalent, meaning they have the same eigenvalues: if $\mathbb F(x\otimes y):=y\otimes x$ is the usual flip operator on vectors $x,y$ (i.e., $\mathbb F$ is the Choi matrix of the transposition map), then $\mathbb F(A\otimes B)\mathbb F=B\otimes A$ for all operators $A,B$ and thus \begin{align*} \mathbb F((\mathcal{N} \otimes {\rm id})(|\phi^+\rangle\langle\phi^+|))\mathbb F&=\frac1N\sum_{i,j}\mathbb F(\mathcal N(|i\rangle\langle j|)\otimes |i\rangle\langle j|)\mathbb F\\ &=\frac1N\sum_{i,j}|i\rangle\langle j|\otimes\mathcal N(|i\rangle\langle j|)=({\rm id} \otimes \mathcal{N})(|\phi^+\rangle\langle\phi^+|)\,. \end{align*} Because the flip operator is unitary (follows from $\mathbb F^\dagger=\mathbb F$ and $\mathbb F^2={\bf1}$), this shows the claimed unitary equivalence.

If you want an explicit example which shows that you get two different operators, consider the qubit reset channel $\mathcal N(X):={\rm tr}(X)|0\rangle\langle 0|$. Then $$ (\mathcal{N} \otimes {\rm id})(|\phi^+\rangle\langle\phi^+|)=\begin{pmatrix} \frac{1}{2} & 0 & 0 & 0 \\ 0 & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$ while $$ ({\rm id} \otimes \mathcal{N})(|\phi^+\rangle\langle\phi^+|)=\begin{pmatrix} \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}. $$