I am interested in a quantum channel from $A^{\otimes n}$ to $B^{\otimes n}$ denoted as $N_{A^{\otimes n} \rightarrow B^{\otimes n}}(\cdot)$. Let $\pi(\cdot)$ be a permutation operation among the $n$ registers. I am given that

$$N_{A^{\otimes n} \rightarrow B^{\otimes n}}\left(\pi_{A^{\otimes n}}(\cdot)\right) = \pi_{B^{\otimes n}} \left(N_{A^{\otimes n} \rightarrow B^{\otimes n}}(\cdot)\right)$$

The Stinespring dilation of $N_{A^{\otimes n}\rightarrow B^{\otimes n}}(\cdot)$ is an isometry $V_{A^{\otimes n}\rightarrow B^{\otimes n}E}$ such that $$\text{Tr}_E\left(V (\cdot) V^\dagger\right) = N(\cdot)$$

Does there exist a Stinespring dilation of $N_{A^{\otimes n}\rightarrow B^{\otimes n}}$ such that

$$V_{A^{\otimes n} \rightarrow B^{\otimes n}E}\pi_{A^{\otimes n}} = \pi_{B^{\otimes n}}V_{A^{\otimes n} \rightarrow B^{\otimes n}E}$$

It is clear that one can achieve this if permutation operations on the $E$ are also allowed but now we are only allowed to permute on $B^n$.


1 Answer 1


No, such a dilation might not exist. For a counter-example that can easily be generalized, let $\Phi$ be any non-isometric channel from $A$ to $B$, let $n=2$, and let $N = \Phi\otimes \Phi$.

By a non-isometric channel I mean one that does not look like $\Phi(\cdot) = V \cdot V^{\dagger}$ for some isometry $V$ from $A$ to $B$, and so it must be that $\Phi$ has Choi rank at least 2. Equivalently, there must be a Kraus representation $$ \Phi(\rho) = \sum_{k=1}^r M_k \rho M_k^{\dagger} $$ where $r\geq 2$ and $\{M_1,\ldots,M_r\}$ are linearly independent. (We can take them to be orthogonal if we wish.)

We can easily obtain a Sinespring dilation of $N = \Phi\otimes\Phi$ by taking $$ V = \sum_{1\leq j,k\leq r} M_j\otimes M_k \otimes v_{j,k} $$ for any orthonormal set $\{v_{j,k}\}\subset E$, and every Stinespring dilation of $N$ is expressible in this way.

The permutation-invariance property suggested by the question implies that we must have $$ \sum_{1\leq j,k\leq r} M_j\otimes M_k \otimes v_{j,k}=\sum_{1\leq j,k\leq r} M_k\otimes M_j \otimes v_{j,k} $$ and therefore $M_j\otimes M_k = M_k\otimes M_j$ for all $j,k$ by the orthonormality of the set $\{v_{j,k}\}$. This, however, contradicts the linear independence of $M_1$ and $M_2$ (for instance).


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