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Let $\mathcal{E}_{A\rightarrow B}$ be a quantum channel and consider its $n-$fold tensor product $\mathcal{E}^{\otimes n}_{A^n\rightarrow B^n}$.

Any isometry $V_{A\rightarrow BE}$ that satisfies $\text{Tr}_E(V\rho V^\dagger) = \mathcal{E}(\rho)$ can be used to construct a Stinespring dilation of $\mathcal{E}^{\otimes n}$. Indeed, a valid Stinespring dilation of $\mathcal{E}^{\otimes n}$ is simply $V^{\otimes n}$.

Is there any other Stinespring dilation of $\mathcal{E}^{\otimes n}$ that has a smaller environment size than $|E|^n$?

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No.

The minimal size of the environment is just the rank of the Choi matrix of $\mathcal E$, call it $J(\mathcal E)$. Since $J(\mathcal E^{\otimes n}) = \big(J(\mathcal E)\big)^{\otimes n}$ and $\text{rank}(A \otimes B) = \text{rank}(A)\text{rank}(B)$, the minimal size of the environment is just $\text{rank}\big(J(\mathcal E)\big)^n$.

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  • $\begingroup$ Thank you for the answer! After I asked the question, I came across Lemma 4.2.2 of arxiv.org/pdf/quant-ph/0512258.pdf but I'm not sure how it fits with your answer. Since $J(\mathcal{E})^{\otimes n}$ is permutation invariant, it has a purification on a smaller Hilbert space $\text{Sym}((\mathcal{H}\otimes\mathcal{H})^{\otimes n})$. Would you be able to comment on this? $\endgroup$
    – user1936752
    Commented Dec 11, 2020 at 11:24
  • $\begingroup$ I guess Renner only means that there is a permutation invariant purification. Otherwise his statement is obviously false, just consider the case of identity. It's hard to say, Renner's notation is notoriously bad. $\endgroup$
    – Mateus Araújo
    Commented Dec 11, 2020 at 12:34
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    $\begingroup$ Lemma 4.2.2 in Renner's thesis is correct, it just isn't applicable to this situation. If $\mathcal{E}$ maps $\operatorname{L}(\mathcal{A})$ to $\operatorname{L}(\mathcal{B})$, then $J(\mathcal{E})\in\operatorname{L}(\mathcal{B}\otimes\mathcal{A})$. The lemma therefore implies the existence of a purification of $J(\mathcal{E})^{\otimes n}$ on the symmetric subspace of $(\mathcal{B}\otimes\mathcal{A}\otimes\mathcal{B}\otimes\mathcal{A})^{\otimes n}$. This does not contradict the answer above, which correctly answers the original question. $\endgroup$
    – John Watrous
    Commented Dec 11, 2020 at 15:21
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    $\begingroup$ I think the point is simply that the dimension of the subspace in which the purification lives has nothing to do with the dimension of the environment you use to purify. You could say that the purification of identity can always be in the one-dimensional subspace spanned by $|\phi^+\rangle$, which does not contradict the fact that the purifying system must be of dimension $d$. $\endgroup$
    – Mateus Araújo
    Commented Dec 11, 2020 at 15:50

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