We study the estimation of an unknown quantum channel $\mathcal{E}$ with input dimension $d_1$, output dimension $d_2$ and Kraus rank at most $r$. We establish a connection between the query complexities in two models: (i) access to $\mathcal{E}$, and (ii) access to a random dilation of $\mathcal{E}$. Specifically, we show that for parallel (possibly coherent) testers, access to dilations does not help. This is proved by constructing a local tester that uses $n$ queries to $\mathcal{E}$ yet faithfully simulates the tester with $n$ queries to a random dilation. As application, we show that: - $O(rd_1d_2/\varepsilon^2)$ queries to $\mathcal{E}$ suffice for channel tomography to within diamond norm error $\varepsilon$. Moreover, when $rd_2=d_1$, we show that the Heisenberg scaling $O(1/\varepsilon)$ can be achieved, even if $\mathcal{E}$ is not a unitary channel: - $O(\min\{d_1^{2.5}/\varepsilon,d_1^2/\varepsilon^2\})$ queries to $\mathcal{E}$ suffice for channel tomography to within diamond norm error $\varepsilon$, and $O(d_1^2/\varepsilon)$ queries suffice for the case of Choi state trace norm error $\varepsilon$. - $O(\min\{d_1^{1.5}/\varepsilon,d_1/\varepsilon^2\})$ queries to $\mathcal{E}$ suffice for tomography of the mixed state $\mathcal{E}(|0\rangle\langle 0|)$ to within trace norm error $\varepsilon$.

翻译：我们研究未知量子信道 $\mathcal{E}$ 的估计问题，其输入维度为 $d_1$，输出维度为 $d_2$，且 Kraus 秩至多为 $r$。我们建立了两种模型下查询复杂度之间的联系：（i）直接访问 $\mathcal{E}$，以及（ii）访问 $\mathcal{E}$ 的随机扩张。具体而言，我们证明对于并行（可能相干）测试者，访问扩张并无助益。这一结论通过构造一个局部测试者来证明，该测试者仅使用 $n$ 次对 $\mathcal{E}$ 的查询，却能忠实模拟使用 $n$ 次对随机扩张查询的测试者。作为应用，我们证明：- 使用 $O(rd_1d_2/\varepsilon^2)$ 次对 $\mathcal{E}$ 的查询足以实现信道层析，其钻石范数误差为 $\varepsilon$。此外，当 $rd_2=d_1$ 时，我们证明即使 $\mathcal{E}$ 不是酉信道，也能达到海森堡标度 $O(1/\varepsilon)$：- 使用 $O(\min\{d_1^{2.5}/\varepsilon,d_1^2/\varepsilon^2\})$ 次对 $\mathcal{E}$ 的查询足以实现信道层析，其钻石范数误差为 $\varepsilon$；而对于 Choi 态迹范数误差 $\varepsilon$ 的情况，$O(d_1^2/\varepsilon)$ 次查询即足够。- 使用 $O(\min\{d_1^{1.5}/\varepsilon,d_1/\varepsilon^2\})$ 次对 $\mathcal{E}$ 的查询足以实现混合态 $\mathcal{E}(|0\rangle\langle 0|)$ 的层析，其迹范数误差为 $\varepsilon$。