论文标题
从其部分重建整体
Reconstructing the whole from its parts
论文作者
论文摘要
量子边缘问题在于确定给定的一组边际减少是否与存在全球量子状态的存在兼容。在这项工作中,我们从动力学系统理论的角度提出了问题,并研究了其在标准方法方面的优势。引入的形式主义使我们能够在任何多部分场景中从一系列自洽的边际减少中分析确定全球量子状态。特别是,我们表明,在通过去极化通道后,任何自洽的多部分边缘递减组都与全球量子状态的存在兼容。该结果表明,在限制对足够混合边缘的注意力时,与边际问题相关的复杂性可以大大降低。我们还以压缩方式提出边际问题,因为标量约束的总数小于标准方法所要求的总数。这一事实表明,在考虑半准编程技术以在经典和量子算法中解决它时,在运行时进行了指数加速。最后,我们从所有$ \ binom {n} {k} $边际减少到$ k $ partions的所有$ \ binom {n} {k} $ n $ qubit量子状态,从随机选择的混合状态产生。数值模拟表明,当$ 5 \ leq n \ leq12 $和$ \ lfloor(n-1)/\ sqrt {2} \ rfloor \ leq k \ leq k \ leq n-1 $,其中$ \ lfloor \ cdot \ cdot \ cdot \ rfloor $ deNOstauctiation。
The quantum marginal problem consists in deciding whether a given set of marginal reductions is compatible with the existence of a global quantum state or not. In this work, we formulate the problem from the perspective of dynamical systems theory and study its advantages with respect to the standard approach. The introduced formalism allows us to analytically determine global quantum states from a wide class of self-consistent marginal reductions in any multipartite scenario. In particular, we show that any self-consistent set of multipartite marginal reductions is compatible with the existence of a global quantum state, after passing through a depolarizing channel. This result reveals that the complexity associated to the marginal problem can be drastically reduced when restricting the attention to sufficiently mixed marginals. We also formulate the marginal problem in a compressed way, in the sense that the total number of scalar constraints is smaller than the one required by the standard approach. This fact suggests an exponential speedup in runtime when considering semi-definite programming techniques to solve it, in both classical and quantum algorithms. Finally, we reconstruct $n$-qubit quantum states from all the $\binom{n}{k}$ marginal reductions to $k$ parties, generated from randomly chosen mixed states. Numerical simulations reveal that the fraction of cases where we can find a global state equals 1 when $5\leq n\leq12$ and $\lfloor(n-1)/\sqrt{2}\rfloor\leq k\leq n-1$, where $\lfloor\cdot\rfloor$ denotes the floor function.