On the continuous Zauner conjecture

D. Yakymenko
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Abstract

In a recent paper by S. Pandey, V. Paulsen, J. Prakash, and M. Rahaman, the authors studied the entanglement breaking quantum channels $\Phi_t:\mbb{C}^{d\times d} \to \mbb{C}^{d \times d}$ for $t \in [-\frac{1}{d^2-1}, \frac{1}{d+1}]$ defined by $\Phi_t(X) = tX+ (1-t)\tr(X) \frac{1}{d}I$. They proved that Zauner's conjecture is equivalent to the statement that entanglement breaking rank of $\Phi_{\frac{1}{d+1}}$ is $d^2$. The authors made the extended conjecture that $\ebr(\Phi_t)=d^2$ for every $t \in [0, \frac{1}{d+1}]$ and proved it in dimensions 2 and 3. In this paper we prove that for any $t \in [-\frac{1}{d^2-1}, \frac{1}{d+1}] \setminus\{0\}$ the equality $\ebr(\Phi_t)=d^2$ is equivalent to the existence of a pair of informationally-complete unit-norm tight frames $\{|x_i\ra\}_{i=1}^{d^2}, \{|y_i\ra\}_{i=1}^{d^2}$ in $\mbb{C}^d $ which are mutually unbiased in the following sense: for any $i\neq j$ it holds that $|\la x_i|y_j\ra|^2 = \frac{1-t}{d}$ and $|\la x_i|y_i\ra|^2 = \frac{t(d^2-1)+1}{d}$. Moreover, it follows that $|\la x_i|x_j\ra\la y_i|y_j\ra|=|t|$ for $i\neq j$. However, our numerical searches for solutions were not successful in dimensions 4 and 5 for values of $t$ other than $0$ or $\frac{1}{d+1}$.
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关于连续Zauner猜想
在S. Pandey, V. Paulsen, J. Prakash和M. Rahaman最近的一篇论文中,作者研究了$\Phi_t(X) = tX+ (1-t)\tr(X) \frac{1}{d}I$定义的$t \in [-\frac{1}{d^2-1}, \frac{1}{d+1}]$的量子通道$\Phi_t:\mbb{C}^{d\times d} \to \mbb{C}^{d \times d}$的纠缠破坏。他们证明了Zauner猜想等价于$\Phi_{\frac{1}{d+1}}$的纠缠破秩为$d^2$的表述。作者对每个$t \in [0, \frac{1}{d+1}]$都提出了扩展猜想$\ebr(\Phi_t)=d^2$,并在2维和3维上进行了证明。在本文中,我们证明了对于任意$t \in [-\frac{1}{d^2-1}, \frac{1}{d+1}] \setminus\{0\}$等式$\ebr(\Phi_t)=d^2$等价于$\mbb{C}^d $中存在一对信息完备的单位范数紧框架$\{|x_i\ra\}_{i=1}^{d^2}, \{|y_i\ra\}_{i=1}^{d^2}$,它们在以下意义上是相互无偏的:对于任意$i\neq j$,它持有$|\la x_i|y_j\ra|^2 = \frac{1-t}{d}$和$|\la x_i|y_i\ra|^2 = \frac{t(d^2-1)+1}{d}$。此外,可以得出$i\neq j$等于$|\la x_i|x_j\ra\la y_i|y_j\ra|=|t|$。然而,我们的数值搜索解决方案是不成功的,在维度4和5的值$t$除了$0$或$\frac{1}{d+1}$。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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