问题描述
我正在寻找采用任意量子态的算法,该量子态由加权的经典态之和组成,这些态由位组成,像这样:
I'm looking for algorithms that take an arbitrary quantum state made up of a sum of weighted classical states made up of bits, like this:
|0000>/2 - |0011>/2 + |0100>/2 - |0111>/2
并使用张量积将其分解为更紧凑的形式,例如:
and factor it into a more compact form using tensor products, like this:
|0> x (|0> + |1>) x (|00> - |11>) / 2
我想将该算法用作可视化/简化(模拟)量子电路状态的一种方式。
I want to use the algorithm as a way of visualizing/simplifying the state of a (simulated) quantum circuit.
对于单个量子位,我知道我可以将所有翻转位,并检查每对在状态之间是否具有相同的x:y关系。在上面的示例中,翻转第二个位始终会为您提供权重为1:1的状态,因此第二个位分解为(1 | 0> + 1 | 1>)。
For individual qubits I know I can just pair all the states with the state where the bit is flipped and check that every pair has the same x:y relation between the states. In the example above, flipping the second bit always gives you a state with a 1:1 weighting, so the second bit factors out as (1|0> + 1|1>).
但是将这种方法扩展为检测纠缠位(例如示例中的第三和第四位)会使它至少花费Ω(n ^ c)
时间(可能更多,我还没有完全考虑过),其中 n
是州的数量, c
是纠缠位数。由于 n
已经随着不理想的位数呈指数增长。
But extending that approach to detect entangled bits (like the third and fourth in the example) causes it to take at least Ω(n^c)
time (probably more, I haven't thought it all the way through), where n
is the number of states and c
is the number of entangled bits. Since n
is already growing exponentially with the number of bits that's... not ideal.
有没有更好的算法?表示更容易从/作为要素?更改基础有多有用?
Are there better algorithms? Representations easier to factor from/to? How useful is changing the basis? Links to papers would be great.
推荐答案
有效的算法似乎很难:
来自:
Gurvits,L.,《埃德蒙兹问题的经典确定性复杂性和量子纠缠》,第35卷ACM计算理论专题讨论会,ACM出版社,纽约,2003年。
Gurvits, L., Classical deterministic complexity of Edmonds’ problem and quantum entanglement, in Proceedings of the 35th ACM Symposium on Theory of Computing, ACM Press, New York, 2003.
Sevag Gharibian,《量子可分离性问题的强NP硬度》,《量子信息与计算》,第一卷。 ,第3号和第4号,第343-360页,2010年。arXiv:0810.4507
Sevag Gharibian, Strong NP-Hardness of the Quantum Separability Problem, Quantum Information and Computation, Vol. 10, No. 3&4, pp. 343-360, 2010. arXiv:0810.4507
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