It is known that computing the permanent $\mathop{\rm Per}(1+A)$, where $A$ is a finite-rank matrix requires a number of operations polynomial in the matrix size. I generalize this result to the expectation values $\left\langle\Psi| P(1+A) |\Psi\right\rangle$, where $P()$ is the multiplicative extension of a single-particle operator and $\left|\Psi\right\rangle$ is a product of a large number of identical finite bosonic states (i.e. bosonic states with a bounded number of bosons). I also improve an earlier polynomial estimate for the fermionic version of the same problem.
翻译:已知的是,计算永久 $\ mathop\ rm Per}(1+A)$, 美元是一个有限基质矩阵, 计算这个基质大小需要多种操作。 我将这一结果概括为预期值 $\ left\ langle\ P ⁇ P(1+A) ⁇ Psi\ right\ rangle$, $P() 是单粒操作员和 $\left\ psi\ right\rangle$ 的倍增效应扩展值, 是大量相同的软骨质状态( 即, 有捆绑数的肉质的肉质状态) 的产物 。 我还改进了同一问题的发酵版本的早期多元估计值 。