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Sep 16, 2023

量子纠缠的数学解释

定义

定义1: 单量子态ψ=a00+a11|\psi\rangle=a_0|0\rangle+a_1|1\rangle,其中0=(10)|0\rangle=\binom{1}{0}, 1=(01)|1\rangle=\binom{0}{1}, a02+a12=1|a_0|^2+|a_1|^2=1.

定义2: 双量子态Ψ=a0000+a0101+a1010+a1111|\Psi\rangle=a_{00}|00\rangle+a_{01}|01\rangle+a_{10}|10\rangle+a_{11}|11\rangle, 其中00=00|00\rangle=|0\rangle\otimes|0\rangle(其余三个同理), a002+a012+a102+a112=1|a_{00}|^2+|a_{01}|^2+|a_{10}|^2+|a_{11}|^2=1.

定义3: 如果Ψ|\Psi\rangle可以写成ψϕ|\psi\rangle\otimes|\phi\rangle的形式,则称之为乘积态,否则称之为纠缠态;

解释

纠缠态存在性: 假设ψ=a00+a11|\psi\rangle=a_0|0\rangle+a_1|1\rangle, ϕ=b00+b11|\phi\rangle=b_0|0\rangle+b_1|1\rangle, 则ψϕ=a0b000+a0b101+a1b010+a1b111|\psi\rangle\otimes|\phi\rangle=a_{0}b_0|00\rangle+a_{0}b_1|01\rangle+a_{1}b_0|10\rangle+a_{1}b_1|11\rangle, 因此得到a00=a0b0a_{00}=a_0b_0等四个方程,不难发现四个方程有解的充要条件是a00a11=a10a01a_{00}a_{11}=a_{10}a_{01}, 除此以外的所有Ψ|\Psi\rangle均为纠缠态

映射解释: 上面的讨论说明,乘积态的坐标集合S={(a0b0,a0b1,a1b0,a1b1)a0,b0,a1,b1C4}S=\{(a_0b_0,a_0b_1,a_1b_0,a_1b_1)|a_0,b_0,a_1,b_1\in\mathbb{C}^4\}不能映满C4\mathbb{C}^4, 实际上由条件a00a11=a10a01a_{00}a_{11}=a_{10}a_{01}可见SS应该是C4\mathbb{C}^4上的一个三维流形(我猜的),可见几乎所有坐标都对应纠缠态

张量积解释: ψ|\psi\rangleϕ|\phi\rangle视作(1,0)形张量VCV^*\to\mathbb{C}, Ψ|\Psi\rangle视作(2,0)形张量V×VCV^*\times V^*\to\mathbb{C}, 乘积态由张量积得到,而纠缠态的存在表明了(2,0)形张量并非都能由两个(1,0)形张量乘积得到,也即(W×W)≇WW(W\times W)^*\not\cong W^*\otimes W^*(实际上(W×W)W×W(W\times W)^*\cong W^*\times W^*,所以说把张量积定义中映射的乘积换成求和就满射了)

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