生活很复杂 – 它具有真实和虚构的部分。 – 匿名
$$ \rho_{m}=\sum_{i} p(i \mid m)\left|\psi_{i}^{m}\right\rangle\left\langle\psi_{i}^{m}\right|=\sum_{i} p(i \mid m) \frac{M_{m}\left|\psi_{i}\right\rangle\left\langle\psi_{i}\right| M_{m}^{\dagger}}{\left\langle\psi_{i}\left|M_{m}^{\dagger} M_{m}\right| \psi_{i}\right\rangle} $$演示隐形传态
下面是演示复杂的公式：
$$ \begin{aligned}\langle v \mid w\rangle &=\left(\sum_{i} v_{i}|i\rangle, \sum_{j} w_{j}|j\rangle\right)=\sum_{i j} v_{i}^{*} w_{j} \delta_{i j}=\sum_{i} v_{i}^{*} w_{i} \\ &=\left[v_{1}^{*} \ldots v_{n}^{*}\right]\left[\begin{array}{c}w_{1} \\ \vdots \\ w_{n}\end{array}\right] . \end{aligned} $$ $$ \left(\sum_{i} a_{i}\left|v_{i}\right\rangle \otimes\left|w_{i}\right\rangle, \sum_{j} b_{j}\left|v_{j}^{\prime}\right\rangle \otimes\left|w_{j}^{\prime}\right\rangle\right) \equiv \sum_{i j} a_{i}^{*} b_{j}\left\langle v_{i} \mid v_{j}^{\prime}\right\rangle\left\langle w_{i} \mid w_{j}^{\prime}\right\rangle $$ $$ \begin{aligned}\left\langle\varphi\left|\left\langle 0\left|U^{\dagger} U\right| \psi\right\rangle\right| 0\right\rangle &=\sum_{m, m^{\prime}}\left\langle\varphi\left|M_{m}^{\dagger} M_{m^{\prime}}\right| \psi\right\rangle\left\langle m \mid m^{\prime}\right\rangle \\ &=\sum_{m}\left\langle\varphi\left|M_{m}^{\dagger} M_{m}\right| \psi\right\rangle \\ &=\langle\varphi \mid \psi\rangle . \end{aligned} $$ $$ \begin{aligned} p(m) &=\left\langle\psi\left|\left\langle 0\left|U^{\dagger} P_{m} U\right| \psi\right\rangle\right| 0\right\rangle \\ &=\sum_{m^{\prime}, m^{\prime \prime}}\left\langle\psi\left|M_{m^{\prime}}^{\dagger}\left\langle m^{\prime}\left|\left(I_{Q} \otimes|m\rangle\langle m|\right) M_{m^{\prime \prime}}\right| \psi\right\rangle\right| m^{\prime \prime}\right\rangle \\ &=\left\langle\psi\left|M_{m}^{\dagger} M_{m}\right| \psi\right\rangle \end{aligned} $$$Exercise 2.2:$
(Matrix representations: example)
Suppose $ V$ is a vector space with basis vectors $|0\rangle$ and $|1\rangle$, and $A$ is a linear operator from $ V$ to $V $ such that $A|0\rangle=|1\rangle$ and $A|1\rangle=|0\rangle$. Give a matrix representation for $A$, with respect to the input basis $|0\rangle , |1\rangle$ , and the output basis$|0\rangle,|1\rangle$. Find input and output bases which give rise to a different matrix representation of $A$.
矩阵表示案例
令$V$是含有基向量$|0\rangle$ 和$|1\rangle$ 的向量空间，$A$是从空间$ V$ 到空间$ V$ 的一个线性算子， $A|0\rangle=|1\rangle ， A\rangle =|0\rangle$.给出输入基是$|0\rangle$ 和$|1\rangle$ ，输出基是$0\rangle，|1\rangle$的矩阵矩阵表示给定一个不同的矩阵矩阵表示出输入基和输出基.
解答
算子表达式为
\begin{array}{l} A|0\rangle=0|0\rangle+1|1\rangle \ A|1\rangle=1|0\rangle+0|1\rangle \end{array}故矩阵表示为
$$ \left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] $$当给定一个矩阵表示为
$$ \left[\begin{array}{ll}a & b \\ c & d\end{array}\right] $$
时，对输入基$|0\rangle$和$|1\rangle$而言，输出基是$(a+c)|0\rangle$和$b+d)|1\rangle$.
