第一章习题答案

1.计算二元对称信道的信道容量

$$\begin{equation} C = \max{I(A,B)} = \max{H(A)-H(A|B)}\\ = \max{H(A)-\sum_{X}P(x_i)\sum{P(y_i|x_i)}\log{\frac{1}{P(y_i|x_i)}}}\\ = \max{H(A) - (p\log{\frac{1}{p}}+(1-p)\log{\frac{1}{1-p}})} \end{equation}$$

2.

$$|\phi_i\rangle = \sum_j |\psi_j\rangle \langle \psi_j| \phi_i\rangle$$

所以这个幺正变换是表象变换U

$$\begin{equation} U_{i\alpha} = \langle \psi_i|\phi_\alpha\rangle\\ U^{-1}_{\alpha i}=\langle\phi_\alpha|\psi_i\rangle \end{equation}$$

3

4

由题有

$$\begin{align} &|\phi\rangle = \frac{1}{\sqrt{2}}|0\rangle_A(\frac{1}{2}|0\rangle_B+\frac{\sqrt{3}}{2}|1\rangle_B)+\frac{1}{\sqrt{2}}|1\rangle_A(\frac{\sqrt{3}}{2}|0\rangle_B+\frac{1}{2}|1\rangle_B)\\ &\rho = |\phi\rangle\langle\phi| \end{align}$$

A的密度算符

$$\begin{align} \rho_A &=tr_B(\rho)=\sum_{\alpha}\langle\alpha|\rho|\alpha\rangle\\ &=\langle0|_B|\phi\rangle\langle\phi|0\rangle_B+\langle1|_B|\phi\rangle\langle\phi|1\rangle_B\\ %%%%%%%%%%%%%%%%%% &=(\frac{\sqrt{2}}{4}|0\rangle_A+\frac{\sqrt{6}}{4}|1\rangle_A)(\frac{\sqrt{2}}{4} \langle 0|_A+\frac{\sqrt{6}}{4}\langle1|_A)+ %%%%%%%%%%%%%%%%%% (\frac{\sqrt{6}}{4}|0\rangle_A+\frac{\sqrt{2}}{4}|1\rangle_A)(\frac{\sqrt{6}}{4} \langle 0|_A+\frac{\sqrt{2}}{4}\langle1|_A) \end{align}$$
约化密度矩阵
$$\begin{gather*} \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{4}\\ \frac{\sqrt{3}}{4} & \frac{1}{2}\\ \end{pmatrix} \end{gather*}$$

B的密度算符

由对称性有
约化密度矩阵

$$\begin{gather*} \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{4}\\ \frac{\sqrt{3}}{4} & \frac{1}{2}\\ \end{pmatrix} \end{gather*}$$

Schimit分解

$$\begin{gather*} \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{4}\\ \frac{\sqrt{3}}{4} & \frac{1}{2}\\ \end{pmatrix}= \begin{pmatrix} \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} &-\frac{1}{\sqrt{2}}\\ \end{pmatrix} \cdot \begin{pmatrix} \frac{2+\sqrt{3}}{4}& 0\\ 0&\frac{2-\sqrt{3}}{4} \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} &-\frac{1}{\sqrt{2}}\\ \end{pmatrix} \end{gather*}$$
$$|\mathbf{\rho_A}-\lambda \mathbf{I}|=0 \Rightarrow \lambda = \frac{1}{2}+\frac{\sqrt{3}}{4} \mathbf{or}\frac{1}{2}-\frac{\sqrt{3}}{4}$$
特征向量
$$(\rho-\lambda I)\mathbf{X}=0 \Rightarrow X=\frac{1}{\sqrt{2}}(1,1)\mathbf{or}\frac{1}{\sqrt{2}}(1,-1)$$
所以Schmidt分解为
$$|\phi\rangle=\frac{1+\sqrt{3}}{2\sqrt{2}}|0'\rangle_A|0'\rangle_B+\frac{1-\sqrt{3}}{2\sqrt{2}}|1'\rangle_A|1'\rangle_B$$

其中

$$|0'\rangle = \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)\\ |1'\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle)$$

5

考虑以下纯态

$$\begin{align} &|\psi_{ABC}\rangle = \psi_{AB}\bigotimes |\omega\rangle \\ &|\psi_{AB}\rangle = \sum_{j}\alpha_j|j\rangle_A |j'\rangle_B , \quad j\ge 2 \end{align}$$

如果能将其写为三体Schmidt分解形式，就有

$$|\Psi\rangle_{ABC}=\sum_{i}\sqrt{p_i}|i\rangle_A |i'\rangle_B |i''\rangle_C ,\forall p_i \neq 0$$
$$\begin{align} &则，当i\ge 2,对粒子C求迹后所剩AB粒子的态是混态而不是原来假定的纯态；\\ &若i=1,则|i''\rangle_C = |\omega\rangle_C，同时A和B是分离的，这也不符合j\ge 2是纠缠的原来假定 \end{align}$$

6

(i)

原式等价于

$$\begin{align} |\langle\phi|\psi\rangle|^2 &= \langle\phi|\psi\rangle\langle\psi|\phi\rangle \\ & = \langle\phi|\rho|\phi\rangle\\ & = \langle\phi|\frac{1}{2}(1+\hat{n}\cdot\vec{\sigma})|\phi\rangle \end{align}$$
由于平均保真度
$$F=\frac{\int_0^{\pi}\int_0^{2\pi}|\langle\omega|\psi\rangle|^2\sin{\theta}d\theta d\phi}{4\pi}=\frac{1}{2}$$

(ii)

7 POVM

i)直和方法

$$|\Psi_{\alpha}\rangle = \sqrt{\frac{2}{3}}\psi_{\alpha} \quad \alpha=1,2,3$$
$$\begin{gather*} \begin{pmatrix} \sqrt{\frac{2}{3}} & -\frac{1}{2}\sqrt{\frac{2}{3}} & -\frac{1}{2}\sqrt{\frac{2}{3}}\\ 0 & \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \end{pmatrix} \end{gather*}$$
补上正交的第三行有
$$\begin{gather*} \begin{pmatrix} \sqrt{\frac{2}{3}} & -\sqrt{\frac{1}{6}} & -\sqrt{\frac{1}{6}}\\ 0 & \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2}\\ \sqrt{\frac{1}{3}} & -\sqrt{\frac{1}{3}} &\sqrt{\frac{1}{3}} \end{pmatrix} \end{gather*}$$
所构造的正交测量为
$$\begin{gather*} |u_1\rangle = \begin{pmatrix} \sqrt{\frac{2}{3}}\\0\\\sqrt{\frac{1}{3}} \end{pmatrix} , |u_2\rangle = \begin{pmatrix} -\sqrt{\frac{1}{6}}\\\frac{\sqrt{2}}{2}\\-\sqrt{\frac{1}{3}} \end{pmatrix} , |u_3\rangle = \begin{pmatrix} -\sqrt{\frac{1}{6}}\\-\frac{\sqrt{2}}{2}\\\sqrt{\frac{1}{3}} \end{pmatrix} \end{gather*}$$

9

$$U_A\bigotimes U_B |\psi^{-}\rangle = \frac{1}{2}(U|0\rangle\bigotimes U|1\rangle-U|1\rangle\bigotimes U|0\rangle)$$

$$\begin{gather*} \sigma_{u}=\vec{u}\cdot\vec{\sigma} = \begin{pmatrix} u_z & u_x-i u_y\\ u_x+i u_y& -u_z \end{pmatrix} \end{gather*}$$

11

构造力学量

$$\begin{align} \xi&=\frac{\vec{\sigma_A\cdot\vec{n}}+I_1}{2}\bigotimes\frac{\vec{\sigma_B}\cdot\vec{m}+I_2}{2}\\ P&=tr(\rho_{AB}\xi)\\ &=\frac{1}{32}(tr(\sigma_n\bigotimes\sigma_m))+2tr(\sigma_n)+2tr(\sigma_m)+\psi) +\\ &\quad\frac{1}{8}\langle{\psi^{-}}|(\sigma_n\bigotimes\sigma_m+\sigma_n\bigotimes I_2+I_1\bigotimes\sigma_m+I_1\bigotimes I_2)|\psi^{-}\rangle\\ &=\frac{2-\cos{\theta}}{8} \end{align}$$

Second Chapter

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