【量子光学】 Quantum Optics 2 Quiz 4.2(9)

Quantum Optics 2课程第九次测验

第1题 (不定项选择)

Which of the following states are entangled?

A) $\frac{1}{\sqrt{2}}\left(\left|x_1,y_2\right\rangle+\left|y_1,x_2\right\rangle\right)$

B) $\frac{1}{\sqrt{2}}\left(\left|x_1,y_2\right\rangle+\left|y_1,y_2\right\rangle\right)$

C) $\frac{1}{2}\left(\left|x_1, x_2\right\rangle+\left|x_1, y_2\right\rangle+\left|y_1, x_2\right\rangle+\left|y_1, y_2\right\rangle\right)$

第2题 (单选)

What probabilities are expected for the state $\frac{1}{\sqrt{2}}\left(\left|x_1,y_2\right\rangle+\left|y_1,x_2\right\rangle\right)$ in the case of parallel polarizers at an angle $\alpha$ from the $\vec{x}$ axis?

A) $ P_{+ +}=P_{- -}=\frac{1}{2}, P_{+ -}=P_{- +}=0$

B) $ P_{+ +}=P_{- -}=0, P_{+ -}=P_{- +}=\frac{1}{2}$

C) $ P_{+ +}=P_{- -}=\frac{1}{2}\cos^2(2\alpha), P_{+ -}=P_{- +}=\frac{1}{2}\sin^2(2\alpha)$

D) $ P_{+ +}=P_{- -}=\frac{1}{2}\sin^2(2\alpha), P_{+ -}=P_{- +}=\frac{1}{2}\cos^2(2\alpha)$

第3题 (不定项选择)

Which angles $\alpha ,\beta$ of the polarizers I and II do lead to total correlation for the entangled state $\frac{1}{\sqrt{2}}\left(\left|x_1,y_2\right\rangle+\left|y_1,x_2\right\rangle\right)$?

A) $\alpha=0, \beta=0$

B) $\alpha=\frac{\pi}{8}, \beta=0$

C) $\alpha=\frac{\pi}{8}, \beta=\frac{\pi}{8}$

D) $\alpha=\frac{\pi}{4}, \beta=0$

E) $\alpha=\frac{\pi}{4}, \beta=\frac{\pi}{4}$

F) $\alpha=\frac{\pi}{2}, \beta=0$

补全下方信息(禁止使用符号‘#’，‘/’)，点击Submit Answer按钮提交答案。
姓名(英语/拼音)：

邮箱(手机/其他)：

备注(英语/不填)：

Submit Answer

---提交后等待口令，若无需刷新后重新提交！---

任何问题请联系：email:happyan@piphipsi.com