名古屋大学 情報学研究科 数理情報学専攻 2021年8月実施 問題6 量子力学

Author

Miyake

Description

Kai

(1)

(2)

\[ \begin{aligned} \left| \psi_\tau \right\rangle &= U(\tau) \left| \psi_0 \right\rangle \\ &= \begin{pmatrix} \cos \tau & - \sin \tau \\ \sin \tau & \cos \tau \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \\ &= \begin{pmatrix} \cos \tau \\ \sin \tau \end{pmatrix} \end{aligned} \]

(3)

\[ \begin{aligned} \left\langle X \right\rangle_\tau &= \left\langle \psi_\tau \right| X \left| \psi_\tau \right\rangle \\ &= \begin{pmatrix} \cos \tau & \sin \tau \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} \cos \tau \\ \sin \tau \end{pmatrix} \\ &= 2 \sin \tau \cos \tau \\ &= \sin 2 \tau \end{aligned} \]

(4)

\(X^2\) は単位行列であるから、

\[ \begin{aligned} \left\langle X^2 \right\rangle_\tau &= \left\langle \psi_\tau | \psi_\tau \right\rangle \\ &= 1 \\ \therefore \ \ \sigma^2_\tau (X) &= \left\langle X^2 \right\rangle_\tau - \left\langle X \right\rangle_\tau^2 \\ &= 1 - \sin^2 2 \tau \end{aligned} \]

(5)

\[ \begin{aligned} \tau^* = \frac{\pi}{4} \end{aligned} \]