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解：
(1) $\because \hat{p} \psi = p\left[e^{\frac{i}{\hbar}(PX - Et)} - e^{-\frac{i}{\hbar}(PX + Et)}\right] \neq p\psi$
$\therefore \psi$ 不是动量算符 $\hat{p}$ 的本征函数。
(2) $\because \hat{p}^2 \psi = p^2\left[e^{\frac{i}{\hbar}(PX - Et)} + e^{-\frac{i}{\hbar}(PX + Et)}\right] = p^2\psi$
$\therefore \psi$ 是动量平方算符 $\hat{p}^2$ 的本征值函数。
解:
$(\hat{p} + \hat{x})\psi = \lambda \psi$
即：$(-i\hbar\frac{\partial}{\partial x} + x)\psi = \lambda \psi \quad \frac{\hbar}{i} \frac{d\psi}{dx} = (\lambda - x)\psi$
$\therefore \frac{\hbar}{i} \frac{d\psi}{\psi} = (\lambda - x)dx$
附注: 本征函数 $\psi = Ae^{\frac{1}{\hbar}(\lambda x - \frac{x^2}{2})}$
本征值 $\lambda$ 为所有实数。
解：
解：
(1) $\because B = A^\dagger A$
$\therefore B^2 = A^\dagger A A^\dagger A = A^\dagger A \left(1 - AA^\dagger \right) = A^\dagger A - A^\dagger A^2 A^\dagger=A^\dagger A=B$
(2)
$\because \quad B^2 = B$
$\therefore B \mathbf{v} = \lambda \mathbf{v}, \quad B^2 \mathbf{v} = \lambda^2 \mathbf{v} \quad \lambda^2 = \lambda = 0, 1$
$\text{而 } B \text{ 的本征值为 } 0, 1 \text{，在 } B \text{ 表示中 } B \text{ 为对角矩阵。}
B = \begin{pmatrix} 0 & 0 \\\\ 0 & 1 \end{pmatrix}$
设
$$A = \begin{pmatrix} a & b \\\\ c & d \end{pmatrix} \text{则}A^\dagger = \begin{pmatrix} a^* & c^* \\\\ b^* & d^* \end{pmatrix}~$$
则有 $$\quad A^ \dagger A^* = \begin{pmatrix} a^*a + c^*c & a^*b + c^*d \\\\ b^*a + d^*c & b^*b + d^*d \end{pmatrix} = B = \begin{pmatrix} 0 & 0 \\\\ 0 & 1 \end{pmatrix} \quad \text{(1)}~$$
$$ A^* A^ \dagger = \begin{pmatrix} a^* a + b b^* & c^* a + d^* b \\\\ a^* c + b^*d & c^* c + d^* d \end{pmatrix} = 1 - B = \begin{pmatrix} 1 & 0 \\\\ 0 & 0 \end{pmatrix} \quad \text{(2)}~$$
$$A^2 = \begin{pmatrix} a^2 + b c & a b + b d \\\\ a c + d c & d^2 + b c \end{pmatrix} = 0 = \begin{pmatrix} 0 & 0 \\\\ 0 & 0 \end{pmatrix} \quad \text{(3)}~$$ 由 (1), (2), (3) 得解: $\quad A = \begin{pmatrix} 0 & e^{i \varphi} \\\\ 0 & 0 \end{pmatrix} \quad \text{或} \quad A = \begin{pmatrix} 0 & 0 \\\\ 0 & 0 \end{pmatrix}$
解：一维无限深势阱中粒子的一维定态波函数： $$\psi_n(x) = \sqrt{\frac{2}{a}} \sin \frac{n\pi x}{a}~,$$能量 $$E_n = n^2 \frac{\pi^2 \hbar^2}{2m a^2}, \quad n = 1, 2, 3, \dots~$$ 将 $\psi(x)$ 按 $\psi_n$ 展开:
解： 电子自旋变 $X(t)$ 满足方程 , $i\hbar \frac{d}{dt} X(t) = \hat{H} X(t) \quad $(1) $$\hat{H} = -\hat{\mu} \cdot \mathbf{B} = \mu_0 B \hat{\sigma}_y = \mu_0 B \begin{pmatrix} 0 & -i \\\\ i & 0 \end{pmatrix}~$$ $$X(t) = \begin{pmatrix} a(t) \\\\ b(t) \end{pmatrix}~$$ 则在 $S_g$ 表中方程(1)可表示为: $$\quad i\hbar \frac{d}{dt} \begin{pmatrix} a(t) \\\\ b(t) \end{pmatrix} = \mu_0 B \begin{pmatrix} 0 & -i \\\\ i & 0 \end{pmatrix} \begin{pmatrix} a(t) \\\\ b(t) \end{pmatrix} \quad (2)~$$
$$\therefore \begin{cases} \frac{da(t)}{dt} = -\frac{\mu_0 B}{\hbar} b(t) = -\omega b(t) \\\\ \frac{db(t)}{dt} = \frac{\mu_0 B}{\hbar} a(t) = \omega a(t) \end{cases} \quad~ \text{其中},\omega = \frac{\mu_0 B}{\hbar}~$$
$\text{应用初始条件} X(0) = \begin{pmatrix} 1 \\\\ 0 \end{pmatrix} \\, \text{有解:} a(t)=\cos \omega t,b(t)=\sin \omega t$ $\therefore \text{t 时刻自旋变为:} X(t) = \begin{pmatrix} \cos \omega t \\\\ \sin \omega t \end{pmatrix}$
$$\text{将} X(t) \text{按} \hat S_0 \text{的本征函数展开：} X(t) = \begin{pmatrix} \cos \omega t \\\\ \sin \omega t \end{pmatrix} = \cos \omega t \begin{pmatrix} 1 \\\\ 0 \end{pmatrix} + \sin \omega t \begin{pmatrix} 0 \\\\ 1 \end{pmatrix}~$$
$$\quad t = 0 \quad \text{时} \quad X(0) = \begin{bmatrix} \cos \omega_0 t \\\\ \sin \omega_0 t \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}~$$
$$\text{设 } t = \tau \text{时}\\, \text{相位反转向下}. \quad X(\tau) = \begin{bmatrix} \cos \omega \tau \\ \sin \omega \tau \end{bmatrix} = \begin{bmatrix} 0 \\\\ 1 \end{bmatrix}~$$
$$\therefore \begin{cases} \cos \omega \tau = 0 \\\\ \sin \omega \tau = 1 \end{cases} \quad \text{解得：} \quad \omega \tau = \left(2k+1\right)-\frac{\pi}{2} \quad \left( k=0,1,2,\ldots \right)~$$ $$\therefore \quad \tau = \frac{\left(2k+1\right)\pi}{2\omega} = \frac{\left(2k+1\right)\pi \hbar}{2 \mu_0 B} \quad (k=0,1,2,\ldots)~$$
解：
转盘转动惯量 $I = \frac{1}{2}MR^2$，取 $\xi$ 轴为转轴。
哈密顿算符：$\hat{H}_0 = \frac{1}{2I}\hat{L}_\xi^2 = -\frac{\hbar^2}{2I}\frac{\partial^2}{\partial\varphi^2}$
本征方程：$-\frac{\hbar^2}{2I}\frac{\partial^2}{\partial\varphi^2}\psi = E\psi$
对应本征函数：$\psi_m^{(0)} = \frac{1}{\sqrt{2\pi}}e^{im\varphi} \quad m = 0, \pm1, \pm2, \ldots $
能量本征值：$E_m^{(0)} = m^2\frac{\hbar^2}{2I} = m^2\frac{\hbar^2}{MR^2} $
除基态($m = 0$)外，第 $\left|m\right|$ 激发态能级为二重简并：$\psi_{1m}^{(0)} = \frac{1}{\sqrt{2\pi}}e^{im\varphi}, \quad \psi_{2m}^{(0)} = \frac{1}{\sqrt{2\pi}}e^{-im\varphi}$
或 $\psi_{1m}^{(0)} = \frac{1}{\sqrt{\pi}}\cos m\varphi, \quad \psi_{2m}^{(0)} = \frac{1}{\sqrt{\pi}}\sin m\varphi $
受微扰后，$H' = F_0 \delta(\varphi-\varphi_0)$。微扰矩阵元：
$$ H'_{11} = \int_0^{2\pi} \frac{1}{\sqrt{\pi}}\cos m_0 \varphi F_0 \delta(\varphi-\varphi_0)\frac{1}{\sqrt{\pi}}\cos m_0 \varphi d\varphi = \frac{F_0}{\pi} \cos^2(m\varphi_0)~$$
$$H'_{22} = \int_0^{2\pi} \frac{1}{\sqrt{\pi}}\sin m_0 \varphi F_0 \delta(\varphi-\varphi_0)\frac{1}{\sqrt{\pi}}\sin m_0 \varphi d\varphi = \frac{F_0}{\pi} \sin^2(m\varphi_0)~$$
$$H'_{12} = H'_{21} = \int_0^{2\pi} \frac{1}{\sqrt{\pi}}\cos m_0 \varphi F_0 \delta(\varphi-\varphi_0)\frac{1}{\sqrt{\pi}}\sin m_0 \varphi d\varphi = \frac{F_0}{2\pi} \sin\left(2m \varphi_0\right) ~$$
由久期方程：
$$\left|
\begin{array}{cc}
\frac{F_0}{\pi} \cos^2(m\varphi_0) - E_m^{(1)} & \frac{F_0}{2\pi} \sin\left(2m\varphi_0\right) \\\\
\frac{F_0}{2\pi} \sin\left(2m\varphi_0\right) & \frac{F_0}{\pi} \sin^2(m\varphi_0) - E_m^{(1)}
\end{array}
\right| = 0~$$
解得:
$E_m^{(1)} =
\begin{cases}
\frac{E_0}{x} \\\\0
\end{cases}$
$$\therefore \\ \text{能量二阶近似为：} E_m = E_m^{(0)} + E_m^{(1)} =
\begin{cases}
m^2\frac{\hbar^2}{M R^2} + \frac{F_0}{\pi}, \\\\
m^2\frac{\hbar^2}{M R^2}
\end{cases}~$$
零级波函数：$\Psi_m^{(0)} = C_1 \frac{1}{\sqrt{\pi}} \cos m \varphi + C_2 \frac{1}{\sqrt{\pi}} \sin m \varphi$
$$\text{由方程} \quad
\begin{pmatrix}
\frac{F_0}{\pi} \cos^2 m \varphi_0 - E_m^{(1)} & \frac{F_0}{\pi} \sin m \varphi_0 \cdot \cos m \varphi_0 \\\\
\frac{F_0}{\pi} \sin m \varphi_0 \cdot \cos m \varphi_0 & \frac{F_0}{\pi} \sin^2 m \varphi_0 - E_m^{(1)}
\end{pmatrix}
\begin{pmatrix}
C_1 \\\\
C_2
\end{pmatrix}
= 0~$$
$\text{当 } E_m^{(1)} = 0 \text{时,解得：} \quad C_1 = -\sin m \varphi_0, \quad C_2 = \cos m \varphi_0$
$\therefore \quad \overline{\Psi}_m^{(1)} = \frac{1}{\sqrt{\pi}} \sin \left[m (\varphi - \varphi_0)\right]$
$\because\text{当} E^{(1)}_m = \frac{F_0}{\pi} \text{ 时：} C_1 = \cos m\varphi_0, \quad C_2 = \sin m\varphi_0$
$\therefore \Phi^{(0)}_{2m} = \frac{1}{\sqrt{\pi}} \cos[m(\varphi-\varphi_0)] $
解：
$t > 0$ 时，总能量算符 $H = H_0 + H'$ 波函数 $\psi(x,t)$ 满足方程。$$i\hbar \frac{\partial}{\partial t}\psi(x,t) = H \psi(x,t)(1)~$$
将 $\psi(x,t)$ 按 $H_0$ 定态波函数：$\psi_n(x) e^{-\frac{i}{\hbar} E_n t}$ 层开:
$$\psi(x,t) = \sum_{n} c_n(t) \psi_n(x) e^{-\frac{i}{\hbar} E_n t}(2)~$$
初始条件为 $t=0$，$\psi(x,0) = \psi_1(x)$，即 $c_n(0) = \delta_{n1}$。
将(2)式化入(1)式得:$$i\hbar \sum_n \frac{d c_n}{dt} \psi_n e^{-\frac{i}{\hbar} E_n t} = \sum_n H' \psi_n c_n e^{-\frac{i}{\hbar} E_n t}~$$
以 $\psi_n^*(x)$ 左乘上式,对全空问积分,并应用正文归一条件：$\int \psi_m^*(x) \psi_n(x) , dx = \delta_{mn}$
得: $$\frac{dC_m}{dt} e^{-\frac{i}{\hbar} E_m t} = \sum_n H'_{mn} C_n(t) e^{-\frac{i}{\hbar} E_n t}(3)~$$
其中绝对矩阵元:$$H'_{mn} = \psi_m | H' | \psi_n \rangle = \int \psi_m^*(x) H' \psi_n(x) , dx~$$
对一级近似(3)式中取 $C_n(t) \approx C_n(0) = \delta_{n1},$
$$\text{得：}i\hbar\frac{dC_n}{dt} e^{-\frac{i}{\hbar} E_n t} = H'_{n1} e^{-\frac{i}{\hbar} E_1 t}~$$
$$\text{即：}i\hbar\frac{dC_n}{dt} = H'_{n1} e^{i\omega_{n1}t}, \quad \omega_{n1} = \frac{E_n - E_1}{\hbar}~$$
$$\text{解得:}C_n(t) = \frac{1}{i\hbar} \int_0^t H'_{n1}(t') e^{i\omega_{n1} t'} dt~$$
其中