贡献者： Entanglement
1.
- 归一化系数 $A=2$，列表如下：
表1：可能的值与相应几率
|
| 几率 | $n$ | $L^{2}=l(l+1)\hbar^{2}$ | $L_{z}=m\hbar$
|
|
$\psi_{210}$ | $ \frac{1}{2}$ | 2 | $2\hbar^2$ | 0
|
|
$\psi_{110}$ | $ \frac{1}{2}$ | 1 | 0 | 0
|
- 不计自旋，守恒量有 $\hat H,\hat {L^2},\hat {L_z} $。
- $\because \quad A^{\dagger} = A,B^{\dagger} = B,(AB)^{\dagger} \neq AB $
$\therefore (1)\quad (AB)^{\dagger} = B^*A^* = BA $，不是厄米算符。
$\quad (2)\quad (AB+BA)^{\dagger} = B^*A^*+A^*+B^* = AB+BA $，是厄米算符。
$\quad (3)\quad (AB-BA)^{\dagger} = B^*A^*-A^*B^* = -(AB-BA) $，不是厄米算符。
$\quad (4)\quad (ABA)^{\dagger} = (ABA)^* = ABA $，是厄米算符。
$\quad (5)\quad [i(AB-BA)]^{\dagger} = -i(BA-AB) = i(AB-BA) $，是厄密算符。
2.
- 由 $\displaystyle \psi (x,0) = \frac{1}{\sqrt{2}}[\varphi_{0}(x) + \varphi_{1}(x)] $ 可得任意时刻的波函数为：
\begin{equation}
\psi(x,t) = \frac{1}{\sqrt{2}}\left[\varphi_{0}(x)e^{-\frac{iE_{0}t}{\hbar}} + \varphi_{1}(x)e^{-\frac{iE_{1}t}{\hbar}} \right]~.
\end{equation}
- 由谐振子的基本性质，有：
\begin{equation}
x\psi_{n} = \frac{1}{\alpha} \left[\sqrt{\frac{n}{2}}\psi_{n-1}(x) + \sqrt{\frac{n+1}{2}}\psi_{n+1}(x) \right]~,
\end{equation}
\begin{equation}
\begin{aligned}
\bar{x} =& \int^{\infty}_{-\infty} \psi^{*}(x)\psi \,\mathrm{d}{x} \\
=& \frac{1}{2\alpha} \int^{\infty}_{-\infty} \left[\varphi^{*}_{0}(x)+\varphi^{*}_{1}(x)\right](x)\left[\varphi_{0}(x)+\varphi_{1}(x)\right] \,\mathrm{d}{x} \\
=& \frac{1}{2\alpha} \left[\sqrt{\frac{1}{2}}\delta_{00}+\sqrt{\frac{1}{2}}\delta_{11} \right] \\
=& \frac{\sqrt{2}}{2} \alpha~,
\end{aligned}
\end{equation}
3.
由一维无限深势阱基本结论得：
\begin{equation}
\begin{aligned}
& E^{0}_{n} = \frac{n^{2} \pi^{2} \hbar^{2}}{2ma^{2}},\quad (n = 1,2,3\cdots) \\
& \psi^{0}_{n} = \sqrt{\frac{2}{a}} \sin{\frac{n \pi x}{a}},\quad (n = 1,2,3\cdots)
\end{aligned}~
\end{equation}
能量的一级修正为：
\begin{equation}
\begin{aligned}
E^{(1)}_{n} =& \int^{a}_{\frac{a}{2}} \psi^{(0)*}_{n} (H') \psi^{(0)}_{n} \,\mathrm{d}{x} \\
=& \frac{2}{a} \int^{a}_{\frac{a}{2}} V_{0} \sin^{2}{\frac{n \pi x}{a}} \,\mathrm{d}{x} \\
=& \frac{2V_{0}}{a} \int^{a}_{\frac{a}{2}} \frac{1}{2} (1-\cos{\frac{2n \pi x}{a}}) \,\mathrm{d}{x} \\
=& \frac{V_{0}}{2}~,
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
H'_{mn} =& \int^{a}_{\frac{a}{2}} \psi^{(0)*}_{m}(H') \psi^{(0)}_{n} \,\mathrm{d}{x} \\
=& \frac{2}{a} \int^{a}_{\frac{a}{2}} \sin{\frac{m \pi x}{a}} (V_{0}) \sin{\frac{n \pi x}{a}} \,\mathrm{d}{x} \\
=& \frac{2}{a} \int^{a}_{\frac{a}{2}} V_{0} \left[ \frac{1}{2} (\cos{\frac{(m-n) \pi x}{a}} - \cos{\frac{(m+n) \pi x}{a}}) \right] \,\mathrm{d}{x} \\
=& \frac{V_{0}}{a} \left[ \frac{a}{(m+n)\pi} \sin{\frac{(m+n)\pi}{a}} - \frac{a}{(m-n)\pi} \sin{\frac{(m-n)}{a}} \right]~.
\end{aligned}
\end{equation}
能量的二级修正为：
\begin{equation}
E^{2}_{n} = \sum_{m}' \frac{\left| H'_{mn} \right|^{2}}{E_{n} - E_{m}}~,
\end{equation}
波函数的一级修正为：
\begin{equation}
\psi^{(1)}_{n} = \sum_{m}' \frac{\left| H'_{mn} \right|}{E_{n} - E_{m}} \psi^{(0)}_{m}~,
\end{equation}
因此：
\begin{equation}
\begin{aligned}
& E = E^{(0)}_{n} + E^{(1)}_{n} + E^{(2)}_{n}~, \\
& \psi = \psi^{(0)}_{n} + \psi^{(1)}_{n}~.
\end{aligned}
\end{equation}
4.
三维无限深势阱的定态薛定谔方程为：
\begin{equation}
\left[ -\frac{\hbar^{2}}{2m} \nabla^{2} +V(x,y,z) \right]\psi(x,y,z) = E\psi(x,y,z)~.
\end{equation}
分离变量 $\psi(x,y,z) = X_(x)Y_(y)Z_(z) $，势阱内 $V=0$。
\begin{equation}
\frac{-\frac{\hbar^{2}}{2m}\frac{d^{2} {x}^{2}}{d{x^{2}}}}{X_(x)} + \frac{-\frac{\hbar^{2}}{2m}\frac{d^{2} {y}^{2}}{d{y^{2}}}}{Y_(y)} + \frac{-\frac{\hbar^{2}}{2m}\frac{d^{2} {z}^{2}}{d{z^{2}}}}{Z_(z)} = E~.
\end{equation}
上述第一项只含 $x$，第二项只含 $y$，第三项只含 $z$，$E$ 与 $x,y,z$ 无关。设第一项为 $E_1$ 利用边界条件 $X_{(0)} = 0,X_{(a)} = 0$ 得：
\begin{equation}
X_{n_1}(x) = \sqrt{\frac{2}{a}} \sin{\frac{n_{1} \pi x}{a}},E_{n_1} = \frac{n^2_1 \pi^2 x^2}{2ma^2}~.
\end{equation}
同理可求得 $Y_{n_2},E_{n_2},Z_{n_3},E_{n_3} $，整理得到：
\begin{equation}
\begin{aligned}
& E_{n_1,n_2,n_3} =\frac{n^2_1 \pi^2 x^2}{2ma^2}+\frac{n^2_2 \pi^2 y^2}{2ma^2}+\frac{n^2_3 \pi^2 z^2}{2ma^2} ~,\\
& \psi(x,y,z) = (\frac{2}{a})^{\frac{3}{2}} \sin{\frac{n_1 \pi x}{a}} \sin{\frac{n_2 \pi y}{a}} \sin{\frac{n_3 \pi z}{a}} ~\\
& (0< x,y,z< a \quad , \quad n_1 n_2 n_3 = 1,2,3 \cdots)~.
\end{aligned}
\end{equation}
当 $n_1 = n_2 = n_3$ 时，能级不简并。$n_1 , n_2 , n_3 $ 中有二者相等时，一般为三重简并，$n_1 , n_2 , n_3 $ 三者都不相等时，一般为 6 度简并，还有偶然简并的情况，如 $n_1 , n_2 , n_3 =(5,6,8) $。简并度满足 $\displaystyle n^2_1 , n^2_2 , n^2_3 = \frac{2ma^2 E}{\hbar^2 \pi^2}$ 条件的正整数。
5.
- 总自旋 $S^2$ 是守恒量，理由为：
(1) $S^2$ 不显含时 $t$。
(2) $S^2$ 与各分量的平方对易，进而与各乘积对易，故：
\begin{equation}
[S^2 ,H^2] = [S^2 , a \boldsymbol{\mathbf{S_1}} \boldsymbol{\mathbf{S_2}} + a \boldsymbol{\mathbf{S_2}} \boldsymbol{\mathbf{S_3}} + a \boldsymbol{\mathbf{S_3}} \boldsymbol{\mathbf{S_1}} ]~,
\end{equation}
- $\displaystyle \boldsymbol{\mathbf{S_1}} \cdot \boldsymbol{\mathbf{S_2}} = \frac{1}{2} ( \boldsymbol{\mathbf{S^2}} - \boldsymbol{\mathbf{S^2_1}} - \boldsymbol{\mathbf{S^2_2}} ) = \frac{1}{2} \boldsymbol{\mathbf{S^2}} - \frac{3}{4} \hbar^2 $
本征函数为：
\begin{equation}
\begin{aligned}
& \chi_{11} = \chi_{\frac{1}{2}}(S_{1z})\chi_{\frac{1}{2}}(S_{2z})~, \\
& \chi_{1-1} = \chi_{-\frac{1}{2}}(S_{1z})\chi_{-\frac{1}{2}}(S_{2z}) ~,\\
& \chi_{10} = \frac{1}{\sqrt{2}} \left[ \chi_{\frac{1}{2}}(S_{1z})\chi_{-\frac{1}{2}}(S_{1z}) + \chi_{\frac{1}{2}}(S_{2z})\chi_{-\frac{1}{2}}(S_{1z}) \right]~, \\
& \chi_{00} = \frac{1}{\sqrt{2}} \left[ \chi_{\frac{1}{2}}(S_{1z})\chi_{-\frac{1}{2}}(S_{1z}) - \chi_{\frac{1}{2}}(S_{2z})\chi_{-\frac{1}{2}}(S_{1z}) \right] ~. \\
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
& \hat{H}\chi_{11} = g(\frac{1}{2}) \cdot 2\hbar - \frac{3}{4} \chi_{11} = \frac{1}{4}g\hbar^{2} \chi_{11}~, \\
& \hat{H}\chi_{1-1} = \frac{1}{4}g\hbar \chi_{1-1} ~, \\
& \hat{H}\chi_{10} = \frac{1}{4}g\hbar \chi_{10} ~, \\
& \hat{H}\chi_{00} = -\frac{3}{4}g\hbar^{2} \chi_{00} ~.\\
\end{aligned}
\end{equation}