贡献者： Entanglement
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1.
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(1)由题可知势能 $\displaystyle \overline{U}=-\frac{e^2}{r}$。
\begin{equation}
\begin{aligned}
\overline{U}=&\iiint \psi^{*} \overline{U} \psi \,\mathrm{d}{\tau} \\
=&-\frac{e^2}{\pi a^{3}_{0}}\int^{\pi}_{0}\int^{2\pi}_{0}\int^{\infty}_{0} \frac{1}{r}e^{-\frac{2r}{a_0}}r^{2}\sin{\theta} \,\mathrm{d}{r} \,\mathrm{d}{\theta} \,\mathrm{d}{\varphi} \\
=&-\frac{e^{2}}{\pi a^{3}_{0}}\int^{\pi}_{0}\int^{2\pi}_{0}\int^{\infty}_{0} e^{-\frac{2r}{a_0}}r\sin{\theta} \,\mathrm{d}{r} \,\mathrm{d}{\theta} \,\mathrm{d}{\varphi} \\
=&-\frac{4 e^2}{a^{3}_{0}}\int^{\infty}_{0}e^{-\frac{2r}{a_0}}r \,\mathrm{d}{r} \\
=&-\frac{4e^{2}}{a^{3}_{0}}(\frac{a_{0}}{2})^2\\
=&-\frac{e^{2}}{a_{0}}~.
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
w(r) \,\mathrm{d}{r} =&\int^{\pi}_{0}\int^{2\pi}_{0} \lvert \psi(r,\theta,\varphi) \rvert \sin{\theta} \,\mathrm{d}{\theta} \,\mathrm{d}{\varphi} \,\mathrm{d}{r} \\
=&\frac{4}{a^{3}}e^{-\frac{2r}{a_0}}r^2 \,\mathrm{d}{r} ~,\\
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
\frac{ \,\mathrm{d}{w(r)} }{ \,\mathrm{d}{r} }=\frac{4}{a^{3}}e^{-\frac{2r}{a_0}}r^2 ~.
\end{aligned}
\end{equation}
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答：康普顿散射是光子与电子做弹性碰撞，在 $X$ 射线通过实物物质发生散射的实验时，除原波长的光外还产生了大于原波长的 $X$ 光，借助光电理论，才可以得到这是由于光子与电子发生碰撞后频率变小的缘故，从而证实了光具有粒子性。
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(1)三维转子的能级为：$\displaystyle E = \frac{l(l+1)}{2I}$，简并度为：$2l+1$。
(2)平面转子设沿 $z$ 轴方向，$\displaystyle \hat{H} = \frac{\hat{l}^{2}_{z}}{2I}$,能级为：$\displaystyle E = \frac{m^{2} \hbar^{2}}{2I} ,m = 0 , \pm 1 , \pm 2$，除了 $m = 0$ 外，能级都是二重简并。
2.
- 对于 $ \boldsymbol{\mathbf{r}} \times \boldsymbol{\mathbf{L}} + \boldsymbol{\mathbf{L}} \times \boldsymbol{\mathbf{r}} $ 有：
\begin{equation}
\begin{aligned}
( \boldsymbol{\mathbf{r}} \times \boldsymbol{\mathbf{L}} + \boldsymbol{\mathbf{L}} \times \boldsymbol{\mathbf{r}} )_{x} =& yL_{z}-zL_{y}+L_{y}z-L_{z}y \\
=& [y,xp_{y}-yp_{x}]+[zp_{x}-xp_{z},z] \\
=& i\hbar x + i\hbar x \\
=&2i\hbar x~.
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
& ( \boldsymbol{\mathbf{r}} \times \boldsymbol{\mathbf{L}} + \boldsymbol{\mathbf{L}} \times \boldsymbol{\mathbf{r}} )_{y} =2i\hbar \boldsymbol{\mathbf{p}} _{y}~, \\
& ( \boldsymbol{\mathbf{r}} \times \boldsymbol{\mathbf{L}} + \boldsymbol{\mathbf{L}} \times \boldsymbol{\mathbf{r}} )_{z} =2i\hbar \boldsymbol{\mathbf{p}} _{z}~.
\end{aligned}
\end{equation}
- 对于 $ \boldsymbol{\mathbf{p}} \times \boldsymbol{\mathbf{L}} + \boldsymbol{\mathbf{L}} \times \boldsymbol{\mathbf{p}} $ 有：
\begin{equation}
\begin{aligned}
( \boldsymbol{\mathbf{p}} \times \boldsymbol{\mathbf{L}} + \boldsymbol{\mathbf{L}} \times \boldsymbol{\mathbf{p}} )_{y} =& \boldsymbol{\mathbf{p}} _{y} \boldsymbol{\mathbf{L}} _{z} - \boldsymbol{\mathbf{p}} _{z} \boldsymbol{\mathbf{L}} _{y} + \boldsymbol{\mathbf{L}} _{y} \boldsymbol{\mathbf{p}} _{z} - \boldsymbol{\mathbf{L}} _{z} \boldsymbol{\mathbf{p}} _{y} \\
=& [ \boldsymbol{\mathbf{p}} _{y}, \boldsymbol{\mathbf{L}} _{z}] + [ \boldsymbol{\mathbf{L}} _{y}, \boldsymbol{\mathbf{p}} _{z}] \\
=& [ \boldsymbol{\mathbf{p}} _{y},x \boldsymbol{\mathbf{p}} _{y} - y \boldsymbol{\mathbf{p}} _{x}] + [z \boldsymbol{\mathbf{p}} _{x}-x \boldsymbol{\mathbf{p}} _{z}, \boldsymbol{\mathbf{p}} _{z}] \\
=& -[ \boldsymbol{\mathbf{p}} _{y},y] \boldsymbol{\mathbf{p}} _{x} + [z, \boldsymbol{\mathbf{p}} _{z}] \boldsymbol{\mathbf{p}} _{x} \\
=& 2i\hbar \boldsymbol{\mathbf{p}} _{x}~.
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
& ( \boldsymbol{\mathbf{p}} \times \boldsymbol{\mathbf{L}} + \boldsymbol{\mathbf{L}} \times \boldsymbol{\mathbf{p}} )_{y} =2i\hbar \boldsymbol{\mathbf{p}} _{y} ~,\\
& ( \boldsymbol{\mathbf{p}} \times \boldsymbol{\mathbf{L}} + \boldsymbol{\mathbf{L}} \times \boldsymbol{\mathbf{p}} )_{z} =2i\hbar \boldsymbol{\mathbf{p}} _{z}~.
\end{aligned}
\end{equation}
3.
在一维无限深势阱 $V_{x} = \left\{\begin{aligned}
& 0 \qquad 0\leqslant x \leqslant a \\
& \infty \qquad x<0,x>a \\
\end{aligned}\right. $ 的基础上，把 $H' = x \quad \frac{a}{2} < x < a $ 看作微扰，一维无限深势阱的本征函数和能量为：
\begin{equation}
\begin{aligned}
\psi^{0}_{n}=& \sqrt{\frac{2}{a}} \sin{\frac{n\pi x}{a}}~,\\
E^{0}_{n}=& \frac{n^{2}\pi{2}\hbar^{2}}{2Ma^2}~.
\end{aligned}
\end{equation}
能量的一级修正为：
\begin{equation}
\begin{aligned}
E^{(1)}_{n}=&\int^{a}_{\frac{a}{2}} \sqrt{\frac{2}{a}} \sin{\frac{n\pi x}{a}}(x)\sqrt{\frac{2}{a}} \sin{\frac{n\pi x}{a}} \,\mathrm{d}{x} \\
=& \frac{2}{a} \int^{a}_{\frac{a}{2}} x\sin^{2}{\frac{n\pi x}{a}} \,\mathrm{d}{x} \\
=& \frac{2}{a}\int^{a}_{\frac{a}{2}} x\Bigl[\frac{1-\cos{\frac{2n \pi}{a}}}{2} \,\mathrm{d}{x} \Bigr] \\
=& \frac{3a}{8}~,
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
H'_{mn} =& \int^{a}_{\frac{a}{2}} \sqrt{\frac{2}{a}} \sin{\frac{m \pi x}{a}} (x) \sqrt{\frac{2}{a} }\sin{\frac{n \pi x}{a}} \,\mathrm{d}{x} \\
=& \frac{2}{a} \int^{a}_{\frac{a}{2}} x\Bigl[\frac{1}{2} \cos{\frac{(m-n)\pi x}{a}} - \frac{1}{2} \cos{\frac{(m+n) \pi x}{a}}\Bigr] \,\mathrm{d}{x} \\
=& \frac{1}{a} \Bigl[\frac{a}{(m-n)\pi} \sin{\frac{(m-n)\pi x}{a}}\Big|^{a}_{\frac{a}{2}} - \frac{a}{(m+n)\pi} \sin{\frac{(m+n)\pi x}{a}}\Big|^{a}_{\frac{a}{2}}\Bigr] \\
=& \frac{1}{(m-n)\pi} \Big[\sin{(m-n)\pi} - \sin{\frac{(m-n)\pi}{2}}\Big]\\
&+\frac{1}{(m+n)\pi} \Big[\sin{(m+n)\pi} -\sin{\frac{(m+n)\pi}{2}} \Big]~.
\end{aligned}
\end{equation}
当 $m-n$ 为偶数时，$H'_{mn}=0$
当 $m-n$ 为奇数时，$\displaystyle H'_{mn} = -\frac{1}{(m-n)\pi}(-1)^{\frac{m-n-1}{2}}+\frac{1}{(m+n)\pi}(-1)^{\frac{m+n-1}{2}} $
能量的二级修正为：$\displaystyle E^{(2)}_{n}= \sum_{m}' \frac{\lvert H'_{mn}\rvert ^{2}}{E^0_{n}-E^0_{m}} $
波函数的一级修正为：$\displaystyle \psi^{(1)}_{n}= \sum_{m}' \frac{H'_{mn}}{E^0_{n}-E^0_{m}} \psi^{0}_{m} $
4.
设整体在 $x$ 轴上可自由运动，谐振子的哈密顿量为：$\displaystyle \hat{H} = -\frac{\hbar^{2}}{2m} \nabla^{2}+\frac{1}{2}m\omega^{2}x^{2} $
薛定谔方程为：$\hat{H}\psi_{x} = E\psi_{x}$
\begin{equation}
\begin{aligned}
& E=(n+\frac{1}{2} \hbar \omega)~, \\
& \psi = Ne^{-\frac{\alpha^{2}x^{2}}{2}}H_{n}(\alpha x)~.
\end{aligned}
\end{equation}
5.
- 由题意可得：
\begin{equation}
\begin{aligned}
H =& aS_{1x}S_{2x}+aS_{1y}S_{2y}+bS_{1z}S_{2z}+aS_{1z}S_{2z}-aS_{1y}S_{2y} \\
=& a(S_{1x}S_{2x}+S_{1y}S_{2y}+S_{1z}S_{2z})+(b-a)S_{z} \\
=& aS_{1}S_{2}+(b-a)S_z \\
=& \frac{a}{2}(s^{2}-\frac{3}{2}\hbar)+(b-a)m\hbar~.
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
\because \quad & s_{1} = \frac{1}{2}~,s_{2}=\frac{1}{2}~, \therefore \quad s=1,0 \\
\therefore \quad & E_{sm} = \frac{a}{2}[s(s+1)\hbar^{2}-\frac{3}{2}\hbar^{2}]+(b-a)m\hbar ~, \\
\therefore \quad & E_{11} = \frac{1}{4}a\hbar^{2}+(b-a)\hbar ~, \\
& E_{1-1} = \frac{1}{4}a\hbar^{2}-(b-a)\hbar ~, \\
& E_{10} = \frac{1}{4}a\hbar^{2} ~, \\
& E_{00} = -\frac{3}{4}a\hbar^{2} ~.\\
\therefore \quad& \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}}[\chi_{\frac{1}{2}}(s_{1z})\chi_{-\frac{1}{2}}(s_{2z})+\chi_{\frac{1}{2}}(s_{2z})\chi_{-\frac{1}{2}}(s_{1z})] ~,\\
& \chi_{00} = \frac{1}{\sqrt{2}}[\chi_{\frac{1}{2}}(s_{1z})\chi_{-\frac{1}{2}}(s_{2z})-\chi_{\frac{1}{2}}(s_{2z})\chi_{-\frac{1}{2}}(s_{1z})] ~.
\end{aligned}
\end{equation}
- 此时：
\begin{equation}
\begin{aligned}
& \left\lvert 1,0 \right\rangle =\chi_{s}=\frac{1}{\sqrt{2}}[\chi_{\frac{1}{2}}(s_{1z})\chi_{-\frac{1}{2}}(s_{2z})+\chi_{\frac{1}{2}}(s_{2z})\chi_{-\frac{1}{2}}(s_{1z})] ~, \\
& \left\lvert 0,0 \right\rangle =\chi_{A}=\frac{1}{\sqrt{2}}[\chi_{\frac{1}{2}}(s_{1z})\chi_{-\frac{1}{2}}(s_{2z})-\chi_{\frac{1}{2}}(s_{2z})\chi_{-\frac{1}{2}}(s_{1z})] ~.
\end{aligned}
\end{equation}
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