天津大学 2014 年考研量子力学答案
贡献者： Entanglement
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1.
- 对于 $\psi(r,\theta,\varphi) = \frac{1}{\sqrt{5}}\psi_{310} + \frac{2}{\sqrt{5}}\psi_{211} $，主量子数 $n$ 可能等于 $3,2$.
表1：$\hat{L}_{z},\hat{L}^{2}$ 的可能值与几率
|
$\hat{L}_z$ 的可能值 | 0 | $\hbar$
|
|
$\hat{L}^2$ 的可能值 | $2\hbar^{2}$ | $2\hbar^{2}$
|
|
相应几率 | $\frac{1}{5}$ | $\frac{4}{5}$
|
\begin{align}
& \overline{\hat{L}_{z}} = 0 \times \frac{1}{5} + \hbar \times \frac{4}{5} = \frac{4\hbar}{5} ~,\\
& \overline{\hat{L}^{2}} = 2\hbar^{2} \times \frac{1}{5} + 2\hbar^{2} \times \frac{4}{5} = 2\hbar^{2}~.
\end{align}
- (1) 光的波动性：光的干涉现象，光的衍射现象。光的粒子性：光电效应，康普顿效应。
(2)戴维孙—革末实验，即电子衍射实验除了证实电子具有粒子性之外也具有波动性。
- 设均匀磁场方向沿 $x$ 方向则：
\begin{equation}
\begin{aligned}
\hat{H}=& - \boldsymbol{\mathbf{\mu}} \boldsymbol\cdot \boldsymbol{\mathbf{B}} \\
=& g_{n} \boldsymbol{\mathbf{S}} _{x} \boldsymbol\cdot \boldsymbol{\mathbf{B}} \\
=& gB\hat{S}_{x}~.
\end{aligned}
\end{equation}
\begin{equation}
\hat{H}=\frac{gB\hbar}{2} \begin{bmatrix}0&1\\1&0\end{bmatrix} ~.
\end{equation}
\begin{equation}
\frac{gB\hbar}{2} \begin{bmatrix}0&1\\1&0\end{bmatrix} \begin{bmatrix}a\\b\end{bmatrix} =E \begin{bmatrix}a\\b\end{bmatrix} ~,
\end{equation}
\begin{equation}
\begin{vmatrix}-E&\frac{gB\hbar}{2}\\\frac{gB\hbar}{2}&-E\end{vmatrix} =0~.
\end{equation}
2.
- 对于二维谐振子，其势能为：
\begin{equation}
V(x,y)=\frac{1}{2}m\Omega^{2}(x^{2}+y^{2})~.
\end{equation}
\begin{equation}
\hat{H}=-\frac{\hbar}{2m}( \frac{\mathrm{d}^{2}}{\mathrm{d}{x}^{2}} + \frac{\mathrm{d}^{2}}{\mathrm{d}{y}^{2}} )+\frac{1}{2}m\Omega^{2}(x^{2}+y^{2})~.
\end{equation}
\begin{equation}
\left[-\frac{\hbar^{2}}{2m}( \frac{\mathrm{d}^{2}}{\mathrm{d}{x}^{2}} + \frac{\mathrm{d}^{2}}{\mathrm{d}{y}^{2}} )+\frac{1}{2}m\Omega^{2}(x^{2}+y^{2}) \right] \psi(x,y)=E\psi(x,y)~.
\end{equation}
\begin{equation}
\frac{ \left[-\frac{\hbar^{2}}{2m} \frac{\mathrm{d}^{2}}{\mathrm{d}{x}^{2}} +\frac{1}{2}m\Omega x^{2} \right] \phi(x)}{\phi(x)}+\frac{ \left[-\frac{\hbar^{2}}{2m} \frac{\mathrm{d}^{2}}{\mathrm{d}{y}^{2}} +\frac{1}{2}m\Omega y^{2} \right] \phi(y)}{\phi(y)}=E~.
\end{equation}
\begin{align}
& \left[-\frac{\hbar^{2}}{2m} \frac{\mathrm{d}^{2}}{\mathrm{d}{x}^{2}} +\frac{1}{2}m\Omega^{2}x^{2} \right] \phi{x}=E_{x}\phi(x)~,\\
& \left[-\frac{\hbar^{2}}{2m} \frac{\mathrm{d}^{2}}{\mathrm{d}{y}^{2}} +\frac{1}{2}m\Omega^{2}y^{2} \right] \phi{y}=E_{y}\phi(y)~.
\end{align}
\begin{align}
&\phi(x)=N_{n_{x}}e^{-\frac{\alpha^{2}x^{2}}{2}}H_{n_{x}}(\alpha x),\quad E_{x}=(n_{x}+\frac{1}{2})\hbar \Omega,\quad n_{x}=0,1,2\cdots~ \\
&\phi(y)=N_{n_{y}}e^{-\frac{\alpha^{2}y^{2}}{2}}H_{n_{y}}(\alpha y),\quad E_{y}=(n_{y}+\frac{1}{2})\hbar \Omega,\quad n_{y}=0,1,2\cdots~
\end{align}
\begin{equation}
\psi(x,y)=\phi(x)\phi(y)=N_{n_{x}}N_{n_{y}}e^{-\frac{\alpha^{2}(x^{2}+y^{2})}{2}}H_{n_{x}}(\alpha x)H_{n_{y}}(\alpha y)~,
\end{equation}
\begin{equation}
E=(n_{x}+n_{y}+1)\hbar \Omega=(N+1)\hbar \Omega,\quad N=0,1,2,\cdots~
\end{equation}
- 当处于基态时，非简并。当处于第 $n$ 激发态时，其简并度为 $n+1$。
- 因为
\begin{align}
&H_{n_x}(-\alpha x)=(-1)^{n_x}H_{n_x}(\alpha x) ~,\\
&H_{n_y}(-\alpha y)=(-1)^{n_y}H_{n_y}(\alpha y)~,
\end{align}
\begin{equation}
H_{n_x}(-\alpha x)H_{n_y}(-\alpha y)=(-1)^{n_x + n_y}H_{n_x}(\alpha x)H_{n_y}(\alpha y)~.
\end{equation}
3.
- $\psi=\sqrt{\frac{1}{3}}\psi_{0}(x)+\sqrt{\frac{2}{3}}\psi_{1}(x)$
对于一维谐振子，其本征能量为：
\begin{equation}
E_{n}=(n+\frac{1}{2})\hbar \omega,\quad n=0,1,2,\cdots~
\end{equation}
\begin{equation}
\psi(x,t)=\sqrt{\frac{1}{3}}\psi_{0}(x)e^{-\frac{i}{\hbar}E_{0}t}+\sqrt{\frac{2}{3}}\psi_{1}(x)e^{-\frac{i}{\hbar}E_{1}t}~.
\end{equation}
\begin{equation}
\begin{aligned}
\bar{x}&=\int^{+\infty}_{-\infty}\psi^{*}(x,t)x\psi(x,t) \,\mathrm{d}{x} \\
&=\int^{+\infty}_{-\infty} \left[\sqrt{\frac{1}{3}}\psi^{*}_{0}(x)e^{\frac{i}{\hbar}E_{0}t}+\sqrt{\frac{2}{3}}\psi^{*}_{1}(x)e^{\frac{i}{\hbar}E_{1}t} \right] x\\
&\qquad \left[\sqrt{\frac{1}{3}}\psi_{0}(x)e^{-\frac{i}{\hbar}E_{0}t}+\sqrt{\frac{2}{3}}\psi_{1}(x)e^{-\frac{i}{\hbar}E_{1}t} \right] \,\mathrm{d}{x} \\
&=\int^{+\infty}_{-\infty} \left[\sqrt{\frac{1}{3}}\psi^{*}_{0}(x)e^{\frac{i}{\hbar}E_{0}t}+\sqrt{\frac{2}{3}}\psi^{*}_{1}(x)e^{\frac{i}{\hbar}E_{1}t} \right] \\
&\qquad \left\{\sqrt{\frac{1}{3}}\frac{1}{\alpha} \left[\sqrt{\frac{1}{2}}\psi_{1}(x) \right] e^{-\frac{i}{\hbar}E_{0}t}+\sqrt{\frac{2}{3}}\frac{1}{\alpha} \left[\sqrt{\frac{1+1}{2}}\psi_{2}(x)+\sqrt{\frac{1}{2}}\psi_{0}(x) \right] e^{-\frac{i}{\hbar}E_{1}t} \right\} \,\mathrm{d}{x} \\
&=\frac{1}{3\alpha}(e^{-i\omega t}+e^{i\omega t})~.
\end{aligned}
\end{equation}
- 能量可能的取值和概率为：
表2：能量与概率
|
能量 | $\frac{1}{2}\hbar\omega$ | $\frac{3}{2}\hbar\omega$
|
|
概率 | $\frac{1}{3}$ | $\frac{2}{3}$
|
\begin{equation}
\bar{E}=\frac{1}{3}\times \frac{1}{2}\hbar\omega + \frac{2}{3}\times \frac{3}{2}\hbar\omega=\frac{7}{6}\hbar\omega~.
\end{equation}
4.
- 微扰项为：
\begin{equation}
H'= \left\{\begin{aligned} &V_{0}\quad (\frac{a}{2} < x< a)\\
&0\quad \text{其他}~.
\end{aligned}\right.
\end{equation}
- 不考虑微扰时，体系的能级和波函数非别为：
\begin{equation}
E^{(0)}_{n}=\frac{n^2 \pi^2 \hbar^2}{2Ma^2}~,
\end{equation}
\begin{equation}
\psi^{(0)}_{n}= \left\{\begin{aligned}
& \sqrt{\frac{2}{a}}\sin{\frac{n\pi x}{a}}\quad 0< x< a\\
& 0 \quad \text{其他}~.\\
\end{aligned}\right.
\end{equation}
\begin{equation}
\begin{aligned}
E^{(1)}_{n}=H'_{nn}=&\int \psi^{(0)*}_{n}H'\psi^{0}_{n} \,\mathrm{d}{x} \\
=&\int^{a}_{\frac{a}{2}} \sqrt{\frac{2}{a}}\sin{\frac{n\pi x}{a}}V_{0}\sqrt{\frac{2}{a}}\sin{\frac{n\pi x}{a}} \,\mathrm{d}{x} \\
=&\frac{2V_{0}}{a}\int^{a}_{\frac{a}{2}} \sin^{2}\left(\frac{n\pi x}{a}\right) \,\mathrm{d}{x} \\
=&\frac{V_{0}}{a}\int^{a}_{\frac{a}{2}} 1- \cos^{2}\left(\frac{2n\pi x}{a}\right) \,\mathrm{d}{x} \\
=&\frac{V_{0}}{2}~.
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
H'_{mn}=&\int \psi^{(0)*}_{m}H'\psi^{(0)}_{n} \,\mathrm{d}{x} \\
=&\int^{a}_{\frac{a}{2}} \sqrt{\frac{2}{a}} \sin{\frac{m\pi x}{a}} V_{0} \sqrt{\frac{2}{a}} \sin{\frac{n\pi x}{a}} \,\mathrm{d}{x} \\
=&\frac{2V_{0}}{a}\int^{a}_{\frac{a}{2}} \sin{\frac{m\pi x}{a}} \sin{\frac{n\pi x}{a}} \,\mathrm{d}{x} \\
=&\frac{2V_{0}}{a}(-\frac{1}{2})\int^{a}_{\frac{a}{2}} \cos{\frac{(m+n)\pi x}{a}}-\cos{\frac{(m-n)\pi x}{a}} \,\mathrm{d}{x} ~.
\end{aligned}
\end{equation}