天津大学 2015 年考研量子力学答案
贡献者： Entanglement
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1.
由归一化条件 $\displaystyle (\sqrt{\frac{1}{3}}A)^{2} + (\sqrt{\frac{2}{3}}A)^{2} = 1 $ 可得到 $A=1$。故归一化函数为：
$\displaystyle \psi(x) = \sqrt{\frac{1}{3}}\phi_{210}(x)+\sqrt{\frac{2}{3}} \phi_{310}(x)$
2.
- 由题可得：
\begin{equation}
\begin{aligned}
\left[L^{2},L_{x}\right] =& [ L^{2}_{x} + L^{2}_{y} + L^{2}_{z} , L_{x}] \\
=& 0 + [L^{2}_{y},L_{x}] + [L^{2}_{z},L_{x}] \\
=& 0 - L_y[L_y,L_x]+[L_y,L_x]L_y \\
=& 0 - i\hbar L_y L_z - i\hbar L_z L_y + i\hbar L_z L_y + i\hbar L_y L_z \\
=& 0~.
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
\left[\hat{L}_+,\hat{L}_z\right]Y_{lm}(\theta ,\phi) =& \left[\hat{L}_{x}+i\hat{L}_{y} ,\hat{L}_z \right]Y_{lm}(\theta ,\phi) \\
=& \left[\hat{L}_x ,\hat{L}_z \right]Y_{lm}(\theta ,\phi) + i\left[\hat{L}_x ,\hat{L}_z \right]Y_{lm}(\theta ,\phi) \\
=& -\hbar \hat{L}_{+}Y_{lm}(\theta ,\phi)~,
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
\left[\hat{L}_{-},\hat{L}_{z}\right]Y_{lm}(\theta ,\phi) =& \left[\hat{L}_{x}-i\hat{L}_{y} ,\hat{L}_z \right]Y_{lm}(\theta ,\phi) \\
=& \left[\hat{L}_x ,\hat{L}_z \right]Y_{lm}(\theta ,\phi) - i\left[\hat{L}_x ,\hat{L}_z \right]Y_{lm}(\theta ,\phi) \\
=& \hbar \hat{L}_{-}Y_{lm}(\theta ,\phi)~,
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
\left[\hat{L}_{+},\hat{L}_{-}\right]Y_{lm}(\theta ,\phi) =& \left[\hat{L}_{x} + i\hat{L}_{y} ,\hat{L}_{x}-i\hat{L}_{y} \right]Y_{lm}(\theta ,\phi) \\
=& 2\hbar \hat{L}_{z}Y_{lm}(\theta ,\phi)~.
\end{aligned}
\end{equation}
- 答：
- 由题可得：
\begin{equation}
\begin{aligned}
e^{i \rho_{y} \partial} =& \sum_{n=0}^{\infty} \frac{(i \rho_y \partial)^{n}}{n!} \\
=& \sum (i\rho_y \partial - \frac{i\rho_y \partial^3}{3!} + \frac{i\rho_y \partial^5}{5!} + \dots) + \sum (1 - \frac{\partial^2}{2!} + \frac{\partial^4}{4!} + \dots) \\
=& i\rho_y \sin{\partial} + \cos{\partial} ~.
\end{aligned}
\end{equation}
3.
由 $E_n = (n+\frac{1}{2})\hbar \omega$ 可得 $E_0 = \frac{1}{2}\hbar \omega ,E_1 = \frac{3}{2}\hbar \omega$ 将波函数归一化可得 $A=\frac{1}{\sqrt{1+x^2}}$。
- $t$ 时刻的波函数为：
\begin{equation}
\psi(x,t)=A\psi_{0}(x)e^{-\frac{iE_{0}t}{\hbar}} + Ax\psi_{1}(x)e^{-\frac{iE_{1}t}{\hbar}}~.
\end{equation}
- 坐标的平均值为：
\begin{equation}
\begin{aligned}
\bar{x} =& \left\langle \psi \middle| x \middle| \psi \right\rangle \\
=& \frac{A^{2}}{\alpha} \left\langle \psi_{0} + x\psi_{1} \middle| \sqrt{\frac{1}{2}}\psi_{1} +\sqrt{\frac{1}{2}}x\psi_{0} + \psi_{2} \right\rangle \\
=& \frac{A^{2}}{\alpha} (\sqrt{\frac{1}{2}}x + \sqrt{\frac{1}{2}}x) \\
=& \frac{\sqrt{2}A^{2}x}{\alpha}~.
\end{aligned}
\end{equation}
- 能量的平均值为：
\begin{equation}
\begin{aligned}
\bar{E} =& \frac{1}{1+x^2} \boldsymbol\cdot \frac{1}{2}\hbar \omega + \frac{x}{1+x^2} \boldsymbol\cdot \hbar \omega \\
=& \frac{\hbar \omega (3x+1)}{2(x+1)}~.
\end{aligned}
\end{equation}
4.
根据题意有：
\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}
将 $H' = -q \epsilon x$ 是为微扰。因此能量的一级修正为：
\begin{equation}
\begin{aligned}
E^{(1)}_{n} =& \int \psi_{n}^{(0)*} (-q\epsilon) \psi_{n}^{(0)} \,\mathrm{d}{x} \\
=& -\frac{2q\epsilon}{a} \int^{a}_{0} x \sin^{2}\left(\frac{n \pi x}{a}\right) \,\mathrm{d}{x} \\
=& -\frac{q\epsilon}{a} \boldsymbol\cdot \frac{a^{2}}{2} \\
=& -\frac{q \epsilon a}{2}~.
\end{aligned}
\end{equation}
又因为：
\begin{equation}
\begin{aligned}
H'_{mn} =& \int \psi_{m}^{(0)*} (-q\epsilon) \psi_{n}^{(0)} \,\mathrm{d}{x} \\
=& -\frac{2q\epsilon}{a} \int^{a}_{0} x \sin{\frac{m \pi x}{a}} \sin{\frac{n \pi x}{a}} \,\mathrm{d}{x} \\
=& -\frac{2q\epsilon}{a} \int^{a}_{0} x \frac{1}{2} \left[\cos{\frac{(m-n)\pi x}{a}} - \cos{\frac{(m+n)\pi x}{a}} \right] \,\mathrm{d}{x} \\
=& -\frac{q\epsilon}{\pi} \left[\frac{1}{m-n}\sin{\frac{(m-n)\pi x}{a}} \Big|^{a}_{0} - \frac{1}{m+n}\sin{\frac{(m+n)\pi x}{a}} \Big|^{a}_{0} \right] \\
=& -\frac{q\epsilon}{\pi} \left[\frac{1}{m-n}\sin{(m-n)\pi x} - \frac{1}{m+n}\sin{(m+n)\pi x} \right]~,
\end{aligned}
\end{equation}
所以能量的二级修正为：
\begin{equation}
E^{(2)}_{n} = \sum_{m}' \frac{\left|H'_{mn} \right|^{2}}{E^{(0)}_{m} - E^{(0)}_{n}}~.
\end{equation}
波函数的一级修正为：
\begin{equation}
\psi^{(1)}_{n} = \sum_{m}'\frac{H'_{mn}}{E^{(0)}_{n} - E^{(0)}_{m}} \psi^{(0)}_{m}~.
\end{equation}
5.
- 由题可得 $\hat{H} = g \boldsymbol{\mathbf{S}} _{1} \boldsymbol\cdot \boldsymbol{\mathbf{S}} _{2} = \frac{1}{2}g(S^{2}-S^{2}_{1}-S^{2}_{2}) = \frac{1}{2}gs(s+1)\hbar^{2}-\frac{3}{2}\hbar^2 $。
自旋波函数为：
\begin{align}
& \chi^{s}_{11} = \chi_{\frac{1}{2}}(s_{1z})\chi_{\frac{1}{2}}(s_{2z})~, \\
& \chi^{s}_{1-1} = \chi_{\frac{1}{2}}(s_{1z})\chi_{-\frac{1}{2}}(s_{2z})~, \\
&\chi^{s}_{10} = \frac{1}{\sqrt{2}} \left[ \chi_{\frac{1}{2}}(s_{1z})\chi_{-\frac{1}{2}}(s_{2z}) + \chi_{-\frac{1}{2}}(s_{1z})\chi_{\frac{1}{2}}(s_{2z}) \right]~, \\
&\chi^{A}_{00} = \frac{1}{\sqrt{2}} \left[ \chi_{\frac{1}{2}}(s_{1z})\chi_{-\frac{1}{2}}(s_{2z}) - \chi_{-\frac{1}{2}}(s_{1z})\chi_{\frac{1}{2}}(s_{2z}) \right]~.
\end{align}
\begin{align}
&E_{11} =\frac{g\hbar^{2}}{2}~, \\
&E_{1-1} =\frac{g\hbar^{2}}{2} ~, \\
&E_{10} =\frac{g\hbar^{2}}{2} ~,\\
&E_{00} =-\frac{3g\hbar^{2}}{2} ~.
\end{align}
- 设磁场 $B$ 沿 $x$ 方向，$ \boldsymbol{\mathbf{\mu}} = g \boldsymbol{\mathbf{S}} _{x} $.在 $S_{z}$ 表象中：$S_{x} = \frac{\hbar}{2} \begin{bmatrix}0&1\\1&0\end{bmatrix} $。
所以：
\begin{equation}
\hat{H} = - \boldsymbol{\mathbf{\mu}} \boldsymbol{\mathbf{B}} = -\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}c_{1}\\c_{2}\end{bmatrix} = E \begin{bmatrix}c_{1}\\c_{2}\end{bmatrix} ~.
\end{equation}
\begin{align}
E = \frac{gB\hbar}{2} ~,\qquad \psi_{1} = \frac{1}{\sqrt{2}} \begin{bmatrix}1\\-1\end{bmatrix} ~,\\
E = -\frac{gB\hbar}{2}~, \qquad \psi_{1} = \frac{1}{\sqrt{2}} \begin{bmatrix}1\\1\end{bmatrix} ~.
\end{align}