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.

  1. 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}
    同理可得:
    \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}
  2. 答:
  3. 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}
    故 $B=\sin{\partial},A=\cos{\partial}$.

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 $A=\frac{1}{\sqrt{1+x^2}}$.

  1. $t$ 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}
  2. 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}
  3. 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}
    系统是定态,力学量平均值不随时间变化.

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.

  1. 由题可得 $\hat{H} = g \vec{S}_{1} \boldsymbol\cdot \vec{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}
    能级为:
    \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}
    $S=1$ 时为三重简并,$S=0$ 时为非简并.
  2. 设磁场 $B$ 沿 $x$ 方向,$\vec{\mu} = g \vec{S}_{x} $.在 $S_{z}$ 表象中:$S_{x} = \frac{\hbar}{2} \begin{bmatrix}0&1\\1&0\end{bmatrix} $.
    所以:
    \begin{equation} \hat{H} = -\vec{\mu}\vec{B} = -\frac{gB\hbar}{2} \begin{bmatrix}0&1\\1&0\end{bmatrix} \end{equation}
    设粒子自旋波函数为:$\psi_{n} = \begin{pmatrix}c_{1}\\c_{2}\end{pmatrix} $,又因为本征方程:$\hat{H}\Psi = E\Psi $,所以可得:
    \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}
    解本征方程得:
    \begin{align} E = \frac{gB\hbar}{2} \quad \psi_{1} = \frac{1}{\sqrt{2}} \begin{bmatrix}1\\-1\end{bmatrix} \\ E = -\frac{gB\hbar}{2} \quad \psi_{1} = \frac{1}{\sqrt{2}} \begin{bmatrix}1\\1\end{bmatrix} \end{align}
    能级简并完全消除.

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