天津大学 2017 年考研量子力学答案

                     

贡献者: Entanglement

  • 本词条处于草稿阶段.

1.

  1. (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}
    (2)电子 $r+dr$ 在球壳内出现的几率为:
    \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}
    令 $\displaystyle \frac{ \,\mathrm{d}{w(r)} }{ \,\mathrm{d}{r} }=0,\Longrightarrow r_1 = 0,r_2 = \infty,r_3 = a_0$,因为 $\displaystyle \frac{dd^{2}{w(r)}}{ \,\mathrm{d}{r^{2}} }|_{r = a_{}} < 0$,所以 $r = a_0$ 为最概然半径.
  2. 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.
  3. (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.

  1. 对于 $\vec{r} \times \vec{L} + \vec{L} \times \vec{r} $ 有:
    \begin{equation} \begin{aligned} (\vec{r} \times \vec{L} + \vec{L} \times \vec{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}
    同理可得:
    \begin{equation} \begin{aligned} & (\vec{r} \times \vec{L} + \vec{L} \times \vec{r})_{y} =2i\hbar \vec{p}_{y} \\ & (\vec{r} \times \vec{L} + \vec{L} \times \vec{r})_{z} =2i\hbar \vec{p}_{z} \end{aligned} \end{equation}
    所以:$\vec{r} \times \vec{L} + \vec{L} \times \vec{r} = 2i\hbar \vec{r} $
  2. 对于 $\vec{p} \times \vec{L} + \vec{L} \times \vec{p}$ 有:
    \begin{equation} \begin{aligned} (\vec{p} \times \vec{L} + \vec{L} \times \vec{p})_{y} =& \vec{p}_{y} \vec{L}_{z} - \vec{p}_{z} \vec{L}_{y} + \vec{L}_{y} \vec{p}_{z} - \vec{L}_{z} \vec{p}_{y} \\ =& [\vec{p}_{y},\vec{L}_{z}] + [\vec{L}_{y},\vec{p}_{z}] \\ =& [\vec{p}_{y},x\vec{p}_{y} - y\vec{p}_{x}] + [z\vec{p}_{x}-x\vec{p}_{z},\vec{p}_{z}] \\ =& -[\vec{p}_{y},y]\vec{p}_{x} + [z,\vec{p}_{z}]\vec{p}_{x} \\ =& 2i\hbar \vec{p}_{x} \end{aligned} \end{equation}
    同理可得:
    \begin{equation} \begin{aligned} & (\vec{p} \times \vec{L} + \vec{L} \times \vec{p})_{y} =2i\hbar \vec{p}_{y} \\ & (\vec{p} \times \vec{L} + \vec{L} \times \vec{p})_{z} =2i\hbar \vec{p}_{z} \end{aligned} \end{equation}
    所以:$\vec{p} \times \vec{L} + \vec{L} \times \vec{p} = 2i\hbar \vec{p} $

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.

  1. 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}
    因为能量本征态是 $S^{2}$ 和 $S_{z}$ 的共同本征函数 $\chi_{sm}$.
    \begin{equation} \begin{aligned} \because & s_{1} = \frac{1}{2},s_{2}=\frac{1}{2},\therefore s=1,0 \\ \therefore & E_{sm} = \frac{a}{2}[s(s+1)\hbar^{2}-\frac{3}{2}\hbar^{2}]+(b-a)m\hbar \\ \therefore & 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 & \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}
  2. 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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