题目
29.试求长度为l的一维势箱中,处于 n=3 状态-|||-的一个粒子的x^2和p^2的平均值x ^2、p^2。
题目解答
答案
解析
步骤 1：写出能量本征态的波函数
对于长度为l的一维势箱，处于n=3状态的粒子的波函数为：
${y}_{3}(x)=\sqrt {\dfrac {2}{l}\sin (\dfrac {3\pi x}{l})}$
步骤 2：计算x^2的平均值
x^2的平均值定义为：
${x}^{2}={\int }_{0}^{l}{\psi }_{3}{(x)}^{2}{x}^{2}dx$
代入波函数：
${x}^{2}=\dfrac {2}{l}{\int }_{0}^{l}{\sin }^{2}(\dfrac {3\pi x}{l}){x}^{2}dx$
利用三角恒等式 ${\sin }^{2}\theta =\dfrac {1-\cos 2\theta }{2}$:
${x}^{2}=\dfrac {2}{l}\times \dfrac {1}{2}{\int }_{0}^{l}(1-\cos (\dfrac {6\pi x}{l})){x}^{2}dx$
${x}^{2}=\dfrac {1}{l}{\int }_{0}^{l}{x}^{2}dx-\dfrac {1}{l}{\int }_{0}^{l}{x}^{2}\cos (\dfrac {6\pi x}{l})dx$
计算第一个积分：
${\int }_{0}^{l}{x}^{2}dx=\dfrac {{l}^{3}}{3}$
计算第二个积分，使用分部积分法：
$\int {x}^{2}\cos (\dfrac {6\pi x}{l})dx=\dfrac {{x}^{2}l}{6\pi }\sin (\dfrac {6\pi x}{l})-\dfrac {2l{x}^{2}}{{(6\pi )}^{2}}\cos (\dfrac {6\pi x}{l})+\dfrac {2{l}^{3}}{{(6\pi )}^{3}}\sin (\dfrac {6\pi x}{l})$
将结果代入：
${x}^{2}=\dfrac {1}{l}\times \dfrac {{l}^{3}}{3}-\dfrac {1}{l}\times (-\dfrac {{l}^{3}}{18{\pi }^{2}})$
${x}^{2}=\dfrac {{l}^{2}}{3}+\dfrac {{l}^{2}}{18{\pi }^{2}}$
${x}^{2}=\dfrac {6{\pi }^{2}+1}{18{\pi }^{2}}{l}^{2}$
步骤 3：计算p^2的平均值
根据能量本征值：
${E}_{3}=\dfrac {9{\pi }^{2}{h}^{2}}{2m{l}^{2}}$
又有：
${p}^{2}=2m{E}_{3}$
代入E3：
${p}^{2}=2m\times \dfrac {9{\pi }^{2}{h}^{2}}{2m{l}^{2}}$
${p}^{2}=\dfrac {9{\pi }^{2}{h}^{2}}{{l}^{2}}$
对于长度为l的一维势箱，处于n=3状态的粒子的波函数为：
${y}_{3}(x)=\sqrt {\dfrac {2}{l}\sin (\dfrac {3\pi x}{l})}$
步骤 2：计算x^2的平均值
x^2的平均值定义为：
${x}^{2}={\int }_{0}^{l}{\psi }_{3}{(x)}^{2}{x}^{2}dx$
代入波函数：
${x}^{2}=\dfrac {2}{l}{\int }_{0}^{l}{\sin }^{2}(\dfrac {3\pi x}{l}){x}^{2}dx$
利用三角恒等式 ${\sin }^{2}\theta =\dfrac {1-\cos 2\theta }{2}$:
${x}^{2}=\dfrac {2}{l}\times \dfrac {1}{2}{\int }_{0}^{l}(1-\cos (\dfrac {6\pi x}{l})){x}^{2}dx$
${x}^{2}=\dfrac {1}{l}{\int }_{0}^{l}{x}^{2}dx-\dfrac {1}{l}{\int }_{0}^{l}{x}^{2}\cos (\dfrac {6\pi x}{l})dx$
计算第一个积分：
${\int }_{0}^{l}{x}^{2}dx=\dfrac {{l}^{3}}{3}$
计算第二个积分，使用分部积分法：
$\int {x}^{2}\cos (\dfrac {6\pi x}{l})dx=\dfrac {{x}^{2}l}{6\pi }\sin (\dfrac {6\pi x}{l})-\dfrac {2l{x}^{2}}{{(6\pi )}^{2}}\cos (\dfrac {6\pi x}{l})+\dfrac {2{l}^{3}}{{(6\pi )}^{3}}\sin (\dfrac {6\pi x}{l})$
将结果代入：
${x}^{2}=\dfrac {1}{l}\times \dfrac {{l}^{3}}{3}-\dfrac {1}{l}\times (-\dfrac {{l}^{3}}{18{\pi }^{2}})$
${x}^{2}=\dfrac {{l}^{2}}{3}+\dfrac {{l}^{2}}{18{\pi }^{2}}$
${x}^{2}=\dfrac {6{\pi }^{2}+1}{18{\pi }^{2}}{l}^{2}$
步骤 3：计算p^2的平均值
根据能量本征值：
${E}_{3}=\dfrac {9{\pi }^{2}{h}^{2}}{2m{l}^{2}}$
又有：
${p}^{2}=2m{E}_{3}$
代入E3：
${p}^{2}=2m\times \dfrac {9{\pi }^{2}{h}^{2}}{2m{l}^{2}}$
${p}^{2}=\dfrac {9{\pi }^{2}{h}^{2}}{{l}^{2}}$