Chapter 34, Problem 54P

To determine

The value of $\langle x\rangle \text{and}\langle x^2\rangle$ .

Answer to Problem 54P

The value of $\langle x\rangle$ is $0$ and value of $\langle x^2\rangle$ is $L^2\left[\frac{1}{12}-\frac{1}{8{\pi }^2}\right]$ .

Explanation of Solution

Given:

The one-dimensional box region is $-\frac{L}{2}\le x\le \frac{L}{2}$ .

The one-dimensional box length is $L$ .

The centeris at origin.

The particle mass is $m$ .

The wave function for $n=1,3,5,\mathrm{...}$ is $\psi \left(x\right)=\sqrt{\frac{2}{L}}\cos \frac{n\pi x}{L}$ .

The wave function for $n=2,4,6,\mathrm{...}$ is $\psi \left(x\right)=\sqrt{\frac{2}{L}}\sin \frac{n\pi x}{L}$ .

State is first excited state $\left(n=2\right)$ .

Formula used:

The expression for $\langle x\rangle$ is given by,

  $\langle x\rangle ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x{\psi }^2\left(x\right)}dx$

The expression for $\langle x^2\rangle$ is given by,

  $\langle x^2\rangle ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{x}^2{\psi }^2\left(x\right)}dx$

The integral formula,

  ${\displaystyle \int {\theta }^2{\sin }^2\theta d\theta }=\frac{{\theta }^3}{6}-\left(\frac{{\theta }^2}{4}-\frac{1}{8}\right)\sin 2\theta -\frac{\theta \cos 2\theta }{4}+c$

Calculation:

The $\langle x\rangle$ is calculated as,

  $\begin{array}{c}\langle x\rangle ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x{\psi }^2\left(x\right)}dx\\ ={{\displaystyle \underset{-L/2}{\overset{L/2}{\int }}x\left(\sqrt{\frac{2}{L}}\sin \frac{2\pi x}{L}\right)}}^{2}dx\end{array}$

The function $x{\left(\sqrt{\frac{2}{L}}\sin \frac{2\pi x}{L}\right)}^2$ is odd function.

Solving further as,

  $\begin{array}{c}\langle x\rangle ={{\displaystyle \underset{-L/2}{\overset{L/2}{\int }}x\left(\sqrt{\frac{2}{L}}\sin \frac{2\pi x}{L}\right)}}^{2}dx\\ =0\end{array}$

The $\langle x^2\rangle$ is calculated as,

  $\begin{array}{c}\langle x^2\rangle ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{x}^2{\psi }^2\left(x\right)}dx\\ ={{\displaystyle \underset{-L/2}{\overset{L/2}{\int }}{x}^2\left(\sqrt{\frac{2}{L}}\sin \frac{2\pi x}{L}\right)}}^{2}dx\end{array}$

The function $x^2{\left(\sqrt{\frac{2}{L}}\sin \frac{2\pi x}{L}\right)}^2$ is an even function.

Solving further as,

  $\begin{array}{c}\langle x^2\rangle ={{\displaystyle \underset{-L/2}{\overset{L/2}{\int }}{x}^2\left(\sqrt{\frac{2}{L}}\sin \frac{2\pi x}{L}\right)}}^{2}dx\\ =2{{\displaystyle \underset{0}{\overset{L/2}{\int }}{x}^2\left(\sqrt{\frac{2}{L}}\sin \frac{2\pi x}{L}\right)}}^{2}dx\end{array}$

Let, $\frac{2\pi x}{L}=\theta$ .So,

  $\frac{2\pi dx}{L}=d\theta$

Solving further as,

  $\begin{array}{c}\langle x^2\rangle =2{{\displaystyle \underset{0}{\overset{L/2}{\int }}{x}^2\left(\sqrt{\frac{2}{L}}\sin \frac{2\pi x}{L}\right)}}^{2}dx\\ =2{{\displaystyle \underset{0}{\overset{\pi }{\int }}{\left(\frac{L\theta }{2\pi }\right)}^2\left(\sqrt{\frac{2}{L}}\sin \theta \right)}}^2\left(\frac{Ld\theta }{2\pi }\right)\\ =\frac{L^2}{2{\pi }^3}{\displaystyle \underset{0}{\overset{\pi }{\int }}{\theta }^2{\sin }^2\theta d\theta }\\ =\frac{L^2}{2{\pi }^3}{\left\{\frac{{\theta }^3}{6}-\left(\frac{{\theta }^2}{4}-\frac{1}{8}\right)\sin 2\theta -\frac{\theta \cos 2\theta }{4}\right\}}_0^{\pi }\end{array}$

Solving further as,

  $\begin{array}{c}\langle x^2\rangle =\frac{L^2}{2{\pi }^3}{\left\{\frac{{\theta }^3}{6}-\left(\frac{{\theta }^2}{4}-\frac{1}{8}\right)\sin 2\theta -\frac{\theta \cos 2\theta }{4}\right\}}_0^{\pi }\\ =\frac{L^2}{2{\pi }^3}\left\{\frac{{\pi }^3}{6}-0-\frac{\pi }{4}\right\}\\ =L^2\left[\frac{1}{12}-\frac{1}{8{\pi }^2}\right]\end{array}$

Conclusion:

Therefore, the value of $\langle x\rangle$ is $0$ and value of $\langle x^2\rangle$ is $L^2\left[\frac{1}{12}-\frac{1}{8{\pi }^2}\right]$ .