Chapter 34, Problem 51P

(a)

To determine

The value of $\langle x\rangle \text{and}\langle x^2\rangle$ for a particle in nth state.

Answer to Problem 51P

The value of $\langle x\rangle$ is $\frac{1}{2}L$ and value of $\langle x^2\rangle$ is $\frac{1}{3}{L}^2-\frac{L^2}{2n^2{\pi }^2}$ for a particle in nth state.

Explanation of Solution

Given:

The one-dimensional box region is $0\le x\le L$ .

The particle is in the nth state.

The wave functionis ${\psi }_n\left(x\right)=\sqrt{\frac{2}{L}}\sin \frac{n\pi x}{L}\text{for},n=1,2,3,\mathrm{...}$ .

Formula used:

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

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

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

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

The integral formula,

  ${\displaystyle \int \theta {\sin }^2\theta d\theta }=\frac{{\theta }^2}{4}-\frac{\theta \sin 2\theta }{4}-\frac{\cos 2\theta }{8}+c$

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:

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

By differentiating both sides,

  $\begin{array}{c}\frac{n\pi dx}{L}=d\theta \\ dx=\frac{Ld\theta }{n\pi }\end{array}$

The limit is $0\text{to}L$ changes to $0\text{to}n\pi$ .

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

  $\begin{array}{c}\langle x\rangle ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x{\left({\psi }_n\left(x\right)\right)}^2}dx\\ ={\displaystyle \underset{0}{\overset{L}{\int }}x{\left(\sqrt{\frac{2}{L}}\sin \frac{n\pi x}{L}\right)}^2}dx\\ ={\displaystyle \underset{0}{\overset{n\pi }{\int }}\left(\frac{L\theta }{n\pi }\right)\frac{2}{L}{\sin }^2\theta }\left(\frac{Ld\theta }{n\pi }\right)\\ =\left(\frac{2L}{n^2{\pi }^2}\right){\displaystyle \underset{0}{\overset{n\pi }{\int }}\theta {\sin }^2\theta d\theta }\end{array}$

Solving further as,

  $\begin{array}{c}\langle x\rangle =\left(\frac{2L}{n^2{\pi }^2}\right){\displaystyle \underset{0}{\overset{n\pi }{\int }}\theta {\sin }^2\theta d\theta }\\ =\left(\frac{2L}{n^2{\pi }^2}\right){\left[\frac{{\theta }^2}{4}-\frac{\theta \sin 2\theta }{4}-\frac{\cos 2\theta }{8}\right]}_0^{n\pi }\\ =\left(\frac{2L}{n^2{\pi }^2}\right)\left[\frac{n^2{\pi }^2}{4}-0-0\right]\\ =\frac{L}{2}\end{array}$

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

  $\begin{array}{c}\langle x^2\rangle ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{x}^2{\left({\psi }_n\left(x\right)\right)}^2}dx\\ ={\displaystyle \underset{0}{\overset{L}{\int }}{x}^2{\left(\sqrt{\frac{2}{L}}\sin \frac{n\pi x}{L}\right)}^2}dx\\ ={\displaystyle \underset{0}{\overset{n\pi }{\int }}{\left(\frac{L\theta }{n\pi }\right)}^2\frac{2}{L}{\sin }^2\theta }\left(\frac{Ld\theta }{n\pi }\right)\\ =\left(\frac{2L^2}{n^3{\pi }^3}\right){\displaystyle \underset{0}{\overset{n\pi }{\int }}\theta {\sin }^2\theta d\theta }\end{array}$

Solving further as,

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

Conclusion:

Therefore, the value of $\langle x\rangle$ is $\frac{1}{2}L$ and value of $\langle x^2\rangle$ is $\frac{1}{3}{L}^2-\frac{L^2}{2n^2{\pi }^2}$ for a particle in nth state.

(b)

To determine

The comparison of $\langle x\rangle \text{and}\langle x^2\rangle$ for $n\gg 1$ with the value of problem $50$ .

Answer to Problem 51P

The value of $\langle x\rangle \text{and}\langle x^2\rangle$ for the classical probability distribution function, $P\left(x\right)=\frac{1}{L}$ and for the wave function ${\psi }_n\left(x\right)=\sqrt{\frac{2}{L}}\sin \frac{n\pi x}{L}$ in the nth state while $n\gg 1$ is same.

Explanation of Solution

Calculation:

The value of $\langle x\rangle$ in both the cases is same and equalto $\frac{L}{2}$ .

The value of $\langle x^2\rangle$ for $n\gg 1$ is calculated as,

  $\begin{array}{c}\langle x^2\rangle =\frac{1}{3}{L}^2-\frac{L^2}{2n^2{\pi }^2}\\ =\frac{1}{3}{L}^2-0\\ =\frac{1}{3}{L}^2\end{array}$

Which is same as in case of problem $50$ .

Conclusion:

Therefore, the value of $\langle x\rangle \text{and}\langle x^2\rangle$ for the classical probability distribution function, $P\left(x\right)=\frac{1}{L}$ and for the wave function ${\psi }_n\left(x\right)=\sqrt{\frac{2}{L}}\sin \frac{n\pi x}{L}$ in the nth state while $n\gg 1$issame.