Chapter 6, Problem 38P

To determine

The expression for average particle position; evaluate the results for mean position; the amplitude of oscillation for an electron in a well; the time to shuttle back and forth in a well; the same time classically for an electron.

Answer to Problem 38P

Average position of particle is $x_0+A\cos \left(\Omega t\right)$ ; the mean position value is $0.5\text{nm}$ ; the amplitude value is $-0.18\text{nm}$ ; time required is equal to $3.68\times {10}^{-15}\text{s}$ ; classical time required is equal to $3.47\times {10}^{-15}\text{s}$ .

Explanation of Solution

Write the expression for average position.

  $\langle x\rangle ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x{\left|\Psi \right|}^{2}dx}$

Substitute $C\left\{{\psi }_1{e}^{-i{E}_{1}t/\hslash }+{\psi }_2{e}^{-i{E}_{2}t/\hslash }\right\}$ for $\text{ψ}$ and $C\left\{{\psi }_1^{*}{e}^{+i{E}_{1}t/\hslash }+{\psi }_2^{*}{e}^{+i{E}_{2}t/\hslash }\right\}$ for $\text{ψ*}$ in above expression.

  $\langle x\rangle =C^2{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x\left\{{\psi }_1^{*}{e}^{+i{E}_{1}t/\hslash }+{\psi }_2^{*}{e}^{+i{E}_{2}t/\hslash }\right\}\left\{{\psi }_1{e}^{-i{E}_{1}t/\hslash }+{\psi }_2{e}^{-i{E}_{2}t/\hslash }\right\}dx}$        (I)

Write the expression for mean position.

  $x_0=\frac{1}{2}\left({\displaystyle \int x{\left|{\psi }_1\right|}^{2}dx+{\displaystyle \int x{\left|{\psi }_2\right|}^{2}dx}}\right)$        (II)

Write the expression for amplitude of oscillation.

  $A={\displaystyle \int x{\psi }_1^{*}{\psi }_{2}dx}$        (III)

Write the expression for energy of particle.

  $E_n=\frac{n^2{h}^2}{2m{L}^2}$        (IV)

Write the expression for time required.

  $T=\frac{h}{E_2-E_1}$        (V)

Write the expression for velocity in terms of kinetic energy.

  $v={\left(\frac{2E}{m}\right)}^{1/2}$        (VI)

Write the expression for classical time required.

  $t=\frac{2L}{v}$        (VII)

Conclusion:

For last two integrals, the integral value of both is same as the functions ${\psi }_1\left(x\right)$ and ${\psi }_2\left(x\right)$ is real.

Substitute $1/\sqrt{2}$ for $C$ and simplify the equation (I).

  $\begin{array}{c}\langle x\rangle =C^2\left\{{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x{\left|{\psi }_1\right|}^{2}dx}+{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x{\left|{\psi }_2\right|}^{2}dx}+e^{-i\Omega t}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x{\psi }_1^{*}{\psi }_{2}dx}+e^{+i\Omega t}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x{\psi }_2^{*}{\psi }_{1}dx}\right\}\\ =\frac{1}{2}\left\{\left({\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x{\left|{\psi }_1\right|}^{2}dx}+{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x{\left|{\psi }_2\right|}^{2}dx}\right)+e^{-i\Omega t}A+e^{+i\Omega t}A\right\}\\ =\frac{1}{2}\left({\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x{\left|{\psi }_1\right|}^{2}dx}+{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x{\left|{\psi }_2\right|}^{2}dx}\right)+\frac{A}{2}\left(e^{-i\Omega t}+e^{+i\Omega t}\right)\\ =x_0+A\cos \left(\Omega t\right)\end{array}$

Thus, average position of particle is $x_0+A\cos \left(\Omega t\right)$ .

The average position of particle for any stationary state of well is $L/2$ .

Substitute $L/2$ for ${\displaystyle \int x{\left|{\psi }_2\right|}^{2}dx}$ in equation (II).

  $\begin{array}{c}{x}_0=\frac{1}{2}\left(\frac{L}{2}+\frac{L}{2}\right)\\ =\frac{L}{2}\\ =0.5\text{nm}\end{array}$

Thus, the mean position value is $0.5\text{nm}$ .

  $\begin{array}{c}A=\frac{2}{L}{\displaystyle \underset{0}{\overset{L}{\int }}x\sin \left(\frac{\pi x}{L}\right)\sin \left(\frac{2\pi x}{L}\right)dx}\\ =\frac{1}{L}{\left[\frac{L}{3\pi }\left\{3\sin \left(\frac{\pi x}{L}\right)-\sin \left(\frac{3\pi x}{L}\right)\right\}\right]}_0^L-\frac{1}{L}\left(\frac{L}{3\pi }\right){\displaystyle \underset{0}{\overset{L}{\int }}\left\{3\sin \left(\frac{\pi x}{L}\right)-\sin \left(\frac{3\pi x}{L}\right)\right\}}\\ =0+\frac{1}{L}{\left(\frac{L}{3\pi }\right)}^2{\left[\left\{9\cos \left(\frac{\pi x}{L}\right)-\cos \left(\frac{3\pi x}{L}\right)\right\}\right]}_0^L\\ =-0.18\text{nm}\end{array}$

Thus, the amplitude value is $-0.18\text{nm}$ .

Substitute $1.24\text{keV}\cdot \text{nm}/c$ for $h$, $511\text{keV}/c^2$ for $m$ and $1\text{nm}$ for $L$ in equation (IV).

      $\begin{array}{c}{E}_1=\frac{{\left(1.24\text{keV}\cdot \text{nm}/c\right)}^2}{8\left(511\text{keV}/c^2\right){\left(1\text{ nm}\right)}^2}\\ =0.376\text{ eV}\end{array}$

Substitute $1.24\text{keV}\cdot \text{nm}/c$ for $h$, $511\text{keV}/c^2$ for $m$, $2$ for $n$ and $1\text{nm}$ for $L$ in equation (IV).

      $\begin{array}{c}{E}_1=\frac{{\left(2\right)}^2{\left(1.24\text{keV}\cdot \text{nm}/c\right)}^2}{8\left(511\text{keV}/c^2\right){\left(1\text{ nm}\right)}^2}\\ =1.51\text{ eV}\end{array}$

Substitute 1.51 eV for $E_2$ , 0.376 eV for $E_1$ and $4.136\times {10}^{-15}\text{ eV}\cdot \text{s}$ for $h$ in equation (V).

  $\begin{array}{c}T=\frac{4.136\times {10}^{-15}\text{ eV}\cdot \text{s}}{\text{1}\text{.124 eV}}\\ =3.68\times {10}^{-15}\text{s}\end{array}$

Thus, time required is equal to $3.68\times {10}^{-15}\text{s}$ .

Calculate the kinetic energy of electron.

  $\begin{array}{c}E=\frac{E_1+E_2}{2}\\ =0.94\text{eV}\end{array}$

Substitute $511\text{keV}/c^2$ for $m$ and 0.94 eV  for $E$ in equation (VI).

  $\begin{array}{c}v={\left(\frac{2\left(0.94\text{eV}\right)}{511\text{keV}/c^2}\right)}^{1/2}\\ =\left(1.92\times {10}^{-3}\right)c\end{array}$

Substitute 1 nm for $L$ and  $\left(1.92\times {10}^{-3}\right)c$  for $v$  in equation (VII).

  $\begin{array}{c}t=\frac{2\left(1\text{nm}\right)}{\left(1.92\times {10}^{-3}\right)c}\\ =3.47\times {10}^{-15}\text{s}\end{array}$

Thus, classical time required is equal to $3.47\times {10}^{-15}\text{s}$ .