Chapter 10, Problem 10.83E

Interpretation Introduction

Interpretation:

The average values of $\langle x^2\rangle$, $\langle y^2\rangle$ and $\langle z^2\rangle$ for ${\Psi }_{111}$ of a $\text{3-D}$ particle-in-a-box are to be stated.

Concept introduction:

The Schrödinger equation is used to find the allowed energy levels for electronic transitions in the quantum mechanics. It is generally expressed as follows.

$\stackrel{⌢}{\text{H}}\Psi =\text{E}\Psi$

Where,

• $\stackrel{⌢}{\text{H}}$is the Hamiltonian operator.

• $\Psi$is the wavefunction.

• $\text{E}$is the energy.

The energy obtained after applying the operator on wavefunction is known as the eigen value for the wavefunction.

Answer to Problem 10.83E

The average values of $\langle x^2\rangle$, $\langle y^2\rangle$ and $\langle z^2\rangle$ for ${\Psi }_{111}$ of a $\text{3-D}$ particle-in-a-box are $\frac{a^2\left(4{\pi }^2-6\right)}{12{\pi }^2}$, $\frac{b^2\left(4{\pi }^2-6\right)}{12{\pi }^2}$ and $\frac{c^2\left(4{\pi }^2-6\right)}{12{\pi }^2}$ respectively.

Explanation of Solution

For particle in $\text{3-D}$ box, ${\Psi }_{111}$ wavefunction is written as follows.

${\Psi }_{111}=\sqrt{\frac{8}{abc}}\sin \frac{\pi x}{a}\sin \frac{\pi y}{b}\sin \frac{\pi z}{c}$

The average value of $\langle x^2\rangle$ is expressed as follows.

$\langle x^2\rangle ={\displaystyle \iiint {\Psi }_{111}{x}^2{\Psi }_{111}^{*}}$

Substitute the value in the above function as follows.

$\begin{array}{rll}\langle x^2\rangle & =& {\displaystyle \iiint {\Psi }_{111}{x}^2{\Psi }_{111}^{*}}\\ & =& {\displaystyle \iiint \sqrt{\frac{8}{abc}}\sin \frac{\pi x}{a}\sin \frac{\pi y}{b}\sin \frac{\pi z}{c}}\left(x^2\right)\sqrt{\frac{8}{abc}}\sin \frac{\pi x}{a}\sin \frac{\pi y}{b}\sin \frac{\pi z}{c}dxdydz\\ & =& \frac{8}{abc}{\displaystyle \underset{0}{\overset{a}{\int }}{x}^2{\sin }^2\frac{\pi x}{a}}dx{\displaystyle \underset{0}{\overset{b}{\int }}{\sin }^2\frac{\pi y}{b}}dy{\displaystyle \underset{0}{\overset{c}{\int }}{\sin }^2\frac{\pi z}{c}dz}\end{array}$

The above equation can be simplified in three parts as follows.

$\begin{array}{l}{\displaystyle \underset{0}{\overset{a}{\int }}{x}^2{\sin }^2\frac{\pi x}{a}}dx\\ ={\left[\frac{x^3}{6}-\left(\frac{a{x}^2}{4\pi }-\frac{a^2}{8{\pi }^2}\right)\sin \left(\frac{2\pi x}{a}\right)-\frac{a^{2}x}{4{\pi }^2}\cos \left(\frac{2\pi x}{a}\right)\right]}_0^a\\ =\frac{a^3}{6}-\frac{a^3}{4{\pi }^2}\\ =\frac{a^3\left(4{\pi }^2-6\right)}{24{\pi }^2}\end{array}$

$\begin{array}{l}{\displaystyle \underset{0}{\overset{b}{\int }}{\sin }^2\frac{\pi y}{a}}dy\\ ={\left[\frac{y}{2}-\frac{b}{4\pi }\sin \left(\frac{2\pi y}{b}\right)\right]}_0^b\\ =\frac{b}{2}\end{array}$

$\begin{array}{l}{\displaystyle \underset{0}{\overset{c}{\int }}{\sin }^2\frac{\pi z}{a}}dz\\ ={\left[\frac{z}{2}-\frac{c}{4\pi }\sin \left(\frac{2\pi z}{c}\right)\right]}_0^c\\ =\frac{c}{2}\end{array}$

Substitute these solved integrals in expression for $\langle x^2\rangle$ as follows.

$\begin{array}{rll}\langle x^2\rangle & =& \frac{8}{abc}\left(\frac{a^3\left(4{\pi }^2-6\right)}{24{\pi }^2}\right)\frac{b}{2}\frac{c}{2}\\ & =& \frac{a^2\left(4{\pi }^2-6\right)}{12{\pi }^2}\end{array}$

Similarly, the average value of $\langle y^2\rangle$ is expressed as follows.

$\langle y^2\rangle ={\displaystyle \iiint {\Psi }_{111}{y}^2{\Psi }_{111}^{*}}$

Substitute the value in the above function as follows.

$\begin{array}{rll}\langle y^2\rangle & =& {\displaystyle \iiint {\Psi }_{111}{y}^2{\Psi }_{111}^{*}}\\ & =& {\displaystyle \iiint \sqrt{\frac{8}{abc}}\sin \frac{\pi x}{a}\sin \frac{\pi y}{b}\sin \frac{\pi z}{c}}\left(y^2\right)\sqrt{\frac{8}{abc}}\sin \frac{\pi x}{a}\sin \frac{\pi y}{b}\sin \frac{\pi z}{c}dxdydz\\ & =& \frac{8}{abc}{\displaystyle \underset{0}{\overset{a}{\int }}{\sin }^2\frac{\pi x}{a}}dx{\displaystyle \underset{0}{\overset{b}{\int }}{y}^2{\sin }^2\frac{\pi y}{b}}dy{\displaystyle \underset{0}{\overset{c}{\int }}{\sin }^2\frac{\pi z}{c}dz}\end{array}$

The above equation can be simplified in three parts as follows.

$\begin{array}{l}{\displaystyle \underset{0}{\overset{a}{\int }}{\sin }^2\frac{\pi x}{a}}dx\\ ={\left[\frac{x}{4}-\frac{a}{4\pi }\sin \left(\frac{2\pi x}{a}\right)\right]}_0^a\\ =\frac{a}{2}\end{array}$

$\begin{array}{l}{\displaystyle \underset{0}{\overset{b}{\int }}{y}^2{\sin }^2\frac{\pi y}{b}}dy\\ ={\left[\frac{y^3}{6}-\left(\frac{b{y}^2}{4\pi }-\frac{b^2}{8{\pi }^2}\right)\sin \left(\frac{2\pi y}{b}\right)-\frac{b^{2}y}{4{\pi }^2}\cos \left(\frac{2\pi y}{b}\right)\right]}_0^b\\ =\frac{b^3}{6}-\frac{b^3}{4{\pi }^2}\\ =\frac{b^3\left(4{\pi }^2-6\right)}{24{\pi }^2}\end{array}$

$\begin{array}{l}{\displaystyle \underset{0}{\overset{c}{\int }}{\sin }^2\frac{\pi z}{a}}dz\\ ={\left[\frac{z}{2}-\frac{c}{4\pi }\sin \left(\frac{2\pi z}{c}\right)\right]}_0^c\\ =\frac{c}{2}\end{array}$

Substitute these solved integrals in expression for $\langle y^2\rangle$ as follows.

$\begin{array}{rll}\langle y^2\rangle & =& \frac{8}{abc}\frac{a}{2}\left(\frac{b^3\left(4{\pi }^2-6\right)}{24{\pi }^2}\right)\frac{c}{2}\\ & =& \frac{b^2\left(4{\pi }^2-6\right)}{12{\pi }^2}\end{array}$

Similarly, the average value of $\langle z^2\rangle$ is expressed as follows.

$\langle z^2\rangle ={\displaystyle \iiint {\Psi }_{111}{z}^2{\Psi }_{111}^{*}}$

Substitute the value in the above function as follows.

$\begin{array}{rll}\langle z^2\rangle & =& {\displaystyle \iiint {\Psi }_{111}{z}^2{\Psi }_{111}^{*}}\\ & =& {\displaystyle \iiint \sqrt{\frac{8}{abc}}\sin \frac{\pi x}{a}\sin \frac{\pi y}{b}\sin \frac{\pi z}{c}}\left(z^2\right)\sqrt{\frac{8}{abc}}\sin \frac{\pi x}{a}\sin \frac{\pi y}{b}\sin \frac{\pi z}{c}dxdydz\\ & =& \frac{8}{abc}{\displaystyle \underset{0}{\overset{a}{\int }}{\sin }^2\frac{\pi x}{a}}dx{\displaystyle \underset{0}{\overset{b}{\int }}{\sin }^2\frac{\pi y}{b}}dy{\displaystyle \underset{0}{\overset{c}{\int }}{z}^2{\sin }^2\frac{\pi z}{c}dz}\end{array}$

The above equation can be simplified in three parts as follows.

$\begin{array}{l}{\displaystyle \underset{0}{\overset{a}{\int }}{\sin }^2\frac{\pi x}{a}}dx\\ ={\left[\frac{x}{4}-\frac{a}{4\pi }\sin \left(\frac{2\pi x}{a}\right)\right]}_0^a\\ =\frac{a}{2}\end{array}$

$\begin{array}{l}{\displaystyle \underset{0}{\overset{b}{\int }}{\sin }^2\frac{\pi y}{b}}dy\\ ={\left[\frac{y}{2}-\frac{b}{4\pi }\sin \left(\frac{2\pi y}{b}\right)\right]}_0^b\\ =\frac{b}{2}\end{array}$

$\begin{array}{l}{\displaystyle \underset{0}{\overset{c}{\int }}{z}^2{\sin }^2\frac{\pi z}{c}}dz\\ ={\left[\frac{z^3}{6}-\left(\frac{c{z}^2}{4\pi }-\frac{c^2}{8{\pi }^2}\right)\sin \left(\frac{2\pi z}{c}\right)-\frac{c^{2}z}{4{\pi }^2}\cos \left(\frac{2\pi z}{c}\right)\right]}_0^c\\ =\frac{c^3}{6}-\frac{c^3}{4{\pi }^2}\\ =\frac{c^3\left(4{\pi }^2-6\right)}{24{\pi }^2}\end{array}$

Substitute these solved integrals in expression for $\langle z^2\rangle$ as follows.

$\begin{array}{rll}\langle z^2\rangle & =& \frac{8}{abc}\frac{a}{2}\frac{b}{2}\left(\frac{c^3\left(4{\pi }^2-6\right)}{24{\pi }^2}\right)\\ & =& \frac{c^2\left(4{\pi }^2-6\right)}{12{\pi }^2}\end{array}$

Conclusion

The average values of $\langle x^2\rangle$, $\langle y^2\rangle$ and $\langle z^2\rangle$ for ${\Psi }_{111}$ of a $\text{3-D}$ particle-in-a-box are $\frac{a^2\left(4{\pi }^2-6\right)}{12{\pi }^2}$, $\frac{b^2\left(4{\pi }^2-6\right)}{12{\pi }^2}$ and $\frac{c^2\left(4{\pi }^2-6\right)}{12{\pi }^2}$ respectively.