Chapter 10, Problem 10.71E

Mathematically, the uncertainty $\Delta A$ in some observable $A$ is given by $\Delta A=\sqrt{\langle A^2\rangle -{\langle A\rangle }^2}$. Use this formula to determine $\Delta x$ and $\Delta p_x$ for $\Psi =\sqrt{\left(2/a\right)}\sin \left(\pi x/a\right)$ and show that the uncertainty principle holds.

Interpretation Introduction

Interpretation:

The value of $\Delta x$ and $\Delta p_x$ for $\Psi$ is to be determined. The validation of uncertainty principle is to be shown.

Concept introduction:

The uncertainty principle deals only with the parameters that cannot be measured simultaneously. Such two parameters are momentum and position in same direction. The uncertainty principle is given by formula as follows.

$\Delta x\Delta p_{\text{x}}\ge \frac{\hslash }{2}$

Where,

• $\Delta x$ represents the uncertainty in the position of the object.

• $\Delta p_{\text{x}}$ represents the uncertainty in the momentum of the object.

• $\hslash$ represents a constant.

Answer to Problem 10.71E

The uncertainty in the position of a particle is $\left(\frac{a}{2\pi }\sqrt{\frac{{\pi }^2-6}{3}}\right)$.

The uncertainty in the momentum of a particle is $\frac{\pi \hslash }{a}$.

The uncertainty principle has been validated.

Explanation of Solution

From Appendix $1$,

${\displaystyle \int x{\sin }^2\left(bx\right)dx}=\frac{x^2}{4}-\frac{x}{4b}\sin \left(2bx\right)-\frac{1}{8b^2}\cos \left(2bx\right)$ …(1)

The average value of $\langle x\rangle$ is calculated as,

$\langle x\rangle ={\displaystyle \underset{0}{\overset{a}{\int }}{\Psi }^{\ast }\widehat{x}\Psi dx}$ …(2)

Where,

• $\widehat{x}$ represents the position operator.

• $\Psi$ represents the wavefunction.

The given wavefunction $\Psi$ is defined as $\sqrt{\frac{2}{a}}\sin \frac{\pi x}{a}$.

The position operator is defined as $x\left(\right)$.

Substitute the value of $\Psi$ and $\widehat{x}$ in the equation (2).

$\begin{array}{rll}\langle x\rangle & =& {\displaystyle \underset{0}{\overset{a}{\int }}{\left(\sqrt{\frac{2}{a}}\sin \frac{\pi x}{a}\right)}^{\ast }x\left(\sqrt{\frac{2}{a}}\sin \frac{\pi x}{a}\right)dx}\\ & =& {\displaystyle \underset{0}{\overset{a}{\int }}x{\left(\sqrt{\frac{2}{a}}\sin \frac{\pi x}{a}\right)}^{2}dx}\\ & =& \frac{2}{a}{\displaystyle \underset{0}{\overset{a}{\int }}x{\left(\sin \frac{\pi x}{a}\right)}^{2}dx}\end{array}$

Assume $b=\frac{\pi }{a}$ and expand the above expression similarly as equation (1).

$\begin{array}{rll}\langle x\rangle & =& \frac{2}{a}{\left[\frac{x^2}{4}-\frac{xa}{4\left(\pi \right)}\sin \frac{2\pi x}{a}-\frac{a^3}{8{\left(\pi \right)}^3}\cos \frac{2\pi x}{a}\right]}_0^a\\ & =& \frac{2}{a}\left[\begin{array}{l}\left(\frac{a^2}{4}-\frac{a^2}{4\pi }\sin \frac{2\pi a}{a}-\frac{a^3}{8{\left(\pi \right)}^3}\cos \frac{2\pi a}{a}\right)\\ -\left(\frac{{\left(0\right)}^2}{4}-\frac{\left(0\right)a}{4\left(\pi \right)}\sin \frac{2\pi \left(0\right)}{a}-\frac{a^3}{8{\left(\pi \right)}^3}\cos \frac{2\pi \left(0\right)}{a}\right)\end{array}\right]\\ & =& \frac{2}{a}\left[\begin{array}{l}\left(\frac{a^2}{4}-\frac{a^2}{4\pi }\sin 2\pi -\frac{a^3}{8{\left(\pi \right)}^3}\cos 2\pi \right)\\ -\left(\frac{{\left(0\right)}^2}{4}-\frac{\left(0\right)a}{4\left(\pi \right)}\sin \frac{2\pi \left(0\right)}{a}-\frac{a^3}{8{\left(\pi \right)}^3}\cos \frac{2\pi \left(0\right)}{a}\right)\end{array}\right]\end{array}$

Substitute the value of $\sin \pi =0$ in the above expression.

$\begin{array}{rllll}\langle x\rangle & =& & =& \frac{2}{a}\left[\begin{array}{l}\left(\frac{a^2}{4}-\frac{a^2}{4\pi }\left(0\right)-\frac{a^3}{8{\left(\pi \right)}^3}\cos 2\pi \right)\\ -\left(\frac{{\left(0\right)}^2}{4}-\frac{\left(0\right)a}{4\left(\pi \right)}\sin \frac{4\pi \left(0\right)}{a}-\frac{a^3}{8{\left(\pi \right)}^3}\cos \frac{2\pi \left(0\right)}{a}\right)\end{array}\right]\\ & =& \left(\frac{2}{a}\right)\left(\frac{a^2}{4}\right)\\ & =& \frac{a}{2}\end{array}$

Therefore, the average value of position, $\langle x\rangle$, for ${\Psi }_1$ of a particle-in-a-box is $\frac{a}{2}$.

From Appendix $1$,

${\displaystyle \int \sin \left(bx\right)\cos \left(bx\right)dx}=\frac{1}{b}{\sin }^2\left(bx\right)$ …(3)

The average value of $\langle p_x\rangle$ is calculated as,

$\langle p_x\rangle ={\displaystyle \underset{0}{\overset{a}{\int }}{\Psi }_1{}^{\ast }\widehat{p_x}{\Psi }_{1}dx}$ …(4)

Where,

• $\widehat{p_x}$ represents an operator.

• ${\Psi }_1$ represents the wavefunction.

The given wavefunction ${\Psi }_1$ is defined as $\sqrt{\frac{2}{a}}\sin \frac{\pi x}{a}$.

The operator $\widehat{p_x}$ is defined as $-i\hslash \frac{d}{dx}\left(\right)$.

Substitute the value of ${\Psi }_1$ and $\widehat{x}$ in the equation (4).

$\begin{array}{rll}\langle p_x\rangle & =& {\displaystyle \underset{0}{\overset{a}{\int }}{\left(\sqrt{\frac{2}{a}}\sin \frac{\pi x}{a}\right)}^{\ast }-i\hslash \frac{d}{dx}\left(\sqrt{\frac{2}{a}}\sin \frac{\pi x}{a}\right)dx}\\ & =& {\displaystyle \underset{0}{\overset{a}{\int }}\left(\sqrt{\frac{2}{a}}\sin \frac{\pi x}{a}\right)\left(-i\hslash \right)\left(\sqrt{\frac{2}{a}}\left(\frac{\pi }{a}\right)\cos \frac{\pi x}{a}\right)dx}\\ & =& \frac{-2i\pi \hslash }{a^2}{\displaystyle \underset{0}{\overset{a}{\int }}\left(\sin \frac{\pi x}{a}\cos \frac{2\pi x}{a}\right)dx}\end{array}$

Assume $b=\frac{\pi }{a}$ and expand the above expression similarly as equation (3).

$\begin{array}{rll}\langle p_x\rangle & =& \frac{-2i\pi \hslash }{a^2}{\left[\left(\frac{1}{\left(\pi /a\right)}\right){\sin }^2\left(\frac{\pi x}{a}\right)\right]}_0^a\\ & =& \frac{-2i\pi \hslash }{a^2}\left[\left(\frac{1}{\left(\pi /a\right)}\right){\sin }^2\left(\frac{\pi \left(a\right)}{a}\right)-\left(\frac{1}{\left(\pi /a\right)}\right){\sin }^2\left(\frac{\pi \left(0\right)}{a}\right)\right]\\ & =& \frac{-2i\pi \hslash }{a^2}\left[\left(\frac{1}{\left(\pi /a\right)}\right){\sin }^2\left(\pi \right)-\left(\frac{1}{\left(\pi /a\right)}\right)\left(0\right)\right]\end{array}$

Substitute the value of $\sin \pi =0$ in the above expression.

$\begin{array}{rll}\langle p_x\rangle & =& \frac{-2i\pi \hslash }{a^2}\left[\left(\frac{1}{\left(\pi /a\right)}\right)\left(0\right)-\left(\frac{1}{\left(\pi /a\right)}\right)\left(0\right)\right]\\ & =& 0\end{array}$

Therefore, the value of $\langle p_x\rangle$, for ${\Psi }_1$ of a particle-in-a-box is $0$.

From Appendix $1$,

${\displaystyle \int x^2{\sin }^2\left(bx\right)dx}=\frac{x^3}{6}-\left(\frac{x^2}{4b}-\frac{1}{8b^3}\right)\sin \left(2bx\right)-\frac{x}{4b^2}\cos \left(2bx\right)$ …(5)

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

$\langle x^2\rangle ={\displaystyle \underset{0}{\overset{a}{\int }}{\Psi }^{\ast }\widehat{x^2}\Psi dx}$ …(6)

Where,

• $\widehat{x^2}$ represents an operator.

• $\Psi$ represents the wavefunction.

The wavefunction at $n=1$, ${\Psi }_1$, is defined as $\sqrt{\frac{2}{a}}\sin \frac{\pi x}{a}$.

The wavefunction at $n=2$, ${\Psi }_2$, is defined as $\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}$.

The operator $\widehat{x^2}$ is defined as $x^2\left(\right)$.

Substitute the value of ${\Psi }_1$ and $\widehat{x^2}$ in the equation (6).

$\begin{array}{rll}\langle x^2\rangle & =& {\displaystyle \underset{0}{\overset{a}{\int }}{\left(\sqrt{\frac{2}{a}}\sin \frac{\pi x}{a}\right)}^{\ast }{x}^2\left(\sqrt{\frac{2}{a}}\sin \frac{\pi x}{a}\right)dx}\\ & =& {\displaystyle \underset{0}{\overset{a}{\int }}{x}^2{\left(\sqrt{\frac{2}{a}}\sin \frac{\pi x}{a}\right)}^{2}dx}\\ & =& \frac{2}{a}{\displaystyle \underset{0}{\overset{a}{\int }}{x}^2{\left(\sin \frac{\pi x}{a}\right)}^{2}dx}\end{array}$

Assume $b=\frac{\pi }{a}$ and expand the above expression similarly as equation (5).

$\begin{array}{rll}\langle x^2\rangle & =& \left(\frac{2}{a}\right){\left[\frac{x^3}{6}-\left(\frac{x^2}{4\left(\pi /a\right)}-\frac{1}{8{\left(\pi /a\right)}^3}\right)\sin \left(2\left(\pi /a\right)x\right)-\frac{x}{4{\left(\pi /a\right)}^2}\cos \left(2\left(\pi /a\right)x\right)\right]}_0^a\\ & =& \left(\frac{2}{a}\right)\left[\begin{array}{l}\left(\frac{{\left(a\right)}^3}{6}-\left(\frac{{\left(a\right)}^2}{4\left(\pi /a\right)}-\frac{1}{8{\left(\pi /a\right)}^3}\right)\sin \left(2\left(\pi /a\right)\left(a\right)\right)-\frac{\left(a\right)}{4{\left(\pi /a\right)}^2}\cos \left(2\left(\pi /a\right)\left(a\right)\right)\right)\\ -\left(\frac{{\left(0\right)}^3}{6}-\left(\frac{{\left(0\right)}^2}{4\left(\pi /a\right)}-\frac{1}{8{\left(\pi /a\right)}^3}\right)\sin \left(2\left(\pi /a\right)\left(0\right)\right)-\frac{\left(0\right)}{4{\left(\pi /a\right)}^2}\cos \left(2\left(\pi /a\right)\left(0\right)\right)\right)\end{array}\right]\\ & =& \left(\frac{2}{a}\right)\left[\left(\frac{{\left(a\right)}^3}{6}-\left(\frac{a^3}{4{\pi }^2}-\frac{a^3}{8{\pi }^3}\right)\sin \left(2\pi \right)-\frac{a^3}{4{\pi }^2}\cos \left(2\pi \right)\right)-0\right]\end{array}$

Substitute the value of $\sin 2\pi =0$ and $\cos 2\pi =1$ in the above expression.

$\begin{array}{rll}\langle x^2\rangle & =& \left(\frac{2}{a}\right)\left[\left(\frac{{\left(a\right)}^3}{6}-\left(\frac{a^3}{4{\pi }^2}-\frac{a^3}{8{\pi }^3}\right)\left(0\right)-\frac{a^3}{4{\pi }^2}\left(1\right)\right)-0\right]\\ & =& \left(\frac{2}{a}\right)\left(\frac{{\left(a\right)}^3}{6}-\frac{a^3}{4{\pi }^2}\right)\\ & =& \frac{a^2}{3}-\frac{a^2}{2{\pi }^2}\end{array}$

Therefore, the value of $\langle x^2\rangle$ for a particle-in-a-box having $n=1$ is $\left(\frac{a^2}{3}-\frac{a^2}{2{\pi }^2}\right)$.

From Appendix $1$,

${\displaystyle \int {\sin }^2\left(bx\right)dx}=\frac{x}{2}-\frac{1}{4b}\sin \left(2bx\right)$ …(7)

The average value of $\langle p_x\rangle$ is calculated as,

$\langle p_x^2\rangle ={\displaystyle \underset{0}{\overset{a}{\int }}{\Psi }^{\ast }\widehat{p_x^2}\Psi dx}$ …(8)

Where,

• $\widehat{p_x^2}$ represents an operator.

• $\Psi$ represents the wavefunction.

The wavefunction at $n=1$, ${\Psi }_1$, is defined as $\sqrt{\frac{2}{a}}\sin \frac{\pi x}{a}$.

The wavefunction at $n=2$, ${\Psi }_2$, is defined as $\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}$.

The operator $\widehat{p_x^2}$ is defined as $-{\hslash }^2\frac{{\partial }^2}{\partial x^2}\left(\right)$.

Substitute the value of ${\Psi }_1$ and $\widehat{p_x^2}$ in the equation (8)

$\begin{array}{rll}\langle p_x^2\rangle & =& {\displaystyle \underset{0}{\overset{a}{\int }}{\left(\sqrt{\frac{2}{a}}\sin \frac{\pi x}{a}\right)}^{\ast }-{\hslash }^2\frac{{\partial }^2}{\partial x^2}\left(\sqrt{\frac{2}{a}}\sin \frac{\pi x}{a}\right)dx}\\ & =& {\displaystyle \underset{0}{\overset{a}{\int }}\left(\sqrt{\frac{2}{a}}\sin \frac{\pi x}{a}\right)\left(-{\hslash }^2\right)\left(\sqrt{\frac{2}{a}}\right)\left(\frac{\pi }{a}\right)\left(\frac{\partial }{\partial x}\left(\cos \frac{\pi x}{a}\right)\right)dx}\\ & =& {\displaystyle \underset{0}{\overset{a}{\int }}\left(\sqrt{\frac{2}{a}}\sin \frac{\pi x}{a}\right)\left(-{\hslash }^2\right)\left(\sqrt{\frac{2}{a}}\right){\left(\frac{\pi }{a}\right)}^2\left(-\sin \frac{\pi x}{a}\right)dx}\\ & =& \frac{2{\pi }^2{\hslash }^2}{a^3}{\displaystyle \underset{0}{\overset{a}{\int }}\left({\sin }^2\frac{\pi x}{a}\right)dx}\end{array}$

Assume $b=\frac{\pi }{a}$ and expand the above expression similarly as equation (7).

$\begin{array}{rll}\langle p_x^2\rangle & =& \left(\frac{2{\pi }^2{\hslash }^2}{a^3}\right){\left[\frac{x}{2}-\frac{1}{4\left(\pi /a\right)}\sin \left(2\left(\pi /a\right)x\right)\right]}_0^a\\ & =& \left(\frac{2{\pi }^2{\hslash }^2}{a^3}\right)\left[\left(\frac{a}{2}-\frac{1}{4\left(\pi /a\right)}\sin \left(2\left(\pi /a\right)\left(a\right)\right)\right)-\left(\frac{\left(0\right)}{2}-\frac{1}{4\left(\pi /a\right)}\sin \left(2\left(\pi /a\right)\left(0\right)\right)\right)\right]\\ & =& \left(\frac{2{\pi }^2{\hslash }^2}{a^3}\right)\left[\left(\frac{a}{2}-\frac{1}{4\left(\pi /a\right)}\sin \left(2\pi \right)\right)-\left(0\right)\right]\end{array}$

Substitute the value of $\sin 2\pi =0$ in the above expression.

$\begin{array}{rll}\langle p_x^2\rangle & =& \left(\frac{2{\pi }^2{\hslash }^2}{a^3}\right)\left(\frac{a}{2}\right)\\ & =& \frac{{\pi }^2{\hslash }^2}{a^2}\end{array}$

Therefore, the value of $\langle p_x^2\rangle$ for a particle-in-a-box having $n=1$ is $\left(\frac{{\pi }^2{\hslash }^2}{a^2}\right)$.

The given relation is represented as,

$\Delta A=\sqrt{\langle A^2\rangle -{\langle A\rangle }^2}$…(9)

Where,

• $\Delta A$ represents the uncertainty $\Delta A$ in an observable.

The equation (9) can be represented for the value of $\Delta x$ as,

$\Delta x=\sqrt{\langle x^2\rangle -{\langle x\rangle }^2}$…(10)

Substitute the value of $\langle x^2\rangle$ and $\langle x\rangle$ in the equation (10).

$\begin{array}{rll}\Delta x& =& \sqrt{\left(\frac{a^2}{3}-\frac{a^2}{2{\pi }^2}\right)-{\left(\frac{a}{2}\right)}^2}\\ & =& a\sqrt{\frac{1}{3}-\frac{1}{2{\pi }^2}-\frac{1}{4}}\\ & =& a\sqrt{\frac{4{\pi }^2-6-3{\pi }^2}{12{\pi }^2}}\\ & =& \frac{a}{2\pi }\sqrt{\frac{{\pi }^2-6}{3}}\end{array}$

Therefore, the uncertainty in the position of a particle is $\left(\frac{a}{2\pi }\sqrt{\frac{{\pi }^2-6}{3}}\right)$.

The equation (9) can be represented for the value of $\Delta p_x$ as,

$\Delta p_x=\sqrt{\langle p_x{}^2\rangle -{\langle p_x\rangle }^2}$…(11)

Substitute the value of $\langle p_x{}^2\rangle$ and ${\langle p_x\rangle }^2$ in the equation (11).

$\begin{array}{rll}\Delta p_x& =& \sqrt{\left(\frac{{\pi }^2{\hslash }^2}{a^2}\right)-0}\\ & =& \frac{\pi \hslash }{a}\end{array}$

Therefore, the uncertainty in the momentum of a particle is $\frac{\pi \hslash }{a}$.

The uncertainty principle is given by formula as follows.

$\Delta x\Delta p_{\text{x}}\ge \frac{\hslash }{2}$…(12)

Where,

• $\Delta x$ represents the uncertainty in the position of the object.

• $\Delta p_{\text{x}}$ represents the uncertainty in the momentum of the object.

• $\hslash$ represents a constant.

Substitute the value of $\Delta x$ and $\Delta p_{\text{x}}$ in the equation (12).

$\begin{array}{c}\left(\frac{a}{2\pi }\sqrt{\frac{{\pi }^2-6}{3}}\right)\left(\frac{\pi \hslash }{a}\right)\ge \frac{\hslash }{2}\\ \left(\frac{\hslash }{2}\sqrt{\frac{{\pi }^2-6}{3}}\right)\ge \frac{\hslash }{2}\end{array}$

Substitute the value of $\pi =3.14$ in the above expression.

$\begin{array}{c}\left(\frac{\hslash }{2}\right)\left(\sqrt{\frac{{\left(3.14\right)}^2-6}{3}}\right)\ge \frac{\hslash }{2}\\ \left(\frac{\hslash }{2}\right)\left(\sqrt{\frac{3.8596}{3}}\right)\ge \frac{\hslash }{2}\\ \left(\frac{\hslash }{2}\right)\left(\sqrt{1.2865}\right)\ge \frac{\hslash }{2}\\ \left(1.1342\right)\left(\frac{\hslash }{2}\right)\ge \frac{\hslash }{2}\end{array}$

The value of left hand side in the above expression is greater than the value of right-hand side. Therefore, the above expression shows that the uncertainty principle is valid.

Conclusion

The uncertainty in the position of a particle is $\left(\frac{a}{2\pi }\sqrt{\frac{{\pi }^2-6}{3}}\right)$.

The uncertainty in the momentum of a particle is $\frac{\pi \hslash }{a}$.

The uncertainty principle has been validated.