Chapter 10, Problem 10.63E

Interpretation Introduction

(a)

Interpretation:

The average value of position, $\langle x\rangle$ for the given wavefunction is to be evaluated.

Concept introduction:

In quantum mechanics, the wavefunction is given by $\Psi$. The wavefunction contains all the information about the state of the system. The wavefunction is the function of the coordinates of particles and time. The square of the probability function, ${\left|\Psi \right|}^2$, relates to the probability density.

Answer to Problem 10.63E

The average value of position, $\langle x\rangle$ for the given wavefunction is $\frac{3}{4}$.

Explanation of Solution

The given wavefunction is $\Psi =x$.

The normalization constant of the given wavefunction is assumed to be $N$. Hence, the normalized wavefunction is $\Psi =Nx$.

The normalization of given wavefunction is done by the formula,

${\displaystyle \underset{0}{\overset{1}{\int }}{\left(\Psi \right)}^{*}\Psi }dx=1$ …(1)

Substitute the value of $\Psi$ in the equation (1).

$\begin{array}{rll}{\displaystyle \underset{0}{\overset{1}{\int }}{\left(Nx\right)}^{*}Nx}dx& =& 1\\ N^2{\displaystyle \underset{0}{\overset{1}{\int }}{x}^2}dx& =& 1\\ N^2{\left[\frac{x^3}{3}\right]}_0^1& =& 1\\ N^2\left[\frac{1}{3}-\frac{0}{3}\right]& =& 1\end{array}$

Solve the above equation.

$\begin{array}{rll}{N}^2\frac{1}{3}& =& 1\\ N^2& =& 3\\ N& =& \pm \sqrt{3}\end{array}$

Only positive square roots are taken for normalization constant. Therefore, the normalized wavefunction is $\Psi =\sqrt{3}x$.

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

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

$\langle x\rangle ={\displaystyle \underset{0}{\overset{1}{\int }}{\Psi }^{\ast }\widehat{x}\Psi dx}$ …(2)

Where,

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

• $\Psi$represents the wavefunction.

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

$\begin{array}{rll}\langle x\rangle & =& {\displaystyle \underset{0}{\overset{1}{\int }}{\left(\sqrt{3}x\right)}^{\ast }x\left(\sqrt{3}x\right)dx}\\ & =& 3{\displaystyle \underset{0}{\overset{1}{\int }}{x}^{3}dx}\\ & =& 3{\left[\frac{x^4}{4}\right]}_0^1\\ & =& 3\left[\frac{{\left(1\right)}^4}{4}-\frac{{\left(0\right)}^4}{4}\right]\end{array}$

Solve the above expression for the value of $\langle x\rangle$.

$\langle x\rangle =\frac{3}{4}$

Therefore, the average value of position, $\langle x\rangle$ for the given wavefunction is $\frac{3}{4}$.

Conclusion

The average value of position, $\langle x\rangle$ for the given wavefunction is $\frac{3}{4}$.

Interpretation Introduction

(b)

Interpretation:

The average value of position, $\langle x\rangle$ for the given wavefunction is to be evaluated.

Concept introduction:

In quantum mechanics, the wavefunction is given by $\Psi$. The wavefunction contains all the information about the state of the system. The wavefunction is the function of the coordinates of particles and time. The square of the probability function, ${\left|\Psi \right|}^2$, relates to the probability density.

Answer to Problem 10.63E

The average value of position, $\langle x\rangle$ for the given wavefunction is $\frac{3}{2}$.

Explanation of Solution

The given wavefunction is $\Psi =x$.

The normalization constant of the given wavefunction is assumed to be $N$. Hence, the normalized wavefunction is $\Psi =Nx$.

The normalization of given wavefunction is done by the formula,

${\displaystyle \underset{0}{\overset{2}{\int }}{\left(\Psi \right)}^{*}\Psi }dx=1$ …(3)

Substitute the value of $\Psi$ in the above formula.

$\begin{array}{rll}{\displaystyle \underset{0}{\overset{2}{\int }}{\left(Nx\right)}^{*}Nxdx}& =& 1\\ N^2{\displaystyle \underset{0}{\overset{2}{\int }}{x}^2}dx& =& 1\\ N^2{\displaystyle \underset{0}{\overset{2}{\int }}\frac{x^{2+1}}{2+1}}& =& 1\\ N^2{\left[\frac{x^3}{3}\right]}_0^2& =& 1\end{array}$

Solve the above equation.

$\begin{array}{rll}{N}^2\left[\frac{8}{3}-\frac{0}{3}\right]& =& 1\\ N^2\left[\frac{8}{3}\right]& =& 1\\ N^2& =& \frac{3}{8}\\ N& =& \pm \sqrt{\frac{3}{8}}\end{array}$

Only positive square root is taken for normalization constant. Therefore, the normalized wavefunction is $\text{ψ}=\sqrt{\frac{3}{8}}x$.

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

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

$\langle x\rangle ={\displaystyle \underset{0}{\overset{2}{\int }}{\Psi }^{\ast }\widehat{x}\Psi dx}$ …(4)

Where,

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

• $\Psi$represents the wavefunction.

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

$\begin{array}{c}\langle x\rangle ={\displaystyle \underset{0}{\overset{2}{\int }}{\left(\sqrt{\frac{3}{8}}x\right)}^{\ast }x\left(\sqrt{\frac{3}{8}}x\right)dx}\\ =\frac{3}{8}{\displaystyle \underset{0}{\overset{2}{\int }}{x}^{3}dx}\\ =\frac{3}{8}{\left[\frac{x^4}{4}\right]}_0^2\\ =\frac{3}{8}\left[\frac{{\left(2\right)}^4}{4}-\frac{{\left(0\right)}^4}{4}\right]\end{array}$

Solve the above expression for the value of $\langle x\rangle$.

$\langle x\rangle =\frac{3}{2}$

Therefore, the average value of position, $\langle x\rangle$ for the given wavefunction is $\frac{3}{2}$.

Conclusion

The average value of position, $\langle x\rangle$ for the given wavefunction is $\frac{3}{2}$.

Interpretation Introduction

(c)

Interpretation:

The average value of position, $\langle x\rangle$ for the given wavefunction is to be evaluated.

Concept introduction:

In quantum mechanics, the wavefunction is given by $\Psi$. The wavefunction contains all the information about the state of the system. The wavefunction is the function of the coordinates of particles and time. The square of the probability function, ${\left|\Psi \right|}^2$, relates to the probability density.

Answer to Problem 10.63E

The average value of position, $\langle x\rangle$ for the given wavefunction is $\frac{\pi }{2}$.

Explanation of Solution

The given wavefunction is $\Psi =\sin 2x$.

The normalization constant of the given wavefunction is assumed to be $N$. Hence, the normalized wavefunction is $\Psi =N\sin 2x$.

From Appendix $1$,

${\displaystyle \int {\sin }^{2}bxdx}=\frac{x}{2}-\frac{1}{4b}\sin \left(2bx\right)$ …(5)

The normalization of given wavefunction is done by the formula,

${\displaystyle \underset{0}{\overset{\pi }{\int }}{\left(\Psi \right)}^{*}\Psi }dx=1$ …(6)

Substitute the value of $\text{ψ}$ in the above formula.

$\begin{array}{rll}{\displaystyle \underset{0}{\overset{\pi }{\int }}{\left(N\sin 2x\right)}^{*}N\sin 2xdx}& =& 1\\ N^2{\displaystyle \underset{0}{\overset{\pi }{\int }}{\sin }^{2}2xdx}& =& 1\end{array}$

Assume $b=2$ and expand the above expression similarly as equation (5).

$\begin{array}{rll}{N}^2{\left[\frac{x}{2}-\frac{1}{8}\sin 4x\right]}_0^{\pi }& =& 1\\ N^2{\left[\left(\frac{\pi }{2}-\frac{1}{8}\sin 4\pi \right)-\left(\frac{0}{2}-\frac{1}{8}\sin \left(0\right)\right)\right]}_0^{\pi }& =& 1\end{array}$

Substitute the value of $\sin 0=0$ and $\sin 4\pi =0$ in the above expression.

$\begin{array}{l}{N}^2{\left[\left(\frac{\pi }{2}-\frac{1}{8}\left(0\right)\right)-\left(\frac{0}{2}-\frac{1}{8}\left(0\right)\right)\right]}_0^{\pi }=1\\ N^2\left(\frac{\pi }{2}\right)=1\end{array}$

Solve the above equation.

$N=\pm \sqrt{\frac{2}{\pi }}$

Only positive square root is taken for normalization constant. Therefore, the normalized wavefunction is $\sqrt{\frac{2}{\pi }}\sin 2x$.

From Appendix $1$,

${\displaystyle \int x{\sin }^2\left(bx\right)dx}=\frac{x^2}{4}-\frac{x}{4b}\sin \left(2bx\right)-\frac{1}{8b^2}\cos \left(2bx\right)$ …(7)

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

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

Where,

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

• $\Psi$represents the wavefunction.

Substitute the value of $\Psi$ and $\widehat{x}$ in the equation (8).

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

Assume $b=2$ and expand the above expression similarly as equation (7).

$\begin{array}{rll}\langle x\rangle & =& \frac{2}{\pi }{\left[\frac{x^2}{4}-\frac{x}{4\left(2\right)}\sin \left(\left(2\right)\left(2\right)x\right)-\frac{1}{\left(8\right){\left(2\right)}^2}\cos \left(\left(2\right)\left(2\right)x\right)\right]}_0^{\pi }\\ & =& \frac{2}{\pi }\left[\begin{array}{l}\left(\frac{{\left(\pi \right)}^2}{4}-\frac{\left(\pi \right)}{8}\sin \left(4\left(\pi \right)\right)-\frac{1}{32}\cos \left(4\left(\pi \right)\right)\right)\\ -\left(\frac{{\left(0\right)}^2}{4}-\frac{\left(0\right)}{8}\sin \left(4\left(0\right)\right)-\frac{1}{32}\cos \left(4\left(0\right)\right)\right)\end{array}\right]\end{array}$

Substitute the value of $\cos 4\pi =1$, $\cos 0=1$, $\sin 0=0$ and $\sin 4\pi =0$ in the above expression.

$\begin{array}{rll}\langle x\rangle & =& \frac{2}{\pi }\left[\left(\frac{{\left(\pi \right)}^2}{4}-\frac{\left(\pi \right)}{8}\left(0\right)-\frac{1}{32}\left(1\right)\right)-\left(\frac{{\left(0\right)}^2}{4}-\frac{\left(0\right)}{8}\left(0\right)-\frac{1}{32}\left(1\right)\right)\right]\\ & =& \frac{2}{\pi }\left(\frac{{\left(\pi \right)}^2}{4}\right)\\ & =& \frac{\pi }{2}\end{array}$

Therefore, the average value of position, $\langle x\rangle$ for the given wavefunction is $\frac{\pi }{2}$.

Conclusion

The average value of position, $\langle x\rangle$ for the given wavefunction is $\frac{\pi }{2}$.