Chapter 10, Problem 10.27E

Interpretation Introduction

(a)

Interpretation:

The probability for the particle having wavefunction $\Psi =\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}$ in the range of $x=0\text{to}0.02a$ is to be calculated.

Concept introduction:

For the normalization of the wavefunction, the wavefunction is integrated as a product of its conjugate over the entire limits. It is expressed by the equation as given below.

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

Where,

• $N$ is the normalization constant.

• ${\Psi }^{*}$ is the conjugate of the wavefunction.

• $\Psi$ is the wavefunction.

Answer to Problem 10.27E

The probability for the particle having wavefunction $\Psi =\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}$ in the range of $x=0\text{to}0.02a$ is $2.10\times {10}^{-4}$.

Explanation of Solution

For the probability of the wavefunction the expression is as follows.

${\displaystyle \underset{0}{\overset{\infty }{\int }}\left(N\Psi \right)\left(N{\Psi }^{*}\right)}=P$

Where,

• $N$ is the normalization constant.

• ${\Psi }^{*}$ is the conjugate of the wavefunction.

• $\Psi$ is the wavefunction.

• $P$ is the probability.

Substitute the values in the above equation as follows.

$\begin{array}{rll}P& =& {\displaystyle \underset{0}{\overset{0.02a}{\int }}\left(\Psi \right)\left({\Psi }^{*}\right)}dx\\ & =& {\displaystyle \underset{0}{\overset{0.02a}{\int }}\left(\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}\right)\left(\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}\right)}dx\\ & =& \frac{2}{a}{\displaystyle \underset{0}{\overset{0.02a}{\int }}\left(\sin \frac{2\pi x}{a}\right)\left(\sin \frac{2\pi x}{a}\right)}dx\\ & =& \frac{2}{a}{\displaystyle \underset{0}{\overset{0.02a}{\int }}{\left(\sin \frac{2\pi x}{a}\right)}^{2}dx}\end{array}$

The above equation is simplified as given below.

$\begin{array}{rll}& =& \frac{2}{a}{\displaystyle \underset{0}{\overset{0.02a}{\int }}\left({\sin }^2\frac{2\pi x}{a}\right)dx}\\ & =& \frac{2}{a}\times {\left[\frac{x}{2}-\frac{a}{8\pi }\sin \frac{4\pi x}{a}\right]}_0^{0.02a}\\ & =& \frac{2}{a}\times \left[\left(\frac{0.02a}{2}-\frac{a}{8\pi }\sin \frac{4\pi \left(0.02a\right)}{a}\right)-0\right]\\ & =& \frac{2}{a}\times \left(\frac{0.02a}{2}-\frac{a}{8\pi }\sin 0.08\pi \right)\end{array}$

$\begin{array}{l}=\frac{2}{a}\times \left(\frac{0.02a}{2}-\frac{a}{25.12}\sin \left(0.2512\right)\right)\\ =\frac{2}{a}\times \left(\frac{0.02a}{2}-\frac{0.24856a}{25.12}\right)\\ =\frac{2}{a}\times \frac{0.0053a}{50.24}\\ =2.10\times {10}^{-4}\end{array}$

Conclusion

The probability for the particle having wavefunction $\Psi =\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}$ in the range of $x=0\text{to}0.02a$ is calculated as $2.10\times {10}^{-4}$.

Interpretation Introduction

(b)

Interpretation:

The probability for the particle having wavefunction $\Psi =\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}$ in the range of $x=0.24a\text{to}0.26a$ is to be calculated.

Concept introduction:

For the normalization of the wavefunction, the wavefunction is integrated as a product of its conjugate over the entire limits. It is expressed by the equation as given below.

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

Where,

• $N$ is the normalization constant.

• ${\Psi }^{*}$ is the conjugate of the wavefunction.

• $\Psi$ is the wavefunction.

Answer to Problem 10.27E

The probability for the particle having wavefunction $\Psi =\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}$ in the range of $x=0.24a\text{to}0.26a$ is $0.075$.

Explanation of Solution

For the probability of the wavefunction the expression is as follows.

${\displaystyle \underset{0}{\overset{\infty }{\int }}\left(N\Psi \right)\left(N{\Psi }^{*}\right)}=P$

Where,

• $N$ is the normalization constant.

• ${\Psi }^{*}$ is the conjugate of the wavefunction.

• $\Psi$ is the wavefuction.

• $P$ is the probability.

Substitute the values in the above equation as follows.

$\begin{array}{rll}P& =& {\displaystyle \underset{0.24a}{\overset{0.26a}{\int }}\left(\Psi \right)\left({\Psi }^{*}\right)}dx\\ & =& {\displaystyle \underset{0.24a}{\overset{0.26a}{\int }}\left(\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}\right)\left(\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}\right)}dx\\ & =& \frac{2}{a}{\displaystyle \underset{0.24a}{\overset{0.26a}{\int }}\left(\sin \frac{2\pi x}{a}\right)\left(\sin \frac{2\pi x}{a}\right)}dx\\ & =& \frac{2}{a}{\displaystyle \underset{0.24a}{\overset{0.26a}{\int }}{\left(\sin \frac{2\pi x}{a}\right)}^{2}dx}\end{array}$

The above equation is simplified as follows.

$\begin{array}{rll}& =& \frac{2}{a}{\displaystyle \underset{0.24a}{\overset{0.26a}{\int }}\left({\sin }^2\frac{2\pi x}{a}\right)dx}\\ & =& \frac{2}{a}\times {\left[\frac{x}{2}-\frac{a}{8\pi }\sin \frac{4\pi x}{a}\right]}_{0.24a}^{0.26a}\\ & =& \frac{2}{a}\times \left[\left(\frac{0.26a}{2}-\frac{a}{8\pi }\sin \frac{4\pi \left(0.26a\right)}{a}\right)-\left(\frac{0.24a}{2}-\frac{a}{8\pi }\sin \frac{4\pi \left(0.24a\right)}{a}\right)\right]\\ & =& \frac{2}{a}\times \left(\frac{0.26a}{2}-\frac{0.24a}{2}-\frac{a}{8\pi }\sin 1.04\pi +\frac{a}{8\pi }\sin 0.81\pi \right)\end{array}$

$\begin{array}{l}=\frac{2}{a}\times \left(\frac{0.26a}{2}-\frac{0.24a}{2}-\frac{a}{8\pi }\sin 1.04\pi +\frac{a}{8\pi }\sin 0.81\pi \right)\\ =\frac{2}{a}\times \left(\frac{0.02a}{2}+\frac{0.68683a}{25.12}\right)\\ =\frac{2}{a}\times \frac{\left(1.87606a\right)}{50.24}\\ =0.075\end{array}$

Conclusion

The probability for the particle having wavefunction $\Psi =\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}$ in the range of $x=0.24a\text{to}0.26a$ is calculated as $0.075$.

Interpretation Introduction

(c)

Interpretation:

The probability for the particle having wavefunction $\Psi =\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}$ in the range of $x=0.49a\text{to}0.51a$ is to be calculated.

Concept introduction:

For the normalization of the wavefunction, the wavefunction is integrated as a product of its conjugate over the entire limits. It is expressed by the equation as given below.

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

Where,

• $N$ is the normalization constant.

• ${\Psi }^{*}$ is the conjugate of the wavefunction.

• $\Psi$ is the wavefunction.

Answer to Problem 10.27E

The probability for the particle having wavefunction $\Psi =\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}$ in the range of $x=0.49a\text{to}0.51a$ is $4.9\times {10}^{-3}$.

Explanation of Solution

For the probability of the wavefunction the expression is as follows.

${\displaystyle \underset{0}{\overset{\infty }{\int }}\left(N\Psi \right)\left(N{\Psi }^{*}\right)}=P$

Where,

• $N$ is the normalization constant.

• ${\Psi }^{*}$ is the conjugate of the wavefunction.

• $\Psi$ is the wavefunction.

• $P$ is the probability.

Substitute the values in the above equation as follows.

$\begin{array}{rll}P& =& {\displaystyle \underset{0.49a}{\overset{0.51a}{\int }}\left(\Psi \right)\left({\Psi }^{*}\right)}dx\\ & =& {\displaystyle \underset{0.49a}{\overset{0.51a}{\int }}\left(\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}\right)\left(\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}\right)}dx\\ & =& \frac{2}{a}{\displaystyle \underset{0.49a}{\overset{0.51a}{\int }}\left(\sin \frac{2\pi x}{a}\right)\left(\sin \frac{2\pi x}{a}\right)}dx\\ & =& \frac{2}{a}{\displaystyle \underset{0.49a}{\overset{0.51a}{\int }}{\left(\sin \frac{2\pi x}{a}\right)}^{2}dx}\end{array}$

The above equation is simplified as follows.

$\begin{array}{rll}& =& \frac{2}{a}{\displaystyle \underset{0.49a}{\overset{0.51a}{\int }}\left({\sin }^2\frac{2\pi x}{a}\right)dx}\\ & =& \frac{2}{a}\times {\left[\frac{x}{2}-\frac{a}{8\pi }\sin \frac{4\pi x}{a}\right]}_{0.49a}^{0.51a}\\ & =& \frac{2}{a}\times \left[\left(\frac{0.51a}{2}-\frac{a}{8\pi }\sin \frac{4\pi \left(0.51a\right)}{a}\right)-\left(\frac{0.49a}{2}-\frac{a}{8\pi }\sin \frac{4\pi \left(0.49a\right)}{a}\right)\right]\\ & =& \frac{2}{a}\times \left(\frac{0.51a}{2}-\frac{0.49a}{2}-\frac{a}{8\pi }\sin 2.04\pi +\frac{a}{8\pi }\sin 1.96\pi \right)\end{array}$

$\begin{array}{l}=\frac{2}{a}\times \left(\frac{0.51a}{2}-\frac{0.49a}{2}-\frac{a}{8\pi }\sin 2.04\pi +\frac{a}{8\pi }\sin 1.96\pi \right)\\ =\frac{2}{a}\times \left(\frac{0.02a}{2}-\frac{0.189605a}{25.12}\right)\\ =\frac{2}{a}\times \frac{0.12319a}{50.24}\\ =4.9\times {10}^{-3}\end{array}$

Conclusion

The probability for the particle having wavefunction $\Psi =\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}$ in the range of $x=0.49a\text{to}0.51a$ is calculated as $4.9\times {10}^{-3}$.

Interpretation Introduction

(d)

Interpretation:

The probability for the particle having wavefunction $\Psi =\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}$ in the range of $x=0.74a\text{to}0.76a$ is to be calculated.

Concept introduction:

For the normalization of the wavefunction, the wavefunction is integrated as a product of its conjugate over the entire limits. It is expressed by the equation as given below.

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

Where,

• $N$ is the normalization constant.

• ${\Psi }^{*}$ is the conjugate of the wavefunction.

• $\Psi$ is the wavefunction.

Answer to Problem 10.27E

The probability for the particle having wavefunction $\Psi =\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}$ in the range of $x=0.74a\text{to}0.76a$ is $0.047$.

Explanation of Solution

For the probability of the wavefunction the expression is as follows.

${\displaystyle \underset{0}{\overset{\infty }{\int }}\left(N\Psi \right)\left(N{\Psi }^{*}\right)}=P$

Where,

• $N$ is the normalization constant.

• ${\Psi }^{*}$ is the conjugate of the wavefunction.

• $\Psi$ is the wavefunction.

• $P$ is the probability.

Substitute the values in the above equation as follows.

$\begin{array}{rll}P& =& {\displaystyle \underset{0.74a}{\overset{0.76a}{\int }}\left(\Psi \right)\left({\Psi }^{*}\right)}dx\\ & =& {\displaystyle \underset{0.74a}{\overset{0.76a}{\int }}\left(\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}\right)\left(\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}\right)}dx\\ & =& \frac{2}{a}{\displaystyle \underset{0.74a}{\overset{0.76a}{\int }}\left(\sin \frac{2\pi x}{a}\right)\left(\sin \frac{2\pi x}{a}\right)}dx\\ & =& \frac{2}{a}{\displaystyle \underset{0.74a}{\overset{0.76a}{\int }}{\left(\sin \frac{2\pi x}{a}\right)}^{2}dx}\end{array}$

The above equation is simplified as given below.

$\begin{array}{rll}& =& \frac{2}{a}{\displaystyle \underset{0.74a}{\overset{0.76a}{\int }}\left({\sin }^2\frac{2\pi x}{a}\right)dx}\\ & =& \frac{2}{a}\times {\left[\frac{\text{x}}{2}-\frac{a}{8\pi }\sin \frac{4\pi x}{a}\right]}_{0.74a}^{0.76a}\\ & =& \frac{2}{a}\times \left[\left(\frac{0.76a}{2}-\frac{a}{8\pi }\sin \frac{4\pi \left(0.76a\right)}{a}\right)-\left(\frac{0.74a}{2}-\frac{a}{8\pi }\sin \frac{4\pi \left(0.74a\right)}{a}\right)\right]\\ & =& \frac{2}{a}\times \left(\frac{0.76a}{2}-\frac{0.74a}{2}-\frac{a}{8\pi }\sin 3.04\pi +\frac{a}{8\pi }\sin 2.96\pi \right)\end{array}$

$\begin{array}{l}=\frac{2}{a}\times \left(\frac{0.76a}{2}-\frac{0.74a}{2}-\frac{a}{8\pi }\sin 3.04\pi +\frac{a}{8\pi }\sin 2.96\pi \right)\\ =\frac{2}{a}\times \left(\frac{0.02a}{2}+\frac{0.337955a}{25.12}\right)\\ =\frac{2}{a}\times \frac{1.17831a}{50.24}\\ =0.047\end{array}$

Conclusion

The probability for the particle having wavefunction $\Psi =\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}$ in the range of $x=0.74a\text{to}0.76a$ is calculated as $0.047$.

Interpretation Introduction

(e)

Interpretation:

The probability for the particle having wavefunction $\Psi =\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}$ in the range of $x=0.98a\text{to}1.00a$ is to be calculated. The graph of probabilities versus $x$ is to be plotted.

Concept introduction:

For the normalization of the wavefunction, the wavefunction is integrated as a product of its conjugate over the entire limits. It is expressed by the equation as given below.

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

Where,

• $N$ is the normalization constant.

• ${\Psi }^{*}$ is the conjugate of the wavefunction.

• $\Psi$ is the wavefunction.

Answer to Problem 10.27E

The probability for the particle having wavefunction $\Psi =\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}$ in the range of $x=0.98a\text{to}1.00a$ is $9.75\times {10}^{-3}$.The graph of probabilities versus $x$ is given below.

Explanation of Solution

For the probability of the wavefunction the expression is as follows.

${\displaystyle \underset{0}{\overset{\infty }{\int }}\left(N\Psi \right)\left(N{\Psi }^{*}\right)}=P$

Where,

• $N$ is the normalization constant.

• ${\Psi }^{*}$ is the conjugate of the wavefunction.

• $\Psi$ is the wavefunction.

• $P$ is the probability.

Substitute the values in the above equation as follows.

$\begin{array}{rll}P& =& {\displaystyle \underset{0.98a}{\overset{1a}{\int }}\left(\Psi \right)\left({\Psi }^{*}\right)}dx\\ & =& {\displaystyle \underset{0.98a}{\overset{1a}{\int }}\left(\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}\right)\left(\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}\right)}dx\\ & =& \frac{2}{a}{\displaystyle \underset{0.98a}{\overset{1a}{\int }}\left(\sin \frac{2\pi x}{a}\right)\left(\sin \frac{2\pi x}{a}\right)}dx\\ & =& \frac{2}{a}{\displaystyle \underset{0.98a}{\overset{1a}{\int }}{\left(\sin \frac{2\pi x}{a}\right)}^{2}dx}\end{array}$

The above equation is simplified as follows.

$\begin{array}{rll}& =& \frac{2}{a}{\displaystyle \underset{0.98a}{\overset{1a}{\int }}\left({\sin }^2\frac{2\pi x}{a}\right)dx}\\ & =& \frac{2}{a}\times {\left[\frac{x}{2}-\frac{a}{8\pi }\sin \frac{4\pi x}{a}\right]}_{0.98a}^{1a}\\ & =& \frac{2}{a}\times \left[\left(\frac{a}{2}-\frac{a}{8\pi }\sin \frac{4\pi \left(a\right)}{a}\right)-\left(\frac{0.98a}{2}-\frac{a}{8\pi }\sin \frac{4\pi \left(0.98a\right)}{a}\right)\right]\\ & =& \frac{2}{a}\times \left(\frac{a}{2}-\frac{0.98a}{2}-\frac{a}{8\pi }\sin 4\pi +\frac{a}{8\pi }\sin 3.92\pi \right)\end{array}$

$\begin{array}{l}=\frac{2}{a}\times \left(\frac{a}{2}-\frac{0.98a}{2}-\frac{a}{8\pi }\sin 4\pi +\frac{a}{8\pi }\sin 3.92\pi \right)\\ =\frac{2}{a}\times \left(\frac{0.02a}{2}-\frac{0.12869a}{25.12}\right)\\ =\frac{2}{a}\times \left(\frac{0.24502a}{50.24}\right)\\ =9.75\times {10}^{-3}\end{array}$

The plot the probabilities versus $x$ is given in figure 1.

Figure 1

The plot shows the probability for the given wave function. According to this plot, the probability of finding the particle is maximum in the range of $x=0.24a\text{to}0.26a$ and $x=0.74a\text{to}0.76a$ which is different from the previous exercise. Also, the plot gives the curve for the sine function as given.

Conclusion

The probability for the particle having wavefunction $\Psi =\sqrt{\frac{2}{a}}\sin \frac{2\pi x}{a}$ in the range of $x=0.98a\text{to}1.00a$ is calculated as $9.75\times {10}^{-3}$.The graph of probabilities versus $x$ is given in figure 1.