Chapter 10, Problem 10.88E

Interpretation Introduction

(a)

Interpretation:

The given integrals of the wavefunctions of particles-in-boxes are to be evaluated by using equation 10.28.

Concept introduction:

For the orthogonality of the two different wave functions, the product of the wave functions is integrated over the entire limits. It is expressed by the equation as given below.

${\displaystyle \underset{0}{\overset{\infty }{\int }}{\Psi }_m{\Psi }_n}=0$

Where, ${\Psi }_m$ and ${\Psi }_n$ are two different wave functions. The essential condition for two wave functions to be orthogonal to each other is $m\ne n$.

Answer to Problem 10.88E

The value of given integral of wavefunction is $1$.

Explanation of Solution

The combined expression for orthogonality and normality of wave functions is given by the equation 10.28 as shown below.

${\displaystyle \int {\Psi }_m^{*}{\Psi }_n}d\tau =\{\begin{array}{l}0\text{if}m\ne n\\ 1\text{if}m=n\end{array}$

Where,

• $\Psi$is the wave function.

The given integrals of the wavefunctions is ${\displaystyle \int {\Psi }_{\text{4}}^{\text{*}}{\Psi }_{\text{4}}d\tau }$. The value of $m$ is equal to $n$. Hence, the integral of the wavefunctions is calculated by using equation 10.28 as follows:

${\displaystyle \int {\Psi }_{\text{4}}^{\text{*}}{\Psi }_{\text{4}}d\tau }=1$

Therefore, the value of given integral of wavefunction is $1$.

Conclusion

The value of given integral of wavefunction is $1$.

Interpretation Introduction

(b)

Interpretation:

The given integrals of the wavefunctions of particles-in-boxes are to be evaluated by using equation 10.28.

Concept introduction:

For the orthogonality of the two different wave functions, the product of the wave functions is integrated over the entire limits. It is expressed by the equation as given below.

${\displaystyle \underset{0}{\overset{\infty }{\int }}{\Psi }_m{\Psi }_n}=0$

Where, ${\Psi }_m$ and ${\Psi }_n$ are two different wave functions. The essential condition for two wave functions to be orthogonal to each other is $m\ne n$.

Answer to Problem 10.88E

The value of given integral of wavefunction is $0$.

Explanation of Solution

The combined expression for orthogonality and normality of wave functions is given by the equation 10.28 as shown below.

${\displaystyle \int {\Psi }_m^{*}{\Psi }_n}d\tau =\{\begin{array}{l}0\text{if}m\ne n\\ 1\text{if}m=n\end{array}$

Where,

• $\Psi$is the wave function.

The given integrals of the wavefunctions is ${\displaystyle \int {\Psi }_3^{\text{*}}{\Psi }_{\text{4}}d\tau }$. The value of $m$ is not equal to $n$. Hence, the integral of the wavefunctions is calculated by using equation 10.28 as follows:

${\displaystyle \int {\Psi }_3^{\text{*}}{\Psi }_{\text{4}}d\tau }=0$

Therefore, the value of given integral of wavefunction is $0$.

Conclusion

The value of given integral of wavefunction is $0$.

Interpretation Introduction

(c)

Interpretation:

The given integrals of the wavefunctions of particles-in-boxes are to be evaluated by using equation 10.28.

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.

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

Where,

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

• $\Psi$is the wave function.

• $\text{E}$ is the energy

Answer to Problem 10.88E

The value of given integral of wavefunction is $\frac{16h^2}{8m{a}^2}$.

Explanation of Solution

The combined expression for orthogonality and normality of wave functions is given by the equation 10.28 as shown below.

${\displaystyle \int {\Psi }_m^{*}{\Psi }_n}d\tau =\{\begin{array}{l}0\text{if}m\ne n\\ 1\text{if}m=n\end{array}$

Where,

• $\Psi$is the wave function.

The given integrals of the wavefunctions is ${\displaystyle \int {\Psi }_{\text{4}}^{\text{*}}\stackrel{⌢}{H}{\Psi }_{\text{4}}d\tau }$. In the given wavefunction, the eigen function of $\stackrel{⌢}{H}$ is ${\Psi }_{\text{4}}$.

Thus, the given wave function can be expressed as follows:

$\begin{array}{rll}\stackrel{⌢}{H}{\Psi }_4& =& E_4{\Psi }_4\\ {\displaystyle \int {\Psi }_{\text{4}}^{\text{*}}\stackrel{⌢}{H}{\Psi }_{\text{4}}d\tau }& =& {\displaystyle \int {\Psi }_{\text{4}}^{\text{*}}{E}_4{\Psi }_{\text{4}}d\tau }\\ {\displaystyle \int {\Psi }_{\text{4}}^{\text{*}}\stackrel{⌢}{H}{\Psi }_{\text{4}}d\tau }& =& E_4{\displaystyle \int {\Psi }_{\text{4}}^{\text{*}}{\Psi }_{\text{4}}d\tau }\end{array}$

Hence, the given wave function is expressed as $E_4{\displaystyle \int {\Psi }_{\text{4}}^{\text{*}}{\Psi }_{\text{4}}d\tau }$. The value of $m$ is equal to $n$. Thus, the integral of the wavefunctions is given by using equation 10.28 as follows:

${\displaystyle \int {\Psi }_{\text{4}}^{\text{*}}{\Psi }_{\text{4}}d\tau }=1$

The value of given wave function is calculated as follows:

$\begin{array}{rll}{E}_4{\displaystyle \int {\Psi }_{\text{4}}^{\text{*}}{\Psi }_{\text{4}}d\tau }& =& E_4\times 1\\ & =& E_4\end{array}$

Substitute the value of $E_n=\frac{n^2{h}^2}{8m{a}^2}$, where $n=4$ in the above expression.

$\begin{array}{rll}{E}_4{\displaystyle \int {\Psi }_{\text{4}}^{\text{*}}{\Psi }_{\text{4}}d\tau }& =& \frac{{\left(4\right)}^2{h}^2}{8m{a}^2}\\ E_4{\displaystyle \int {\Psi }_{\text{4}}^{\text{*}}{\Psi }_{\text{4}}d\tau }& =& \frac{16h^2}{8m{a}^2}\end{array}$

Therefore, the value of given integral of wavefunction is $\frac{16h^2}{8m{a}^2}$.

Conclusion

The value of given integral of wavefunction is $\frac{16h^2}{8m{a}^2}$.

Interpretation Introduction

(d)

Interpretation:

The given integrals of the wavefunctions of particles-in-boxes are to be evaluated by using equation 10.28.

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.

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

Where,

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

• $\Psi$is the wave function.

• $\text{E}$ is the energy

Answer to Problem 10.88E

The value of given integral of wavefunction is $0$.

Explanation of Solution

The combined expression for orthogonality and normality of wave functions is given by the equation 10.28 as shown below.

${\displaystyle \int {\Psi }_m^{*}{\Psi }_n}d\tau =\{\begin{array}{l}0\text{if}m\ne n\\ 1\text{if}m=n\end{array}$

Where,

• $\Psi$is the wave function.

The given integrals of the wavefunctions is ${\displaystyle \int {\Psi }_{\text{4}}^{\text{*}}\stackrel{⌢}{H}{\Psi }_{2}d\tau }$. In the given wavefunction, the eigen function of $\stackrel{⌢}{H}$ is ${\Psi }_2$.

Thus, the given wave function can be expressed as follows:

$\begin{array}{rll}\stackrel{⌢}{H}{\Psi }_2& =& E_2{\Psi }_2\\ {\displaystyle \int {\Psi }_{\text{4}}^{\text{*}}\stackrel{⌢}{H}{\Psi }_{2}d\tau }& =& {\displaystyle \int {\Psi }_{\text{4}}^{\text{*}}{E}_2{\Psi }_{2}d\tau }\\ {\displaystyle \int {\Psi }_{\text{4}}^{\text{*}}\stackrel{⌢}{H}{\Psi }_{2}d\tau }& =& E_2{\displaystyle \int {\Psi }_{\text{4}}^{\text{*}}{\Psi }_{2}d\tau }\end{array}$

Hence, the given wave function is expressed as $E_2{\displaystyle \int {\Psi }_{\text{4}}^{\text{*}}{\Psi }_{2}d\tau }$. The $m$ is not equal to $n$. thus, the integral of the wavefunctions is given by using equation 10.28 as follows:

${\displaystyle \int {\Psi }_{\text{4}}^{\text{*}}{\Psi }_{2}d\tau }=0$

The value of given wave function is calculated as follows:

$\begin{array}{rll}{E}_2{\displaystyle \int {\Psi }_{\text{4}}^{\text{*}}{\Psi }_{2}d\tau }& =& E_2\times 0\\ & =& 0\end{array}$

Therefore, the value of given integral of wavefunction is $0$.

Conclusion

The value of given integral of wavefunction is $0$.

Interpretation Introduction

(e)

Interpretation:

The given integrals of the wavefunctions of particles-in-boxes are to be evaluated by using equation 10.28.

Concept introduction:

For the orthogonality of the two different wave functions, the product of the wave functions is integrated over the entire limits. It is expressed by the equation as given below.

${\displaystyle \underset{0}{\overset{\infty }{\int }}{\Psi }_m{\Psi }_n}=0$

Where, ${\Psi }_m$ and ${\Psi }_n$ are two different wave functions. The essential condition for two wave functions to be orthogonal to each other is $m\ne n$.

Answer to Problem 10.88E

The value of given integral of wavefunction is $1$.

Explanation of Solution

The combined expression for orthogonality and normality of wave functions is given by the equation 10.28 as shown below.

${\displaystyle \int {\Psi }_m^{*}{\Psi }_n}d\tau =\{\begin{array}{l}0\text{if}m\ne n\\ 1\text{if}m=n\end{array}$

Where,

• $\Psi$is the wave function.

The given integrals of the wavefunctions is ${\displaystyle \iiint {\Psi }_{111}^{\text{*}}{\Psi }_{111}d\tau }$. The $m$ is equal to $n$. Hence, the integral of the wavefunctions is calculated by using equation 10.28 as follows:

${\displaystyle \iiint {\Psi }_{111}^{\text{*}}{\Psi }_{111}d\tau }=1$

Therefore, the value of given integral of wavefunction is $1$.

Conclusion

The value of given integral of wavefunction is $1$.

Interpretation Introduction

(f)

Interpretation:

The given integrals of the wavefunctions of particles-in-boxes are to be evaluated by using equation 10.28.

Concept introduction:

For the orthogonality of the two different wave functions, the product of the wave functions is integrated over the entire limits. It is expressed by the equation as given below.

${\displaystyle \underset{0}{\overset{\infty }{\int }}{\Psi }_m{\Psi }_n}=0$

Where, ${\Psi }_m$ and ${\Psi }_n$ are two different wave functions. The essential condition for two wave functions to be orthogonal to each other is $m\ne n$.

Answer to Problem 10.88E

The value of given integral of wavefunction is $0$.

Explanation of Solution

The combined expression for orthogonality and normality of wave functions is given by the equation 10.28 as shown below.

${\displaystyle \int {\Psi }_m^{*}{\Psi }_n}d\tau =\{\begin{array}{l}0\text{if}m\ne n\\ 1\text{if}m=n\end{array}$

Where,

• $\Psi$is the wave function.

The given integrals of the wavefunctions is ${\displaystyle \iiint {\Psi }_{111}^{\text{*}}{\Psi }_{121}d\tau }$. The $m$ is not equal to $n$. Hence, the integral of the wavefunctions is calculated by using equation 10.28 as follows:

${\displaystyle \iiint {\Psi }_{111}^{\text{*}}{\Psi }_{121}d\tau }=0$

Therefore, the value of given integral of wavefunction is $0$.

Conclusion

The value of given integral of wavefunction is $0$.

Interpretation Introduction

(g)

Interpretation:

The given integrals of the wavefunctions of particles-in-boxes are to be evaluated by using equation 10.28.

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.

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

Where,

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

• $\Psi$is the wave function.

• $\text{E}$ is the energy

Answer to Problem 10.88E

The value of given integral of wavefunction is $\frac{h^2}{8m}\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)$.

Explanation of Solution

The combined expression for orthogonality and normality of wave functions is given by the equation 10.28 as shown below.

${\displaystyle \int {\Psi }_m^{*}{\Psi }_n}d\tau =\{\begin{array}{l}0\text{if}m\ne n\\ 1\text{if}m=n\end{array}$

Where,

• $\Psi$is the wave function.

The given integral of the wavefunctions is ${\displaystyle \iiint {\Psi }_{111}^{\text{*}}\stackrel{⌢}{H}{\Psi }_{111}d\tau }$. In the given wavefunction, the eigen function of $\stackrel{⌢}{H}$ is ${\Psi }_{111}$.

Thus, the given wave function can be expressed as follows:

$\begin{array}{rll}\stackrel{⌢}{H}{\Psi }_{111}& =& E_{111}{\Psi }_{111}\\ {\displaystyle \iiint {\Psi }_{111}^{\text{*}}\stackrel{⌢}{H}{\Psi }_{111}d\tau }& =& {\displaystyle \iiint {\Psi }_{111}^{\text{*}}{E}_{111}{\Psi }_{111}d\tau }\\ {\displaystyle \iiint {\Psi }_{111}^{\text{*}}\stackrel{⌢}{H}{\Psi }_{111}d\tau }& =& E_{111}{\displaystyle \iiint {\Psi }_{111}^{\text{*}}{\Psi }_{111}d\tau }\end{array}$

Hence, the given wave function is expressed as $E_{111}{\displaystyle \iiint {\Psi }_{111}^{\text{*}}{\Psi }_{111}d\tau }$. The value of $m$ is equal to $n$. Thus, the integral of the wavefunctions is given by using equation 10.28 as follows:

${\displaystyle \iiint {\Psi }_{111}^{\text{*}}{\Psi }_{111}d\tau }=1$

The value of given wave function is calculated as follows:

$\begin{array}{rll}{E}_{111}{\displaystyle \iiint {\Psi }_{111}^{\text{*}}{\Psi }_{111}d\tau }& =& E_{111}\times 1\\ & =& E_{111}\end{array}$

Substitute the value of $E_n=\frac{h^2}{8m}\left(\frac{n_x^2}{a^2}+\frac{n_y^2}{b^2}+\frac{n_z^2}{c^2}\right)$, where $n_x=n_y=n_z=1$ in the above expression.

$E_{111}{\displaystyle \iiint {\Psi }_{111}^{\text{*}}{\Psi }_{111}d\tau }=\frac{h^2}{8m}\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)$

Therefore, the value of given integral of wavefunction is $\frac{h^2}{8m}\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)$.

Conclusion

The value of given integral of wavefunction is $\frac{h^2}{8m}\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)$.

Interpretation Introduction

(h)

Interpretation:

The given integrals of the wavefunctions of particles-in-boxes are to be evaluated by using equation 10.28.

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.

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

Where,

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

• $\Psi$is the wave function.

• $\text{E}$ is the energy

Answer to Problem 10.88E

The value of given integral of wavefunction is $0$.

Explanation of Solution

The combined expression for orthogonality and normality of wave functions is given by the equation 10.28 as shown below.

${\displaystyle \int {\Psi }_m^{*}{\Psi }_n}d\tau =\{\begin{array}{l}0\text{if}m\ne n\\ 1\text{if}m=n\end{array}$

Where,

• $\Psi$is the wave function.

The given integrals of the wavefunctions is ${\displaystyle \iiint {\Psi }_{223}^{\text{*}}\stackrel{⌢}{H}{\Psi }_{322}d\tau }$. In the given wavefunction, the eigen function of $\widehat{H}$ is ${\Psi }_{322}$.

Thus, the given wave function can be expressed as follows:

$\begin{array}{rll}\stackrel{⌢}{H}{\Psi }_{322}& =& E_{322}{\Psi }_{322}\\ {\displaystyle \iiint {\Psi }_{223}^{\text{*}}\stackrel{⌢}{H}{\Psi }_{322}d\tau }& =& {\displaystyle \iiint {\Psi }_{223}^{\text{*}}{E}_{322}{\Psi }_{322}d\tau }\\ {\displaystyle \iiint {\Psi }_{223}^{\text{*}}\stackrel{⌢}{H}{\Psi }_{322}d\tau }& =& E_{322}{\displaystyle \iiint {\Psi }_{223}^{\text{*}}{\Psi }_{322}d\tau }\end{array}$

Hence, the given wave function is expressed as $E_{322}{\displaystyle \iiint {\Psi }_{223}^{\text{*}}{\Psi }_{322}d\tau }$. The value of $m$ is not equal to $n$. Thus, the integral of the wavefunctions is given by using equation 10.28 as follows:

${\displaystyle \iiint {\Psi }_{223}^{\text{*}}{\Psi }_{322}d\tau }=0$

The value of given wave function is calculated as follows:

$\begin{array}{rll}{E}_{322}{\displaystyle \iiint {\Psi }_{223}^{\text{*}}{\Psi }_{322}d\tau }& =& E_{322}\times 0\\ & =& 0\end{array}$

Therefore, the value of given integral of wavefunction is $0$.

Conclusion

The value of given integral of wavefunction is $0$.