Chapter 11, Problem 11.54E

Interpretation Introduction

(a)

Interpretation:

The eigenvalues of total angular momentum is to be evaluated using the complete forms of given wavefunction ${\Psi }_{3,-2}$ and operators.

Concept introduction:

The eigenvalues of the wavefunction that are obtained when an operator is applied are the only possible values of observables. The expression for the eigenvalue is given by,

$\widehat{A}\Psi =a\Psi$

The total angular momentum does not depend on the mass of the particle, radius of the rotation and also the magnetic quantum number.

Answer to Problem 11.54E

The eigenvalues of total angular momentum is $12{\hslash }^2$.

Explanation of Solution

Explanation:

The general equation for the wavefunction in the 3-dimensional rotation is,

${\Psi }_{l,m_l}=\frac{1}{\sqrt{2\pi }}{e}^{im\varphi }\cdot {\theta }_{l,m_l}$

The complete form of ${\Psi }_{3,-2}$ using Table 11.3 is,

${\Psi }_{3,-2}=\frac{\sqrt{105}}{4\sqrt{2\pi }}{\sin }^2\theta \cos \theta e^{-2i\varphi }$

The total angular momentum using the complete forms of operators is,

${\widehat{L}}^2\Psi =-{\hslash }^2\left(\frac{{\partial }^2}{\partial {\theta }^2}+\cot \theta \frac{\partial }{\partial \theta }+\frac{1}{{\sin }^2\theta }\frac{{\partial }^2}{\partial {\varphi }^2}\right)\Psi$

The first derivative of the given wavefunction with respect to $\theta$ is,

$\frac{\partial }{\partial \theta }\left(\frac{\sqrt{105}}{4\sqrt{2\pi }}{\sin }^2\theta \cos \theta e^{-2i\varphi }\right)$

$=\frac{\sqrt{105}}{4\sqrt{2\pi }}\left(2\sin \theta {\cos }^2\theta -{\sin }^3\theta \right)e^{-2i\varphi }$…(1)

The second derivative of the given wavefunction with respect to $\theta$ is,

$\frac{{\partial }^2}{\partial {\theta }^2}=\frac{\partial }{\partial \theta }\left(\frac{\sqrt{105}}{4\sqrt{2\pi }}\left(2\sin \theta {\cos }^2\theta -{\sin }^3\theta \right)e^{-2i\varphi }\right)$

$=\frac{\sqrt{105}}{4\sqrt{2\pi }}\left(2{\cos }^3\theta -4{\sin }^2\theta \cos \theta -3{\sin }^2\theta \cos \theta \right)e^{-2i\varphi }$…(2)

The second derivative of the given wavefunction with respect to $\varphi$ is,

$\begin{array}{l}\frac{{\partial }^2}{\partial {\varphi }^2}\left(\frac{\sqrt{105}}{4\sqrt{2\pi }}{\sin }^2\theta \cos \theta e^{-2i\varphi }\right)\\ ={\left(-2\right)}^2{i}^2\times \frac{\sqrt{105}}{4\sqrt{2\pi }}{\sin }^2\theta \cos \theta e^{-2i\varphi }\\ =-4\times \frac{\sqrt{105}}{4\sqrt{2\pi }}{\sin }^2\theta \cos \theta e^{-2i\varphi }\end{array}$

$=-\frac{\sqrt{105}}{\sqrt{2\pi }}{\sin }^2\theta \cos \theta e^{-2i\varphi }$…(3)

Substitute equation (1), (2) and (3) in the equation of total angular momentum as shown below.

${\widehat{L}}^2{\Psi }_{3,-2}=-{\hslash }^2\left(\begin{array}{l}\frac{\sqrt{105}}{4\sqrt{2\pi }}\left(2{\cos }^3\theta -4{\sin }^2\theta \cos \theta -3{\sin }^2\theta \cos \theta \right)e^{-2i\varphi }\\ +\cot \theta \frac{\sqrt{105}}{4\sqrt{2\pi }}\left(2\sin \theta {\cos }^2\theta -{\sin }^3\theta \right)e^{-2i\varphi }-\frac{1}{{\sin }^2\theta }\frac{\sqrt{105}}{\sqrt{2\pi }}{\sin }^2\theta \cos \theta e^{-2i\varphi }\end{array}\right)$

Take common terms together and rearrange the given equation as shown below.

${\widehat{L}}^2{\Psi }_{3,-2}=-{\hslash }^2\frac{\sqrt{105}}{4\sqrt{2\pi }}{e}^{-2i\varphi }\left(\begin{array}{l}\left(2{\cos }^3\theta -7{\sin }^2\theta \cos \theta \right)\\ +\cot \theta \left(2\sin \theta {\cos }^2\theta -{\sin }^3\theta \right)e^{-2i\varphi }-4\cos \theta e^{-2i\varphi }\end{array}\right)$

Substitute the value of $\cot \theta =\frac{\cos \theta }{\sin \theta }$ in the given equation.

$\begin{array}{l}{\widehat{L}}^2{\Psi }_{3,-2}=-{\hslash }^2\frac{\sqrt{105}}{4\sqrt{2\pi }}{e}^{-2i\varphi }\left(\begin{array}{l}\left(2{\cos }^3\theta -7{\sin }^2\theta \cos \theta \right)\\ +\frac{\cos \theta }{\sin \theta }\left(2\sin \theta {\cos }^2\theta -{\sin }^3\theta \right)e^{-2i\varphi }-4\cos \theta e^{-2i\varphi }\end{array}\right)\\ {\widehat{L}}^2{\Psi }_{3,-2}=-{\hslash }^2\frac{\sqrt{105}}{4\sqrt{2\pi }}{e}^{-2i\varphi }\left(\left(2{\cos }^3\theta -7{\sin }^2\theta \cos \theta \right)+\left(2{\cos }^3\theta -\cos \theta {\sin }^2\theta \right)-4\cos \theta \right)\\ {\widehat{L}}^2{\Psi }_{3,-2}=-{\hslash }^2\frac{\sqrt{105}}{4\sqrt{2\pi }}{e}^{-2i\varphi }\left(4{\cos }^3\theta -8{\sin }^2\theta \cos \theta -4\cos \theta \right)\end{array}$

Substitute ${\cos }^2\theta =1-{\sin }^2\theta$ in the given equation.

$\begin{array}{l}{\widehat{L}}^2{\Psi }_{3,-2}=-{\hslash }^2\frac{\sqrt{105}}{4\sqrt{2\pi }}{e}^{-2i\varphi }\left(4\cos \theta \left(1-{\sin }^2\theta \right)-8{\sin }^2\theta \cos \theta -4\cos \theta \right)\\ {\widehat{L}}^2{\Psi }_{3,-2}=-{\hslash }^2\frac{\sqrt{105}}{4\sqrt{2\pi }}{e}^{-2i\varphi }\left(4\cos \theta -4{\sin }^2\theta \cos \theta -8{\sin }^2\theta \cos \theta -4\cos \theta \right)\\ {\widehat{L}}^2{\Psi }_{3,-2}=-{\hslash }^2\frac{\sqrt{105}}{4\sqrt{2\pi }}{e}^{-2i\varphi }\left(-12{\sin }^2\theta \cos \theta \right)\\ {\widehat{L}}^2{\Psi }_{3,-2}=12{\hslash }^2\frac{\sqrt{105}}{4\sqrt{2\pi }}{e}^{-2i\varphi }\left({\sin }^2\theta \cos \theta \right)\end{array}$

Thus, the total angular momentum is represented as,

${\widehat{L}}^2{\Psi }_{3,-2}=12{\hslash }^2{\Psi }_{3,-2}$

The eigenvalues of total angular momentum is $12{\hslash }^2$.

Conclusion

The eigenvalues of total angular momentum is $12{\hslash }^2$.

Interpretation Introduction

(b)

Interpretation:

The eigenvalues of z-component of angular momentum is to be evaluated using the complete forms of given wavefunction ${\Psi }_{3,-2}$ and operators.

Concept introduction:

The eigenvalues of the wavefunction that are obtained when an operator is applied are the only possible values of observables. The expression for the eigenvalue is given by,

$\widehat{A}\Psi =a\Psi$

The z-component of the three dimensional angular momentum that has components in x, y and z direction is quantized.

Answer to Problem 11.54E

Explanation of Solution

The general equation for the wavefunction in the 3-dimensional rotation is,

${\Psi }_{l,m_l}=\frac{1}{\sqrt{2\pi }}{e}^{im\varphi }\cdot {\theta }_{l,m_l}$

The complete form of ${\Psi }_{3,-2}$ using Table 11.3 is,

${\Psi }_{3,-2}=\frac{\sqrt{105}}{4\sqrt{2\pi }}{\sin }^2\theta \cos \theta e^{-2i\varphi }$

The z-component of angular momentum using the complete forms of operators is,

${\widehat{L}}_z{\Psi }_{3,-2}=-i\hslash \frac{\partial }{\partial \varphi }{\Psi }_{3,-2}$

The first derivative of the given wavefunction with respect to $\varphi$ is,

$\frac{\partial }{\partial \varphi }\left(\frac{\sqrt{105}}{4\sqrt{2\pi }}{\sin }^2\theta \cos \theta e^{-2i\varphi }\right)$

$=\left(-2\right)i\times \frac{\sqrt{105}}{4\sqrt{2\pi }}{\sin }^2\theta \cos \theta e^{-2i\varphi }$…(4)

Substitute equation (4) in the equation of z-component of angular momentum as shown below.

$\begin{array}{l}{\widehat{L}}_z{\Psi }_{3,-2}=-i\hslash \left(\left(-2\right)i\times \frac{\sqrt{105}}{4\sqrt{2\pi }}{\sin }^2\theta \cos \theta e^{-2i\varphi }\right)\\ {\widehat{L}}_z{\Psi }_{3,-2}=-2\hslash \left(\frac{\sqrt{105}}{4\sqrt{2\pi }}{\sin }^2\theta \cos \theta e^{-2i\varphi }\right)\\ {\widehat{L}}_z{\Psi }_{3,-2}=-2\hslash {\Psi }_{3,-2}\end{array}$

The eigenvalues of z-component of angular momentum is $-2\hslash$.

Conclusion

The eigenvalues of z-component of angular momentum is $-2\hslash$.

Interpretation Introduction

(c)

Interpretation:

The eigenvalue of energy is to be evaluated using the complete forms of given wavefunction ${\Psi }_{3,-2}$ and operators.

Concept introduction:

The eigenvalues of the wavefunction that are obtained when an operator is applied are the only possible values of observables. The expression for the eigenvalue is given by,

$\widehat{A}\Psi =a\Psi$

The energy of the particle depends on the moment of inertia, quantum number and Planck’s constant. The total energy is quantized.

Answer to Problem 11.54E

The eigenvalue of energy for the given wavefunctionis $\frac{6{\hslash }^2}{I}$.

Explanation of Solution

The general equation for the wavefunction in the 3-dimensional rotation is,

${\Psi }_{l,m_l}=\frac{1}{\sqrt{2\pi }}{e}^{im\varphi }\cdot {\theta }_{l,m_l}$

The complete form of ${\Psi }_{3,-2}$ using Table 11.3 is,

${\Psi }_{3,-2}=\frac{\sqrt{105}}{4\sqrt{2\pi }}{\sin }^2\theta \cos \theta e^{-2i\varphi }$

The eigen equation for the Hamiltonian operator is,

$\widehat{H}{\Psi }_{3,-2}=E{\Psi }_{3,-2}$

The Hamiltonian operator for energy applied on the given wavefunction is also represented in the form of total angular momentum.

$\widehat{H}{\Psi }_{3,-2}=\frac{{\widehat{L}}^2{\Psi }_{3,-2}}{2I}$

The value of total angular momentum is $12{\hslash }^2$.

$\begin{array}{l}\widehat{H}{\Psi }_{3,-2}=\frac{12{\hslash }^2{\Psi }_{3,-2}}{2I}\\ \widehat{H}{\Psi }_{3,-2}=\frac{6{\hslash }^2{\Psi }_{3,-2}}{I}\end{array}$

The eigenvalue of energy $E$ for the given wavefunction is $\frac{6{\hslash }^2}{I}$.

Conclusion

The eigenvalue of energy $E$ for the given wavefunction is $\frac{6{\hslash }^2}{I}$.