Chapter 7, Problem 7F.8P

(a)

Interpretation Introduction

Interpretation:

The wavefunction, $\psi =c_1{Y}_{l,m_l}+c_2{Y}_{l,m_{l\text{'}}}$ is an eigen function of ${\Lambda }^2$ with an eigen value $-l\left(l+1\right)$ has to be shown.

Concept introduction:

When an operator operates on a function and gives back the function along with a constant, then the function is known as eigen function of the operator and the constant is known as eigen value.  A general eigenvalue equation is represented as shown below.

  $\widehat{\text{O}}\psi =\text{K}\psi$

Answer to Problem 7F.8P

The wavefunction, $\psi =c_1{Y}_{l,m_l}+c_2{Y}_{l,m_{l\text{'}}}$ has been shown as the eigen function of ${\Lambda }^2$ with an eigen value $-l\left(l+1\right)$.

Explanation of Solution

The wavefunction given is as follows.

    $\psi =c_1{Y}_{l,m_l}+c_2{Y}_{l,m_{l\text{'}}}$

Apply ${\Lambda }^2$ on the wavefunction as shown below.

    $\begin{array}{c}{\Lambda }^2\psi ={\Lambda }^2\left(c_1{Y}_{l,m_l}+c_2{Y}_{l,m_{l\text{'}}}\right)\\ ={\Lambda }^2{c}_1{Y}_{l,m_l}+{\Lambda }^2{c}_2{Y}_{l,m_{l\text{'}}}\\ =-l\left(l+1\right)c_1{Y}_{l,m_l}+-l\left(l+1\right)c_2{Y}_{l,m_{l\text{'}}}\\ =-l\left(l+1\right)\left(c_1{Y}_{l,m_l}+c_2{Y}_{l,m_{l\text{'}}}\right)\end{array}$

The above equation shows that wavefunction, $\psi =c_1{Y}_{l,m_l}+c_2{Y}_{l,m_{l\text{'}}}$ is the eigen function of ${\Lambda }^2$ with an eigen value $-l\left(l+1\right)$.

(b)

Interpretation Introduction

Interpretation:

The wavefunctions ${\psi }_{\text{a}}=-Y_{1,+1}+Y_{1,-1}$ and ${\psi }_{\text{b}}=i\left(Y_{1,+1}+Y_{1,-1}\right)$ have to be shown real.

Concept introduction:

Same as mentioned in part (a).

Answer to Problem 7F.8P

The wavefunctions, ${\psi }_{\text{a}}=-Y_{1,+1}+Y_{1,-1}$ and ${\psi }_{\text{b}}=i\left(Y_{1,+1}+Y_{1,-1}\right)$ have been shown real.

Explanation of Solution

The wavefunctions given are as follows.

    $\begin{array}{c}{\psi }_{\text{a}}=-Y_{1,+1}+Y_{1,-1}\\ {\psi }_{\text{b}}=i\left(Y_{1,+1}+Y_{1,-1}\right)\end{array}$

In order to show a wavefunction real, the product of that wavefunction with its conjugate is estimated as shown below.

    $\begin{array}{c}{\psi }_{\text{a}}{\psi }_{\text{a}}^{*}=\left(-Y_{1,+1}+Y_{1,-1}\right){\left(-Y_{1,+1}+Y_{1,-1}\right)}^{*}\\ =\left(-Y_{1,+1}+Y_{1,-1}\right)\left(-Y_{1,+1}+Y_{1,-1}\right)\\ =-Y_{1,+1}\cdot -Y_{1,+1}+\left(-Y_{1,+1}\cdot Y_{1,-1}\right)+\left(Y_{1,-1}\cdot -Y_{1,+1}\right)+Y_{1,-1}\cdot Y_{1,-1}\\ =Y_{1,+1}^2-Y_{1,+1}\cdot Y_{1,-1}-Y_{1,+1}\cdot Y_{1,-1}+Y_{1,-1}^2\end{array}$

The value of $Y_{1,+1}\cdot Y_{1,-1}=0$, $Y_{1,+1}^2=1$ and $Y_{1,-1}^2=1$ as the wavefunctions $Y_{1,+1}$ and $Y_{1,-1}$ are orthonormal.

Substitute $Y_{1,+1}\cdot Y_{1,-1}=0$, $Y_{1,+1}^2=1$ and $Y_{1,-1}^2=1$ in the above equation.

    $\begin{array}{c}{\psi }_{\text{a}}{\psi }_{\text{a}}^{*}=Y_{1,+1}^2-Y_{1,+1}\cdot Y_{1,-1}-Y_{1,+1}\cdot Y_{1,-1}+Y_{1,-1}^2\\ =1-0-0+1\\ =2\end{array}$

Therefore, ${\psi }_{\text{a}}=-Y_{1,+1}+Y_{1,-1}$ is a real wavefunction.

The product of wavefunction ${\psi }_{\text{b}}=i\left(Y_{1,+1}+Y_{1,-1}\right)$ with its conjugate is estimated as shown below.

    $\begin{array}{c}{\psi }_{\text{b}}{\psi }_{\text{b}}^{*}=i\left(Y_{1,+1}+Y_{1,-1}\right)\left(-i\right){\left(Y_{1,+1}+Y_{1,-1}\right)}^{*}\\ =-i^2\left(Y_{1,+1}+Y_{1,-1}\right)\left(Y_{1,+1}+Y_{1,-1}\right)\\ =-i^2\left(Y_{1,+1}\cdot Y_{1,+1}+\left(Y_{1,+1}\cdot Y_{1,-1}\right)+\left(Y_{1,-1}\cdot Y_{1,+1}\right)+Y_{1,-1}\cdot Y_{1,-1}\right)\\ =-\left(-1\right)\left(Y_{1,+1}^2+Y_{1,+1}\cdot Y_{1,-1}+Y_{1,+1}\cdot Y_{1,-1}+Y_{1,-1}^2\right)\because i^2=-1\end{array}$

Substitute $Y_{1,+1}\cdot Y_{1,-1}=0$, $Y_{1,+1}^2=1$ and $Y_{1,-1}^2=1$ in the above equation.

    $\begin{array}{c}{\psi }_{\text{b}}{\psi }_{\text{b}}^{*}=Y_{1,+1}^2-Y_{1,+1}\cdot Y_{1,-1}-Y_{1,+1}\cdot Y_{1,-1}+Y_{1,-1}^2\\ =1-0-0+1\\ =2\end{array}$

Therefore, ${\psi }_{\text{b}}=i\left(Y_{1,+1}+Y_{1,-1}\right)$ is a real wavefunction.

(c)

Interpretation Introduction

Interpretation:

The wavefunctions ${\psi }_{\text{a}}=-Y_{1,+1}+Y_{1,-1}$ and ${\psi }_{\text{b}}=i\left(Y_{1,+1}+Y_{1,-1}\right)$ have to be shown orthogonal.

Concept introduction:

For the orthogonality of the wavefunctions, one of the wavefunction is integrated with the conjugate of the other wavefunction over the entire limits.  It is expressed by the equation as given below.

  ${\displaystyle \underset{0}{\overset{\infty }{\int }}\left(N{\psi }_1\right)\left(N{\psi }_2^{*}\right)}=0$

Answer to Problem 7F.8P

The wavefunctions ${\psi }_{\text{a}}=-Y_{1,+1}+Y_{1,-1}$ and ${\psi }_{\text{b}}=i\left(Y_{1,+1}+Y_{1,-1}\right)$ have been shown orthogonal.

Explanation of Solution

In order to show orthogonality, the product wavefunction, ${\psi }_{\text{a}}$ with conjugate of ${\psi }_{\text{b}}$ is estimated as shown below.

    $\begin{array}{c}{\psi }_{\text{a}}{\psi }_{\text{b}}^{*}=\left(-Y_{1,+1}+Y_{1,-1}\right)\left(-i\right){\left(Y_{1,+1}+Y_{1,-1}\right)}^{*}\\ =\left(-Y_{1,+1}+Y_{1,-1}\right)\left(-i\right)\left(Y_{1,+1}+Y_{1,-1}\right)\\ =\left(-i\right)\left(-Y_{1,+1}\cdot Y_{1,+1}+\left(-Y_{1,+1}\cdot Y_{1,-1}\right)+Y_{1,-1}\cdot Y_{1,+1}+Y_{1,-1}\cdot Y_{1,-1}\right)\\ =\left(-i\right)\left(-Y_{1,+1}^2-Y_{1,+1}\cdot Y_{1,-1}+Y_{1,+1}\cdot Y_{1,-1}+Y_{1,-1}^2\right)\end{array}$

Substitute $Y_{1,+1}\cdot Y_{1,-1}=0$, $Y_{1,+1}^2=1$ and $Y_{1,-1}^2=1$ in the above equation.

    $\begin{array}{c}{\psi }_{\text{a}}{\psi }_{\text{b}}^{*}=\left(-i\right)\left(-Y_{1,+1}^2-Y_{1,+1}\cdot Y_{1,-1}+Y_{1,+1}\cdot Y_{1,-1}+Y_{1,-1}^2\right)\\ =\left(-i\right)\left(-1-0+0+1\right)\\ =0\end{array}$

Therefore, the wavefunctions ${\psi }_{\text{a}}$ and ${\psi }_{\text{b}}$ are orthogonal to each other.

(d)

Interpretation Introduction

Interpretation:

The normalization of ${\psi }_{\text{a}}$ and ${\psi }_{\text{b}}$ have to be shown.

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.

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

Answer to Problem 7F.8P

The normalization of ${\psi }_{\text{a}}$ and ${\psi }_{\text{b}}$ have been shown.

Explanation of Solution

In order to show normalization, the product wavefunction, ${\psi }_{\text{a}}$ with its conjugate is estimated as shown below.

    $\begin{array}{c}{\psi }_{\text{a}}{\psi }_{\text{a}}^{*}=1\\ c_{\text{a}}\left(-Y_{1,+1}+Y_{1,-1}\right)c_{\text{a}}{\left(-Y_{1,+1}+Y_{1,-1}\right)}^{*}=1\\ c_{\text{a}}^2\left(-Y_{1,+1}+Y_{1,-1}\right)\left(-Y_{1,+1}+Y_{1,-1}\right)=1\\ c_{\text{a}}^2\left(-Y_{1,+1}\cdot -Y_{1,+1}+\left(-Y_{1,+1}\cdot Y_{1,-1}\right)+\left(Y_{1,-1}\cdot -Y_{1,+1}\right)+Y_{1,-1}\cdot Y_{1,-1}\right)=1\end{array}$

Substitute $Y_{1,+1}\cdot Y_{1,-1}=0$, $Y_{1,+1}^2=1$ and $Y_{1,-1}^2=1$ in the above equation.

    $\begin{array}{c}{c}_{\text{a}}^2\left(Y_{1,+1}^2-Y_{1,+1}\cdot Y_{1,-1}-Y_{1,+1}\cdot Y_{1,-1}+Y_{1,-1}^2\right)=1\\ c_{\text{a}}^2\left(1-0-0+1\right)=1\\ 2c_{\text{a}}^2=1\\ c_{\text{a}}=\frac{1}{\sqrt{2}}\end{array}$

Therefore, ${\psi }_{\text{a}}=-Y_{1,+1}+Y_{1,-1}$ is a normalized wavefunction with normalization constant $\frac{1}{\sqrt{2}}$.

The product of wavefunction ${\psi }_{\text{b}}=i\left(Y_{1,+1}+Y_{1,-1}\right)$ with its conjugate is estimated as shown below.

    $\begin{array}{c}{\psi }_{\text{b}}{\psi }_{\text{b}}^{*}=1\\ c_{\text{b}}^{2}i\left(Y_{1,+1}+Y_{1,-1}\right)\left(-i\right){\left(Y_{1,+1}+Y_{1,-1}\right)}^{*}=1\\ -i^2{c}_{\text{b}}^2\left(Y_{1,+1}+Y_{1,-1}\right)\left(Y_{1,+1}+Y_{1,-1}\right)=1\\ -\left(-1\right)c_{\text{b}}^2\left(Y_{1,+1}^2+Y_{1,+1}\cdot Y_{1,-1}+Y_{1,+1}\cdot Y_{1,-1}+Y_{1,-1}^2\right)=1\because i^2=-1\end{array}$

Substitute $Y_{1,+1}\cdot Y_{1,-1}=0$, $Y_{1,+1}^2=1$ and $Y_{1,-1}^2=1$ in the above equation.

    $\begin{array}{c}-\left(-1\right)c_{\text{b}}^2\left(Y_{1,+1}^2+Y_{1,+1}\cdot Y_{1,-1}+Y_{1,+1}\cdot Y_{1,-1}+Y_{1,-1}^2\right)=1\\ c_{\text{b}}^2\left(1-0-0+1\right)=1\\ 2c_{\text{b}}^2=1\\ c_{\text{b}}=\frac{1}{\sqrt{2}}\end{array}$

Therefore, ${\psi }_{\text{b}}=i\left(Y_{1,+1}+Y_{1,-1}\right)$ is a normalized wavefunction with normalization constant $\frac{1}{\sqrt{2}}$.

(e)

Interpretation Introduction

Interpretation:

The angular nodes of the wavefunctions ${\psi }_{\text{a}}$ and ${\psi }_{\text{b}}$ have to be stated and the plane to which these angular nodes correspond has to be stated.

Concept introduction:

Same as mentioned in part (a).

Answer to Problem 7F.8P

The angular nodes in the wavefunctions ${\psi }_{\text{a}}$ and ${\psi }_{\text{b}}$ is one each.  The plane to which the angular nodes of wavefunctions ${\psi }_{\text{a}}$ and ${\psi }_{\text{b}}$ correspond is x-y plane.

Explanation of Solution

The angular node is equal to the value of $l$.  For the wavefunctions ${\psi }_{\text{a}}$ and ${\psi }_{\text{b}}$, the value of $l$ is $1$.

Therefore, the numbers of angular nodes present in the wavefunctions ${\psi }_{\text{a}}$ and ${\psi }_{\text{b}}$ are one each.

The plane to which the angular nodes of wavefunctions ${\psi }_{\text{a}}$ and ${\psi }_{\text{b}}$ correspond is x-y plane.

(f)

Interpretation Introduction

Interpretation:

Whether the wavefunctions ${\psi }_{\text{a}}$ is an eigen function of ${\widehat{l}}_{\text{z}}$ or not has to be stated. The significance of this answer has to be explained.

Concept introduction:

Same as mentioned in part (a).

Answer to Problem 7F.8P

The wavefunction ${\psi }_{\text{a}}$ is not an eigen function of ${\widehat{l}}_{\text{z}}$.  The significance of this answer is that ${\psi }_{\text{a}}$ does not give exact value of ${\widehat{l}}_{\text{z}}$.

Explanation of Solution

The value of operator ${\widehat{l}}_{\text{z}}$ is $m_l\hslash$.

Apply ${\widehat{l}}_{\text{z}}$ on the wavefunction as shown below.

    $\begin{array}{c}{\widehat{l}}_{\text{z}}{\psi }_{\text{a}}={\widehat{l}}_{\text{z}}\frac{1}{\sqrt{2}}\left(-Y_{1,+1}+Y_{1,-1}\right)\\ =\frac{\hslash }{\sqrt{2}}\left(1\times \left(-Y_{1,+1}\right)+\left(-1\right)\times \left(Y_{1,-1}\right)\right)\\ =\frac{\hslash }{\sqrt{2}}\left(-Y_{1,+1}-Y_{1,-1}\right)\end{array}$

Since on applying ${\widehat{l}}_{\text{z}}$, the same wavefunction is not obtained, therefore, the wavefunction ${\psi }_{\text{a}}$ is not an eigen function of ${\widehat{l}}_{\text{z}}$.

This means the value of ${\widehat{l}}_{\text{z}}$ operator cannot be estimated exactly for the wavefunction ${\psi }_{\text{a}}$.