Chapter 9, Problem 14P

To determine

The expression for the scalar product of $L\cdot S$.

Answer to Problem 14P

The expression for the scalar product of $L\cdot S$ is $\frac{{\hslash }^2}{2}\left(j\left(j+1\right)-l\left(l+1\right)-s\left(s+1\right)\right)$.

Explanation of Solution

The existence of magnetic moment and orbital angular momentum leads to the spin obit coupling which further leads to a net moment. The net moment is the sum of orbital angular moment and spin moment. The spin orbit coupling implies that the spin angular momentum and orbital angular momentum cannot be conserved separately.

The total angular momentum of any system is conserved which is defined as the sum of angular momentum and spin angular momentum.

Write the expression for the total angular momentum.

  $J=L+S$        (I)

Here, $J$ is the total angular mo0metum, $L$ is the orbital angular momentum and $S$ is spin angular momentum.

Take scalar product of equation (I) with $J=L+S$.

  $J\cdot J=\left(L+S\right)\cdot \left(L+S\right)$

Simplify the above expression.

  $J^2=L^2+S^2+2L\cdot S$

Rearrange the above expression in terms of $L\cdot S$.

  $L\cdot S=\frac{1}{2}\left(J^2-L^2-S^2\right)$

Substitute $\sqrt{j\left(j+1\right)}\hslash$ for $J$, $\sqrt{l\left(l+1\right)}\hslash$ for $L$ and $\sqrt{s\left(s+1\right)}\hslash$ in above expression and simplify.

  $L\cdot S=\frac{{\hslash }^2}{2}\left(j\left(j+1\right)-l\left(l+1\right)-s\left(s+1\right)\right)$

Here, $j$ is the total quantum number, $l$ is orbital quantum number and $s$ is spin quantum number.

Conclusion:

Thus, the expression for the scalar product of $L\cdot S$ is $\frac{{\hslash }^2}{2}\left(j\left(j+1\right)-l\left(l+1\right)-s\left(s+1\right)\right)$.

(b)

To determine

The angle between the orbital angular momentum and spin angular momentum.

Answer to Problem 14P

The angle between the orbital angular momentum and spin angular momentum is Case 1(i) $144.7°$, (ii)$65.9°$, Case 2(i)$129.2°$ ,(ii) $58.2°$.

Explanation of Solution

Write the expression for the scalar product of $L$ and $S$.

  $L\cdot S=\left|L\right|\left|S\right|\cos \theta$

Rearrange the above expression.

  $\cos \theta =\frac{L\cdot S}{\left|L\right|\left|S\right|}$

Substitute $\frac{{\hslash }^2}{2}\left(j\left(j+1\right)-l\left(l+1\right)-s\left(s+1\right)\right)$ for $L\cdot S$, $\sqrt{l\left(l+1\right)}\hslash$ for $L$ and $\sqrt{s\left(s+1\right)}\hslash$ in above expression.

  $\cos \theta =\frac{\frac{{\hslash }^2}{2}\left(j\left(j+1\right)-l\left(l+1\right)-s\left(s+1\right)\right)}{\left(\sqrt{l\left(l+1\right)}\hslash \right)\left(\sqrt{s\left(s+1\right)}\hslash \right)}$

Simplify the above expression.

  $\cos \theta =\frac{1}{2}\frac{\left(j\left(j+1\right)-l\left(l+1\right)-s\left(s+1\right)\right)}{\left(\sqrt{l\left(l+1\right)}\right)\left(\sqrt{s\left(s+1\right)}\right)}$        (II)

Here, $\theta$ is the angle between $L$ and $S$.

Conclusion:

Case 1:

(i)$P_{1/2}$

Substitute $1$ for $l$, $\frac{1}{2}$ for $s$ and $\frac{1}{2}$ for $j$ in equation (II).

   $\begin{array}{c}\cos \theta =\frac{1}{2}\frac{\left(\frac{1}{2}\left(\frac{1}{2}+1\right)-1\left(1+1\right)-\frac{1}{2}\left(\frac{1}{2}+1\right)\right)}{\left(\sqrt{1\left(1+1\right)}\right)\left(\sqrt{\frac{1}{2}\left(\frac{1}{2}+1\right)}\right)}\\ =-\sqrt{\frac{2}{3}}\\ \theta =144.7°\end{array}$

(ii)$P_{3/2}$

Substitute $1$ for $l$, $\frac{1}{2}$ for $s$ and $\frac{3}{2}$ for $j$ in equation (II).

   $\begin{array}{c}\cos \theta =\frac{1}{2}\frac{\left(\frac{3}{2}\left(\frac{3}{2}+1\right)-1\left(1+1\right)-\frac{1}{2}\left(\frac{1}{2}+1\right)\right)}{\left(\sqrt{1\left(1+1\right)}\right)\left(\sqrt{\frac{1}{2}\left(\frac{1}{2}+1\right)}\right)}\\ =\sqrt{\frac{1}{6}}\\ \theta =65.9°\end{array}$

Case 2:

(i)$H_{9/2}$

Substitute $5$ for $l$, $\frac{1}{2}$ for $s$ and $\frac{9}{2}$ for $j$ in equation (II).

   $\begin{array}{c}\cos \theta =\frac{1}{2}\frac{\left(\frac{9}{2}\left(\frac{9}{2}+1\right)-5\left(5+1\right)-\frac{1}{2}\left(\frac{1}{2}+1\right)\right)}{\left(\sqrt{5\left(5+1\right)}\right)\left(\sqrt{\frac{1}{2}\left(\frac{1}{2}+1\right)}\right)}\\ =-\frac{6}{\sqrt{90}}\\ \theta =129.2°\end{array}$

(ii)$H_{11/2}$

Substitute $5$ for $l$, $\frac{1}{2}$ for $s$ and $\frac{11}{2}$ for $j$ in equation (II).

     $\begin{array}{c}\cos \theta =\frac{1}{2}\frac{\left(\frac{11}{2}\left(\frac{11}{2}+1\right)-5\left(5+1\right)-\frac{1}{2}\left(\frac{1}{2}+1\right)\right)}{\left(\sqrt{5\left(5+1\right)}\right)\left(\sqrt{\frac{1}{2}\left(\frac{1}{2}+1\right)}\right)}\\ =\frac{5}{\sqrt{90}}\\ \theta =58.2°\end{array}$

Thus, the angle between the orbital angular momentum and spin angular momentum is Case 1(i) $144.7°$, (ii)$65.9°$, Case 2(i)$129.2°$ ,(ii) $58.2°$.