Question

Exercises with the nuclear shell model, spin and parity .(8) Define the azimuthal angle e associated to each azimuthal quantum number m in the cases below (j-1/2, 1, 2) can you show me step by step how to get to these answers 1/2 n/2 2n #54.7,-54.7 deg 45., 90, -45 deg 35.3, 65.9, 90,-65.9,-35.3 deg

Answer 1

For every value of  $j$ , we will have $2j+1$ different value of $m_j$, ranging from $-j\ldots,0,\ldots,+j$

Total angular momentum will be $\sqrt{j(j+1)}\,\hbar$

The Z-component of total angular momentum will be $m_j\,\hbar$

The different azimuthal angles are given by $\theta=\cos^{-1}\left ( \frac{m_j \hbar}{\sqrt{j(j+1)}\hbar} \right )=\cos^{-1}\left ( \frac{m_j }{\sqrt{j(j+1)}} \right )$

i) For $j=\frac{1}{2}$ ,  $m_j=\frac{1}{2}$,   $\theta=\cos^{-1}\left ( \frac{\frac{1}{2} }{\sqrt{\frac{1}{2}(\frac{1}{2}+1)}} \right )=\cos^{-1}\left ( \frac{1}{\sqrt{3}} \right )=54.7\degree$

For $j=\frac{1}{2}$ ,  $m_j=-\frac{1}{2}$ , azimuthal angle is $\theta=-54.7\degree$

ii)

For $j=1$  ,  $m_j=1$ ,   $\theta=\cos^{-1}\left ( \frac{1 }{\sqrt{1(1+1)}} \right )=\cos^{-1}\left ( \frac{1}{\sqrt{2}} \right )=45\degree$

For $j=0$  ,  $m_j=0$ , $\theta=\cos^{-1}\left ( \frac{0 }{\sqrt{1(1+1)}} \right )=90\degree$

For $j=1$  ,  $m_j=-1$ ,  $\theta=-45\degree$

iii)

For $j=2$  ,  $m_j=2$ ,   $\theta=\cos^{-1}\left ( \frac{2 }{\sqrt{2(2+1)}} \right )=\cos^{-1}\left ( \frac{2}{\sqrt{6}} \right )=35.3\degree$

For $j=2$  ,  $m_j=1$ ,   $\theta=\cos^{-1}\left ( \frac{1 }{\sqrt{2(2+1)}} \right )=\cos^{-1}\left ( \frac{1}{\sqrt{6}} \right )=65.9\degree$

For $j=2$  ,  $m_j=0$ ,   $\theta=\cos^{-1}\left ( \frac{0 }{\sqrt{2(2+1)}} \right )=\cos^{-1}\left ( \frac{0}{\sqrt{6}} \right )=90\degree$

For $j=2$  ,  $m_j=-1$ ,   $\theta=-65.9\degree$

For $j=2$  ,  $m_j=-2$ ,   $\theta=-35.3\degree$​​​​​​​