A nitrogen atom has in its ground state a spin with a quantum number s=3/2: . Determine the possible results of measurement of the size and the angle with respect to the given z-axis for the nitrogen atoms spin!
Solution
A nitrogen atom in its ground state has a spin with a quantum number $s = 3/2$. The magnitude of the spin angular momentum (|s|) can be calculated as follows:
\begin{equation} |s| = \sqrt{s(s+1)} = \sqrt{3/2 * (3/2 + 1)} = \sqrt{\frac{3}{2} * \frac{5}{2}} = \sqrt{\frac{15}{4}} = \sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{2} \end{equation}
The possible results of measurement for the spin angular momentum projection along a given $z$-axis ($m_z$) are: $m_z = -3/2, -1/2, 1/2,$ and $3/2$.
The corresponding angles ($\theta$) with respect to the $z$-axis can be calculated as follows:
\begin{equation} \theta = \cos^{-1}\left(\frac{m_z}{|s|}\right) \end{equation}
So for each possible value of $m_z$, the corresponding angle $\theta$ can be found using the above expression: \begin{align*} m_z &= -3/2 &\rightarrow &\quad \theta &= \cos^{-1}\left(-\frac{3/2}{\frac{\sqrt{15}}{2}}\right) \\ m_z &= -1/2 &\rightarrow &\quad \theta &= \cos^{-1}\left(-\frac{1/2}{\frac{\sqrt{15}}{2}}\right) \\ m_z &= 1/2 &\rightarrow &\quad \theta &= \cos^{-1}\left(\frac{1/2}{\frac{\sqrt{15}}{2}}\right) \\ m_z &= 3/2 &\rightarrow &\quad \theta &= \cos^{-1}\left(\frac{3/2}{\frac{\sqrt{15}}{2}}\right) \end{align*}
Note: The angles $\theta$ are unique up to $360^\circ$, so they can take on multiple values depending on the reference frame.