Atomic $J$ vs $B_1$ Angle Calculation
This solution calculates the angle $ \theta$ between the total angular momentum $ \vec{J}$ and the magnetic field $ \vec{B}_1$. The calculation uses the state corresponding to the maximum possible values of $j$ and $j_z$, with $ \vec{B}_1$ defining the quantization axis.
Maximum $j$ and $j_z$ Values
- Given orbital angular momentum quantum number $ l=2 $ and spin quantum number $ s=1/2 $.
- The maximum total angular momentum quantum number is $ j_{max} = l+s = 2 + \frac{1}{2} = \frac{5}{2} $.
- For the state with $ j=5/2 $, the maximum $z$-component quantum number $m_j$ (corresponding to the largest $j_z$) is $ m_{j, max} = +j = +\frac{5}{2} $.
Relation: $ \theta$ and Angular Momentum
The angle $ \theta$ between $ \vec{J}$ and the $z$-axis (direction of $ \vec{B}_1$) is found using the projection of $ \vec{J}$ onto the $z$-axis: $ \langle J_z \rangle = |\vec{J}| \cos \theta $ Substituting $ \langle J_z \rangle = m_j \hbar $ and $ |\vec{J}| = \sqrt{j(j+1)}\hbar $, we derive: $ \cos \theta = \frac{m_j \hbar}{\sqrt{j(j+1)}\hbar} = \frac{m_j}{\sqrt{j(j+1)}} $
Numerical Calculation of $ \theta$
Using the maximum values $ j = 5/2 $ and $ m_j = 5/2 $: $ \cos \theta = \frac{5/2}{\sqrt{\frac{5}{2}(\frac{5}{2}+1)}} = \frac{5/2}{\sqrt{\frac{5}{2} \cdot \frac{7}{2}}} = \frac{5/2}{\sqrt{35/4}} = \frac{5/2}{\sqrt{35}/2} = \frac{5}{\sqrt{35}} $ Simplify and find $ \theta$: $ \cos \theta = \frac{5\sqrt{35}}{35} = \frac{\sqrt{35}}{7} $ $ \theta = \arccos\left(\frac{\sqrt{35}}{7}\right) $ Using a calculator, $ \frac{\sqrt{35}}{7} \approx 0.84515 $. $ \theta \approx \arccos(0.84515) \approx 32.32^\circ $
Final Rounded Angle
Rounding to the nearest integer, the angle is $ \theta \approx 32^\circ $. This result is consistent with the range provided.
Important Questions from Angular Momentum Operators Eigenvalues Clebsch Gordan
- Consider two particles with angular momenta $j_1 = 2\hbar$ and $j_2 = \hbar/2$. If the expression
$$|j = 5/2, m = 3/2\rangle = \begin{cases} c_1|j_1 = 2, m_1 = 1\rangle|j_2 = 1/2, m_2 = 1/2\rangle + \\ c_2|j_1 = 2, m_1 = 2\rangle|j_2 = 1/2, m_2 = -1/2\rangle \end{cases}$$
gives an eigenstate of the total angular momentum of the two particles, using standard notation. Which of the following is true?
(Hint: $\hat{J}_{\pm}|j, m\rangle = \sqrt{j(j + 1) - m(m \pm 1)} |j, m \pm 1\rangle$) - A system of three non-identical spin $\frac{1}{2}$ particles has the Hamiltonian $H = \frac{A}{\hbar^2} (\vec{S}_1 + \vec{S}_2) \cdot \vec{S}_3$, where $\vec{S}_1, \vec{S}_2$ and $\vec{S}_3$ are the spin operators of particles labelled $1,2$ and $3$ respectively and $A$ is a constant with appropriate dimensions. The set of possible energy eigenvalues of the system is
- $H$ is the Hamiltonian, $\vec{L}$ the orbital angular momentum and $L_z$ is the $z$-component of $\vec{L}$. The $1s$ state of the hydrogen atom in the non-relativistic formalism is an eigen function of which one of the following sets of operators?
- An atom with non-zero magnetic moment has an angular momentum of magnitude $\sqrt{12}\hbar$. When a beam of such atoms is passed through a Stern-Gerlach apparatus, how many beams does it split into?
- In the vector model of angular momentum applied to atoms, what is the minimum angle in degrees (in integer) made by the orbital angular momentum vector and the positive $z$ axis for a $2p$ electron?
