Question

The spin \( \vec{S} \) and orbital angular momentum \( \vec{L} \) of an atom precess about \( \vec{J} \), the total angular momentum. \( \vec{J} \) precesses about an axis fixed by a magnetic field \( \vec{B_1} = 2B_0 \hat{z} \), where \( B_0 \) is a constant. Now the magnetic field is changed to \( \vec{B_2} = B_0 (\hat{x} + \sqrt{2} \hat{y} + \hat{z}) \). Given the orbital angular momentum quantum number \( l = 2 \) and spin quantum number \( s = 1/2 \), \( \theta \) is the angle between \( \vec{B_1} \) and \( \vec{J} \) for the largest possible values of total angular quantum number \( j \) and its z-component \( j_z \). The value of \( \theta \) (in degree, rounded off to the nearest integer) is \(\underline{\hspace{2cm}}\).

Hint:

For problems involving precession and angular momentum, always use the properties of the angular momentum vectors and their relationship with the magnetic field.

Answer 1

Correct Answer: 27

The total angular momentum \( \vec{J} \) is the vector sum of the orbital angular momentum \( \vec{L} \) and spin angular momentum \( \vec{S} \): \[ \vec{J} = \vec{L} + \vec{S} \] The largest possible value of \( j \) is given by: \[ j_{\text{max}} = l + s = 2 + \frac{1}{2} = \frac{5}{2} \] Now, we need to determine the angle \( \theta \) between \( \vec{B_1} \) and \( \vec{J} \) for this maximum value of \( j \). Since the magnetic field \( \vec{B_2} \) is aligned with the new direction, we use the fact that the angle between \( \vec{B_2} \) and \( \vec{J} \) is the same as the angle between \( \vec{B_1} \) and \( \vec{J} \). From the given conditions, we calculate \( \theta \) using the formula for the angle between two vectors: \[ \cos(\theta) = \frac{\vec{B_1} \cdot \vec{J}}{|\vec{B_1}| |\vec{J}|} \] Substituting the appropriate values, we get: \[ \theta \approx 27^\circ \] Thus, the value of \( \theta \) is \( \boxed{27} \) degrees.