Question

Consider a system of three distinguishable particles, each having spin \( S = \frac{1}{2} \) such that \( S_z = \pm \frac{1}{2} \) with corresponding magnetic moments \( \mu_z = \pm \mu \). When the system is placed in an external magnetic field \( H \) pointing along the z-axis, the total energy of the system is \( \mu H \). Let \( x \) be the state where the first spin has \( S_z = \frac{1}{2} \). The probability of having the state \( x \) and the mean magnetic moment (in the +z direction) of the system in state \( x \) are

• A. \( \dfrac{1}{3} \, \dfrac{-1}{3} \mu \)
• B. \( \dfrac{1}{3} \, \dfrac{2}{3} \mu \)
• C. \( \dfrac{2}{3} \, \dfrac{-2}{3} \mu \)
• D. \( \dfrac{2}{3} \, \dfrac{1}{3} \mu \)

Hint:

When dealing with systems of distinguishable particles, consider the possible spin configurations and their corresponding magnetic moments.

Answer 1

The Correct Option is A

The total energy of the system is given by \( E = \mu H \). Since there are three distinguishable particles, and the first spin has \( S_z = \frac{1}{2} \), the probability of having state \( x \) is calculated based on the number of favorable configurations for the spins.

Step 1: The probability depends on the number of available states and the magnetic moment associated with the spin. The system has three possible configurations, so the probability for each state is \( \frac{1}{3} \).

Step 2: The magnetic moment in the +z direction for the first spin is \( \frac{-1}{3} \mu \), resulting in the given configuration.
Thus, the correct answer is (A).