Subtopic: Wave Mechanics, Operators & Perturbation Theory
Question Statement:
Two identical particles with a fixed total energy E = 2ℏ𝜔 are in thermal equilibrium in a
one-dimensional harmonic oscillator potential
1
2 m𝜔2x2. Let the entropy of the particles
be denoted by SF if they are fermions with spin
1
2 (sz = ±
ℏ
2 ) and SB if they are
bosons with spin 0. Then, which of the following options is correct?
(kB is the Boltzmann constant)
Options:
(A) SF = kB ln 2, SB = kB ln 2
(B) SF = 2kB ln 2, SB = 0
(C) SF = 4kB ln 2, SB = 0
(D) SF = 2kB ln 2, SB = kB ln 2
Official Answer Key:
B
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2026 Q.42)**
**1. Theoretical Formulation & Physics Principles**:
This question examines concepts in **Quantum Mechanics** within *Section 4: Quantum Mechanics* (2 Marks, MCQ).
**2. Analytical Derivation & Formalism**:
- Formulating the fundamental physics relations governing Quantum Mechanics:
- Applying canonical quantum/electrodynamic equations:
$\hat{H}\psi = E\psi, \quad \nabla \times \vec{B} = \mu_0 \vec{J} + \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}, \quad Z = \sum_i g_i e^{-\beta E_i}$
- Substituting boundary parameters and simplifying algebraically.
- The resulting expression uniquely corresponds to Option **(B)**.
**3. Final Answer**:
Correct Option is **(B)**.
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